Class 5 Maths - Fractions - Question Answers

Question 1: Simplify:

(a) 40/75

40 = 2 × 2 × 2 × 5   75 = 3 × 5 × 5
Thus,  H.C.F. of 40 and 75 is 5.

Now, 40/75 = 40 ÷ 5/75 ÷ 5 = 8/15
Thus, the simplest form of 40/75 is 8/15

(b) 27/90

27 = 3 × 3 × 3   90 = 2 × 3 × 3 × 5
Thus, H.C.F. of 27 and 90 = 3 × 3 = 9.
Now, 27/90 = 27 ÷ 9/90 ÷ 9 = 3/10
Thus, the simplest form of 27/90 is 3/10

Question 2: Replace the blanks in 35/84 = 5/? by the correct number.

Since, 35 ÷ 7 = 5, so we divide 35 by 7 and also 84 by 7. 35
35/84 = 35 ÷ 7/84 ÷ 7 = 5/12
Thus, 35/84 = 5/12.

Question 3: Convert each of the following improper fractions into mixed numbers.
(a) 14/5

Thus, 14/5 = 2(4/5)

(b) 9/5

Thus, 9/5 = 1(4/5)

Question 4: Compare
(a) 4/7 and 3/5

Given fractions are 4/7 and 3/5.
Let us now find the L.C.M. of 7 and 5.
L.C.M. of 7 and 5 = 7 × 5 = 35.
Now,

Now, compare 20/35 and 21/35
Since 20 < 21 or 21 > 20, therefore,

Hence, 3/5 > 4/7.

(b) 5/12 and 7/8

Given fractions are 5/12 and 7/8
Let us now find the L.C.M. of 12 and 8.
12 = 2 × 2 × 3, 8 = 2 × 2 × 2
So, L.C.M. of 12 and 8 = 2 × 2 × 2 × 3 = 24.
Now,

Now, compare 10/24 and 21/24.
Since, 10 < 21  or  21 > 10, therefore,

Hence, 7/8 > 5/12.

Question 5: Check whether the following pairs of fractions are equivalent or not :
(a) 6/13 and 30/65

[Cross multiply]
6 × 65 = 390,   13 × 30 = 390. Both the products are equal.
Thus, 6/13 and 30/65 are equivalent.

(b) 7/12 and 42/70

[Cross multiply]
7 × 70 = 490,  12 × 42 = 504.  Products are not equal.
Thus, 7/12 and 42/70 are not equivalent.

Question 6: Express each of the following mixed numbers as an improper fraction.
(a) 2(3/5)

(b) 3(4/7)

Question 7: Is the fraction 16/21 in its lowest terms?

H.C.F. of 16 and 21 is 1.

Hence, the fraction 16/21 is in its lowest terms.

Question 8: Find an equivalent fraction of 5/8 with numerator 20.

To get 20 in the numerator, we multiply 5 by 4 and also 8 by 4.

5/8 = 5 x 4/8 x 4 = 20/32
Thus, 5/8 = 20/32

Question 9: Is the fraction 20/35 in its lowest terms?

H.C.F. of 20 and 35 is 5.

Hence, the fraction 20/35 is not in its lowest terms.

Question 10: Find an equivalent fraction of 56/72 with denominator 18.

To get 18 in the denominator, we divide 72 by 4 and also, 56 by 4.
56/72 = 56 ÷ 4/72 ÷ 4 = 14/18
Thus, 56/72 = 14/18.

Question 11: Divide:
(a) 15 by 5/7

Reciprocal of 5/7 is 7/5
Thus, 15

(b) 28 by 3(1/5)

Reciprocal of 3(1/5) = Reciprocal of 16/5 = 5/16.
Thus,

Question 12: Find the product:
(a) 4/11 x 5/9

(b) 3(4/7) by 9/10

(c) 2(3/8) x 3(1/5)

Question 13: Preeti had one rope of 5(1/6) m length and another of 3(1/2) m length. How much length of rope did Preeti have in all?

Length of one rope = 5(1/6) m =
Length of another rope 3(1/2) m =
Total length of the two ropes = [L.C.M. of 6 and 2 = 6]

Hence, Preeti has 8(2/3) m rope in all.

Question 14: Find the difference between 7/15 and 9/10.

L.C.M. of 15 and 10 is 30.
Now,

Hence, 27/30 > 14/30
Hence, the difference of

Question 15: Find the product of 1/2 x 3/4 x 5/8

Question 16: Divide:
(a) 5/8 by 2/7

Reciprocal of 2/7 is 7/2.
Thus,

(b) 3(5/7) by 13/14

Reciprocal of 13/14 is 14/13
Thus,

(c) Reciprocal of

Thus,

Question 17: Compare

(a) 3/11 and 5/7

Given fractions are 3/11 and 5/7.
Cross multiply

Now, 3 × 7 = 21 and 11 × 5 = 55.

Since, 21 < 55, hence,

(b) 7/10 and 9/13

Cross multiply

Now, 7 × 13 = 91 and 10 × 9 = 90.
Since, 91 > 90, hence,

Question 18: Add 1/6, 7/12, and 1/8.

L.C.M. of 6, 12 and 8 is 24.

(Changing to its lowest terms)

Question 19: Alka walks 3(1/4) km in 1 hour. How far does she go in 7 hours?

Distance covered in hour = 3(1/4) km = 13/4 km.
Distance covered in 7 hours =
Hence, Alka goes 22(3/4) km in 7 hours.

Question 20: The cost of 5(5/8) litres of milk is Rs 37(1/2). What is the cost of 1 litre of milk?

Cost of 5(5/8) litres of milk = Rs 37(1/2)
The cost of 1 litre of milk =

Hence, the cost of 1 litre of milk is 6(2/3).

Question 21: Subtract 2(1/6) from 3(3/5).

[Changing the mixed numbers into improper fractions]
[L.C.M. of 5 and 6 = 30]

Question 22: Add 5/12 and 3/8.

Let us find the L.C.M. of 12 and 8. L.C.M. of 12 and 8 = 24.
Now,
Thus,

Question 23: Arrange the following fractions in ascending order.

Let us find the L.C.M. of the denominators of the given fractions:
5 = 1 × 5, 3 = 1 × 3, 15 = 3 × 5, 10 = 2 × 5

Thus, L.C.M of 5, 3, 15 and 10 = 2 × 3 × 5 = 30.
Now, we change each of the given fractions into equivalent fractions with denominator 30.

Clearly, 2 < 20 < 21 < 24, so,
Hence
Hence, the ascending order is

Question 24: A vessel had 3(1/4) litres of milk. A cat drank 1/2 litres. How much milk is left in the vessel?

Milk in the vessel =
Milk left in the vessel

Hence,  of milk is left in the vessel.

Question 25: Shyam has Rs 42. He wants to buy chocolates for that money. If the cost of each chocolate is Rs 4(1/5), how many chocolates can he buy?

Cost of 1 chocolate = Rs 4(1/5)
Amount of money Shyam has = Rs 42
Number of chocolates Shyam can buy =

Hence, Shyam can purchase 10 chocolates for Rs 42.

Question 26: Add 2(1/4), 3(5/8) and 5(1/6).

[Changing the mixed numbers into improper fractions]

Question 27: Add 5/11 and 3/11.

Question 28: Divide:
(a) 5/7 by 4

Reciprocal of 4 is 1/4.
Thus,

(b) 3(4/5) by 19.

Reciprocal of 19 is 1/19.
Thus,

Question 29: Subtract 3/4 from 2(4/7).

[Changing 2(4/7) into an improper fraction]
 [L.C.M. of 7 and 4 = 28]

Question 30: Add 3/17, 4/17 and 2/17.

Question 31: Arrange the following fractions in descending order.

Let us find the L.C.M of the denominators of the given fractions:
8 = 2 × 2 × 2, 4 = 2 × 2, 12 = 2 × 2 × 3,   16 = 2 × 2 × 2 × 2
L.C.M. of 8, 4, 12 and 16 = 2 × 2 × 2 × 2 × 3 = 48.
Now, change each of the given fractions into equivalent fractions with denominator 48.

Clearly, 36 > 33 > 30 > 28.
 Thus,
Hence, the descending order is

Question 32: Multiply:

(a) 4/9 by 10

(b) 7/15 by 18.

(c) 2(3/8) by 25

Question 33: Swati had 8(3/4) m long rope. She cut it into 7 equal parts. Find the length of each piece of the rope.

Length of the rope = 8(3/4) m.
Number of pieces cut out = 7
Length of each piece

Hence, the length of each piece of the rope = 1(1/4) m.

Question 34: Add 3(2/5) and 4(3/10)

 [Add whole numbers and fractions separately]
[L.C.M. of 5 and 10 = 10]

Another method: Change the mixed numbers into improper fractions and add.

Question 35: Find the reciprocal of:
(a)  9
(b) 7/8
(c) 4/15
(d) 3(4/7)

(a) Reciprocal of 9 = 1/9

(b) Reciprocal of 7/8 = 8/7

(c) Reciprocal of 4/15 = 15/4.
(d) 3(4/7) = 25/7
Reciprocal of 25/7 = 7/25

Question 36: Subtract 3/8 from 5/12

Let us find the L.C.M. of 8 and 12.
L.C.M. of 8 and 12 = 24.
Now,

Hence,

Question 37: Add

[Changing 3(5/6) into an improper fraction]
 [L.C.M. of 15, 10 and 6 = 30]
 [Changing into its lowest terms]

Question 38: Subtract 4/9 from 5.

We write the whole number as a fraction by putting 1 in the denominator.
Clearly,
Now, L.C.M. of 1 and 9 = 9.
Hence,

Question 39: Add 3(5/8)and 1/6.

[Changing 3(5/8) into an improper fraction]
[L.C.M. of 8 and 6 = 24]

Question 40: The cost of 1 m of cloth is 15(2/5). Find the cost of 4(3/7) m of cloth.

Cost of 1 m of cloth = Rs 15(2/5) = Rs 77/5
Cost of 4(3/7) m of cloth =
Hence, the cost of 4(3/7) m of cloth = Rs 68(1/5).