Question 1

In a polling booth, total number of voters is 1575 out of which 0.4 part are male voters. If a candidate gets 0.4 part of female votes, then find how many votes did the candidate get?

Accepted Answer

Number of male voters = 4/10 of 1575 = 4/10 x 1575 = 630 Number of female voters=1575 - 630 = 945 Now, total votes that candidate got = (6/10) x 630 + (4/10) x 945 = 378 + 378 = 756

Question 2

In a village, (5/8)th of the population are adults,half of the adults are male,(4/5)th of adult females are illiterate. if 400 females are illiterate, then the population of the village is?

Accepted Answer

Let the population of village = N Then, adult's population = 5N/8 and adult females = 1/2 x 5N/8 = 5N/16 ∴ illiterate females = 4/5 of adult female population ⇒ 400 = 4/5 x 5N/16 ⇒ N = (400 x 16)/4 ∴ N = 1600

Question 3

(1/4)th of number of boys and (3/8)th of number of girls participated in annual sport days in the school. What fractional part of total number of students participated?

Accepted Answer

Total number of students participated = (B/4 + 3G/4) ∴ Required percentage= (B/4 + 3G/4) / (B+G) x 100% Clearly, given data is inadequate.

Question 4

A scooter before overhauling requires 2/3 h service time every 45 days,while after overhauling it requires 2/3 h service time every 60 days. what fraction of pre-overhauling service time is saved in the latter case?

Accepted Answer

Time required for service before overhauling = 2/3h = 40 min in 45 days and time required for service after overhualing =2/3h= 40 min in 60 days Now,LCM of 45 and 60 = 180 days time required for service before overhauling is 180 days = 40 x 4 = 160 min and time required for service before overhauling is 180 days =40 x 3 = 120 min ∴ fraction of time saved = 160 - 120/160 =40/160 = 1/4

Question 5

From a rope 30 m long, a person cuts off as many pieces as possible, each 3 1/ 4 m long. What fraction of the whole will be left?

Accepted Answer

Length of each piece = 31/4 = 13/4 m Maximum pieces that can be cut = 30/(13/4) = 9 ∴ Rope left = 30 - 13/4 x 9 = 120 - 117/4 = 3/4 m ∴ Required fraction = 3/4/30 = 1/4

Question 6

If 0. 3+ 0. 6 + 0. 7 + 0. 8 = x, then x is equal to?

Accepted Answer

0.3+ 0.6 + 0.7 + 0.8 = x, ⇒ x = 3/9 + 6/9 + 7/9 + 8/9 = 24/9 = 8/3 = 22/3

Question 7

A boy on being asked 13/14 of a certain fraction had made the mistake of dividing the fraction by 14/13 and so got an answer that exceeded the correct answer by 3/65. The correct fraction is?

Accepted Answer

Let the fraction be N. Then, N / (13/14) - (13/14) x N = 3/65 ⇒ 14N/13 - 13N/14 = 3/65 ⇒ (196N - 169N)/182 = 3/65 ⇒ 27N/182 = 3/65 ∴ N = (3 x 182) / (27 x 65) = 14/45 So, correct answer will be (13/14) x (14/45) = 13/45

Question 8

A man left (1/7)th of his property to his daughter and the remaining to his sons to be equally divided among them. If the share of the each son be double than that of the daughter. Then, find the number of sons?

Accepted Answer

Let the total property = N Then, property of his daughter = P/7 ∴ Remaining property = P - P/7 = 6P/7 Now, this property is divided among his sons, so that each son gets twice the property of his daughter. Let number of sons be K. Then, 6N/7 x K = 2N/7 ⇒ K = 3

Question 9

A vessel, full of water, weighs 16.5 kg. When the vessel is (1/4)th full, it weight 5.25 kg. Then, weight of the empty vessel (in kg) is?

Accepted Answer

Let the weight of vesel = x kg Then, 1/4 of volume of water = (5.25 - x)kg Full volume of water = (16.5 - x) ∴ 4(5.25 - x) = (16.5 - x) ⇒ 21 - 4x = 16.5 - x ⇒ 21 - 16.5 = 3x ∴ x = 4.5/3 = 1.5 kg

Question 10

A has twice as much money as B. They play together and at the end of the first game B wins one-third of A's money from A, what fraction of the sum that B now has, must A win black in the second game, so that they may have exactly equal money?

Accepted Answer

Let A has ₹ 200 at the starting of game, then B has ₹ 100 at the starting of game. After the game, money left with A = 200 - 200/3 = ₹ 400/3 ∴ Total money with B = 100 + 200/3 = ₹ 500/3 Let the fraction of money lost by B = N Then, 500/3 - N x 500/3 = 400/3 + N x 500/3 ⇒ 500/3 - 400/3 = 500N/3 + 500N/3 ⇒ 100/3 = 1000N/3 ∴ N = (100/3) x (3/1000) = 1/10 so, the fraction of money lost is 1/10.

Question 11

Sum of three fractions is 2 11/ 24. If the greatest fraction is divided by the smallest fraction, the result is 7/6, which is greater than the middle fraction by 1/3. find all the three fractions?

Accepted Answer

Let the fraction be x, y and z respectively in decreasing order. Then, according to the question, x/z = 7/6 ⇒ x = 7z/6 And y = 7/6 - 1/3 = 7-2/6 = 5/6 Now, x + y + z = 211/24 ⇒ 7z/6 + 5/6 + z = 59/24 ⇒ (13z + 5)/ 6 = 59/24 ⇒ 13z = 59/4 - 5 =39/4 ⇒ z = 3/4 ∴ x =7z/6 = 7/8 Hence, the fraction are 7/8, 5/6, 3/4.

Question 12

Sum of three fraction is 2 11/ 24 . If the greatest fraction is divided by the smallest fraction the result is 7/6, which is greater than the middle fraction by 1/3. Find all the three fractions.

Accepted Answer

Let the greatest, middle and smallest fraction be P, Q and R respectively in decreasing order. According to the question. (Greatest fraction) / (Smallest fraction) = 7/6 ⇒ P/R = 7/6 ⇒ P = 7R/6 ....(i) and Q = 7/6 - 1/3 = 7 - 2/6 = 5/6 (ii) Now, P + Q + R = 211/24 (iii ) Substitute the value of P and Q from Eqs. (i), (ii) and put in Eqs (iii), we get ⇒ 7R/6 + 5/6 + R = 59/24 ⇒ (7R + 5 + 6R)/6 = 59/24 ⇒ (13R + 5) / 6 = 59/24 ⇒ 13R = (59/4) - 5 = 39/4 ⇒ R = 39 / (4 x 13) ∴ R = 3/4 On putting the value of R in Eq. (i), we get P = 7R/6 = (7/6) x (3/4) = 7/8

Question 13

In the year 2011, Shantanu gets ₹ 3832.5 as his pocket allowance. Find his pocket allowance per day.

Accepted Answer

Shantanu's pocket allowance = ₹ 3832.50 Total days in 2011 (general year) = 365 days Allowance per day = 3832.5/365 = ₹ 10.5

Question 14

1/8 part of a pencil is black and 1/2 part of the remaining is white. If the remaining part is blue and length of this blue part is 31/2 cm, then find the length of the pencil.

Accepted Answer

Let total length of pencil be x cm. Then, black part = x/8 Remaining part pencil after black part = x - x/8 = 7x/8 then White part = 1/2 (7x/8) = 7x/16 Now Remaining part of pencil is blue = total length of pencil - (length of black part + length of white part) length of blue part of pencil = x - (x/8 + 7x/16) length of blue part of pencil = x - (2x + 7x)/16 length of blue part of pencil = x - 9x/16 length of blue part of pencil = (16 x - 9x)/16 ∴ Length of blue part = 7x/16; According to question Length of blue part = 31/2 7x/16 = 31/2 ⇒ 7x/16 = 7/2 ⇒ x/16 = 1/2 ⇒ x = 16/2 ∴ x = 8 cm

Question 15

A man reads 3/8 of a book on a day and 4/5 of the remainder on the second day. If the number of pages still unread are 40, then how many pages did the book contain?

Accepted Answer

Let the number of pages be p. Number of pages read on first day = 3p/8 Remaining pages = p - 3p/8 = (8p - 3p)/8 = 5p/8 Number of pages read on second day = 4/5 x 5p/8 = p/2 Now , remaining pages = p - [ 3p/8 + p/2] = p - [(3p + 4p)/8] = [p- 7p/ 8] = [(8p - 7p)/ 8] = p/8 According to the question p/8 = 40 ⇒ p = 40 x 8 ∴ p = 320

Question 16

If the numerator of a fraction is increased by 200% and the denominator of the fraction is increased by 150%, the resultant fraction is 9/35. What is the original fraction?

Accepted Answer

Let the original fraction be p/q. Numerator is increased by 200%. ∴ Numerator = p + 200% of p = p + 200p/100 = p + 2p = 3p and denominator of the fraction is increased by 150%. Denominator = q + 150q/100 = (100q + 150q)/100 = 250q/100 = 5q/2 Then, according to the question, 3p/5q/2 = 9/35 3p x 2/5q = 9/35 6p/5q = 9/35 ∴ p/q = 9 x 5/35 x 6 p/q = 3/14

Question 17

A, B, C and D purchase a gift worth ₹ 60. A pays 1/2 of what others are paying, B pays 1/3rd of what other are paying and C pays 1/4th of what others are paying. What is the amount paid by D?

Accepted Answer

Let A, B, C and D pay ₹x, ₹y, ₹z and ₹a According to the question, x = 1/2(y + z + a) 2x = (y + z + a) ..........(i) y = 1/3(x + z + a) 3y = (x + z + a) ...........(ii) z = 1/4(x + y + a) 4z = (x + y + a) ..........(iii) Also, x + y + z + a = 60 ...............(iv) Now, put the value of x + y + a = 4z from (iii) in (iv) Then, 4z + z = 60 ⇒ 5z = 60 ∴ z = 12 Similarly, on putting the value of x + z + a = 3y from (ii) in (iv), we get 3y + y = 60 = ⇒ 4y = 60 ∴ y = 15 Again, on putting the value of (y + z + a) = 2x from (i) in (iv), we get 2x + x = 60 ⇒ 3x = 60 &ther4; x = 20 Now, x + y + z + a = 60 On putting the value of x, y and z, we get 12 + 15 + 20 + a = 60 ∴ a = 60 - 47 = ₹ 13

Question 18

The Numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator, then find the fraction.

Accepted Answer

Let denominator of fraction = x Then, numerator = x - 4 ∴ Fraction = x - 4/x Now, according to the question, 8[(x - 4) - 2] = (x + 1) (x -4) - 2= (x + 1)/8 ⇒ x - 6 = x + 1/8 ⇒ 8x - 48 = x + 1 ⇒ 8x - x = 48 + 1 ⇒ 7x = 49 ⇒ x = 49/7 ∴ x = 7 ∴ Fraction = (x - 4)/x Put the value of x. ∴ Fraction = (7 - 4)/7 =3/7

Question 19

783 ÷ 9 ÷ 0.75 =?

Accepted Answer

? = 783/9 x 0.75 = 116

Question 20

4/7 of a pole is in the mud. When 1/3 of it is pulled out, 250 cm of the pole is still in the mud. Find the full length of the pole .

Accepted Answer

Let us assume length of pole is L. Length of the pole in Mud is = L x 4/7 = 4L/7 Length of the Pole which is pulled out = L x 1/3 = L/3 Length of the pole is still in mud = Length of the pole in Mud - Length of the Pole which is pulled out Length of the pole is still in mud = 4L/7 - L/3 According to question, Length of the pole is still in mud = 250 4L/7 - L/3 = 250 (4L x 3 - L x 7)/7 x 3 = 250 12L - 7L = 250 x 21 5L = 250 x 21 L = 250 x 21/5 L = 50 x 21 L = 1050

Decimal Fraction Questions