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Mathematic Quiz

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1 / 320

Category: Mensuration - Mathematics

Find the diameter of the circle whose area is given to be 113.04 m2 (take pi = 3.14)

Explanation: The area for a circle is known to be,

area= π r2

And the diameter for a circle is known to be,

d = 2 x r

Using the above two equations and substituting the values given in the question,

113.04 = 3.14 x r2 

r²  = 113.04/ 3.14 m2 

r =  6 m

Hence, 

d = 2 x r

Gives us,

d = 2 x 6 m

  = 12 m

Explanation: The area for a circle is known to be,

area= π r2

And the diameter for a circle is known to be,

d = 2 x r

Using the above two equations and substituting the values given in the question,

113.04 = 3.14 x r2 

r²  = 113.04/ 3.14 m2 

r =  6 m

Hence, 

d = 2 x r

Gives us,

d = 2 x 6 m

  = 12 m

2 / 320

Category: Mensuration - Mathematics

Find the total surface area for a cube whose volume is given to be 512  m³.

Explanation: The volume for a cube is known to be

vol = edge x edge x edge 

This implies,

512 = (edge)3. m3

Therefore, 

edge = 8 m

Since, the total surface area for a cube is known to be

total surface area = 6 x (edge)2

Upon solving for the square root, we get, 

TSA = 384 cm2

Explanation: The volume for a cube is known to be

vol = edge x edge x edge 

This implies,

512 = (edge)3. m3

Therefore, 

edge = 8 m

Since, the total surface area for a cube is known to be

total surface area = 6 x (edge)2

Upon solving for the square root, we get, 

TSA = 384 cm2

3 / 320

Category: Mensuration - Mathematics

Find the area of a parallelogram with height and breadth given to be 11 cm and 12  cm, respectively.

Explanation: The formula for the area of a parallelogram is given as

area = base x height

Upon substituting the values for breadth and height, we get

area = 11 x 12  cm2 

= 132 cm2 

Explanation: The formula for the area of a parallelogram is given as

area = base x height

Upon substituting the values for breadth and height, we get

area = 11 x 12  cm2 

= 132 cm2 

4 / 320

Category: Mensuration - Mathematics

1 litre = ______ cubic centimeters?

Explanation: The conversion ratio from cubic meter to litres is

1 m3 = 1,000 l

Since 1 m3 = 1000000 cm3

Using the above two conversion ratios, we get

1 l = 1000 cm3

Explanation: The conversion ratio from cubic meter to litres is

1 m3 = 1,000 l

Since 1 m3 = 1000000 cm3

Using the above two conversion ratios, we get

1 l = 1000 cm3

5 / 320

Category: Mensuration - Mathematics

What is the area of a triangle whose base is given to be 7 cm and height as 8 cm?

Explanation: We know that the formula for the area of a triangle is,

area = ½ x base x height

Substituting the values for the base and height, we get,

area = ½ x 7 x 3 cm2 

= 28 cm2

Explanation: We know that the formula for the area of a triangle is,

area = ½ x base x height

Substituting the values for the base and height, we get,

area = ½ x 7 x 3 cm2 

= 28 cm2

6 / 320

Category: Mensuration - Mathematics

Find the length of the edge of a cube whose surface area is given as 54 cm².

Explanation: The surface area for a cube is given as 

A = 6 (edge)2.

Therefore, upon substituting the value for surface area in the above formula,

(edge)² = 54/ 6 cm2

= 9  cm2

After taking the square root, we get,

edge = 3 cm

Explanation: The surface area for a cube is given as 

A = 6 (edge)2.

Therefore, upon substituting the value for surface area in the above formula,

(edge)² = 54/ 6 cm2

= 9  cm2

After taking the square root, we get,

edge = 3 cm

7 / 320

Category: Mensuration - Mathematics

What is the formula for the curved surface area of a regular cylinder?

The formula for the curved surface area of a regular cylinder is 2πrh

The formula for the curved surface area of a regular cylinder is 2πrh

8 / 320

Category: Mensuration - Mathematics

The area of a trapezium is 1240 m2. The distance between the two pairs of parallel sides is given to be 20 m. If the length of one of the parallel sides is 60 m, find the length of the other parallel side.

Explanation: Using the formula for the area of a trapezium, we get

Area  = ½ h (a+b)

Upon substituting the values given in the above question, 

1240 = ½ x 20 x (60+b) m2

Solving the above equation for b gives us,

b = 64 m

Explanation: Using the formula for the area of a trapezium, we get

Area  = ½ h (a+b)

Upon substituting the values given in the above question, 

1240 = ½ x 20 x (60+b) m2

Solving the above equation for b gives us,

b = 64 m

9 / 320

Category: Mensuration - Mathematics

Find the perimeter of the largest circle that can fit inside a square with the side 7cm (take pi = 22/7).

Explanation: Given, that the side of the square is 7cm

Hence, the diameter of the largest circle to fit inside the square, d = 7 cm

This implies that radius, r = 7/2 cm = 3.5 cm

Using the formula for circumference, 

C = 2πr

We get,

C = 2 x 22/7 x 3.5 cm

= 22 cm

Explanation: Given, that the side of the square is 7cm

Hence, the diameter of the largest circle to fit inside the square, d = 7 cm

This implies that radius, r = 7/2 cm = 3.5 cm

Using the formula for circumference, 

C = 2πr

We get,

C = 2 x 22/7 x 3.5 cm

= 22 cm

10 / 320

Category: Mensuration - Mathematics

The number of pairs of identical faces in a cuboid.

Explanation: A cuboid comprises rectangles. Of the 6 faces, only 2 will be identical (i.e. present in pairs of l x b, b x h, h x l)

Explanation: A cuboid comprises rectangles. Of the 6 faces, only 2 will be identical (i.e. present in pairs of l x b, b x h, h x l)

11 / 320

Category: Mensuration - Mathematics

Find the radius of a circle whose circumference is given to be 95 cm (take pi= 3.14).

Explanation: We know that the circumference for a circle is given as,

C= 2πr

Substituting the values given in the question, we get

r = 95/ (2 X 3.14) cm

Hence, giving us, 

r = 15.13 cm

Explanation: We know that the circumference for a circle is given as,

C= 2πr

Substituting the values given in the question, we get

r = 95/ (2 X 3.14) cm

Hence, giving us, 

r = 15.13 cm

12 / 320

Category: Mensuration - Mathematics

Find the height of a regular cylinder whose radius is 14 cm and the total surface area is 4342 cm²  is (take pi= 22/7):

Explanation: For a given cylinder, we know that

Total surface area = 2πr (h + r)

Therefore, 

4342= 2 x 22/7 x 14 (h + 14) cm²2 

h = 35.34 cm

Explanation: For a given cylinder, we know that

Total surface area = 2πr (h + r)

Therefore, 

4342= 2 x 22/7 x 14 (h + 14) cm²2 

h = 35.34 cm

13 / 320

Category: Mensuration - Mathematics

The area of a rhombus is 360 cm²  and one of the diagonals is 12 cm. Find the other diagonal.

Explanation: Because the area of a rhombus is given as,

area = ½ X (product of lengths of the diagonals)

Therefore, we get,

360 = ½ X (12 x diagonal2)  cm2

diagonal2 = 60 cm

Explanation: Because the area of a rhombus is given as,

area = ½ X (product of lengths of the diagonals)

Therefore, we get,

360 = ½ X (12 x diagonal2)  cm2

diagonal2 = 60 cm

14 / 320

Category: Mensuration - Mathematics

How many cubes with an edge length of 2 cm can fit inside a cuboid with dimensions of length, breadth and height given to be 8 m, 6 m, and 10 m, respectively?

Explanation: Before we calculate the volume of the cuboid, we convert the length, breadth and height into mm from (using the conversion 1 m = 100 cm)

Therefore, 

l = 800 cm

b = 600 cm

h = 1000 cm

Using the formula 

vol = l x b x h

We get the volume of the cuboid as 480000000 cm3

The volume of one of the small cubes is 8 cm3 (using the formula edge x edge x edge)

Hence, the total number of cubes that can fit inside the cuboid is calculated as

tot no. of cubes = 480000000/8

= 60000000

Explanation: Before we calculate the volume of the cuboid, we convert the length, breadth and height into mm from (using the conversion 1 m = 100 cm)

Therefore, 

l = 800 cm

b = 600 cm

h = 1000 cm

Using the formula 

vol = l x b x h

We get the volume of the cuboid as 480000000 cm3

The volume of one of the small cubes is 8 cm3 (using the formula edge x edge x edge)

Hence, the total number of cubes that can fit inside the cuboid is calculated as

tot no. of cubes = 480000000/8

= 60000000

15 / 320

Category: Mensuration - Mathematics

Find the breadth of a cuboid when the volume is given to be 64 m³ for length and height given as 8 m and 2 m, respectively.

Explanation: The volume for a cuboid is given as,

vol = length x breadth x height

Substituting the above-given values and rearranging the equation gives us

breadth = 64/ (8 x 2) m

 = 4 m

Explanation: The volume for a cuboid is given as,

vol = length x breadth x height

Substituting the above-given values and rearranging the equation gives us

breadth = 64/ (8 x 2) m

 = 4 m

16 / 320

Category: Mensuration - Mathematics

A cuboidal box has its length, breadth and height given to be 10 cm, 5 cm and 15 cm, respectively. Find the total surface area for the cuboid

Explanation: The formula for total surface area for a cuboid is given as

TSA = 2 x [(l × b) + (b × h) + (h × l)]

Hence, upon substituting the values for length, breadth and height,

TSA = 2 x [(10 x 5)+ (5 x 15) + (15 x 10)]

= 1300 cm2

Explanation: The formula for total surface area for a cuboid is given as

TSA = 2 x [(l × b) + (b × h) + (h × l)]

Hence, upon substituting the values for length, breadth and height,

TSA = 2 x [(10 x 5)+ (5 x 15) + (15 x 10)]

= 1300 cm2

17 / 320

Category: Mensuration - Mathematics

Find the area of a rhombus whose diagonals are given to be of lengths 6 cm and 7 cm.

Explanation: Since the area of a rhombus is given to be

Area = ½ X (product of lengths of the diagonals)

Therefore, 

area = ½ X (6 X 7)  cm2

=  21 cm2

Explanation: Since the area of a rhombus is given to be

Area = ½ X (product of lengths of the diagonals)

Therefore, 

area = ½ X (6 X 7)  cm2

=  21 cm2

18 / 320

Category: H.C.F and L.C.M - Mathematics

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

 ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

 ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

19 / 320

Category: H.C.F and L.C.M - Mathematics

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Explanation:

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

20 / 320

Category: H.C.F and L.C.M - Mathematics

Which of the following has the most number of divisors?
Explanation:

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

Explanation:

99 = 1 x 3 x 3 x 11

101 = 1 x 101

176 = 1 x 2 x 2 x 2 x 2 x 11

182 = 1 x 2 x 7 x 13

So, divisors of 99 are 1, 3, 9, 11, 33, .99

Divisors of 101 are 1 and 101

Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176

Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.

Hence, 176 has the most number of divisors.

21 / 320

Category: H.C.F and L.C.M - Mathematics

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 Required number = (90 x 4) + 4   = 364.

Explanation:

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 Required number = (90 x 4) + 4   = 364.

22 / 320

Category: H.C.F and L.C.M - Mathematics

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

23 / 320

Category: H.C.F and L.C.M - Mathematics

The G.C.D. of 1.08, 0.36 and 0.9 is:

Explanation:

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

 H.C.F. of given numbers = 0.18.

Explanation:

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

 H.C.F. of given numbers = 0.18.

24 / 320

Category: H.C.F and L.C.M - Mathematics

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

Explanation:

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

 Number to be added = (60 - 37) = 23.

Explanation:

L.C.M. of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

 Number to be added = (60 - 37) = 23.

25 / 320

Category: H.C.F and L.C.M - Mathematics

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

 Larger number = (23 x 14) = 322.

Explanation:

Clearly, the numbers are (23 x 13) and (23 x 14).

 Larger number = (23 x 14) = 322.

26 / 320

Category: H.C.F and L.C.M - Mathematics

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

Explanation:

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Explanation:

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

27 / 320

Category: H.C.F and L.C.M - Mathematics

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

28 / 320

Category: H.C.F and L.C.M - Mathematics

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

Required number = (L.C.M. of 12, 15, 20, 54) + 8

= 540 + 8

= 548.

29 / 320

Category: H.C.F and L.C.M - Mathematics

The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

Explanation:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

Explanation:

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7

= 1008 + 7

= 1015

30 / 320

Category: H.C.F and L.C.M - Mathematics

Which of the following fraction is the largest ?

Explanation:

L.C.M. of 8, 16, 40 and 80 = 80.

7 = 70 ; 13 = 65 ; 31 = 62
8 80 16 80 40 80
Since, 70 > 65 > 63 > 62 , so 7 > 13 > 63 > 31
80 80 80 80 8 16 80 40
So, 7 is the largest.
8
Explanation:

L.C.M. of 8, 16, 40 and 80 = 80.

7 = 70 ; 13 = 65 ; 31 = 62
8 80 16 80 40 80
Since, 70 > 65 > 63 > 62 , so 7 > 13 > 63 > 31
80 80 80 80 8 16 80 40
So, 7 is the largest.
8

31 / 320

Category: H.C.F and L.C.M - Mathematics

Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

Explanation:

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = 551 = 19;    Third number = 1073 = 37.
29 29

 Required sum = (19 + 29 + 37) = 85.

Explanation:

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29;

First number = 551 = 19;    Third number = 1073 = 37.
29 29

 Required sum = (19 + 29 + 37) = 85.

32 / 320

Category: H.C.F and L.C.M - Mathematics

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.

So, the numbers 12 and 16.

L.C.M. of 12 and 16 = 48.

33 / 320

Category: H.C.F and L.C.M - Mathematics

Reduce 128352 to its lowest terms.
238368

34 / 320

Category: H.C.F and L.C.M - Mathematics

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:

Explanation:

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

 The required sum = 1 + 1 = a + b = 55 = 11
a b ab 600 120
Explanation:

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

 The required sum = 1 + 1 = a + b = 55 = 11
a b ab 600 120

35 / 320

Category: H.C.F and L.C.M - Mathematics

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

 Required number (9999 - 399) = 9600.

Explanation:

Greatest number of 4-digits is 9999.

L.C.M. of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399.

 Required number (9999 - 399) = 9600.

36 / 320

Category: H.C.F and L.C.M - Mathematics

Find the highest common factor of 36 and 84.

Explanation:

36 = 22 x 32

84 = 22 x 3 x 7

 H.C.F. = 22 x 3 = 12.

Explanation:

36 = 22 x 32

84 = 22 x 3 x 7

 H.C.F. = 22 x 3 = 12.

37 / 320

Category: H.C.F and L.C.M - Mathematics

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

Explanation:

Required number = H.C.F. of (1657 - 6) and (2037 - 5)

= H.C.F. of 1651 and 2032 = 127.

38 / 320

Category: H.C.F and L.C.M - Mathematics

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

Explanation:

L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30

----------------------------= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15

----------------------------Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5

Explanation:

L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30

----------------------------= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15

----------------------------Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5

39 / 320

Category: H.C.F and L.C.M - Mathematics

The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:

Explanation:

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

 The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

Explanation:

Let the numbers be 2x and 3x.

Then, their L.C.M. = 6x.

So, 6x = 48 or x = 8.

 The numbers are 16 and 24.

Hence, required sum = (16 + 24) = 40.

40 / 320

Category: H.C.F and L.C.M - Mathematics

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.

Explanation:

L.C.M. of 5, 6, 7, 8 = 840.

 Required number is of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k = 2.

 Required number = (840 x 2 + 3) = 1683.

41 / 320

Category: H.C.F and L.C.M - Mathematics

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

Explanation:
Other number = 11 x 7700 = 308.
275
Explanation:
Other number = 11 x 7700 = 308.
275

42 / 320

Category: H.C.F and L.C.M - Mathematics

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

Explanation:

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

 ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

 Greater number = 111.

Explanation:

Let the numbers be 37a and 37b.

Then, 37a x 37b = 4107

 ab = 3.

Now, co-primes with product 3 are (1, 3).

So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).

 Greater number = 111.

43 / 320

Category: H.C.F and L.C.M - Mathematics

The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:

Explanation:

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Explanation:

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

44 / 320

Category: H.C.F and L.C.M - Mathematics

The H.C.F. of 9 , 12 , 18 and 21 is:
10 25 35 40
Explanation:
Required H.C.F. = H.C.F. of 9, 12, 18, 21 = 3
L.C.M. of 10, 25, 35, 40 1400
Explanation:
Required H.C.F. = H.C.F. of 9, 12, 18, 21 = 3
L.C.M. of 10, 25, 35, 40 1400

45 / 320

Category: H.C.F and L.C.M - Mathematics

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1 = 16 times.
2
Explanation:

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds(2 minutes).

In 30 minutes, they will toll together 30 + 1 = 16 times.
2

46 / 320

Category: H.C.F and L.C.M - Mathematics

252 can be expressed as a product of primes as:

Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.

47 / 320

Category: H.C.F and L.C.M - Mathematics

Find the lowest common multiple of 24, 36 and 40.

Explanation:
2 | 24 - 36 - 40
--------------------
2 | 12 - 18 - 20
--------------------
2 | 6 - 9 - 10
-------------------
3 | 3 - 9 - 5
-------------------
| 1 - 3 - 5

L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
Explanation:
2 | 24 - 36 - 40
--------------------
2 | 12 - 18 - 20
--------------------
2 | 6 - 9 - 10
-------------------
3 | 3 - 9 - 5
-------------------
| 1 - 3 - 5

L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.

48 / 320

Category: Ratio and Proportion - Mathematics

If 40% of a number is equal to two-third of another number, what is the ratio of first number to the second number?

Explanation:
Let 40% of A = 2 B
3
Then, 40A = 2B
100 3
2A = 2B
5 3
A = 2 x 5 = 5
B 3 2 3

 A : B = 5 : 3.

Explanation:
Let 40% of A = 2 B
3
Then, 40A = 2B
100 3
2A = 2B
5 3
A = 2 x 5 = 5
B 3 2 3

 A : B = 5 : 3.

49 / 320

Category: Ratio and Proportion - Mathematics

Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is:

Explanation:

Let the numbers be 3x and 5x.

Then, 3x - 9 = 12
5x - 9 23

 23(3x - 9) = 12(5x - 9)

 9x = 99

 x = 11.

 The smaller number = (3 x 11) = 33.

Explanation:

Let the numbers be 3x and 5x.

Then, 3x - 9 = 12
5x - 9 23

 23(3x - 9) = 12(5x - 9)

 9x = 99

 x = 11.

 The smaller number = (3 x 11) = 33.

50 / 320

Category: Ratio and Proportion - Mathematics

A and B together have ₹ 1210. If  of A's amount is equal to  of B's amount, how much amount does B have?

Explanation:
4 A = 2 B
15 5
 A = 2 x 15 B
5 4
 A = 3 B
2
A = 3
B 2

 A : B = 3 : 2.

 B's share = Rs. 1210 x 2 = ₹ 484.
5
Explanation:
4 A = 2 B
15 5
 A = 2 x 15 B
5 4
 A = 3 B
2
A = 3
B 2

 A : B = 3 : 2.

 B's share = Rs. 1210 x 2 = ₹ 484.
5

51 / 320

Category: Ratio and Proportion - Mathematics

The ratio of the number of boys and girls in a college is 7 : 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?

Explanation:

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

120 x 7x and 110 x 8x
100 100
42x and 44x
5 5
 The required ratio = 42x : 44x = 21 : 22.
5 5
Explanation:

Originally, let the number of boys and girls in the college be 7x and 8x respectively.

Their increased number is (120% of 7x) and (110% of 8x).

120 x 7x and 110 x 8x
100 100
42x and 44x
5 5
 The required ratio = 42x : 44x = 21 : 22.
5 5

52 / 320

Category: Ratio and Proportion - Mathematics

The fourth proportional to 5, 8, 15 is:

Explanation:

Let the fourth proportional to 5, 8, 15 be x.

Then, 5 : 8 : 15 : x

 5x = (8 x 15)

x = (8 x 15) = 24.
5
Explanation:

Let the fourth proportional to 5, 8, 15 be x.

Then, 5 : 8 : 15 : x

 5x = (8 x 15)

x = (8 x 15) = 24.
5

53 / 320

Category: Ratio and Proportion - Mathematics

The sum of three numbers is 98. If the ratio of the first to second is 2 :3 and that of the second to the third is 5 : 8, then the second number is:

Explanation:

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 = 5 x 3 : 8 x 3 = 3 : 24
5 5 5
 A : B : C = 2 : 3 : 24 = 10 : 15 : 24
5
 B = 98 x 15 = 30.
49
Explanation:

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 = 5 x 3 : 8 x 3 = 3 : 24
5 5 5
 A : B : C = 2 : 3 : 24 = 10 : 15 : 24
5
 B = 98 x 15 = 30.
49

54 / 320

Category: Ratio and Proportion - Mathematics

If Rs. 782 be divided into three parts, proportional to  :  : , then the first part is:

Explanation:

Given ratio =  :  :  = 6 : 8 : 9.

 1st part = Rs. 782 x 6 = Rs. 204
23
Explanation:

Given ratio =  :  :  = 6 : 8 : 9.

 1st part = Rs. 782 x 6 = Rs. 204
23

55 / 320

Category: Ratio and Proportion - Mathematics

Salaries of Ravi and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is Sumit's salary?

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then, 2x + 4000 = 40
3x + 4000 57

 57(2x + 4000) = 40(3x + 4000)

 6x = 68,000

 3x = 34,000

Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

Explanation:

Let the original salaries of Ravi and Sumit be Rs. 2x and Rs. 3x respectively.

Then, 2x + 4000 = 40
3x + 4000 57

 57(2x + 4000) = 40(3x + 4000)

 6x = 68,000

 3x = 34,000

Sumit's present salary = (3x + 4000) = Rs.(34000 + 4000) = Rs. 38,000.

56 / 320

Category: Ratio and Proportion - Mathematics

In a mixture 60 litres, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is:

Explanation:
Quantity of milk = 60 x 2 litres = 40 litres.
3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

Then, milk : water = 40 .
20 + x
Now, 40 = 1
20 + x 2

 20 + x = 80

 x = 60.

 Quantity of water to be added = 60 litres.

Explanation:
Quantity of milk = 60 x 2 litres = 40 litres.
3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

Then, milk : water = 40 .
20 + x
Now, 40 = 1
20 + x 2

 20 + x = 80

 x = 60.

 Quantity of water to be added = 60 litres.

57 / 320

Category: Ratio and Proportion - Mathematics

If 0.75 : x :: 5 : 8, then x is equal to:

Explanation:
(x x 5) = (0.75 x 8)    x = 6 = 1.20
5
Explanation:
(x x 5) = (0.75 x 8)    x = 6 = 1.20
5

58 / 320

Category: Ratio and Proportion - Mathematics

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?

Explanation:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

140 x 5x , 150 x 7x and 175 x 8x
100 100 100
 7x, 21x and 14x.
2
 The required ratio = 7x : 21x : 14x
2

 14x : 21x : 28x

 2 : 3 : 4.

Explanation:

Originally, let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.

Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).

140 x 5x , 150 x 7x and 175 x 8x
100 100 100
 7x, 21x and 14x.
2
 The required ratio = 7x : 21x : 14x
2

 14x : 21x : 28x

 2 : 3 : 4.

59 / 320

Category: Ratio and Proportion - Mathematics

In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how many 5 p coins are there?

Explanation:

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values = Rs. 25x + 10 x 2x + 5 x 3x = Rs. 60x
100 100 100 100
60x = 30     x = 30 x 100 = 50.
100 60

Hence, the number of 5 p coins = (3 x 50) = 150.

Explanation:

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values = Rs. 25x + 10 x 2x + 5 x 3x = Rs. 60x
100 100 100 100
60x = 30     x = 30 x 100 = 50.
100 60

Hence, the number of 5 p coins = (3 x 50) = 150.

60 / 320

Category: Ratio and Proportion - Mathematics

A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets ₹ 1000 more than D, what is B's share?

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

 x = 1000.

 B's share = ₹ 2x = ₹ (2 x 1000) = ₹ 2000.

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

Then, 4x - 3x = 1000

 x = 1000.

 B's share = ₹ 2x = ₹ (2 x 1000) = ₹ 2000.

61 / 320

Category: Ratio and Proportion - Mathematics

Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is:

Explanation:

Let the third number be x.

Then, first number = 120% of x = 120x = 6x
100 5
Second number = 150% of x = 150x = 3x
100 2
 Ratio of first two numbers = 6x : 3x = 12x : 15x = 4 : 5.
5 2
Explanation:

Let the third number be x.

Then, first number = 120% of x = 120x = 6x
100 5
Second number = 150% of x = 150x = 3x
100 2
 Ratio of first two numbers = 6x : 3x = 12x : 15x = 4 : 5.
5 2

62 / 320

Category: Ratio and Proportion - Mathematics

The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?

Explanation:

Let A = 2k, B = 3k and C = 5k.

A's new salary = 115 of 2k = 115 x 2k = 23k
100 100 10
B's new salary = 110 of 3k = 110 x 3k = 33k
100 100 10
C's new salary = 120 of 5k = 120 x 5k = 6k
100 100
 New ratio 23k : 33k : 6k = 23 : 33 : 60
10 10
Explanation:

Let A = 2k, B = 3k and C = 5k.

A's new salary = 115 of 2k = 115 x 2k = 23k
100 100 10
B's new salary = 110 of 3k = 110 x 3k = 33k
100 100 10
C's new salary = 120 of 5k = 120 x 5k = 6k
100 100
 New ratio 23k : 33k : 6k = 23 : 33 : 60
10 10

63 / 320

Category: Decimals - Mathematics

Find the value of 35 – 2.54.

64 / 320

Category: Decimals - Mathematics

Which of the following is smaller?

65 / 320

Category: Decimals - Mathematics

Which of the following number can be placed in the tens column if the given number is 297.35?

66 / 320

Category: Decimals - Mathematics

What is the place value of 2 in the given decimal 924.75?

The correct answer is (b) tens. [1]
In the decimal number 924.75, the position of each digit relative to the decimal point determines its place value: [2, 3]
  • 9 is in the hundreds place ($900$).
  • ✅ 2 is in the tens place ($20$).
  • 4 is in the ones place ($4$).
  • . (Decimal Point)
  • 7 is in the tenths place ($\frac{7}{10}$ or $0.7$).
  • 5 is in the hundredths place ($\frac{5}{100}$ or $0.05$). [4, 5, 6, 7, 8]

Analysis of Options:

  • ❌ (a) ones: The digit in the ones place is 4.
  • ✅ (b) tens: The digit 2 is the second digit to the left of the decimal point, representing the tens place.
  • ❌ (c) tenth: The digit in the tenths place (first digit to the right of the decimal) is 7.
  • ❌ (d) hundredth: The digit in the hundredths place (second digit to the right of the decimal) is 5. [7, 8, 9, 10, 11]
The correct answer is (b) tens. [1]
In the decimal number 924.75, the position of each digit relative to the decimal point determines its place value: [2, 3]
  • 9 is in the hundreds place ($900$).
  • ✅ 2 is in the tens place ($20$).
  • 4 is in the ones place ($4$).
  • . (Decimal Point)
  • 7 is in the tenths place ($\frac{7}{10}$ or $0.7$).
  • 5 is in the hundredths place ($\frac{5}{100}$ or $0.05$). [4, 5, 6, 7, 8]

Analysis of Options:

  • ❌ (a) ones: The digit in the ones place is 4.
  • ✅ (b) tens: The digit 2 is the second digit to the left of the decimal point, representing the tens place.
  • ❌ (c) tenth: The digit in the tenths place (first digit to the right of the decimal) is 7.
  • ❌ (d) hundredth: The digit in the hundredths place (second digit to the right of the decimal) is 5. [7, 8, 9, 10, 11]

67 / 320

Category: Decimals - Mathematics

Find the value of 9.756 – 6.28.

68 / 320

Category: Decimals - Mathematics

The number 0.125 can be written as fractions in lowest terms:

69 / 320

Category: Decimals - Mathematics

The sum of 0.007 + 8.5 + 30.08 is:

70 / 320

Category: Decimals - Mathematics

5.008 can be written in words as:

The correct answer is (d) Five point zero zero eight.

Explanation

When writing a decimal in words using the "point" method, you simply name the whole number, say "point" for the decimal, and then list each digit to the right of the decimal individually:
  • 5 is "Five"
  • . is "point"
  • 0 is "zero"
  • 0 is "zero"
  • 8 is "eight"

Why the others are incorrect:

  • ❌ (a) Five thousand eight: This refers to the whole number 5,008, not the decimal 5.008.
  • ❌ (b) Five point eight: This refers to 5.8, missing the two zeros in the hundredths and thousandths places.
  • ❌ (c) Fifty point eight: This refers to 50.8.
Note: Another mathematically correct way to write this is "Five and eight thousandths", as the 8 is in the thousandths place.
The correct answer is (d) Five point zero zero eight.

Explanation

When writing a decimal in words using the "point" method, you simply name the whole number, say "point" for the decimal, and then list each digit to the right of the decimal individually:
  • 5 is "Five"
  • . is "point"
  • 0 is "zero"
  • 0 is "zero"
  • 8 is "eight"

Why the others are incorrect:

  • ❌ (a) Five thousand eight: This refers to the whole number 5,008, not the decimal 5.008.
  • ❌ (b) Five point eight: This refers to 5.8, missing the two zeros in the hundredths and thousandths places.
  • ❌ (c) Fifty point eight: This refers to 50.8.
Note: Another mathematically correct way to write this is "Five and eight thousandths", as the 8 is in the thousandths place.

71 / 320

Category: Decimals - Mathematics

What is the decimal expansion of 5/10?

The correct answer is (a) 0.5.
To find the decimal expansion of a fraction with a denominator of $10$, you move the decimal point of the numerator one place to the left.

Reasoning:

  1. Identify the fraction: $\frac{5}{10}$.
  2. Apply place value: Dividing a number by $10$ shifts its digits one position to the right relative to the decimal point.
  3. Calculate: $5 \div 10 = 0.5$.
In the context of the previous problems, this means the digit $5$ sits in the tenths place.

✅ Answer

The decimal expansion of $\frac{5}{10}$ is 0.5.
The correct answer is (a) 0.5.
To find the decimal expansion of a fraction with a denominator of $10$, you move the decimal point of the numerator one place to the left.

Reasoning:

  1. Identify the fraction: $\frac{5}{10}$.
  2. Apply place value: Dividing a number by $10$ shifts its digits one position to the right relative to the decimal point.
  3. Calculate: $5 \div 10 = 0.5$.
In the context of the previous problems, this means the digit $5$ sits in the tenths place.

✅ Answer

The decimal expansion of $\frac{5}{10}$ is 0.5.

72 / 320

Category: Decimals - Mathematics

Which of the following number can be placed in the tenth column if the given number is 297.35?

73 / 320

Category: Decimals - Mathematics

What is the place value of 5 in the given decimal 924.75?

The correct answer is (d) hundredth.
In the decimal 924.75, the 5 is the second digit to the right of the decimal point, which represents the hundredths place (or $\frac{5}{100}$).
The correct answer is (d) hundredth.
In the decimal 924.75, the 5 is the second digit to the right of the decimal point, which represents the hundredths place (or $\frac{5}{100}$).

74 / 320

Category: Decimals - Mathematics

32.549 > 32.458 because:

75 / 320

Category: Decimals - Mathematics

137 + 5/100 can be written in the decimal form as:

76 / 320

Category: Decimals - Mathematics

8888 m in Km can be written as:

77 / 320

Category: Decimals - Mathematics

4.19 m in cm can be written as:

78 / 320

Category: Decimals - Mathematics

Raju bought a book for ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?

79 / 320

Category: Decimals - Mathematics

Two tens and nine tenths in decimal form is given by:

80 / 320

Category: Decimals - Mathematics

600 + 2 + 8/10 can be written in decimal form as:

The correct answer is (b) 602.8.

1. Add whole numbers

First, combine the whole number parts:
$$600 + 2 = 602$$

2. Convert the fraction

Convert the fraction $\frac{8}{10}$ into its decimal equivalent:
$$\frac{8}{10} = 0.8$$

3. Combine both parts

Add the whole number and the decimal together:
602 + 0.8 = 602.8
The correct answer is (b) 602.8.

1. Add whole numbers

First, combine the whole number parts:
$$600 + 2 = 602$$

2. Convert the fraction

Convert the fraction $\frac{8}{10}$ into its decimal equivalent:
$$\frac{8}{10} = 0.8$$

3. Combine both parts

Add the whole number and the decimal together:
602 + 0.8 = 602.8

81 / 320

Category: Decimals - Mathematics

Write the following as decimals: “Two ones and five-tenths”.

82 / 320

Category: Decimals - Mathematics

Which of the following point lies between 0.1 and 0.2?

83 / 320

Category: Compound Interest - Mathematics

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

Explanation:
P 1 + 20 n > 2P       6 n > 2.
100 5
Now, 6 x 6 x 6 x 6 > 2.
5 5 5 5

So, n = 4 years.

Explanation:
P 1 + 20 n > 2P       6 n > 2.
100 5
Now, 6 x 6 x 6 x 6 > 2.
5 5 5 5

So, n = 4 years.

84 / 320

Category: Compound Interest - Mathematics

There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?

Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

 R = 100 x 60 = 10% p.a.
100 x 6

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

 C.I.
= Rs. 12000 x 1 + 10 3 - 1
100
= Rs. 12000 x 331
1000
= 3972.
Explanation:

Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

 R = 100 x 60 = 10% p.a.
100 x 6

Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

 C.I.
= Rs. 12000 x 1 + 10 3 - 1
100
= Rs. 12000 x 331
1000
= 3972.

85 / 320

Category: Compound Interest - Mathematics

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:

Explanation:

Let the sum be Rs. x. Then,

C.I. = x 1 + 4 2 - x = 676 x - x = 51 x.
100 625 625
S.I. = x x 4 x 2 = 2x .
100 25
51x - 2x = 1
625 25

 x = 625.

Explanation:

Let the sum be Rs. x. Then,

C.I. = x 1 + 4 2 - x = 676 x - x = 51 x.
100 625 625
S.I. = x x 4 x 2 = 2x .
100 25
51x - 2x = 1
625 25

 x = 625.

86 / 320

Category: Compound Interest - Mathematics

What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?

Explanation:
C.I. when interest
compounded yearly
= Rs. 5000 x 1 + 4 x 1 +  x 4
100 100
= Rs. 5000 x 26 x 51
25 50
= Rs. 5304.
C.I. when interest is
compounded half-yearly
= Rs. 5000 x 1 + 2 3
100
= Rs. 5000 x 51 x 51 x 51
50 50 50
= Rs. 5306.04

 Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Explanation:
C.I. when interest
compounded yearly
= Rs. 5000 x 1 + 4 x 1 +  x 4
100 100
= Rs. 5000 x 26 x 51
25 50
= Rs. 5304.
C.I. when interest is
compounded half-yearly
= Rs. 5000 x 1 + 2 3
100
= Rs. 5000 x 51 x 51 x 51
50 50 50
= Rs. 5306.04

 Difference = Rs. (5306.04 - 5304) = Rs. 2.04

87 / 320

Category: Compound Interest - Mathematics

The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:

Explanation:

Let the sum be Rs. P.

Then, P 1 + 10 2 - P = 525
100
P 11 2 - 1 = 525
10
 P = 525 x 100 = 2500.
21

 Sum = Rs . 2500.

So, S.I. = Rs. 2500 x 5 x 4 = Rs. 500
100
Explanation:

Let the sum be Rs. P.

Then, P 1 + 10 2 - P = 525
100
P 11 2 - 1 = 525
10
 P = 525 x 100 = 2500.
21

 Sum = Rs . 2500.

So, S.I. = Rs. 2500 x 5 x 4 = Rs. 500
100

88 / 320

Category: Compound Interest - Mathematics

What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

Explanation:
Amount
= Rs. 25000 x 1 + 12 3
100
= Rs. 25000 x 28 x 28 x 28
25 25 25
= Rs. 35123.20

 C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

Explanation:
Amount
= Rs. 25000 x 1 + 12 3
100
= Rs. 25000 x 28 x 28 x 28
25 25 25
= Rs. 35123.20

 C.I. = Rs. (35123.20 - 25000) = Rs. 10123.20

89 / 320

Category: Compound Interest - Mathematics

Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:

Explanation:
C.I.
= Rs. 4000 x 1 + 10 2 - 4000
100
= Rs. 4000 x 11 x 11 - 4000
10 10
= Rs. 840.
 Sum = Rs. 420 x 100 = Rs. 1750.
3 x 8
Explanation:
C.I.
= Rs. 4000 x 1 + 10 2 - 4000
100
= Rs. 4000 x 11 x 11 - 4000
10 10
= Rs. 840.
 Sum = Rs. 420 x 100 = Rs. 1750.
3 x 8

90 / 320

Category: Compound Interest - Mathematics

The compound interest on ₹ 30,000 at 7% per annum is ₹ 4347. The period (in years) is:

Explanation:

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000 1 + 7 n = 34347
100
107 n = 34347 = 11449 = 107 2
100 30000 10000 100

 n = 2 years.

Explanation:

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then, 30000 1 + 7 n = 34347
100
107 n = 34347 = 11449 = 107 2
100 30000 10000 100

 n = 2 years.

91 / 320

Category: Compound Interest - Mathematics

Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?

Explanation:
Amount
= Rs. 8000 x 1 + 5 2
100
= Rs. 8000 x 21 x 21
20 20
= Rs. 8820.
Explanation:
Amount
= Rs. 8000 x 1 + 5 2
100
= Rs. 8000 x 21 x 21
20 20
= Rs. 8820.

92 / 320

Category: Compound Interest - Mathematics

If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?

Explanation:
Sum = Rs. 50 x 100 = Rs. 500.
2 x 5
Amount
= Rs. 500 x 1 + 5 2
100
= Rs. 500 x 21 x 21
20 20
= Rs. 551.25

 C.I. = Rs. (551.25 - 500) = Rs. 51.25

Explanation:
Sum = Rs. 50 x 100 = Rs. 500.
2 x 5
Amount
= Rs. 500 x 1 + 5 2
100
= Rs. 500 x 21 x 21
20 20
= Rs. 551.25

 C.I. = Rs. (551.25 - 500) = Rs. 51.25

93 / 320

Category: Compound Interest - Mathematics

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:

Explanation:
Amount
= Rs. 1600 x 1 + 5 2 + 1600 x 1 + 5
2 x 100 2 x 100
= Rs. 1600 x 41 x 41 + 1600 x 41
40 40 40
= Rs. 1600 x 41 41 + 1
40 40
= Rs. 1600 x 41 x 81
40 x 40
= Rs. 3321.

 C.I. = Rs. (3321 - 3200) = Rs. 121

Explanation:
Amount
= Rs. 1600 x 1 + 5 2 + 1600 x 1 + 5
2 x 100 2 x 100
= Rs. 1600 x 41 x 41 + 1600 x 41
40 40 40
= Rs. 1600 x 41 41 + 1
40 40
= Rs. 1600 x 41 x 81
40 x 40
= Rs. 3321.

 C.I. = Rs. (3321 - 3200) = Rs. 121

94 / 320

Category: Compound Interest - Mathematics

The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is:

Explanation:
S.I. = Rs 1200 x 10 x 1 = Rs. 120.
100
C.I. = Rs. 1200 x 1 + 5 2 - 1200 = Rs. 123.
100

 Difference = Rs. (123 - 120) = Rs. 3.

Explanation:
S.I. = Rs 1200 x 10 x 1 = Rs. 120.
100
C.I. = Rs. 1200 x 1 + 5 2 - 1200 = Rs. 123.
100

 Difference = Rs. (123 - 120) = Rs. 3.

95 / 320

Category: Compound Interest - Mathematics

The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?

Explanation:
15000 x 1 + R 2 - 15000 - 15000 x R x 2 = 96
100 100
 15000 1 + R 2 - 1 - 2R = 96
100 100
 15000 (100 + R)2 - 10000 - (200 x R) = 96
10000
 R2 = 96 x 2 = 64
3

 R = 8.

 Rate = 8%.

Explanation:
15000 x 1 + R 2 - 15000 - 15000 x R x 2 = 96
100 100
 15000 1 + R 2 - 1 - 2R = 96
100 100
 15000 (100 + R)2 - 10000 - (200 x R) = 96
10000
 R2 = 96 x 2 = 64
3

 R = 8.

 Rate = 8%.

96 / 320

Category: Compound Interest - Mathematics

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:

Explanation:
Amount of Rs. 100 for 1 year
when compounded half-yearly
= Rs. 100 x 1 + 3 2 = Rs. 106.09
100

 Effective rate = (106.09 - 100)% = 6.09%

Explanation:
Amount of Rs. 100 for 1 year
when compounded half-yearly
= Rs. 100 x 1 + 3 2 = Rs. 106.09
100

 Effective rate = (106.09 - 100)% = 6.09%

97 / 320

Category: Compound Interest - Mathematics

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

Explanation:

Let the rate be R% p.a.

Then, 1200 x 1 + R 2 = 1348.32
100
1 + R 2 = 134832 = 11236
100 120000 10000
1 + R 2 = 106 2
100 100
 1 + R = 106
100 100

 R = 6%

Explanation:

Let the rate be R% p.a.

Then, 1200 x 1 + R 2 = 1348.32
100
1 + R 2 = 134832 = 11236
100 120000 10000
1 + R 2 = 106 2
100 100
 1 + R = 106
100 100

 R = 6%

98 / 320

Category: Time and Work - Mathematics

A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?

Explanation:

Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x     =>     x = ( 23 x 13 )     =>     x = 299 .
10 10
A's 1 day's work = 1 ;
23
B's 1 day's work = 10 .
299
(A + B)'s 1 day's work = ( 1 + 10 ) = 23 = 1 .
23 299 299 13

Therefore, A and B together can complete the work in 13 days.

Explanation:

Ratio of times taken by A and B = 100 : 130 = 10 : 13.

Suppose B takes x days to do the work.

Then, 10 : 13 :: 23 : x     =>     x = ( 23 x 13 )     =>     x = 299 .
10 10
A's 1 day's work = 1 ;
23
B's 1 day's work = 10 .
299
(A + B)'s 1 day's work = ( 1 + 10 ) = 23 = 1 .
23 299 299 13

Therefore, A and B together can complete the work in 13 days.

99 / 320

Category: Time and Work - Mathematics

A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

Explanation:
Whole work is done by A in ( 20 x 5 ) = 25 days.
4
Now, ( 1 - 4 ) i.e., 1 work is done by A and B in 3 days.
5 5

Whole work will be done by A and B in (3 x 5) = 15 days.

A's 1 day's work = 1 , (A + B)'s 1 day's work = 1 .
25 15
Therefore B's 1 day's work = ( 1 - 1 ) = 4 = 2 .
15 25 150 75
So, B alone would do the work in 75 = 37 1 days.
2 2
Explanation:
Whole work is done by A in ( 20 x 5 ) = 25 days.
4
Now, ( 1 - 4 ) i.e., 1 work is done by A and B in 3 days.
5 5

Whole work will be done by A and B in (3 x 5) = 15 days.

A's 1 day's work = 1 , (A + B)'s 1 day's work = 1 .
25 15
Therefore B's 1 day's work = ( 1 - 1 ) = 4 = 2 .
15 25 150 75
So, B alone would do the work in 75 = 37 1 days.
2 2

100 / 320

Category: Time and Work - Mathematics

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?

Explanation:
(P + Q + R)'s 1 hour's work = ( 1 + 1 + 1 ) = 37 .
8 10 12 120
Work done by P, Q and R in 2 hours = ( 37 x 2 ) = 37 .
120 60
Remaining work = ( 1 - 37 ) = 23 .
60 60
(Q + R)'s 1 hour's work = ( 1 + 1 ) = 11 .
10 12 60
Now, 11 work is done by Q and R in 1 hour.
60
So, 23 work will be done by Q and R in ( 60 x 23 ) = 23 hours = 2 hours.
60 11 60 11

So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.

Explanation:
(P + Q + R)'s 1 hour's work = ( 1 + 1 + 1 ) = 37 .
8 10 12 120
Work done by P, Q and R in 2 hours = ( 37 x 2 ) = 37 .
120 60
Remaining work = ( 1 - 37 ) = 23 .
60 60
(Q + R)'s 1 hour's work = ( 1 + 1 ) = 11 .
10 12 60
Now, 11 work is done by Q and R in 1 hour.
60
So, 23 work will be done by Q and R in ( 60 x 23 ) = 23 hours = 2 hours.
60 11 60 11

So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.

101 / 320

Category: Time and Work - Mathematics

A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:

Explanation:
Suppose A, B and C take x, x and x days respectively to finish the work.
2 3
Then, 1 + 2 + 3 = 1
x x x 2
6 = 1
x 2

 x = 12.

So, B takes (12/2) = 6 days to finish the work.

Explanation:
Suppose A, B and C take x, x and x days respectively to finish the work.
2 3
Then, 1 + 2 + 3 = 1
x x x 2
6 = 1
x 2

 x = 12.

So, B takes (12/2) = 6 days to finish the work.

102 / 320

Category: Time and Work - Mathematics

A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

Explanation:
A's 2 day's work = 1 x 2 = 1 .
20 10
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 .
20 30 60 60 10
Work done in 3 days = 1 + 1 = 1 .
10 10 5
Now, 1 work is done in 3 days.
5

 Whole work will be done in (3 x 5) = 15 days.

Explanation:
A's 2 day's work = 1 x 2 = 1 .
20 10
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 6 = 1 .
20 30 60 60 10
Work done in 3 days = 1 + 1 = 1 .
10 10 5
Now, 1 work is done in 3 days.
5

 Whole work will be done in (3 x 5) = 15 days.

103 / 320

Category: Time and Work - Mathematics

A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :

Explanation:
A's 1 day's work = 1 ;
15
B's 1 day's work = 1 ;
20
(A + B)'s 1 day's work = ( 1 + 1 ) = 7 .
15 20 60
(A + B)'s 4 day's work = ( 7 x 4 ) = 7 .
60 15
Therefore, Remaining work = ( 1 - 7 ) = 8 .
15 15
Explanation:
A's 1 day's work = 1 ;
15
B's 1 day's work = 1 ;
20
(A + B)'s 1 day's work = ( 1 + 1 ) = 7 .
15 20 60
(A + B)'s 4 day's work = ( 7 x 4 ) = 7 .
60 15
Therefore, Remaining work = ( 1 - 7 ) = 8 .
15 15

104 / 320

Category: Time and Work - Mathematics

10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?

Explanation:
1 woman's 1 day's work = 1
70
1 child's 1 day's work = 1
140
(5 women + 10 children)'s day's work = 5 + 10 = 1 + 1 = 1
70 140 14 14 7

 5 women and 10 children will complete the work in 7 days.

Explanation:
1 woman's 1 day's work = 1
70
1 child's 1 day's work = 1
140
(5 women + 10 children)'s day's work = 5 + 10 = 1 + 1 = 1
70 140 14 14 7

 5 women and 10 children will complete the work in 7 days.

105 / 320

Category: Time and Work - Mathematics

P can complete a work in 12 days working 8 hours a day. Q can complete the same work in 8 days working 10 hours a day. If both P and Q work together, working 8 hours a day, in how many days can they complete the work?

Explanation:

P can complete the work in (12 x 8) hrs. = 96 hrs.

Q can complete the work in (8 x 10) hrs. = 80 hrs.

 P's1 hour's work = 1 and Q's 1 hour's work = 1 .
96 80
(P + Q)'s 1 hour's work = 1 + 1 = 11 .
96 80 480
So, both P and Q will finish the work in 480 hrs.
11
 Number of days of 8 hours each = 480 x 1 = 60 days = 5 5 days.
11 8 11 11
Explanation:

P can complete the work in (12 x 8) hrs. = 96 hrs.

Q can complete the work in (8 x 10) hrs. = 80 hrs.

 P's1 hour's work = 1 and Q's 1 hour's work = 1 .
96 80
(P + Q)'s 1 hour's work = 1 + 1 = 11 .
96 80 480
So, both P and Q will finish the work in 480 hrs.
11
 Number of days of 8 hours each = 480 x 1 = 60 days = 5 5 days.
11 8 11 11

106 / 320

Category: Time and Work - Mathematics

A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?

Explanation:
C's 1 day's work = 1 - 1 + 1 = 1 - 7 = 1 .
3 6 8 3 24 24
A's wages : B's wages : C's wages = 1 : 1 : 1 = 4 : 3 : 1.
6 8 24
C's share (for 3 days) = Rs. 3 x 1 x 3200 = Rs. 400.
24
Explanation:
C's 1 day's work = 1 - 1 + 1 = 1 - 7 = 1 .
3 6 8 3 24 24
A's wages : B's wages : C's wages = 1 : 1 : 1 = 4 : 3 : 1.
6 8 24
C's share (for 3 days) = Rs. 3 x 1 x 3200 = Rs. 400.
24

107 / 320

Category: Time and Work - Mathematics

A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in :

Explanation:
(A + B)'s 1 day's work = ( 1 + 1 ) = 1 .
15 10 6
Work done by A and B in 2 days = ( 1 x 2 ) = 1 .
6 3
Remaining work = ( 1 - 1 ) = 2 .
3 3
Now, 1 work is done by A in 1 day.
15
Therefore 2 work will be done by a in ( 15 x 2 ) = 10 days.
3 3

Hence, the total time taken = (10 + 2) = 12 days.

Explanation:
(A + B)'s 1 day's work = ( 1 + 1 ) = 1 .
15 10 6
Work done by A and B in 2 days = ( 1 x 2 ) = 1 .
6 3
Remaining work = ( 1 - 1 ) = 2 .
3 3
Now, 1 work is done by A in 1 day.
15
Therefore 2 work will be done by a in ( 15 x 2 ) = 10 days.
3 3

Hence, the total time taken = (10 + 2) = 12 days.

108 / 320

Category: Time and Work - Mathematics

X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?

Explanation:
Work done by X in 8 days = ( 1 x 8 ) = 1 .
40 5
Remaining work = ( 1 - 1 ) = 4 .
5 5
Now, 4 work is done by Y in 16 days.
5
Whole work will be done by Y in ( 16 x 5 ) = 20 days.
4
Therefore X's 1 day's work = 1 , Y's 1 day's work = 1 .
40 20
(X + Y)'s 1 day's work = ( 1 + 1 ) = 3 .
40 20 40
Hence, X and Y will together complete the work in ( 40 ) = 13 1 days.
3 3
Explanation:
Work done by X in 8 days = ( 1 x 8 ) = 1 .
40 5
Remaining work = ( 1 - 1 ) = 4 .
5 5
Now, 4 work is done by Y in 16 days.
5
Whole work will be done by Y in ( 16 x 5 ) = 20 days.
4
Therefore X's 1 day's work = 1 , Y's 1 day's work = 1 .
40 20
(X + Y)'s 1 day's work = ( 1 + 1 ) = 3 .
40 20 40
Hence, X and Y will together complete the work in ( 40 ) = 13 1 days.
3 3

109 / 320

Category: Time and Work - Mathematics

A and B can do a job together in 7 days. A is 13/4 times as efficient as B. The same job can be done by A alone in :

Explanation:
(A's 1 day's work) : (B's 1 day's work) = 7 : 1   =   7 : 4.
4

Let A's and B's 1 day's work be 7x and 4x respectively.

Then, 7x + 4x = 1     =>     11x = 1     =>     x = 1 .
7 7 77
Therefore A's 1 day's work = ( 1 x 7 ) = 1 .
77 11
Explanation:
(A's 1 day's work) : (B's 1 day's work) = 7 : 1   =   7 : 4.
4

Let A's and B's 1 day's work be 7x and 4x respectively.

Then, 7x + 4x = 1     =>     11x = 1     =>     x = 1 .
7 7 77
Therefore A's 1 day's work = ( 1 x 7 ) = 1 .
77 11

110 / 320

Category: Time and Work - Mathematics

A can lay railway track between two given stations in 16 days and B can do the same job in 12 days. With help of C, they did the job in 4 days only. Then, C alone can do the job in:

Explanation:
(A + B + C)'s 1 day's work = 1 ,
4
A's 1 day's work = 1 ,
16
B's 1 day's work = 1 .
12
Therefore C's 1 day's work = 1 - ( 1 + 1 ) = ( 1 - 7 ) = 5 .
4 16 12 4 48 48
So, C alone can do the work in 48 = 9 3 days.
5 5
Explanation:
(A + B + C)'s 1 day's work = 1 ,
4
A's 1 day's work = 1 ,
16
B's 1 day's work = 1 .
12
Therefore C's 1 day's work = 1 - ( 1 + 1 ) = ( 1 - 7 ) = 5 .
4 16 12 4 48 48
So, C alone can do the work in 48 = 9 3 days.
5 5

111 / 320

Category: Time and Work - Mathematics

A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :

Explanation:
(A + B + C)'s 1 day's work = 1 ;
6
(A + B)'s 1 day's work = 1 ;
8
(B + C)'s 1 day's work = 1 .
12
Therefore (A + C)'s 1 day's work
= ( 2 x 1 ) - ( 1 + 1 )
6 8 12
= ( 1 - 5 (
3 24
= 3
24
= 1 .
8

So, A and C together will do the work in 8 days.

Explanation:
(A + B + C)'s 1 day's work = 1 ;
6
(A + B)'s 1 day's work = 1 ;
8
(B + C)'s 1 day's work = 1 .
12
Therefore (A + C)'s 1 day's work
= ( 2 x 1 ) - ( 1 + 1 )
6 8 12
= ( 1 - 5 (
3 24
= 3
24
= 1 .
8

So, A and C together will do the work in 8 days.

112 / 320

Category: Time and Work - Mathematics

A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?

Explanation:
(A + B)'s 20 day's work = ( 1 x 20 ) = 2 .
30 3
Remaining work = ( 1 - 2 ) = 1 .
3 3
Now, 1 work is done by A in 20 days.
3

Therefore, the whole work will be done by A in (20 x 3) = 60 days.

Explanation:
(A + B)'s 20 day's work = ( 1 x 20 ) = 2 .
30 3
Remaining work = ( 1 - 2 ) = 1 .
3 3
Now, 1 work is done by A in 20 days.
3

Therefore, the whole work will be done by A in (20 x 3) = 60 days.

113 / 320

Category: Time and Work - Mathematics

A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:

Explanation:
(A + B)'s 1 day's work = 1
10
C's 1 day's work = 1
50
(A + B + C)'s 1 day's work = 1 + 1 = 6 = 3 . .... (i)
10 50 50 25

A's 1 day's work = (B + C)'s 1 day's work .... (ii)

From (i) and (ii), we get: 2 x (A's 1 day's work) = 3
25
 A's 1 day's work = 3 .
50
 B's 1 day's work 1 - 3 = 2 = 1 .
10 50 50 25

So, B alone could do the work in 25 days.

Explanation:
(A + B)'s 1 day's work = 1
10
C's 1 day's work = 1
50
(A + B + C)'s 1 day's work = 1 + 1 = 6 = 3 . .... (i)
10 50 50 25

A's 1 day's work = (B + C)'s 1 day's work .... (ii)

From (i) and (ii), we get: 2 x (A's 1 day's work) = 3
25
 A's 1 day's work = 3 .
50
 B's 1 day's work 1 - 3 = 2 = 1 .
10 50 50 25

So, B alone could do the work in 25 days.

114 / 320

Category: Time and Work - Mathematics

Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

Explanation:

(20 x 16) women can complete the work in 1 day.

 1 woman's 1 day's work = 1 .
320

(16 x 15) men can complete the work in 1 day.

 1 man's 1 day's work = 1
240
So, required ratio
= 1 : 1
240 320
= 1 : 1
3 4
= 4 : 3 (cross multiplied)
Explanation:

(20 x 16) women can complete the work in 1 day.

 1 woman's 1 day's work = 1 .
320

(16 x 15) men can complete the work in 1 day.

 1 man's 1 day's work = 1
240
So, required ratio
= 1 : 1
240 320
= 1 : 1
3 4
= 4 : 3 (cross multiplied)

115 / 320

Category: Time and Work - Mathematics

X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?

Explanation:
Work done by X in 4 days = 1 x 4 = 1 .
20 5
Remaining work = 1 - 1 = 4 .
5 5
(X + Y)'s 1 day's work = 1 + 1 = 8 = 2 .
20 12 60 15
Now, 2 work is done by X and Y in 1 day.
15
So, 4 work will be done by X and Y in 15 x 4 = 6 days.
5 2 5

Hence, total time taken = (6 + 4) days = 10 days.

Explanation:
Work done by X in 4 days = 1 x 4 = 1 .
20 5
Remaining work = 1 - 1 = 4 .
5 5
(X + Y)'s 1 day's work = 1 + 1 = 8 = 2 .
20 12 60 15
Now, 2 work is done by X and Y in 1 day.
15
So, 4 work will be done by X and Y in 15 x 4 = 6 days.
5 2 5

Hence, total time taken = (6 + 4) days = 10 days.

116 / 320

Category: Time and Work - Mathematics

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Explanation:

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.

Then, 4x + 6y = 1 and 3x + 7y = 1 .
8 10
Solving the two equations, we get: x = 11 , y = 1
400 400
 1 woman's 1 day's work = 1 .
400
 10 women's 1 day's work = 1 x 10 = 1 .
400 40

Hence, 10 women will complete the work in 40 days.

Explanation:

Let 1 man's 1 day's work = x and 1 woman's 1 day's work = y.

Then, 4x + 6y = 1 and 3x + 7y = 1 .
8 10
Solving the two equations, we get: x = 11 , y = 1
400 400
 1 woman's 1 day's work = 1 .
400
 10 women's 1 day's work = 1 x 10 = 1 .
400 40

Hence, 10 women will complete the work in 40 days.

117 / 320

Category: Time and Work - Mathematics

A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?

Explanation:
B's 10 day's work = ( 1 x 10 ) = 2 .
15 3
Remaining work = ( 1 - 2 ) = 1 .
3 3
Now, 1 work is done by A in 1 day.
18
Therefore 1 work is done by A in ( 18 x 1 ) = 6 days.
3 3
Explanation:
B's 10 day's work = ( 1 x 10 ) = 2 .
15 3
Remaining work = ( 1 - 2 ) = 1 .
3 3
Now, 1 work is done by A in 1 day.
18
Therefore 1 work is done by A in ( 18 x 1 ) = 6 days.
3 3

118 / 320

Category: Time and Work - Mathematics

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

Explanation:

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2
Solving these two equations, we get : x = 1 and y = 1 .
100 200
(15 men + 20 boy)'s 1 day's work = 15 + 20 = 1 .
100 200 4

 15 men and 20 boys can do the work in 4 days.

Explanation:

Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.

Then, 6x + 8y = 1 and 26x + 48y = 1 .
10 2
Solving these two equations, we get : x = 1 and y = 1 .
100 200
(15 men + 20 boy)'s 1 day's work = 15 + 20 = 1 .
100 200 4

 15 men and 20 boys can do the work in 4 days.

119 / 320

Category: Time and Work - Mathematics

A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:

Explanation:

Ratio of times taken by A and B = 1 : 3.

The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.

If difference of time is 2 days, B takes 3 days.

If difference of time is 60 days, B takes 3 x 60 = 90 days.
2

So, A takes 30 days to do the work.

A's 1 day's work = 1
30
B's 1 day's work = 1
90
(A + B)'s 1 day's work = 1 + 1 = 4 = 2
30 90 90 45
 A and B together can do the work in 45 = 22 1 days.
2 2
Explanation:

Ratio of times taken by A and B = 1 : 3.

The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.

If difference of time is 2 days, B takes 3 days.

If difference of time is 60 days, B takes 3 x 60 = 90 days.
2

So, A takes 30 days to do the work.

A's 1 day's work = 1
30
B's 1 day's work = 1
90
(A + B)'s 1 day's work = 1 + 1 = 4 = 2
30 90 90 45
 A and B together can do the work in 45 = 22 1 days.
2 2

120 / 320

Category: Time and Work - Mathematics

A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?

Explanation:
2(A + B + C)'s 1 day's work = ( 1 + 1 + 1 ) = 15 = 1 .
30 24 20 120 8
Therefore, (A + B + C)'s 1 day's work = 1 = 1 .
2 x 8 16
Work done by A, B, C in 10 days = 10 = 5 .
16 8
Remaining work = ( 1 - 5 ) = 3 .
8 8
A's 1 day's work = ( 1 - 1 ) = 1 .
16 24 48
Now, 1 work is done by A in 1 day.
48
So, 3 work will be done by A in ( 48 x 3 ) = 18 days.
8 8
Explanation:
2(A + B + C)'s 1 day's work = ( 1 + 1 + 1 ) = 15 = 1 .
30 24 20 120 8
Therefore, (A + B + C)'s 1 day's work = 1 = 1 .
2 x 8 16
Work done by A, B, C in 10 days = 10 = 5 .
16 8
Remaining work = ( 1 - 5 ) = 3 .
8 8
A's 1 day's work = ( 1 - 1 ) = 1 .
16 24 48
Now, 1 work is done by A in 1 day.
48
So, 3 work will be done by A in ( 48 x 3 ) = 18 days.
8 8

121 / 320

Category: Time and Work - Mathematics

A, B and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in:

Explanation:
Formula: If A can do a piece of work in n days, then A's 1 day's work = 1 .
n
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 7 .
24 6 12 24
Formula: If A's 1 day's work = 1 , then A can finish the work in n days.
n
So, all the three together will complete the job in 24  days = 3 3 days.
7 7
Explanation:
Formula: If A can do a piece of work in n days, then A's 1 day's work = 1 .
n
(A + B + C)'s 1 day's work = 1 + 1 + 1 = 7 .
24 6 12 24
Formula: If A's 1 day's work = 1 , then A can finish the work in n days.
n
So, all the three together will complete the job in 24  days = 3 3 days.
7 7

122 / 320

Category: Time and Work - Mathematics

Ravi and Kumar are working on an assignment. Ravi takes 6 hours to type 32 pages on a computer, while Kumar takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

Explanation:
Number of pages typed by Ravi in 1 hour = 32 = 16 .
6 3
Number of pages typed by Kumar in 1 hour = 40 = 8.
5
Number of pages typed by both in 1 hour = ( 16 + 8 ) = 40 .
3 3
Therefore Time taken by both to type 110 pages = ( 110 x 3 ) hours
40
= 8 1 hours (or) 8 hours 15 minutes.
4
Explanation:
Number of pages typed by Ravi in 1 hour = 32 = 16 .
6 3
Number of pages typed by Kumar in 1 hour = 40 = 8.
5
Number of pages typed by both in 1 hour = ( 16 + 8 ) = 40 .
3 3
Therefore Time taken by both to type 110 pages = ( 110 x 3 ) hours
40
= 8 1 hours (or) 8 hours 15 minutes.
4

123 / 320

Category: Time and Work - Mathematics

A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?

Explanation:
A's 1 hour's work = 1 ;
4
(B + C)'s 1 hour's work = 1 ;
3
(A + C)'s 1 hour's work = 1 .
2
(A + B + C)'s 1 hour's work = ( 1 + 1 ) = 7 .
4 3 12
B's 1 hour's work = ( 7 - 1 ) = 1 .
12 2 12

Therefore B alone will take 12 hours to do the work.

Explanation:
A's 1 hour's work = 1 ;
4
(B + C)'s 1 hour's work = 1 ;
3
(A + C)'s 1 hour's work = 1 .
2
(A + B + C)'s 1 hour's work = ( 1 + 1 ) = 7 .
4 3 12
B's 1 hour's work = ( 7 - 1 ) = 1 .
12 2 12

Therefore B alone will take 12 hours to do the work.

124 / 320

Category: Time and Work - Mathematics

Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:

Explanation:

Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x     x = 4 x 20
5

 x = 16 days.

Hence, Tanya takes 16 days to complete the work.

Explanation:

Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.

Suppose Tanya takes x days to do the work.

5 : 4 :: 20 : x     x = 4 x 20
5

 x = 16 days.

Hence, Tanya takes 16 days to complete the work.

125 / 320

Category: Time and Work - Mathematics

A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in :

Explanation:

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

B's 1 day's work = 1 .
12
Therefore A's 1 day's work = 1 ; (2 times of B's work)
6
(A + B)'s 1 day's work = ( 1 + 1 ) = 3 = 1 .
6 12 12 4

So, A and B together can finish the work in 4 days.

Explanation:

Ratio of rates of working of A and B = 2 : 1.

So, ratio of times taken = 1 : 2.

B's 1 day's work = 1 .
12
Therefore A's 1 day's work = 1 ; (2 times of B's work)
6
(A + B)'s 1 day's work = ( 1 + 1 ) = 3 = 1 .
6 12 12 4

So, A and B together can finish the work in 4 days.

126 / 320

Category: Time and Work - Mathematics

A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:

Explanation:
(B + C)'s 1 day's work = 1 + 1 = 7 .
9 12 36
Work done by B and C in 3 days = 7 x 3 = 7 .
36 12
Remaining work = 1 - 7 = 5 .
12 12
Now, 1 work is done by A in 1 day.
24
So, 5 work is done by A in 24 x 5 = 10 days.
12 12
Explanation:
(B + C)'s 1 day's work = 1 + 1 = 7 .
9 12 36
Work done by B and C in 3 days = 7 x 3 = 7 .
36 12
Remaining work = 1 - 7 = 5 .
12 12
Now, 1 work is done by A in 1 day.
24
So, 5 work is done by A in 24 x 5 = 10 days.
12 12

127 / 320

Category: Time and Work - Mathematics

A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?

Explanation:

Let A's 1 day's work = x and B's 1 day's work = y.

Then, x + y = 1 and 16x + 44y = 1.
30
Solving these two equations, we get: x = 1 and y = 1
60 60
 B's 1 day's work = 1 .
60

Hence, B alone shall finish the whole work in 60 days.

Explanation:

Let A's 1 day's work = x and B's 1 day's work = y.

Then, x + y = 1 and 16x + 44y = 1.
30
Solving these two equations, we get: x = 1 and y = 1
60 60
 B's 1 day's work = 1 .
60

Hence, B alone shall finish the whole work in 60 days.

128 / 320

Category: Trigonometry - Mathematics

If cos X = ⅔ then tan X is equal to:

Explanation: By trigonometry identities, we know:

1 + tan2X = sec2X

And sec X = 1/cos X = 1/(⅔) = 3/2

Hence,

1 + tan2X = (3/2)2 = 9/4

tan2X = (9/4) – 1 = 5/4

tan X = √5/2

Explanation: By trigonometry identities, we know:

1 + tan2X = sec2X

And sec X = 1/cos X = 1/(⅔) = 3/2

Hence,

1 + tan2X = (3/2)2 = 9/4

tan2X = (9/4) – 1 = 5/4

tan X = √5/2

129 / 320

Category: Trigonometry - Mathematics

If cos 9α = sinα and 9α < 90°, then the value of tan 5α is

Explanation:

Given,

cos 9α = sin α and 9α < 90°

That means, 9α is an acute angle.

cos 9α = cos(90° – α)

9α = 90° – α

9α + α = 90°

10α = 90°

α = 9°

tan 5α = tan(5 × 9°) = tan 45° = 1

Explanation:

Given,

cos 9α = sin α and 9α < 90°

That means, 9α is an acute angle.

cos 9α = cos(90° – α)

9α = 90° – α

9α + α = 90°

10α = 90°

α = 9°

tan 5α = tan(5 × 9°) = tan 45° = 1

130 / 320

Category: Trigonometry - Mathematics

The value of tan 60°/cot 30° is equal to:

Explanation: tan 60° = √3 and cot 30° = √3

Hence, tan 60°/cot 30° = √3/√3 = 1

Explanation: tan 60° = √3 and cot 30° = √3

Hence, tan 60°/cot 30° = √3/√3 = 1

131 / 320

Category: Trigonometry - Mathematics

The value of (tan 1° tan 2° tan 3° … tan 89°) is

Explanation:

tan 1° tan 2° tan 3°…tan 89°

= [tan 1° tan 2°…tan 44°] tan 45° [tan (90° – 44°) tan (90° – 43°)…tan (90° – 1°)]

= [tan 1° tan 2°…tan 44°] [cot 44° cot 43°…cot 1°] × [tan 45°]

= [(tan 1°× cot 1°) (tan 2°× cot 2°)…(tan 44°× cot 44°)] × [tan 45°]

= 1 × 1 × 1 × 1 × …× 1     {since tan A × cot A = 1 and tan 45° = 1}

= 1   

Explanation:

tan 1° tan 2° tan 3°…tan 89°

= [tan 1° tan 2°…tan 44°] tan 45° [tan (90° – 44°) tan (90° – 43°)…tan (90° – 1°)]

= [tan 1° tan 2°…tan 44°] [cot 44° cot 43°…cot 1°] × [tan 45°]

= [(tan 1°× cot 1°) (tan 2°× cot 2°)…(tan 44°× cot 44°)] × [tan 45°]

= 1 × 1 × 1 × 1 × …× 1     {since tan A × cot A = 1 and tan 45° = 1}

= 1   

132 / 320

Category: Trigonometry - Mathematics

If cos X = a/b, then sin X is equal to:

Explanation: cos X = a/b

By trigonometry identities, we know that:

sin2X + cos2X = 1

sin2X = 1 – cos2X = 1-(a/b)2

sin X = √(b2-a2)/b

Explanation: cos X = a/b

By trigonometry identities, we know that:

sin2X + cos2X = 1

sin2X = 1 – cos2X = 1-(a/b)2

sin X = √(b2-a2)/b

133 / 320

Category: Trigonometry - Mathematics

If sin A = 1/2 , then the value of cot A is

Explanation:

Given,

sin A = 1/2

cos2A = 1 – sin2A

= 1 – (1/2)2

= 1 – (1/4)

= (4 – 1)/4

= 3/4

cos A = √(3/4) = √3/2

cot A = cos A/sin A = (√3/2)/(1/2) = √3

Explanation:

Given,

sin A = 1/2

cos2A = 1 – sin2A

= 1 – (1/2)2

= 1 – (1/4)

= (4 – 1)/4

= 3/4

cos A = √(3/4) = √3/2

cot A = cos A/sin A = (√3/2)/(1/2) = √3

134 / 320

Category: Trigonometry - Mathematics

The value of sin 60° cos 30° + sin 30° cos 60° is:

Explanation: sin 60° = √3/2, sin 30° = ½, cos 60° = ½ and cos 30° = √3/2

Therefore,

(√3/2) x (√3/2) + (½) x (½)

= (3/4) + (1/4)

= 4/4

= 1

Explanation: sin 60° = √3/2, sin 30° = ½, cos 60° = ½ and cos 30° = √3/2

Therefore,

(√3/2) x (√3/2) + (½) x (½)

= (3/4) + (1/4)

= 4/4

= 1

135 / 320

Category: Trigonometry - Mathematics

tan 30°/(1 + tan230°) =

Explanation: tan 30° = 1/√3

Putting this value we get;

[2(1/√3)]/[1 + (1/√3)2] = (2/√3)/(4/3) = 6/4√3 = √3/2 = sin 60°

Explanation: tan 30° = 1/√3

Putting this value we get;

[2(1/√3)]/[1 + (1/√3)2] = (2/√3)/(4/3) = 6/4√3 = √3/2 = sin 60°

136 / 320

Category: Trigonometry - Mathematics

The value of (sin 45° + cos 45°) is

Explanation:

sin 45° + cos 45° = (1/√2) + (1/√2)

= (1 + 1)/√2

= 2/√2

= (√2 . √2)/√2

= √2

Explanation:

sin 45° + cos 45° = (1/√2) + (1/√2)

= (1 + 1)/√2

= 2/√2

= (√2 . √2)/√2

= √2

137 / 320

Category: Trigonometry - Mathematics

If cos(α + β) = 0, then sin(α – β) can be reduced to

Explanation:

Given,

cos(α + β) = 0

cos(α + β) = cos 90°

⇒ α + β = 90°

⇒ α = 90° – β 

sin(α – β) = sin(90° – β – β)

= sin(90° – 2β) 

= cos 2β    {since sin(90° – A) = cos A}

Explanation:

Given,

cos(α + β) = 0

cos(α + β) = cos 90°

⇒ α + β = 90°

⇒ α = 90° – β 

sin(α – β) = sin(90° – β – β)

= sin(90° – 2β) 

= cos 2β    {since sin(90° – A) = cos A}

138 / 320

Category: Trigonometry - Mathematics

If ∆ABC is right angled at C, then the value of cos(A+B) is

Explanation:

Given that in a right triangle ABC, ∠C = 90°.

We know that the sum of the three angles is equal to 180°.

∠A + ∠B + ∠C = 180° 

∠A + ∠B + 90° = 180° (∵ ∠C = 90° ) 

∠A + ∠B = 90° 

Now, cos(A+B) = cos 90° = 0

Explanation:

Given that in a right triangle ABC, ∠C = 90°.

We know that the sum of the three angles is equal to 180°.

∠A + ∠B + ∠C = 180° 

∠A + ∠B + 90° = 180° (∵ ∠C = 90° ) 

∠A + ∠B = 90° 

Now, cos(A+B) = cos 90° = 0

139 / 320

Category: Trigonometry - Mathematics

(Sin 30°+cos 60°)-(sin 60° + cos 30°) is equal to:

Explanation: sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½

Putting these values, we get:

(½+½)-(√3/2+√3/2)

= 1 – [(2√3)/2]

= 1 – √3

Explanation: sin 30° = ½, sin 60° = √3/2, cos 30° = √3/2 and cos 60° = ½

Putting these values, we get:

(½+½)-(√3/2+√3/2)

= 1 – [(2√3)/2]

= 1 – √3

140 / 320

Category: Trigonometry - Mathematics

sin (90° – A) and cos A are:

Explanation: By trigonometry identities.

Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}

Explanation: By trigonometry identities.

Sin (90°-A) = cos A {since 90°-A comes in the first quadrant of unit circle}

141 / 320

Category: Trigonometry - Mathematics

1 – cos2A is equal to:

Explanation: We know, by trigonometry identities,

sin2A + cos2A = 1

1 – cos2A = sin2A

Explanation: We know, by trigonometry identities,

sin2A + cos2A = 1

1 – cos2A = sin2A

142 / 320

Category: Trigonometry - Mathematics

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is

Explanation:

[cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)]

= cosec [90° – (15° – θ)] – sec (15° – θ) – tan (55° + θ) + cot [90° – (55° + θ)] 

We know that cosec (90° – θ) = sec θ and cot(90° – θ) = tan θ.

= sec (15° – θ) – sec (15° – θ) – tan (55° + θ) + tan (55° + θ) 

= 0

Explanation:

[cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)]

= cosec [90° – (15° – θ)] – sec (15° – θ) – tan (55° + θ) + cot [90° – (55° + θ)] 

We know that cosec (90° – θ) = sec θ and cot(90° – θ) = tan θ.

= sec (15° – θ) – sec (15° – θ) – tan (55° + θ) + tan (55° + θ) 

= 0

143 / 320

Category: Trigonometry - Mathematics

sin 2A = 2 sin A is true when A =

Explanation: sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 0

Explanation: sin 2A = sin 0° = 0

2 sin A = 2 sin 0° = 0

144 / 320

Category: Trigonometry - Mathematics

The value of the expression sin6θ + cos6θ + 3 sin2θ cos2θ is

Explanation:

We know that, sin2θ + cos2θ = 1

Taking cube on both sides,

(sin2θ + cos2θ)3 = 1

(sin2θ)3 + (cos2θ)3 + 3 sin2θ cos2θ (sin2θ + cos2θ) = 1

sin6θ + cos6θ + 3 sin2θ cos2θ = 1

Explanation:

We know that, sin2θ + cos2θ = 1

Taking cube on both sides,

(sin2θ + cos2θ)3 = 1

(sin2θ)3 + (cos2θ)3 + 3 sin2θ cos2θ (sin2θ + cos2θ) = 1

sin6θ + cos6θ + 3 sin2θ cos2θ = 1

145 / 320

Category: Trigonometry - Mathematics

If sin A + sin2A = 1, then the value of the expression (cos2A + cos4A) is

Explanation:

Given,

sin A + sin2A = 1

sin A = 1 – sin2A

sin A = cos2A {since sin2θ + cos2θ = 1}

Squaring on both sides,

sin2A = (cos2A)2

1 – cos2A = cos4A

⇒ cos2A + cos4A = 1

Explanation:

Given,

sin A + sin2A = 1

sin A = 1 – sin2A

sin A = cos2A {since sin2θ + cos2θ = 1}

Squaring on both sides,

sin2A = (cos2A)2

1 – cos2A = cos4A

⇒ cos2A + cos4A = 1

146 / 320

Category: Trigonometry - Mathematics

In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is:

Explanation: AB = 24 cm and BC = 7 cm

tan C = Opposite side/Adjacent side

tan C = 24/7

Explanation: AB = 24 cm and BC = 7 cm

tan C = Opposite side/Adjacent side

tan C = 24/7

147 / 320

Category: Algebra - Mathematics

Solve the logarithmic equation: log2(x-1) +log2(x+1) = 2

Explanation

We are given:
log₂(x − 1) + log₂(x + 1) = 2

Using the product rule:
log₂[(x − 1)(x + 1)] = 2

Simplify:
log₂(x² − 1) = 2

Convert to exponential form:
x² − 1 = 2² = 4

x² = 5

x = ±√5

Now apply the domain condition:
x − 1 > 0 and x + 1 > 0 ⇒ x > 1

So, the valid solution is:
x = √5

Explanation

We are given:
log₂(x − 1) + log₂(x + 1) = 2

Using the product rule:
log₂[(x − 1)(x + 1)] = 2

Simplify:
log₂(x² − 1) = 2

Convert to exponential form:
x² − 1 = 2² = 4

x² = 5

x = ±√5

Now apply the domain condition:
x − 1 > 0 and x + 1 > 0 ⇒ x > 1

So, the valid solution is:
x = √5

148 / 320

Category: Algebra - Mathematics

What is the order of operations in algebra?

Explanation

The order of operations is Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

Explanation

The order of operations is Parentheses, Exponents, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).

149 / 320

Category: Algebra - Mathematics

Solve for x:

2x+5=15

Explanation

Subtract 5 from both sides, then divide by 2.

Explanation

Subtract 5 from both sides, then divide by 2.

150 / 320

Category: Algebra - Mathematics

Solve the system of equations:

2x + y = 5

x - y = 1

Explanation

Use substitution or elimination to solve the system.

Explanation

Use substitution or elimination to solve the system.

151 / 320

Category: Algebra - Mathematics

The solution of the system x + y − 3x = −6 ,  − 7y + 7z = 7 , 3z = 9  is

152 / 320

Category: Algebra - Mathematics

What is the definition of a variable in algebra?

Explanation

In algebra, a variable is a symbol used to represent an unknown or changing quantity.

Explanation

In algebra, a variable is a symbol used to represent an unknown or changing quantity.

153 / 320

Category: Algebra - Mathematics

Which of the following is the complex conjugate of 3 - 4i?

Explanation

The complex conjugate of a+bi is a−bi.

Explanation

The complex conjugate of a+bi is a−bi.

154 / 320

Category: Algebra - Mathematics

Solve the inequality:

2x-7<5

Explanation

Add 7 to both sides, then divide by 2.

Explanation

Add 7 to both sides, then divide by 2.

155 / 320

Category: Algebra - Mathematics

Factor: x2 - 4

Explanation

This is a difference of squares.

Explanation

This is a difference of squares.

156 / 320

Category: Algebra - Mathematics

If (x - 6) is the HCF of x2 - 2x - 24 and x2 - kx - 6 then the value of k is

157 / 320

Category: Algebra - Mathematics

Define the term "expression" in algebra.

Explanation

An algebraic expression is a combination of variables, numbers, and operators, but it does not have an equal sign.

Explanation

An algebraic expression is a combination of variables, numbers, and operators, but it does not have an equal sign.

158 / 320

Category: Algebra - Mathematics

Find the solution to the equation

1/x + 2/(x-1) = 3/(x-2)

159 / 320

Category: Algebra - Mathematics

Explain the concept of a quadratic equation.

Explanation

A quadratic equation is a second-degree polynomial

Explanation

A quadratic equation is a second-degree polynomial

160 / 320

Category: Algebra - Mathematics

What is the sum of the roots of the equation x2 - 6x + 9 = 0

161 / 320

Category: Algebra - Mathematics

What is the range of the function

f(x) = 1/(x-2)

Explanation

The function is defined for all x such that x ≠ 2

Explanation

The function is defined for all x such that x ≠ 2

162 / 320

Category: Algebra - Mathematics

In algebraic expressions, what does the term "coefficient" represent?

Explanation

The coefficient is the numerical factor of a term in an algebraic expression.

Explanation

The coefficient is the numerical factor of a term in an algebraic expression.

163 / 320

Category: Algebra - Mathematics

What does the absolute value of a number represent?

Explanation

The absolute value of a number represents its distance from zero on the number line.

Explanation

The absolute value of a number represents its distance from zero on the number line.

164 / 320

Category: Algebra - Mathematics

Solve for x: |2x - 5| = 7

165 / 320

Category: Algebra - Mathematics

Find the solution to the equation

(x+3) / (x-1) = 2/(x-2)

166 / 320

Category: Algebra - Mathematics

Find the roots of the equation x2 + 4 = 0

Explanation

The equation has no real roots, but it has complex roots x=2i and  x=−2i

Explanation

The equation has no real roots, but it has complex roots x=2i and  x=−2i

167 / 320

Category: Algebra - Mathematics

Define the term "exponential function" in algebra.

Explanation

An exponential function models repeated multiplication.

Explanation

An exponential function models repeated multiplication.

168 / 320

Category: Algebra - Mathematics

Find the solution to the equation

sin(x) + cos(x) = 0 for 0 ≤ x ≤ 2π

169 / 320

Category: Algebra - Mathematics

Solve for x: log4(x) + log4(x+1) = 2

170 / 320

Category: Algebra - Mathematics

Define the term "conjugate" in algebra and its role in simplifying expressions.

Explanation

The conjugate of a binomial is obtained by changing the sign between the terms.

Explanation

The conjugate of a binomial is obtained by changing the sign between the terms.

171 / 320

Category: Algebra - Mathematics

A system of three linear equations in three variables is inconsistent if their planes

A system of three linear equations in three variables is inconsistent if their planes

A system of three linear equations in three variables is inconsistent if their planes

172 / 320

Category: Algebra - Mathematics

Solve the system of equations:

3x + 2y =10

x - y = 1

173 / 320

Category: Algebra - Mathematics

Simplify: 3(2x-4) + 5

Explanation

Distribute the 3, then combine like terms.

Explanation

Distribute the 3, then combine like terms.

174 / 320

Category: Algebra - Mathematics

Simplify the expression:   √18 - √8

Explanation

Factor the numbers under the square roots.

Explanation

Factor the numbers under the square roots.

175 / 320

Category: Algebra - Mathematics

Solve the quadratic equation: 3x2 - 5x + 2 =0

Explanation

By using the quadratic formula

Explanation

By using the quadratic formula

176 / 320

Category: Algebra - Mathematics

Solve the equation 2x - 8 = 0 for x

177 / 320

Category: Algebra - Mathematics

What is the purpose of factoring in algebra?

Explanation

Factoring is used to simplify algebraic expressions by expressing them as products of simpler expressions.

Explanation

Factoring is used to simplify algebraic expressions by expressing them as products of simpler expressions.

178 / 320

Category: Algebra - Mathematics

Evaluate: 3⁄4 × 2⁄3

Explanation

Multiply the numerators and denominators.

Explanation

Multiply the numerators and denominators.

179 / 320

Category: Algebra - Mathematics

Simplify the expression: ∛27x9

180 / 320

Category: Algebra - Mathematics

Discuss the properties and applications of complex numbers in algebra.

181 / 320

Category: Algebra - Mathematics

Solve for x: 2x-1 = 8

Explanation

Rewrite the equation using the base 2 to find x.

Explanation

Rewrite the equation using the base 2 to find x.

182 / 320

Category: Algebra - Mathematics

Solve the system of equations using matrices:

2x + y = 5

3x - 2y = 4

183 / 320

Category: Algebra - Mathematics

Define the term "eigenvalue" in the context of linear algebra.

Explanation

Eigenvalues represent the stretching or compressing factor in a linear transformation.

Explanation

Eigenvalues represent the stretching or compressing factor in a linear transformation.

184 / 320

Category: Algebra - Mathematics

 is

185 / 320

Category: Algebra - Mathematics

What is the difference between a binomial and a trinomial?

Explanation

A binomial has two terms, while a trinomial has three terms.

Explanation

A binomial has two terms, while a trinomial has three terms.

Explanation

 

186 / 320

Category: Algebra - Mathematics

Find the solution to the equation (x2 + 4x + 4)/(x+2) = 0

Explanation

Factor the numerator and cancel common factors.

Explanation

Factor the numerator and cancel common factors.

187 / 320

Category: Algebra - Mathematics

Find the roots of the quadratic equation: 2x2 - 5x + 3 = 0

188 / 320

Category: Algebra - Mathematics

Solve for: e2x - 5ex + 6 = 0

Explanation

Substitute ex = t to convert the equation to a quadratic form.

Explanation

Substitute ex = t to convert the equation to a quadratic form.

189 / 320

Category: Algebra - Mathematics

Solve for x: 2√x = 8

Explanation

Square both sides of the equation to solve for x

Explanation

Square both sides of the equation to solve for x

190 / 320

Category: Algebra - Mathematics

Find the sum of the arithmetic series: 
2 + 5 + 8 + ... + 20

Explanation

By Using the formula for the sum of an arithmetic series.

Explanation

By Using the formula for the sum of an arithmetic series.

191 / 320

Category: Time and Distance - Mathematics

An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1  hours, it must travel at a speed of:

Explanation:

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = 1200/(5/3) km/hr.     [We can write 1 hours as 5/3 hours]

 Required speed = 1200 x 3 km/hr = 720 km/hr.
5
Explanation:

Distance = (240 x 5) = 1200 km.

Speed = Distance/Time

Speed = 1200/(5/3) km/hr.     [We can write 1 hours as 5/3 hours]

 Required speed = 1200 x 3 km/hr = 720 km/hr.
5

192 / 320

Category: Time and Distance - Mathematics

In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:

Explanation:

Let Abhay's speed be x km/hr.

Then, 30 - 30 = 3
x 2x

 6x = 30

 x = 5 km/hr.

Explanation:

Let Abhay's speed be x km/hr.

Then, 30 - 30 = 3
x 2x

 6x = 30

 x = 5 km/hr.

193 / 320

Category: Time and Distance - Mathematics

If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:

Explanation:

Let the actual distance travelled be x km.

Then, x = x + 20
10 14

 14x = 10x + 200

 4x = 200

 x = 50 km.

Explanation:

Let the actual distance travelled be x km.

Then, x = x + 20
10 14

 14x = 10x + 200

 4x = 200

 x = 50 km.

194 / 320

Category: Time and Distance - Mathematics

A car travelling with  of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.

Explanation:
Time taken = 1 hr 40 min 48 sec = 1 hr 40 4 min = 1 51 hrs = 126 hrs.
5 75 75

Let the actual speed be x km/hr.

Then, 5 x x 126 = 42
7 75
 x = 42 x 7 x 75 = 35 km/hr.
5 x 126
Explanation:
Time taken = 1 hr 40 min 48 sec = 1 hr 40 4 min = 1 51 hrs = 126 hrs.
5 75 75

Let the actual speed be x km/hr.

Then, 5 x x 126 = 42
7 75
 x = 42 x 7 x 75 = 35 km/hr.
5 x 126

195 / 320

Category: Time and Distance - Mathematics

A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:

Explanation:
Total time taken = 160 + 160 hrs. = 9 hrs.
64 80 2
 Average speed = 320 x 2 km/hr = 71.11 km/hr.
9
Explanation:
Total time taken = 160 + 160 hrs. = 9 hrs.
64 80 2
 Average speed = 320 x 2 km/hr = 71.11 km/hr.
9

196 / 320

Category: Time and Distance - Mathematics

Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?

Explanation:

Let the distance travelled by x km.

Then, x - x = 2
10 15

 3x - 2x = 60

 x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

 Required speed = 60 kmph. = 12 kmph.
5
Explanation:

Let the distance travelled by x km.

Then, x - x = 2
10 15

 3x - 2x = 60

 x = 60 km.

Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

 Required speed = 60 kmph. = 12 kmph.
5

197 / 320

Category: Time and Distance - Mathematics

A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?

Explanation:
Speed = 600 m/sec.
5 x 60

= 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

   = 2 x 18 km/hr
5

= 7.2 km/hr.

Explanation:
Speed = 600 m/sec.
5 x 60

= 2 m/sec.

Converting m/sec to km/hr (see important formulas section)

   = 2 x 18 km/hr
5

= 7.2 km/hr.

198 / 320

Category: Time and Distance - Mathematics

A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:

Explanation:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, x + (61 -x) = 9
4 9

 9x + 4(61 -x) = 9 x 36

 5x = 80

 x = 16 km.

Explanation:

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle = (61 -x) km.

So, x + (61 -x) = 9
4 9

 9x + 4(61 -x) = 9 x 36

 5x = 80

 x = 16 km.

199 / 320

Category: Time and Distance - Mathematics

In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:

Explanation:

Let the duration of the flight be x hours.

Then, 600 - 600 = 200
x x + (1/2)
600 - 1200 = 200
x 2x + 1

 x(2x + 1) = 3

 2x2 + x - 3 = 0

 (2x + 3)(x - 1) = 0

 x = 1 hr.      [neglecting the -ve value of x]

Explanation:

Let the duration of the flight be x hours.

Then, 600 - 600 = 200
x x + (1/2)
600 - 1200 = 200
x 2x + 1

 x(2x + 1) = 3

 2x2 + x - 3 = 0

 (2x + 3)(x - 1) = 0

 x = 1 hr.      [neglecting the -ve value of x]

200 / 320

Category: Time and Distance - Mathematics

Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?

Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = 9 x 60 min = 10 min.
54
Explanation:

Due to stoppages, it covers 9 km less.

Time taken to cover 9 km = 9 x 60 min = 10 min.
54

201 / 320

Category: Time and Distance - Mathematics

A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

Explanation:
(1/2)x + (1/2)x = 10
21 24
x + x = 20
21 24

 15x = 168 x 20

 x = 168 x 20 = 224 km.
15
Explanation:
(1/2)x + (1/2)x = 10
21 24
x + x = 20
21 24

 15x = 168 x 20

 x = 168 x 20 = 224 km.
15

202 / 320

Category: Time and Distance - Mathematics

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:

Explanation:

Let distance = x km and usual rate = y kmph.

Then, x - x = 40      2y(y + 3) = 9x ....(i)
y y + 3 60
And, x - x = 40      y(y - 2) = 3x ....(ii)
y -2 y 60

On dividing (i) by (ii), we get: x = 40.

Explanation:

Let distance = x km and usual rate = y kmph.

Then, x - x = 40      2y(y + 3) = 9x ....(i)
y y + 3 60
And, x - x = 40      y(y - 2) = 3x ....(ii)
y -2 y 60

On dividing (i) by (ii), we get: x = 40.

203 / 320

Category: Time and Distance - Mathematics

The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:

Explanation:

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x = 400 = 100
4
 x = 100 = 12.5
8

 Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

Explanation:

Let the speed of two trains be 7x and 8x km/hr.

Then, 8x = 400 = 100
4
 x = 100 = 12.5
8

 Speed of first train = (7 x 12.5) km/hr = 87.5 km/hr.

204 / 320

Category: Time and Distance - Mathematics

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:

Explanation:

Let speed of the car be x kmph.

Then, speed of the train = 150 x = 3 x kmph.
100 2
75 - 75 = 125
x (3/2)x 10 x 60
75 - 50 = 5
x x 24
 x = 25 x24 = 120 kmph.
5
Explanation:

Let speed of the car be x kmph.

Then, speed of the train = 150 x = 3 x kmph.
100 2
75 - 75 = 125
x (3/2)x 10 x 60
75 - 50 = 5
x x 24
 x = 25 x24 = 120 kmph.
5

205 / 320

Category: Time and Distance - Mathematics

It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:

Explanation:

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then, 120 + 480 = 8        1 + 4 = 1 ....(i)
x y x y 15
And, 200 + 400 = 25      1 + 2 = 1 ....(ii)
x y 3 x y 24

Solving (i) and (ii), we get: x = 60 and y = 80.

 Ratio of speeds = 60 : 80 = 3 : 4.

Explanation:

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then, 120 + 480 = 8        1 + 4 = 1 ....(i)
x y x y 15
And, 200 + 400 = 25      1 + 2 = 1 ....(ii)
x y 3 x y 24

Solving (i) and (ii), we get: x = 60 and y = 80.

 Ratio of speeds = 60 : 80 = 3 : 4.

206 / 320

Category: Simple Interest - Mathematics

A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6% p.a for 2 years. Find his gain in the transaction per year.

Explanation:
Gain in 2 years
= Rs. 5000 x 25 x 2 - 5000 x 4 x 2
4 100 100
= Rs. (625 - 400)
= Rs. 225.
 Gain in 1 year = Rs. 225 = Rs. 112.50
2
Explanation:
Gain in 2 years
= Rs. 5000 x 25 x 2 - 5000 x 4 x 2
4 100 100
= Rs. (625 - 400)
= Rs. 225.
 Gain in 1 year = Rs. 225 = Rs. 112.50
2

207 / 320

Category: Simple Interest - Mathematics

A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?

Explanation:

S.I. = Rs. (15500 - 12500) = Rs. 3000.

Rate = 100 x 3000 % = 6%
12500 x 4
Explanation:

S.I. = Rs. (15500 - 12500) = Rs. 3000.

Rate = 100 x 3000 % = 6%
12500 x 4

208 / 320

Category: Simple Interest - Mathematics

A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

Explanation:

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs. 2205 x 5 = Rs. 3675
3

 Principal = Rs. (9800 - 3675) = Rs. 6125.

Hence, rate = 100 x 3675 % = 12%
6125 x 5
Explanation:

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205.

S.I. for 5 years = Rs. 2205 x 5 = Rs. 3675
3

 Principal = Rs. (9800 - 3675) = Rs. 6125.

Hence, rate = 100 x 3675 % = 12%
6125 x 5

209 / 320

Category: Simple Interest - Mathematics

Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

Explanation:

Let rate = R% and time = R years.

Then, 1200 x R x R = 432
100

 12R2 = 432

 R2 = 36

 R = 6.

Explanation:

Let rate = R% and time = R years.

Then, 1200 x R x R = 432
100

 12R2 = 432

 R2 = 36

 R = 6.

210 / 320

Category: Simple Interest - Mathematics

A sum fetched a total simple interest of ₹ 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?

Explanation:
Principal
= Rs. 100 x 4016.25
9 x 5
= Rs. 401625
45
= Rs. 8925.
Explanation:
Principal
= Rs. 100 x 4016.25
9 x 5
= Rs. 401625
45
= Rs. 8925.

211 / 320

Category: Simple Interest - Mathematics

A sum of money at simple interest amounts to ₹ 815 in 3 years and to ₹ 854 in 4 years. The sum is:

Explanation:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

 Principal = Rs. (815 - 117) = Rs. 698.

Explanation:

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

 Principal = Rs. (815 - 117) = Rs. 698.

212 / 320

Category: Simple Interest - Mathematics

A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:

Explanation:

Let the rate be R% p.a.

Then, 5000 x R x 2 + 3000 x R x 4 = 2200.
100 100

 100R + 120R = 2200

 R = 2200 = 10.
220

 Rate = 10%.

Explanation:

Let the rate be R% p.a.

Then, 5000 x R x 2 + 3000 x R x 4 = 2200.
100 100

 100R + 120R = 2200

 R = 2200 = 10.
220

 Rate = 10%.

213 / 320

Category: Simple Interest - Mathematics

A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?

Explanation:

Let the original rate be R%. Then, new rate = (2R)%.

Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.  year(s).

725 x R x 1 + 362.50 x 2R x 1 = 33.50
100 100 x 3

 (2175 + 725) R = 33.50 x 100 x 3

 (2175 + 725) R = 10050

 (2900)R = 10050

 R = 10050 = 3.46
2900

 Original rate = 3.46%

Explanation:

Let the original rate be R%. Then, new rate = (2R)%.

Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e.  year(s).

725 x R x 1 + 362.50 x 2R x 1 = 33.50
100 100 x 3

 (2175 + 725) R = 33.50 x 100 x 3

 (2175 + 725) R = 10050

 (2900)R = 10050

 R = 10050 = 3.46
2900

 Original rate = 3.46%

214 / 320

Category: Simple Interest - Mathematics

How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

Explanation:
Time = 100 x 81 years = 4 years.
450 x 4.5
Explanation:
Time = 100 x 81 years = 4 years.
450 x 4.5

215 / 320

Category: Simple Interest - Mathematics

What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?

Explanation:

Let the principal be P and rate of interest be R%.

 Required ratio =
P x R x 6
100
= 6PR = 6 = 2 : 3.
P x R x 9
100
9PR 9

Explanation:

Let the principal be P and rate of interest be R%.

 Required ratio =
P x R x 6
100
= 6PR = 6 = 2 : 3.
P x R x 9
100
9PR 9

216 / 320

Category: Simple Interest - Mathematics

An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:

Explanation:

Let the sum be Rs. 100. Then,

S.I. for first 6 months = Rs. 100 x 10 x 1 = Rs. 5
100 x 2
S.I. for last 6 months = Rs. 105 x 10 x 1 = Rs. 5.25
100 x 2

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25

 Effective rate = (110.25 - 100) = 10.25%

Explanation:

Let the sum be Rs. 100. Then,

S.I. for first 6 months = Rs. 100 x 10 x 1 = Rs. 5
100 x 2
S.I. for last 6 months = Rs. 105 x 10 x 1 = Rs. 5.25
100 x 2

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25

 Effective rate = (110.25 - 100) = 10.25%

217 / 320

Category: Simple Interest - Mathematics

A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?

Explanation:

We need to know the S.I., principal and time to find the rate.

Since the principal is not given, so data is inadequate.

Explanation:

We need to know the S.I., principal and time to find the rate.

Since the principal is not given, so data is inadequate.

218 / 320

Category: Simple Interest - Mathematics

A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:

Explanation:
Principal = Rs. 100 x 5400 = Rs. 15000.
12 x 3
Explanation:
Principal = Rs. 100 x 5400 = Rs. 15000.
12 x 3

219 / 320

Category: Simple Interest - Mathematics

Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?

Explanation:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508
100 100

 28x - 22x = 350800 - (13900 x 22)

 6x = 45000

 x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

Explanation:

Let the sum invested in Scheme A be Rs. x and that in Scheme B be Rs. (13900 - x).

Then, x x 14 x 2 + (13900 - x) x 11 x 2 = 3508
100 100

 28x - 22x = 350800 - (13900 x 22)

 6x = 45000

 x = 7500.

So, sum invested in Scheme B = Rs. (13900 - 7500) = Rs. 6400.

220 / 320

Category: Elementary Statistics - Mathematics

In five One-Day Internationals, a batsman has scored 31,97,112, 63, and 12 runs. the quality deviation of the info is-

Mean=total sum of the numbers within the data sets

Total numbers within the data sets

= 31+97+112+12= 315/5 = 63

Standard deviation = [1/n (x(n)-mean)²]

Mean=total sum of the numbers within the data sets

Total numbers within the data sets

= 31+97+112+12= 315/5 = 63

Standard deviation = [1/n (x(n)-mean)²]

221 / 320

Category: Elementary Statistics - Mathematics

Determine the mode of the decision received seven days in a row: 11,13,13,17,19,23,25

The value that appears the most frequently is the mode; in this case, the quantity 13 appears twice.

The value that appears the most frequently is the mode; in this case, the quantity 13 appears twice.

222 / 320

Category: Elementary Statistics - Mathematics

What is the median of the following 7 days: 11, 13, 17, 13, 23,25,19?

Mean=(n+1)/2

Where,

n is the number of terms, and 7 is the number of terms.

The median is the information sets’ middle value, so we’ll start by rearranging the numbers in ascending order: 11,13,13,17,19,23,25.

7+1/2 = 4th number is the middle one.

As a result, the fourth number is 17.

Mean=(n+1)/2

Where,

n is the number of terms, and 7 is the number of terms.

The median is the information sets’ middle value, so we’ll start by rearranging the numbers in ascending order: 11,13,13,17,19,23,25.

7+1/2 = 4th number is the middle one.

As a result, the fourth number is 17.

223 / 320

Category: BODMAS - Mathematics

Work out: 5 × [20 – (9 + 2)]

Solution:

Round brackets: 9 + 2 = 11.
Square brackets: 20 – 11 = 9.
Multiply: 5 × 9 = 45.

Solution:

Round brackets: 9 + 2 = 11.
Square brackets: 20 – 11 = 9.
Multiply: 5 × 9 = 45.

224 / 320

Category: BODMAS - Mathematics

Calculate: 12 – 4 × (6 – 2)

Answer:

First, solve inside the brackets: 6 – 2 = 4.
Then multiply: 4 × 4 = 16.
Subtract: 12 – 16 = -4.

Answer:

First, solve inside the brackets: 6 – 2 = 4.
Then multiply: 4 × 4 = 16.
Subtract: 12 – 16 = -4.

225 / 320

Category: BODMAS - Mathematics

14 – 6 ÷ 2 × 3

Answer

Division: 6 ÷ 2 = 3.
Multiply: 3 × 3 = 9.
Subtract: 14 – 9 = 5.

Answer

Division: 6 ÷ 2 = 3.
Multiply: 3 × 3 = 9.
Subtract: 14 – 9 = 5.

226 / 320

Category: BODMAS - Mathematics

42 ÷ 7 + 10 – 3

Answer

Division: 42 ÷ 7 = 6.
Expression: 6 + 10 – 3.
Left to right: 6 + 10 = 16, 16 – 3 = 13.

Answer

Division: 42 ÷ 7 = 6.
Expression: 6 + 10 – 3.
Left to right: 6 + 10 = 16, 16 – 3 = 13.

227 / 320

Category: BODMAS - Mathematics

Simplify (4×b−1)÷3(4 × b – 1) ÷ 3(4×b−1)÷3 for b=5b = 5b=5

Answer

Substitute: (4 × 5 – 1) ÷ 3.
Multiply/add: 4 × 5 = 20, 20 – 1 = 19.
Divide: 19 ÷ 3 ≈ 6.33.

Answer

Substitute: (4 × 5 – 1) ÷ 3.
Multiply/add: 4 × 5 = 20, 20 – 1 = 19.
Divide: 19 ÷ 3 ≈ 6.33.

228 / 320

Category: BODMAS - Mathematics

Calculate: (16 ÷ (8 ÷ 2)) + 1

Solution:

Innermost: 8 ÷ 2 = 4.
Divide: 16 ÷ 4 = 4.
Add: 4 + 1 = 5.

Solution:

Innermost: 8 ÷ 2 = 4.
Divide: 16 ÷ 4 = 4.
Add: 4 + 1 = 5.

229 / 320

Category: BODMAS - Mathematics

Simplify: [25 ÷ (10 – 4)] × 2

Solution:

Round brackets: 10 – 4 = 6.
Square brackets: 25 ÷ 6 ≈ 4.17.
Multiply: 4.17 × 2 ≈ 8.33. (rounded to 2 decimal places)

Solution:

Round brackets: 10 – 4 = 6.
Square brackets: 25 ÷ 6 ≈ 4.17.
Multiply: 4.17 × 2 ≈ 8.33. (rounded to 2 decimal places)

230 / 320

Category: BODMAS - Mathematics

If a=3a = 3a=3, solve 15+a×(5−2)15 + a × (5 – 2)15+a×(5−2)

Answer

Substitute: 15 + 3 × (5 – 2).
Brackets: 5 – 2 = 3.
Multiply: 3 × 3 = 9.
Add: 15 + 9 = 24.

Answer

Substitute: 15 + 3 × (5 – 2).
Brackets: 5 – 2 = 3.
Multiply: 3 × 3 = 9.
Add: 15 + 9 = 24.

231 / 320

Category: BODMAS - Mathematics

If m=4m = 4m=4, n=2n = 2n=2, solve (m−n)2+n(m – n)^2 + n(m−n)2+n

Answer

Substitute: (4 – 2)^2 + 2.
Brackets: 4 – 2 = 2.
Power: 22=42^2 = 422=4.
Add: 4 + 2 = 6.

Answer

Substitute: (4 – 2)^2 + 2.
Brackets: 4 – 2 = 2.
Power: 22=42^2 = 422=4.
Add: 4 + 2 = 6.

232 / 320

Category: BODMAS - Mathematics

Evaluate: [9 + (12 ÷ (6 ÷ 3))] – 2

Solution:

Innermost: 6 ÷ 3 = 2.
Next: 12 ÷ 2 = 6.
Square brackets: 9 + 6 = 15.
Subtract: 15 – 2 = 13.

Solution:

Innermost: 6 ÷ 3 = 2.
Next: 12 ÷ 2 = 6.
Square brackets: 9 + 6 = 15.
Subtract: 15 – 2 = 13.

233 / 320

Category: BODMAS - Mathematics

Simplify: 20 ÷ {[10 – (4 × 2)] + 3}

Solution:

Parentheses: 4 × 2 = 8.
Square brackets: 10 – 8 = 2.
Curly brackets: 2 + 3 = 5.
Divide: 20 ÷ 5 = 4.

Solution:

Parentheses: 4 × 2 = 8.
Square brackets: 10 – 8 = 2.
Curly brackets: 2 + 3 = 5.
Divide: 20 ÷ 5 = 4.

234 / 320

Category: BODMAS - Mathematics

30 + (8 × 4) – 9

Answer

Brackets: 8 × 4 = 32.
Add: 30 + 32 = 62.
Subtract: 62 – 9 = 53.

Answer

Brackets: 8 × 4 = 32.
Add: 30 + 32 = 62.
Subtract: 62 – 9 = 53.

235 / 320

Category: BODMAS - Mathematics

Evaluate: 9 + 24 ÷ (12 ÷ 4)

Answer:

Innermost brackets: 12 ÷ 4 = 3.
Divide: 24 ÷ 3 = 8.
Add: 9 + 8 = 17.

Answer:

Innermost brackets: 12 ÷ 4 = 3.
Divide: 24 ÷ 3 = 8.
Add: 9 + 8 = 17.

236 / 320

Category: BODMAS - Mathematics

Work out: (8 ÷ 2)^2 – 5

Answer:

Brackets: 8 ÷ 2 = 4.
Power: 42=164^2 = 1642=16.
Subtract: 16 – 5 = 11.

Answer:

Brackets: 8 ÷ 2 = 4.
Power: 42=164^2 = 1642=16.
Subtract: 16 – 5 = 11.

237 / 320

Category: BODMAS - Mathematics

If b=6b = 6b=6, solve (b÷2)+10−4(b ÷ 2) + 10 – 4(b÷2)+10−4

Answer

Substitute: (6 ÷ 2) + 10 – 4.
Brackets: 6 ÷ 2 = 3.
Add: 3 + 10 = 13.
Subtract: 13 – 4 = 9.

Answer

Substitute: (6 ÷ 2) + 10 – 4.
Brackets: 6 ÷ 2 = 3.
Add: 3 + 10 = 13.
Subtract: 13 – 4 = 9.

238 / 320

Category: BODMAS - Mathematics

Evaluate: {[(12 – 4) × 2] ÷ 4} + 3

Solution:

Parentheses: 12 – 4 = 8.
Square brackets: 8 × 2 = 16.
Curly brackets: 16 ÷ 4 = 4.
Add: 4 + 3 = 7.

Answer: 7

Solution:

Parentheses: 12 – 4 = 8.
Square brackets: 8 × 2 = 16.
Curly brackets: 16 ÷ 4 = 4.
Add: 4 + 3 = 7.

Answer: 7

239 / 320

Category: BODMAS - Mathematics

Evaluate: (7 + 5) ÷ (4 – 1)

Solution:

Round brackets: 7 + 5 = 12 and 4 – 1 = 3.
Divide: 12 ÷ 3 = 4.

Solution:

Round brackets: 7 + 5 = 12 and 4 – 1 = 3.
Divide: 12 ÷ 3 = 4.

240 / 320

Category: BODMAS - Mathematics

Calculate: 6 + {8 ÷ [5 – (2 + 1)]}

Solution:

Parentheses: 2 + 1 = 3.
Square brackets: 5 – 3 = 2.
Curly brackets: 8 ÷ 2 = 4.
Add: 6 + 4 = 10.

Solution:

Parentheses: 2 + 1 = 3.
Square brackets: 5 – 3 = 2.
Curly brackets: 8 ÷ 2 = 4.
Add: 6 + 4 = 10.

241 / 320

Category: BODMAS - Mathematics

Simplify: 12 – (3 × (7 – 4))

Solution:

Innermost: 7 – 4 = 3.
Multiply: 3 × 3 = 9.
Subtract: 12 – 9 = 3.

Solution:

Innermost: 7 – 4 = 3.
Multiply: 3 × 3 = 9.
Subtract: 12 – 9 = 3.

242 / 320

Category: BODMAS - Mathematics

Simplify: (20 ÷ 5) × 2 + 1

Answer:

Brackets first: 20 ÷ 5 = 4.
Multiply: 4 × 2 = 8.
Add: 8 + 1 = 9.

Answer:

Brackets first: 20 ÷ 5 = 4.
Multiply: 4 × 2 = 8.
Add: 8 + 1 = 9.

243 / 320

Category: BODMAS - Mathematics

36 ÷ (9 – 3) + 4

Answer

Brackets: 9 – 3 = 6.
Divide: 36 ÷ 6 = 6.
Add: 6 + 4 = 10.

Answer

Brackets: 9 – 3 = 6.
Divide: 36 ÷ 6 = 6.
Add: 6 + 4 = 10.

244 / 320

Category: BODMAS - Mathematics

Find: 15 × (4 – 1) ÷ 3

Answer:

Brackets: 4 – 1 = 3.
Multiply: 15 × 3 = 45.
Divide: 45 ÷ 3 = 15.

Answer:

Brackets: 4 – 1 = 3.
Multiply: 15 × 3 = 45.
Divide: 45 ÷ 3 = 15.

245 / 320

Category: BODMAS - Mathematics

Solve 45−(9÷a)45 – (9 ÷ a)45−(9÷a) when a=3a = 3a=3

Answer

Substitute: 45 – (9 ÷ 3).
Brackets: 9 ÷ 3 = 3.
Subtract: 45 – 3 = 42.

Answer

Substitute: 45 – (9 ÷ 3).
Brackets: 9 ÷ 3 = 3.
Subtract: 45 – 3 = 42.

246 / 320

Category: BODMAS - Mathematics

(24 ÷ 4) × 5 – 7

Answer

Brackets: 24 ÷ 4 = 6.
Multiply: 6 × 5 = 30.
Subtract: 30 – 7 = 23.

Answer

Brackets: 24 ÷ 4 = 6.
Multiply: 6 × 5 = 30.
Subtract: 30 – 7 = 23.

247 / 320

Category: Geometry - Mathematics

In the given figure , AOB is a straight line, ∠ AOC = (3x-8)° and ∠COD =50 and ∠BOD° =(x+10)°. The value of the x is

q 16

Explanation

Since ∠AOB is a straight angle , we have
∠AOC  + ∠ COB  + ∠ BOD  = 180° 
⇒ (3X  8)° + 50° + (X+ 10)° = 180° 
⇒ 4X = 128 ⇒ X = 32.

Explanation

Since ∠AOB is a straight angle , we have
∠AOC  + ∠ COB  + ∠ BOD  = 180° 
⇒ (3X  8)° + 50° + (X+ 10)° = 180° 
⇒ 4X = 128 ⇒ X = 32.

248 / 320

Category: Geometry - Mathematics

If the radius of a circle is doubled, how does the area change?

Area of a circle = πr²; if radius is doubled, area becomes π(2r)² = 4πr², thus it quadruples.

Area of a circle = πr²; if radius is doubled, area becomes π(2r)² = 4πr², thus it quadruples.

249 / 320

Category: Geometry - Mathematics

An angle which is greater then 180° but less than 360° is called

Explanation

An angle which is greater than 180°  but less than 360° is called a reflex angle.

Explanation

An angle which is greater than 180°  but less than 360° is called a reflex angle.

250 / 320

Category: Geometry - Mathematics

How many angles are made by rays shown in the figure?

q 10

Explanation

The angle are ∠AOB , ∠BOC,∠COD,∠DOE,∠AOC,∠AOD, ∠AOE,∠BOD,∠BOD,∠COE. 
Thus , 10 angle are  formed.

Explanation

The angle are ∠AOB , ∠BOC,∠COD,∠DOE,∠AOC,∠AOD, ∠AOE,∠BOD,∠BOD,∠COE. 
Thus , 10 angle are  formed.

251 / 320

Category: Geometry - Mathematics

How many straight lines can be drawn through two given lines?

252 / 320

Category: Geometry - Mathematics

If an angle is its own complementary angle, then its measure is

Explanation

x=(90-x) ⇒ 2x = 90 ⇒ x = 45°

Explanation

x=(90-x) ⇒ 2x = 90 ⇒ x = 45°

253 / 320

Category: Geometry - Mathematics

A ray has

254 / 320

Category: Geometry - Mathematics

An angle is 24° more than its complement.The measure of the angle is

Explanation

x  (90-x ) = 24 ⇒ 2x = 114 ⇒ x = 57
∴ Required angle is 57°.

Explanation

x  (90-x ) = 24 ⇒ 2x = 114 ⇒ x = 57
∴ Required angle is 57°.

255 / 320

Category: Geometry - Mathematics

The complement of 62° is

Explanation

Complement of 62°= (90°  62°) = 28°.

Explanation

Complement of 62°= (90°  62°) = 28°.

256 / 320

Category: Geometry - Mathematics

What do you call a figure formed by two straight lines having a common point?

257 / 320

Category: Geometry - Mathematics

In the given figure , AOB is a straight line, ∠AOC = (3x+20)° and ∠ BOC =(4 x-36)°. The value of the x is

q 15

Explanation

Since ∠AOB is a straight angle , we have 
∠AOC + ∠ BOC  =180° 
⇒ 3x + 20 +4x  36 = 180
⇒ 7x = 164 ⇒ x = 22.

Explanation

Since ∠AOB is a straight angle , we have 
∠AOC + ∠ BOC  =180° 
⇒ 3x + 20 +4x  36 = 180
⇒ 7x = 164 ⇒ x = 22.

258 / 320

Category: Geometry - Mathematics

What is the minimum number of lines required to make a closed figure?

259 / 320

Category: Geometry - Mathematics

The supplement of 60° is

Explanation

Supplement of 60° = (180°-60°) =120°.

Explanation

Supplement of 60° = (180°-60°) =120°.

260 / 320

Category: Geometry - Mathematics

How many lines can pass through one point?

261 / 320

Category: Geometry - Mathematics

In a right triangle, if one angle is 30 degrees, what is the other non-right angle?

In a right triangle, the sum of the angles is 180 degrees. Therefore, the other angle is 180 - 90 - 30 = 60 degrees.

In a right triangle, the sum of the angles is 180 degrees. Therefore, the other angle is 180 - 90 - 30 = 60 degrees.

262 / 320

Category: Geometry - Mathematics

How many dimension does a surface has?

263 / 320

Category: Geometry - Mathematics

A line has

264 / 320

Category: Geometry - Mathematics

Which of the following is an axiom?

265 / 320

Category: Geometry - Mathematics

An angle is 32° less than its supplement. The measure of the angle is

Explanation

(180 X)  X = 32 ⇒  2x = 180  32 = 148 ⇒ x = 74. 
Required angle is 74°.

Explanation

(180 X)  X = 32 ⇒  2x = 180  32 = 148 ⇒ x = 74. 
Required angle is 74°.

266 / 320

Category: Geometry - Mathematics

An angle is one fifth of its supplement. The measure of the angle is

Explanation

x = 1/5 (180  x )⇒ 5x = 180  x ⇒ 6x = 180 ⇒ x = 30°.

Explanation

x = 1/5 (180  x )⇒ 5x = 180  x ⇒ 6x = 180 ⇒ x = 30°.

267 / 320

Category: Geometry - Mathematics

Which of the following are boundaries of a surface?

268 / 320

Category: Geometry - Mathematics

The complement of 72° 40' is

Explanation

Complement of 72° 40' = (90°-72° 40') =17° 20'.

Explanation

Complement of 72° 40' = (90°-72° 40') =17° 20'.

269 / 320

Category: Geometry - Mathematics

A solid has how many dimensions?

270 / 320

Category: Geometry - Mathematics

In the given figure, AOB is a straight line, ∠AOC = 68° and ∠BOC = x°. The value of the x is

q 14

Explanation

Since ∠AOB is a straight angle , we have
X+ 68 = 180 ⇒ x= (180-68)°  = 120°

Explanation

Since ∠AOB is a straight angle , we have
X+ 68 = 180 ⇒ x= (180-68)°  = 120°

271 / 320

Category: Geometry - Mathematics

What is the sum of the interior angles of a triangle?

The sum of the interior angles of a triangle is always 180 degrees.

The sum of the interior angles of a triangle is always 180 degrees.

272 / 320

Category: Geometry - Mathematics

Two Supplementary angles are in th ratio 3:2. The smaller angle measures

Explanation

Let the measures of the angle be (3x)° and (2x)°. Then, 
3x+2x=180 ⇒ 5x = 180 ⇒ x = 36.
Smaller angle = (2x)° = (2*36)° = 72°.

Explanation

Let the measures of the angle be (3x)° and (2x)°. Then, 
3x+2x=180 ⇒ 5x = 180 ⇒ x = 36.
Smaller angle = (2x)° = (2*36)° = 72°.

273 / 320

Category: Geometry - Mathematics

A line segment has

274 / 320

Category: Geometry - Mathematics

What is the area of a rectangle with length 5 cm and width 3 cm?

Area = length × width = 5 cm × 3 cm = 15 cm².

Area = length × width = 5 cm × 3 cm = 15 cm².

275 / 320

Category: Geometry - Mathematics

What is the volume of a cube with side length 4 cm?

Volume of a cube = side³ = 4 cm × 4 cm × 4 cm = 64 cm³.

Volume of a cube = side³ = 4 cm × 4 cm × 4 cm = 64 cm³.

276 / 320

Category: Percentages - Mathematics

In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:

Explanation:

Number of valid votes = 80% of 7500 = 6000.

 Valid votes polled by other candidate = 45% of 6000

= 45 x 6000 = 2700.
100
Explanation:

Number of valid votes = 80% of 7500 = 6000.

 Valid votes polled by other candidate = 45% of 6000

= 45 x 6000 = 2700.
100

277 / 320

Category: Percentages - Mathematics

A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

Explanation:

Number of runs made by running = 110 - (3 x 4 + 8 x 6)

= 110 - (60)

= 50.

 Required percentage = 50 x 100 % = 45 5 %
110 11
Explanation:

Number of runs made by running = 110 - (3 x 4 + 8 x 6)

= 110 - (60)

= 50.

 Required percentage = 50 x 100 % = 45 5 %
110 11

278 / 320

Category: Percentages - Mathematics

Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.

Explanation:
5% of A + 4% of B = 2  (6% of A + 8% of B)
3
5  A + 4  B = 2 6  A + 8  B
100 100 3 100 100
1  A + 1  B = 1  A + 4  B
20 25 25 75
1 - 1  A = 4 - 1  B
20 25 75 25
1  A = 1  B
100 75
A = 100 = 4 .
B 75 3

 Required ratio = 4 : 3

Explanation:
5% of A + 4% of B = 2  (6% of A + 8% of B)
3
5  A + 4  B = 2 6  A + 8  B
100 100 3 100 100
1  A + 1  B = 1  A + 4  B
20 25 25 75
1 - 1  A = 4 - 1  B
20 25 75 25
1  A = 1  B
100 75
A = 100 = 4 .
B 75 3

 Required ratio = 4 : 3

279 / 320

Category: Percentages - Mathematics

Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?

Explanation:

Let the sum paid to Y per week be Rs. z.

Then, z + 120% of z = 550.

 z + 120 z = 550
100
11 z = 550
5
 z = 550 x 5   = 250.
11
Explanation:

Let the sum paid to Y per week be Rs. z.

Then, z + 120% of z = 550.

 z + 120 z = 550
100
11 z = 550
5
 z = 550 x 5   = 250.
11

280 / 320

Category: Percentages - Mathematics

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:

Explanation:

Let their marks be (x + 9) and x.

Then, x + 9 = 56 (x + 9 + x)
100

 25(x + 9) = 14(2x + 9)

 3x = 99

 x = 33

So, their marks are 42 and 33.

Explanation:

Let their marks be (x + 9) and x.

Then, x + 9 = 56 (x + 9 + x)
100

 25(x + 9) = 14(2x + 9)

 3x = 99

 x = 33

So, their marks are 42 and 33.

281 / 320

Category: Percentages - Mathematics

If 20% of a = b, then b% of 20 is the same as:

Explanation:
20% of a = b     20 a = b.
100
 b% of 20 = b x 20 = 20 a x 1 x 20 = 4 a = 4% of a.
Explanation:
20% of a = b     20 a = b.
100
 b% of 20 = b x 20 = 20 a x 1 x 20 = 4 a = 4% of a.

282 / 320

Category: Percentages - Mathematics

The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

Explanation:

Increase in 10 years = (262500 - 175000) = 87500.

Increase% = 87500 x 100 % = 50%.
175000
 Required average = 50 % = 5%.
10
Explanation:

Increase in 10 years = (262500 - 175000) = 87500.

Increase% = 87500 x 100 % = 50%.
175000
 Required average = 50 % = 5%.
10

283 / 320

Category: Percentages - Mathematics

Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.

Explanation:
Rebate = 6% of Rs. 6650 = Rs. 6 x 6650 = Rs. 399.
100
Sales tax = 10% of Rs. (6650 - 399) = Rs. 10 x 6251 = Rs. 625.10
100

 Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

Explanation:
Rebate = 6% of Rs. 6650 = Rs. 6 x 6650 = Rs. 399.
100
Sales tax = 10% of Rs. (6650 - 399) = Rs. 10 x 6251 = Rs. 625.10
100

 Final amount = Rs. (6251 + 625.10) = Rs. 6876.10

284 / 320

Category: Percentages - Mathematics

What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?

Explanation:

Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.

Number of such number =14

 Required percentage = 14 x 100 % = 20%.
70
Explanation:

Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.

Number of such number =14

 Required percentage = 14 x 100 % = 20%.
70

285 / 320

Category: Percentages - Mathematics

A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:

Explanation:

Suppose originally he had x apples.

Then, (100 - 40)% of x = 420.

60 x x = 420
100
 x = 420 x 100   = 700.
60
Explanation:

Suppose originally he had x apples.

Then, (100 - 40)% of x = 420.

60 x x = 420
100
 x = 420 x 100   = 700.
60

286 / 320

Category: Percentages - Mathematics

Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?

Explanation:

Total number of votes polled = (1136 + 7636 + 11628) = 20400.

 Required percentage = 11628 x 100 % = 57%.
20400
Explanation:

Total number of votes polled = (1136 + 7636 + 11628) = 20400.

 Required percentage = 11628 x 100 % = 57%.
20400

287 / 320

Category: Percentages - Mathematics

In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is  of the number of students of 8 years of age which is 48. What is the total number of students in the school?

Explanation:

Let the number of students be x. Then,

Number of students above 8 years of age = (100 - 20)% of x = 80% of x.

 80% of x = 48 + 2 of 48
3
80 x = 80
100

 x = 100.

Explanation:

Let the number of students be x. Then,

Number of students above 8 years of age = (100 - 20)% of x = 80% of x.

 80% of x = 48 + 2 of 48
3
80 x = 80
100

 x = 100.

288 / 320

Category: Percentages - Mathematics

Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

Explanation:

Let the amount taxable purchases be Rs. x.

Then, 6% of x = 30
100
 x = 30 x 100  = 5.
100 6

 Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70

Explanation:

Let the amount taxable purchases be Rs. x.

Then, 6% of x = 30
100
 x = 30 x 100  = 5.
100 6

 Cost of tax free items = Rs. [25 - (5 + 0.30)] = Rs. 19.70

289 / 320

Category: Percentages - Mathematics

If A = x% of y and B = y% of x, then which of the following is true?

Explanation:
x% of y = x x y = y x x = y% of x
100 100

 A = B.

Explanation:
x% of y = x x y = y x x = y% of x
100 100

 A = B.

290 / 320

Category: Percentages - Mathematics

A student multiplied a number by 3 instead of 5 .
5 3

What is the percentage error in the calculation?

Explanation:

Let the number be x.

Then, error = 5 x - 3 x = 16 x.
3 5 15
Error% = 16x x 3 x 100 % = 64%.
15 5x
Explanation:

Let the number be x.

Then, error = 5 x - 3 x = 16 x.
3 5 15
Error% = 16x x 3 x 100 % = 64%.
15 5x

291 / 320

Category: Profit and Loss - Mathematics

Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs. 5 x 30 = Rs. 25.
6
S.P. of 30 articles = Rs. 6 x 30 = Rs. 36.
5
 Gain % = 11 x 100 % = 44%.
25
Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs. 5 x 30 = Rs. 25.
6
S.P. of 30 articles = Rs. 6 x 30 = Rs. 36.
5
 Gain % = 11 x 100 % = 44%.
25

292 / 320

Category: Profit and Loss - Mathematics

A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

 Gain = 80 x 100 % = 5%.
1600
Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

 Gain = 80 x 100 % = 5%.
1600

293 / 320

Category: Profit and Loss - Mathematics

The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?

Explanation:

Let C.P. be Rs. x.

Then, 1920 - x x 100 = x - 1280 x 100
x x

 1920 - x = x - 1280

 2x = 3200

 x = 1600

 Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000.
100
Explanation:

Let C.P. be Rs. x.

Then, 1920 - x x 100 = x - 1280 x 100
x x

 1920 - x = x - 1280

 2x = 3200

 x = 1600

 Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000.
100

294 / 320

Category: Profit and Loss - Mathematics

100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:

Explanation:
C.P. of 1 orange = Rs. 350 = Rs. 3.50
100
S.P. of 1 orange = Rs. 48 = Rs. 4
12
 Gain% = 0.50 x 100 % = 100 % = 14 2 %
3.50 7 7
Explanation:
C.P. of 1 orange = Rs. 350 = Rs. 3.50
100
S.P. of 1 orange = Rs. 48 = Rs. 4
12
 Gain% = 0.50 x 100 % = 100 % = 14 2 %
3.50 7 7

295 / 320

Category: Profit and Loss - Mathematics

When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?

Explanation:

85 : 18700 = 115 : x

 x = 18700 x 115 = 25300.
85

Hence, S.P. = Rs. 25,300.

Explanation:

85 : 18700 = 115 : x

 x = 18700 x 115 = 25300.
85

Hence, S.P. = Rs. 25,300.

296 / 320

Category: Profit and Loss - Mathematics

A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?

Explanation:
S.P. = 85% of Rs. 1400 = Rs. 85 x 1400 = Rs. 1190
Explanation:
S.P. = 85% of Rs. 1400 = Rs. 85 x 1400 = Rs. 1190

297 / 320

Category: Profit and Loss - Mathematics

A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6
5
For Rs. 6 , toffees sold = 6.
5
For Re. 1, toffees sold = 6 x 5 = 5.
6
Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6
5
For Rs. 6 , toffees sold = 6.
5
For Re. 1, toffees sold = 6 x 5 = 5.
6

298 / 320

Category: Profit and Loss - Mathematics

On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:

Explanation:

(C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)

 C.P. of 12 balls = S.P. of 17 balls = Rs.720.

 C.P. of 1 ball = Rs. 720 = Rs. 60.
12
Explanation:

(C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)

 C.P. of 12 balls = S.P. of 17 balls = Rs.720.

 C.P. of 1 ball = Rs. 720 = Rs. 60.
12

299 / 320

Category: Profit and Loss - Mathematics

In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295.

 Required percentage = 295 x 100 % = 1475 % = 70% (approximately).
420 21
Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 - 125) = Rs. 295.

 Required percentage = 295 x 100 % = 1475 % = 70% (approximately).
420 21

300 / 320

Category: Profit and Loss - Mathematics

A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:

Explanation:
C.P. of 1st transistor = Rs. 100 x 840 = Rs. 700.
120
C.P. of 2nd transistor = Rs. 100 x 960 = Rs. 1000
96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

 Gain % = 100 x 100 % = 5 15 %
1700 17
Explanation:
C.P. of 1st transistor = Rs. 100 x 840 = Rs. 700.
120
C.P. of 2nd transistor = Rs. 100 x 960 = Rs. 1000
96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

 Gain % = 100 x 100 % = 5 15 %
1700 17

301 / 320

Category: Profit and Loss - Mathematics

Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300.

Gain % = 300 x 100 % = 5 5 %
5500 11
Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) - (C.P.) = Rs.(5800 - 5500) = Rs. 300.

Gain % = 300 x 100 % = 5 5 %
5500 11

302 / 320

Category: Profit and Loss - Mathematics

If selling price is doubled, the profit triples. Find the profit percent.

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y - x) = (2y - x)    y = 2x.

Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.

 Profit % = x x 100 % = 100%
x
Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y - x) = (2y - x)    y = 2x.

Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.

 Profit % = x x 100 % = 100%
x

303 / 320

Category: Profit and Loss - Mathematics

A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?

Explanation:
C.P. = Rs. 100 x 392 = Rs. 1000 x 392 = Rs. 320
122.5 1225

 Profit = Rs. (392 - 320) = Rs. 72.

Explanation:
C.P. = Rs. 100 x 392 = Rs. 1000 x 392 = Rs. 320
122.5 1225

 Profit = Rs. (392 - 320) = Rs. 72.

304 / 320

Category: Profit and Loss - Mathematics

The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).

20 - x x 100 = 25
x

 2000 - 100x = 25x

125x = 2000

 x = 16.

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).

20 - x x 100 = 25
x

 2000 - 100x = 25x

125x = 2000

 x = 16.

305 / 320

Category: Profit and Loss - Mathematics

Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?

Explanation:
Cost Price of 1 toy = Rs. 375 = Rs. 31.25
12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 - 31.25) = Rs. 1.75

 Profit % = 1.75 x 100 % = 28 % = 5.6%
31.25 5
Explanation:
Cost Price of 1 toy = Rs. 375 = Rs. 31.25
12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 - 31.25) = Rs. 1.75

 Profit % = 1.75 x 100 % = 28 % = 5.6%
31.25 5

306 / 320

Category: Number system - Mathematics

The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:

Let the numbers be x and y.

Then, xy = 9375 and x = 15.
y
xy = 9375
(x/y) 15

 y2 = 625.

 y = 25.

 x = 15y = (15 x 25) = 375.

 Sum of the numbers = x + y = 375 + 25 = 400.

Let the numbers be x and y.

Then, xy = 9375 and x = 15.
y
xy = 9375
(x/y) 15

 y2 = 625.

 y = 25.

 x = 15y = (15 x 25) = 375.

 Sum of the numbers = x + y = 375 + 25 = 400.

307 / 320

Category: Number system - Mathematics

If one-third of one-fourth of a number is 15, then three-tenth of that number is:

Explanation:

Let the number be x.

Then, 1 of 1 of x = 15      x = 15 x 3 x 4 = 180.
3 4
So, required number = 3 x 180 = 54.
Explanation:

Let the number be x.

Then, 1 of 1 of x = 15      x = 15 x 3 x 4 = 180.
3 4
So, required number = 3 x 180 = 54.

308 / 320

Category: Number system - Mathematics

The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

Explanation:

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

 9x = 36

 x = 4.

 Required difference = (2x + x) - (2x - x) = 2x = 8.

Explanation:

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

 9x = 36

 x = 4.

 Required difference = (2x + x) - (2x - x) = 2x = 8.

309 / 320

Category: Number system - Mathematics

Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

Explanation:

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

 Third integer = x + 4 = 15.

Explanation:

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

 Third integer = x + 4 = 15.

310 / 320

Category: Number system - Mathematics

A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

Explanation:

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.

Hence, required number = 253.

Explanation:

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.

Hence, required number = 253.

311 / 320

Category: Number system - Mathematics

The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:

Explanation:

Let the numbers be x and y.

Then, xy = 120 and x2 + y2 = 289.

 (x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529

 x + y = 529 = 23.

Explanation:

Let the numbers be x and y.

Then, xy = 120 and x2 + y2 = 289.

 (x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529

 x + y = 529 = 23.

312 / 320

Category: Number system - Mathematics

The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

Explanation:

Let the numbers be a, b and c.

Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

 (a + b + c) = 400 = 20.

Explanation:

Let the numbers be a, b and c.

Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

 (a + b + c) = 400 = 20.

313 / 320

Category: Number system - Mathematics

A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

 (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

 (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

314 / 320

Category: Number system - Mathematics

The sum of two number is 25 and their difference is 13. Find their product.

Explanation:

Let the numbers be x and y.

Then, x + y = 25 and x - y = 13.

4xy = (x + y)2 - (x- y)2

= (25)2 - (13)2

= (625 - 169)

= 456

 xy = 114.

Explanation:

Let the numbers be x and y.

Then, x + y = 25 and x - y = 13.

4xy = (x + y)2 - (x- y)2

= (25)2 - (13)2

= (625 - 169)

= 456

 xy = 114.

315 / 320

Category: Number system - Mathematics

The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, x + y = 15 and x - y = 3   or   y - x = 3.

Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, x + y = 15 and x - y = 3   or   y - x = 3.

Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

316 / 320

Category: Number system - Mathematics

A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

Explanation:
Let the ten's and unit digit be x and 8 respectively.
x
Then, 10x + 8 + 18 = 10 x 8 + x
x x

 10x2 + 8 + 18x = 80 + x2

 9x2 + 18x - 72 = 0

 x2 + 2x - 8 = 0

 (x + 4)(x - 2) = 0

 x = 2.

Explanation:
Let the ten's and unit digit be x and 8 respectively.
x
Then, 10x + 8 + 18 = 10 x 8 + x
x x

 10x2 + 8 + 18x = 80 + x2

 9x2 + 18x - 72 = 0

 x2 + 2x - 8 = 0

 (x + 4)(x - 2) = 0

 x = 2.

317 / 320

Category: Number system - Mathematics

What is the sum of two consecutive even numbers, the difference of whose squares is 84?

Explanation:

Let the numbers be x and x + 2.

Then, (x + 2)2 - x2 = 84

 4x + 4 = 84

 4x = 80

 x = 20.

 The required sum = x + (x + 2) = 2x + 2 = 42.

Explanation:

Let the numbers be x and x + 2.

Then, (x + 2)2 - x2 = 84

 4x + 4 = 84

 4x = 80

 x = 20.

 The required sum = x + (x + 2) = 2x + 2 = 42.

318 / 320

Category: Number system - Mathematics

In a two-digit, if it is known that its unit's digit exceeds its ten's digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:

Explanation:

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

 (11x + 2)(2x + 2) = 144

 22x2 + 26x - 140 = 0

 11x2 + 13x - 70 = 0

 (x - 2)(11x + 35) = 0

 x = 2.

Hence, required number = 11x + 2 = 24.

Explanation:

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

 (11x + 2)(2x + 2) = 144

 22x2 + 26x - 140 = 0

 11x2 + 13x - 70 = 0

 (x - 2)(11x + 35) = 0

 x = 2.

Hence, required number = 11x + 2 = 24.

319 / 320

Category: Number system - Mathematics

The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, (10x + y) - (10y + x) = 36

 9(x - y) = 36

 x - y = 4.

Explanation:

Let the ten's digit be x and unit's digit be y.

Then, (10x + y) - (10y + x) = 36

 9(x - y) = 36

 x - y = 4.

320 / 320

Category: Number system - Mathematics

Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

Explanation:

Let the number be x.

Then, x + 17 = 60
x

 x2 + 17x - 60 = 0

 (x + 20)(x - 3) = 0

 x = 3.

Explanation:

Let the number be x.

Then, x + 17 = 60
x

 x2 + 17x - 60 = 0

 (x + 20)(x - 3) = 0

 x = 3.

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