Class 5

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.9

Question 1.
Find the product of the following decimal numbers :
(a) 5.15 × 6
(b) 52.4 × 2
(c) 0.31 × 5
(d) 9.05 × 0.2
(e) 7.24 × 2.3.
Solution:
(a) 5.15 × 6

5.15 × 6 = 30.90

(b) 52.4 × 2

52.4 × 2 = 104.8

(c) 0.31 × 5

0.31 × 5 = 1.55

(d) 9.05 × 0.2

9.05 × 0.2 = 1.810

(e) 7.24 ×2.3

7.24 × 2.3 = 16.652

Question 2.
Find the division of the following decimal numbers :
(a) 18.24 ÷ 3
(b) 8.64 ÷ 4
(c) 2.48 ÷ 8
(d) 16.5 ÷ 15
(e) 34.3 ÷ 7.
Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 5 Money (Currency) Ex 5.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 5 Money (Currency) Ex 5.4

1. Find the value of the following :

Question 1.
₹ 258 × 17
Solution:

Question 2.
₹ 410 × 20
Solution:

Question 3.
₹ 518 × 18
Solution:

Question 4.
₹ 220 × 14
Solution:

Question 5.
₹ 206 × 25.
Solution:

2. Evaluate :

Question 1.
₹ 3120 ÷ 10
Solution:
₹ 3120 ÷ 10 = ₹ 312

Question 2.
₹ 1590 ÷ 15
Solution:
₹ 1590 ÷ 15 = ₹ 106

Question 3.
₹ 4272 ÷ 16
Solution:
₹ 4272 ÷ 16 = ₹ 267

Question 4.
₹ 4200 ÷ 20
Solution:
₹ 4200 ÷ 20 = ₹ 210

Question 5.
₹ 6500 ÷ 25.
Solution:
₹ 6500 ÷ 25 = ₹ 260

Question 3.
The cost of a calculator is ₹ 415. Find the cost of 17 such calculators.
Solution:
The cost price of 1 calculator = ₹ 415
The cost price of 17 calculators = ₹ 415 × 17 = ₹ 7055

The cost price of 17 calculators = ₹ 7055

Question 4.
The price of 1 kg ghee is ₹ 435. What is the price of 18 kg ghee ?
Solution:
The cost price of 1 kg ghee = ₹ 435
The cost price of 18 kg ghee = ₹ 435 × 18

Thus, the cost price of 18 kg ghee = ₹ 7830

Question 5.
The cost of 24 glasses is ? 2880. Find the cost of a glass.
Sol. The cost price of 24 glasses = ₹ 2880
The cost price of 1 glass = ₹ 2880 ÷ 24 = ₹ 120

The cost price of 1 glass = ₹ 120

Question 6.
The cost of 19 geometry boxes is ₹ 2850. Find the cost of a geometry box.
Solution:
The cost price of 19 geometry boxes = ₹ 2850
The cost price of 1 geometry box = ₹ 2850 ÷ 19 = ₹ 150

The cost price of 1 geometry box = ₹ 1150

Question 7.
The price of 1 litre petrol is ₹ 73. What is the cost of 12 litres petrol.
Solution:
Cost of 1 litre of petrol = ₹ 73
Cost of 12 litres of petrol = ₹ 73 × 12 = ₹ 876

Cost of 12 litres of petrol = ₹ 876

Question 8.
The cost of 25 kg rice is ₹ 2000. Find the cost of 1 kg rice.
Solution:
Cost of 25 kg of rice = ₹ 2000
Cost of 1 kg of rice = ₹ 2000 ÷ 25 = ₹ 80

Cost of 1 kg of rice = ₹ 80

Question 9.
The price of 1 m cloth is ₹ 500. Find the cost of 18 m cloth.
Solution:
Cost of 1 m cloth = ₹ 500
Cost of 18 m cloth = ₹ 500 × 18 = ₹ 9000

Cost of 18 m cloth = ₹ 9000

10. Fill in the blanks :

Question 1.
₹ 13 × 8 = ____________
Solution:
₹ 104

Question 2.
₹ 24 × 5 = ___________
Solution:
₹ 120

Question 3.
₹ 24 ÷ 3 = __________
Solution:
₹ 8

Question 4.
₹ 72 ÷ 8 = ____________
Solution:
₹ 9

Question 5.
₹ 25 × 6 = __________
Solution:
₹ 150

Question 6.
₹ 100 ÷ 10 = __________
Solution:
₹ 10

Question 7.
There are __________ ₹ 100 notes in ₹ 1000.
Solution:
10

Question 8.
There are ___________ ₹ 50 notes in ₹ 300.
Solution:
6

Question 9.
There are ___________ ₹ 20 notes in ₹ 500.
Solution:
25

Question 10.
There are _________ ₹ 500 notes in ₹ 2000.
Solution:
4

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 5 Money (Currency) Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 5 Money (Currency) Ex 5.3

Question 1.
Ravi bought a notebook for ₹ 50, a book for ₹ 125 and bought a pen for ₹ 150. Blow much money did he spend ?
Solution:
Ravi bouight a notebook for = ₹ 50
Ravi bought a book for = ₹ 25
Ravi ibought a pen for = ₹ 150
The amount of money he spent =

Question 2.
Manveet Kaur has ₹ 148.50 with her. Her father gave ₹ 116.50 to her. How much amount does she has?
Solution:

Question 3.
Paras purchased a bag for ₹ 450 and gave ₹ 500 to the shopkeeper. How much amount would he get back from the shopkeeper?
Solution:
Paras bought a bag for = ₹ 450
The amount given to the shopkeeper = ₹ 500
The amount he will get back from the shopkeeper

Paras will get ₹ 50 from the shopkeeper

Question 4.
Gurdeep has ₹ 1000 with him. He purchases shoes for ₹ 742. How much amount is left with him ?
Solution:

Question 5.
Prabhjot has ₹ 2168.50 and her brother Simarjeet has ₹ 1248.50. How much amount they both have ?
Solution:
The amount of money Prabhjot has = ₹ 2168.50
The amount of money her brother has = ₹ 1248.50
The amount of money they together have

Question 6.
A shopkeeper has ₹ 1000. He bought a radio for ₹ 650. How much amount was left with him?
Solution:
The amount of money shopkeeper had = ₹ 1000
He bought a radio for = ₹ 650
The amount of money left with him = ₹ 350
He has now ₹ 350 with him

Question 7.
Harleen went to the market with her friend. She bought goods worth ₹ 3467.50 and her friend bought goods, worth ₹ 3350.25. How much Harleen spent more than her friend ?
Solution:
Harleen bought goods for = ₹ 3467.50
Her friend bought goods for = ₹ 3350.25
The amount Harleen spent more than her friend = ₹ 117.25
Harleen has spent ₹ 117.25 more than her friend

Question 8.
Avneet has purchased a shirt for ₹ 1865.90, pants for ₹ 1060.30 and a pair of shoes for ₹ 990.10 from a shop. How much total amount has he spent ?
Solution:
The amount for which Avneet bought a shirt = ₹ 1865.90
The amount for which Avneet bought a trouser = ₹ 1060.30
The amount for which Avneet bought a pair of shoes = ₹ 990.10
The total amount spent by Avneet = ₹ 3916.30

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.8

Addition and Subtraction of Decimals

Question 1.
Add the following decimal numbers:
(a) 2.4, 5.3 and 4.1
(b) 6.25, 5.65 and 3.01
(c) 4.32, 2.320 and 7.038
(d) 8.4, 7.03 and 2.432
(e) 12, 13.8 and 8.120.
Solution:

Question 2.
Find the difference of the following decimals numbers :
(a) 8.82 and 7.31
(b) 6.9 and 3.43
(c) 25.750 and 15.375
(d) 45 and 13.220.
(e) 13.752 and 9.27.
Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 5 Money (Currency) Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 5 Money (Currency) Ex 5.2

1. Solve the following :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

2. Solve the following:

Question 1.
₹ 3138.65 + ₹ 2124.15
Solution:

Question 2.
₹ 4472.85 + ₹ 5200.32
Solution:

Question 3.
₹ 5245.18 + ₹ 4216.27
Solution:

Question 4.
₹ 4580.42 – ₹ 2292.18
Solution:

Question 5.
₹ 8314.24 – ₹ 5218.16.
Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.7

Question 1.
Convert the following fractions into decimals :

(a) \(\frac{9}{10}\)
(b) \(\frac{35}{1000}\)
(c) \(\frac{31}{1000}\)
(d) \(\frac{117}{100}\)
(e) \(\frac{37}{10}\)
Solution:
(a) 0.9
(b) 0.35
(c) 0.031
(d) 1.17
(e) 3.7.

Question 2.
Represent the following fractions into decimals:
(a) \(\frac{3}{5}\)
(b) \(\frac{15}{20}\)
(c) \(\frac{4}{25}\)
(d) \(\frac{5}{4}\)
(e) \(\frac{7}{40}\)
Solution:
(a) \(\frac{3}{5}\) = \(\frac{3 \times 20}{5 \times 20}\) = \(\frac{60}{100}\) = 0.60
(b) \(\frac{15}{20}\) = \(\frac{15 \times 4}{25 \times 4}\) = \(\frac{60}{100}\) = 0.60
(c) \(\frac{4}{25}\) = \(\frac{4 \times 4}{25 \times 4}\) = \(\frac{16}{100}\) = 0.16
(d) \(\frac{5}{4}\) = \(\frac{5 \times 25}{4 \times 25}\) = \(\frac{125}{100}\) = 1.25
(e) \(\frac{7}{40}\) = \(\frac{7 \times 25}{40 \times 25}\) = \(\frac{175}{1000}\) = 0.175

Question 3.
Represent the following decimals into fractions :
(a) 1.3
(b) 1.75
(c) 4.5
(d) 0.35
(e) 0.8
(f) 3.84
(g) 8.345
(h) 0.024
(i) 3.001
(j) 0.98.
Solution:
(a) 1.3 = \(\frac{13}{10}\)
(b) 1.75 = \(\frac{175}{100}\)
(c) 4.5 = \(\frac{45}{10}\)
(d) 0.35 = \(\frac{35}{100}\)
(e) 0.8 = \(\frac{8}{10}\)
(f) 3.84 = \(\frac{384}{100}\)
(g) 8.345 = \(\frac{384}{100}\)
(h) 0.024 = \(\frac{24}{1000}\)
(i) 3.001 = \(\frac{3001}{1000}\)
(j) 0.98 = \(\frac{98}{100}\)

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 5 Money (Currency) Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 5 Money (Currency) Ex 5.1

1. Convert the following Rupees into Paise :

Question 1.
₹ 15
Solution:
₹ 15 = (15 × 100) Paise
= 1500 Paise

Question 2.
₹ 8.13
Solution:
₹ 8.13 = (8.13 × 100) Paise
= 813 Paise

Question 3.
₹ 12.63
Solution:
₹ 12.63 = (12.63 × 100) Paise
= 1263 Paise

Question 4.
₹ 13.50
Solution:
₹ 13.50 = (13.50 × 100) Paise
= 1350 Paise

Question 5.
₹ 98.75.
Solution:
₹ 98.75 = (98.75 × 100) Paise
= 9875 Paise

2. Convert the following paise into rupees :

Question 1.
700 Paise
Solution:
700 Paise = ₹ (700 ÷ 100)
= ₹ 7

Question 2.
925 Paise
Solution:
925 Paise = ₹ (925 ÷ 100)
= ₹ 9.25

Question 3.
1972 Paise
Solution:
1972 Paise = ₹ (1972 ÷ 100)
= ₹ 19.72

Question 4.
2816 Paise
Solution:
2816 Paise = ₹ (2816 ÷ 100)
= ₹ 28.16

Question 5.
3650 Paise
Solution:
3650 Paise = ₹ (3650 ÷ 100)
= ₹ 36.50

3. Fill in the blanks :

Question 1.
There are …………….., 50 paise coins in ₹ 1.
Solution:
2

Question 2.
There are ……………… ₹ 2 coins in ₹ 10.
Solution:
5

Question 3.
There are …………………, 50 paise coins in ₹ 1.50.
Solution:
3

Question 4.
There is a need of ……………….., ₹ 10 notes to make ₹ 100.
Solution:
10

Question 5.
There is a need of …………………, ₹ 5 notes to make ₹ 20.
Solution:
4

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.6

Question 1.
Find the greater fraction for each pair of the following :
(a) \(\frac{2}{5}\), \(\frac{2}{3}\)
(b) \(\frac{7}{9}\), \(\frac{7}{12}\)
(c) \(\frac{1}{8}\), \(\frac{1}{4}\)
(d) \(\frac{4}{6}\), \(\frac{4}{8}\)
(e) \(\frac{3}{7}\), \(\frac{3}{11}\)
(f) \(\frac{7}{9}\), \(\frac{4}{9}\)
(g) \(\frac{3}{4}\), \(\frac{1}{4}\)
(h) \(\frac{5}{8}\), \(\frac{7}{8}\)
Solution:
(a) \(\frac{2}{5}\), \(\frac{2}{3}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, \(\frac{2}{3}\) is greater than \(\frac{2}{5}\).

(b) \(\frac{7}{9}\), \(\frac{7}{12}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, \(\frac{7}{9}\) is greater than \(\frac{7}{12}\).

(c) \(\frac{1}{8}\), \(\frac{1}{4}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, \(\frac{1}{4}\) is greater than \(\frac{1}{8}\)

(d) \(\frac{4}{6}\), \(\frac{4}{8}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, \(\frac{4}{6}\) is greater than \(\frac{4}{8}\)

(e) \(\frac{3}{7}\), \(\frac{3}{11}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is smaller will be greater.
Hence, \(\frac{3}{7}\) is greater than \(\frac{3}{11}\).

(f) \(\frac{7}{9}\), \(\frac{4}{9}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, \(\frac{7}{9}\) is greater than \(\frac{4}{9}\)

(g) \(\frac{3}{4}\), \(\frac{1}{4}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, \(\frac{3}{4}\) is greater than \(\frac{1}{4}\).

(h) \(\frac{5}{8}\), \(\frac{7}{8}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be greater.
Hence, \(\frac{7}{8}\) is greater than \(\frac{5}{8}\)

Question 2.
Find the smaller fraction for each part of the following :
(a) \(\frac{3}{5}\), \(\frac{3}{4}\)
(b) \(\frac{5}{8}\), \(\frac{5}{12}\)
(c) \(\frac{7}{9}\), \(\frac{4}{9}\)
(d) \(\frac{3}{6}\), \(\frac{3}{8}\)
(e) \(\frac{5}{7}\), \(\frac{5}{11}\)
(f) \(\frac{8}{12}\), \(\frac{5}{12}\)
(g) \(\frac{9}{4}\), \(\frac{7}{4}\)
(h) \(\frac{9}{8}\), \(\frac{7}{8}\)
Solution:
(a) \(\frac{3}{5}\), \(\frac{3}{4}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, \(\frac{3}{5}\) is Smaller than \(\frac{3}{4}\)

(b) \(\frac{5}{8}\), \(\frac{5}{12}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, \(\frac{5}{12}\) is Smaller than \(\frac{5}{8}\)

(c) \(\frac{7}{9}\), \(\frac{4}{9}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, \(\frac{3}{5}\) is Smaller than \(\frac{3}{4}\)

(d) \(\frac{3}{6}\), \(\frac{3}{8}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, \(\frac{3}{8}\) is Smaller than \(\frac{3}{6}\)

(e) \(\frac{5}{7}\), \(\frac{5}{11}\)
The numerator of both the fractions are equal. So, the fraction whose denominator is greater will be smaller.
Hence, \(\frac{5}{11}\) is Smaller than \(\frac{5}{7}\)

(f) \(\frac{8}{12}\), \(\frac{5}{12}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, \(\frac{5}{12}\) is Smaller than \(\frac{8}{12}\)

(g) \(\frac{9}{4}\), \(\frac{7}{4}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, \(\frac{7}{4}\) is Smaller than \(\frac{9}{4}\)

(h) \(\frac{9}{8}\), \(\frac{7}{8}\)
The denominator of both the fractions are equal. So, the fraction whose numerator is greater will be smaller.
Hence, \(\frac{7}{8}\) is Smaller than \(\frac{9}{8}\)

Question 3.
Write the following in increasing (ascending) order :

(a) \(\frac{7}{12}\), \(\frac{4}{12}\), \(\frac{1}{12}\), \(\frac{5}{12}\)
(b) \(\frac{5}{12}\), \(\frac{5}{9}\), \(\frac{5}{7}\), \(\frac{5}{4}\)
(c) \(\frac{6}{11}\), \(\frac{4}{11}\), \(\frac{9}{11}\), \(\frac{3}{11}\)
(d) \(\frac{7}{8}\), \(\frac{7}{12}\), \(\frac{7}{4}\), \(\frac{7}{2}\)
(e) \(\frac{12}{15}\), \(\frac{12}{13}\), \(\frac{12}{17}\), \(\frac{12}{10}\)
Solution:
(a) \(\frac{7}{12}\), \(\frac{4}{12}\), \(\frac{1}{12}\), \(\frac{5}{12}\)
The denominators of all of these fractions are equal. Therefore, the fraction which has the lowest numerator, will be the smallest.
Among these fractions, \(\frac{1}{12}\) is the smallest.
Ascending order is \(\frac{1}{12}\), \(\frac{4}{12}\), \(\frac{5}{12}\) and \(\frac{7}{12}\).

(b) \(\frac{5}{12}\), \(\frac{5}{9}\), \(\frac{5}{7}\), \(\frac{5}{4}\)
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, \(\frac{5}{12}\) is the smallest.
Ascending order is \(\frac{5}{12}\), \(\frac{5}{9}\), \(\frac{5}{7}\) and \(\frac{5}{4}\)

(c) \(\frac{6}{11}\), \(\frac{4}{11}\), \(\frac{9}{11}\), \(\frac{3}{11}\)
The denominators of all of these fractions are equal. Therefore, the fraction which has the lowest numerator, will be the smallest fraction.
Among these fractions, \(\frac{3}{11}\) is the smallest.
Ascending order is \(\frac{3}{11}\), \(\frac{4}{11}\), \(\frac{6}{11}\) and \(\frac{9}{11}\).

(d) \(\frac{7}{8}\), \(\frac{7}{12}\), \(\frac{7}{4}\), \(\frac{7}{2}\)
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, \(\frac{7}{12}\) is the smallest.
Ascending order is \(\frac{7}{12}\), \(\frac{7}{8}\), \(\frac{7}{4}\) and \(\frac{7}{2}\)

(e) \(\frac{12}{15}\), \(\frac{12}{13}\), \(\frac{12}{17}\), \(\frac{12}{10}\)
The numerator of all of these fractions are equal. Therefore, the fraction whose denominator is the largest, is the smallest fraction.
Among these fractions, \(\frac{12}{17}\) is the smallest.
Ascending order is \(\frac{12}{17}\), \(\frac{12}{15}\), \(\frac{12}{13}\) and \(\frac{12}{10}\)

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.5

Question 1.
Write the like and unlike fractions for the following groups :
(a) \(\frac{3}{7}\), \(\frac{5}{7}\), \(\frac{1}{7}\), ……………….
(b) \(\frac{6}{9}\), \(\frac{4}{9}\), \(\frac{1}{9}\), ……………….
(c) \(\frac{9}{12}\), \(\frac{7}{11}\), \(\frac{7}{10}\), ……………….
(d) \(\frac{7}{10}\), \(\frac{6}{10}\), \(\frac{8}{10}\), ……………….
(e) \(\frac{5}{3}\), \(\frac{5}{7}\), \(\frac{5}{9}\), ……………….
Solution:
(a) Like fractions,
(b) Like fractions,
(c) Unlike fractions,
(d) Like fractions,
(e) Unlike fractions.

Question 2.
Write two like fractions for the following :
(a) \(\frac{1}{5}\), \(\frac{4}{5}\), \(\frac{3}{4}\), -, –
(b) \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{7}{9}\), -, –
(c) \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{9}{7}\), -, –
Solution:
(a) \(\frac{2}{5}\) and \(\frac{6}{5}\)
(b) \(\frac{1}{9}\) and \(\frac{5}{9}\)
(c) \(\frac{1}{7}\) and \(\frac{4}{7}\)

Question 3.
Write the unit fraction, whose denominator is as follows :
(a) 7
(b) 5
(c) 8
(d) 3
(e) 15.
Solution:
(a) \(\frac{1}{7}\),
(b) \(\frac{1}{5}\),
(c) \(\frac{1}{8}\),
(d) \(\frac{1}{3}\),
(e) \(\frac{7}{9}\)

Question 4.
Which of the following fractions are proper and improper fractions :
(a) \(\frac{7}{12}\)
(b) \(\frac{8}{3}\)
(c) \(\frac{12}{18}\)
(d) \(\frac{3}{5}\)
(e) \(\frac{7}{9}\)
Solution:
(a) Proper fraction,
(b) Improper fraction,
(c) Proper fraction,
(d) Proper fraction,
(e) Proper fraction.

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 4 Fractions Ex 4.4

Question 1.
Check whether the following fractions are in its lowest form or not :
(a) \(\frac{12}{14}\)
(b) \(\frac{21}{3}\)
(c) \(\frac{13}{17}\)
(d) \(\frac{25}{50}\)
(e) \(\frac{14}{21}\)
(f) \(\frac{8}{13}\)
(g) \(\frac{7}{15}\)
(h) \(\frac{14}{27}\)
(i) \(\frac{25}{35}\)
(j) \(\frac{18}{23}\)
Solution:
(a) To find whether the fraction \(\frac{12}{14}\) is in its lowest term or not, we will find the HCF of 12 and 14.
Now, HCF of 12 and 14 = 2
Because the HCF of numerator and denominator is not 1.
∴ The fraction is not in its lowest term. To change the fraction into its lowest term, we will divide the numerator (12) and denominator (14) by HCF (2)
Fraction = \(\frac{12}{14}\) = \(\frac{12 \div 2}{14 \div 2}\) = \(\frac{6}{7}\)
∴ The lowest form of \(\frac{12}{14}\) is \(\frac{6}{7}\).

(b) To find whether the fraction \(\frac{21}{35}\) in its lowest term or not, we will find the HCF of 21 and 35 Now, HCF of 21 and 35 = 7
Now, HCF of 21 and 35 = 7

Because the HCF of numerator and denominator is not 1
∴ The fraction is not in its lowest terms. To change the fraction into its lowest terms, we will divide the numerator (21) and denominator (35) by their HCF (7)
Fraction = \(\frac{21}{35}\) = \(\frac{21 \div 7}{35 \div 7}\) = \(\frac{3}{5}\)
∴ The lowest form of \(\frac{21}{35}\) is \(\frac{3}{5}\)

(c) To find whether the fraction \(\frac{13}{17}\) is in its lowest terms or not, we will find the HCF of 13 and 17 Now, HCF of 13 and 17 = 1
Therefore, the fraction \(\frac{13}{17}\) is in its lowest terms.

(d) To find whether the fraction \(\frac{25}{50}\) is in its lowest terms or not, we will find the HCF of 25 and 50.
Now, HCF of 25 and 50 = 25

Because, the HCF of numerator and denominator is not 1, therefore, the fraction is not in its lowest terms. To change the fraction into its lowest terms, we will divide the numerator (25) and denominator (50) by their HCF (25)
Fraction = \(\frac{25}{50}\) = \(\frac{25 \div 25}{50 \div 25}\) = \(\frac{1}{2}\)
∴The lowest form of \(\frac{25}{50}\) is \(\frac{1}{2}\)

(e) In \(\frac{14}{21}\), HCF of numerator (14) and denominator (21) = 7

The fraction \(\frac{14}{21}\) is not in its lowest terms because HCF of numerator (14) and denominator (21) is not 1. To change, the fraction, into its lowest terms we will divide the numerator (14) and denominator (21) by their HCF (7)
\(\frac{14}{21}\) = \(\frac{14 \div 7}{21 \div 7}\) = \(\frac{2}{3}\)
∴ The lowest form of fraction \(\frac{14}{21}\) is \(\frac{2}{3}\).

(f) To find whether the fraction \(\frac{8}{13}\) is in its lowest terms or not, we will find the HCF of numerator (8) and denominator (13)
Now, HCF of 8 and 13 = 1

Therefore, the fraction \(\frac{8}{13}\) is in its lowest form.

(g) To find whether the fraction \(\frac{7}{15}\) is in its lowest terms or not, we will find the HCF of numerator 7 and denominator 15.
Now, HCF of 7 and 15 = 1

Therefore, the fraction is in its lowest form.

(h) To find whether the fraction \(\frac{14}{27}\) is in its lowest terms or not, we will find the HCF of numerator (14) and denominator (27).
Now, HCF of 14 and 27 = 1

Therefore, the fraction \(\frac{14}{27}\) is in its lowest terms.

(i) To find whether the fraction \(\frac{25}{35}\) is in its lowest term or not, we will find the HCF of numerator (25) and denominator (35).
Now, the HCF of 25 and 35 = 5

Because, the HCF of numerator and denominator is not 1, therefore, the fraction \(\frac{25}{35}\) is not m its lowest terms.
To change the fraction into its lowest terms, we will divide the numerator and denominator by their HCF (5).
\(\frac{25}{35}\) = \(\frac{25 \div 5}{35 \div 5}\) = \(\frac{5}{7}\)
Therefore, the lowest form of \(\frac{25}{35}\) is \(\frac{5}{7}\).

(j) To find whether the fraction \(\frac{18}{23}\) is in its lowest terms or not, we will find the HCF of numerator (18) and denominator (23).
Now, HCF of 18 and 23 = 1

Therefore, the fraction \(\frac{18}{23}\) is in its lowest terms.

Question 2.
Write the lowest form of following fractions :
(a) \(\frac{4}{8}\)
(b) \(\frac{12}{18}\)
(c) \(\frac{15}{20}\)
(d) \(\frac{35}{45}\)
(e) \(\frac{24}{36}\)
(f) \(\frac{8}{12}\)
(g) \(\frac{18}{21}\)
(h) \(\frac{25}{45}\)
(i) \(\frac{6}{12}\)
(j) \(\frac{9}{27}\)
Solution:
(a) HCF of 4 and 8 = 4
Therefore, the lowest form of \(\frac{4}{8}\)
= \(\frac{4 \div 4}{8 \div 4}\) = \(\frac{1}{2}\)

(b) HCF of 12 and 18 = 6
Therefore, the lowest form of \(\frac{12}{18}\)
= \(\frac{12 \div 6}{18 \div 6}\) = \(\frac{2}{3}\)

(c) HCF of 15 and 20 = 5
Therefore, the lowest form of \(\frac{15}{20}\)
= \(\frac{15 \div 5}{20 \div 5}\) = \(\frac{3}{4}\)

(d) HCF of 35 and 45 = 5
Therefore, the lowest form of \(\frac{35}{45}\)
= \(\frac{35 \div 5}{45 \div 5}\) = \(\frac{7}{9}\)

(e) HCF of 24 and 36 = 12
Therefore, the lowest form of \(\frac{24}{36}\)
= \(\frac{24 \div 12}{36 \div 12}\) = \(\frac{2}{3}\)

(f) HCF of 8 and 12 = 4
Therefore, the lowest form of \(\frac{8}{12}\)
= \(\frac{8 \div 4}{12 \div 4}\) = \(\frac{2}{3}\)

(g) HCF of 18 and 21 = 3
Therefore, the lowest form of \(\frac{18}{21}\)
= \(\frac{18 \div 3}{21 \div 3}\) = \(\frac{6}{7}\)

(h) HCF of 25 and 45 = 5
Therefore, the lowest form of \(\frac{25}{45}\)
= \(\frac{25 \div 5}{45 \div 5}\) = \(\frac{5}{9}\)

(i) HCF of 6 and 12 = 6
Therefore, the lowest form of \(\frac{6}{12}\)
= \(\frac{6 \div 6}{12 \div 6}\) = \(\frac{1}{2}\)

(j) HCF of 9 and 27 = 9
Therefore, the lowest form of \(\frac{9}{27}\)
= \(\frac{9 \div 9}{27 \div 9}\) = \(\frac{1}{3}\)