Fractions Extra Questions And Answers - Class 5 Mathematics | Quick Revision Study Guide

Q1: Simplify:

(a) 40/75

40 = 2 × 2 × 2 × 5  75 = 3 × 5 × 5
H.C.F. of 40 and 75 is 5.
Divide numerator and denominator by 5:
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
H.C.F. of 27 and 90 is 3 × 3 = 9.
Divide numerator and denominator by 9:
27/90 = (27 ÷ 9)/(90 ÷ 9) = 3/10
Thus, the simplest form of 27/90 is 3/10.

Q2: Replace the blanks in 35/84 = 5/? by the correct number.

35 ÷ 7 = 5, so divide both numerator and denominator by 7.
35/84 = (35 ÷ 7)/(84 ÷ 7) = 5/12
Thus, 35/84 = 5/12.

Q3: Convert each of the following improper fractions into mixed numbers.
(a) 14/5

Divide 14 by 5: 14 ÷ 5 = 2 remainder 4.

(b) 9/5

Divide 9 by 5: 9 ÷ 5 = 1 remainder 4.

Q4: 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.

Q5: 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.

Q6: Express each of the following mixed numbers as an improper fraction.
(a) 

Method to convert a mixed number to an improper fraction:
Multiply the whole number by the denominator, add the numerator, and write the result over the same denominator.
Hence, Mixed number = (Whole × Denominator + Numerator)/Denominator.

(b) 

Apply the same method:
Multiply the whole part by the denominator, add the numerator, and place over the denominator.

Q7: Is the fraction 16/21 in its lowest terms?

Find prime factors:
16 = 2 × 2 × 2 × 2 (2^4),  21 = 3 × 7.
No common prime factor, so H.C.F. of 16 and 21 is 1.
Hence, the fraction 16/21 is in its lowest terms.

Q8: Find an equivalent fraction of 5/8 with numerator 20.

To get 20 in the numerator multiply numerator and denominator by 4:
5/8 = (5 × 4)/(8 × 4) = 20/32.
Thus, 5/8 = 20/32.

Q9: Is the fraction 20/35 in its lowest terms?

H.C.F. of 20 and 35 is 5.
Divide numerator and denominator by 5:
20/35 = (20 ÷ 5)/(35 ÷ 5) = 4/7.
Since it can be simplified, 20/35 is not in its lowest terms.
Its simplest form is 4/7.

Q10: Find an equivalent fraction of 56/72 with denominator 18.

To change the denominator from 72 to 18 divide numerator and denominator by 4:
56/72 = (56 ÷ 4)/(72 ÷ 4) = 14/18.
We can simplify further by dividing by 2:
14/18 = 7/9.
Thus, 56/72 = 14/18 = 7/9.

Q11: 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,

Q12: Find the product:
(a) 4/11 x 5/9

Thus, the product is 20/99.

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

Thus, the product is 45/14 or 3 (3/14).

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

Thus, the product is 38/5 or 7 (3/5).

Q13: 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.

Q14: 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

Q15:Find the product of 1/2 x 3/4 x 5/8

Thus, the product is 15/64.

Q16:Divide:
(a) 5/8 by 2/7

Reciprocal of 2/7 is 7/2.

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

Reciprocal of 13/14 is 14/13.

(c) Reciprocal of

Q17: 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,

Q18: Add 1/6, 7/12, and 1/8.

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

(Changing to its lowest terms)

Q19: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.

Q20: 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).

Q21: Subtract 2(1/6) from 3(3/5).

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

Q22: Add 5/12 and 3/8.

Find L.C.M. of 12 and 8: L.C.M. = 24.
Convert:
5/12 = (5 × 2)/(12 × 2) = 10/24
3/8 = (3 × 3)/(8 × 3) = 9/24
Add: 10/24 + 9/24 = 19/24.
Thus, the sum is 19/24.

Q23: 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

Q24: A vessel had  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.

Q25: Shyam has Rs 42. He wants to buy chocolates from that money. If the cost of each chocolate is Rs , 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.

Q26: Add 2(1/4), 3(5/8) and 5(1/6).

[Changing the mixed numbers into improper fractions]

Q27: Add 5/11 and 3/11.

Thus, the sum is 8/11.

Q28: Divide:
(a) 5/7 by 4

Reciprocal of 4 is 1/4.

(b) by 19

Reciprocal of 19 is 1/19.

Q29: Subtract 3/4 from.

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

Q30: Add 3/17, 4/17 and 2/17.

Thus, the sum is 9/17.

Q31: 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

Q32: Multiply:

(a) 4/9 by 10

Thus, the product is 40/9 or 4 (4/9).

(b) 7/15 by 18.

Thus, the product is 42/5 or 8 (2/5).

(c)  by 25

Q33: Swati had m long rope. She cut it into 7 equal parts. Find the length of each piece of the rope.

Length of the rope =  m.
Number of pieces cut out = 7Length of each piece

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

Q34: Add  and

[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.

Q35: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) = (3 × 7 + 4)/7 = 25/7, so its reciprocal = 7/25.

Q36: 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,

Q37: Add

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

Q 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,

Q39: Add and 1/6.

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

Q40: 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)