Class 5

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 5 Money (Currency) MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 5 Money (Currency) MCQ Questions

Multiple Choice Questions

Tick (✓) the right answer :

Question 1.
What is the standard method of writing ₹ 13 and 50 paise ?
(a) ₹ 1350
(b) ₹ 13.50
(c) ₹ 1350 paise
(d) None.
Answer:
(b) ₹ 13.50

Question 2.
How many ₹ 2 coins are there in ₹ 26?
(a) 52
(b) 26
(c) 13
(d) 20.
Answer:
(c) 13

Question 3.
What is the standard symbol of Indian currency ?
(a) ₹
(b) $
(c) £
(d) None.
Answer:
(a) ₹

Question 4.
If the cost of a pen is ₹ 12. What is the cost of 11 pens ?
(a) ₹ 120
(b) ₹ 23
(c) ₹ 1
(d) ₹ 132.
Answer:
(d) ₹ 132

Question 5.
1 kg apples cost ₹ 80. What is the cost of half kg apples ?
(a) ₹ 20
(b) ₹ 160
(c) ₹ 40
(d) ₹ 80.
Answer:
(c) ₹ 40

Question 6.
The cost of a dozen pencils is ₹ 60. Find the cost of one pencil.
(a) ₹ 12
(b) ₹ 5
(c) ₹ 60
(d) ₹ 30.
Answer:
(b) ₹ 5

Question 7.
What is the amount of 7 notes of ₹ 20 ?
(a) ₹ 27
(b) ₹ 14
(c) ₹ 140
(d) ₹ 13.
Answer:
(c) ₹ 140

Question 8.
Convert 480 paise into rupees.
(a) ₹ 4.80
(b) ₹ 48.00
(c) ₹ 480
(d) ₹ 8.40.
Answer:
(a) ₹ 4.80

Question 9.
Sukhdev had spent ₹ 25 in a market and he was left with ₹ 25. How many rupees had he in the beginning ?
(a) ₹ 25
(b) ₹ 5.00
(c) ₹ 50
(d) ₹ 40.
Answer:
(c) ₹ 50

Question 10.
Find the value of ₹ 10.40 + ₹ 15.30 + ₹ 8.20.
(a) ₹ 33.90
(b) ₹ 34.00
(c) ₹ 30.90
(d) ₹ 339.
Answer:
(a) ₹ 33.90

Question 11.
The cost of a shirt is ₹ 999.90. How much estimate cost shopkeeper will take ?
(a) ₹ 990
(b) ₹ 999
(c) ₹ 1000
(d) ₹ 999.95
Answer:
(c) ₹ 1000

Question 12.
The cost of a daily newspaper is ₹ 4. What is the cost of newspaper in the month of January ?
(a) ₹ 124
(b) ₹ 12
(c) ₹ 35
(d) ₹ 25.
Answer:
(a) ₹ 124

Question 13.
Anmol saves ₹ 5 daily from his pocket expenses. How many rupees has he saved in the month of March ?
(a) ₹ 36
(b) ₹ 31
(c) ₹ 155
(d) ₹ 150.
Answer:
(c) ₹ 155

Question 14.
The cost of 8 meter cloth is ₹ 680. What is the cost of 1 meter cloth ?
(a) ₹ 80
(b) ₹ 85
(c) ₹ 70
(d) ₹ 90.
Answer:
(b) ₹ 85

Question 15.
How many coins of 50 paise will be there in ₹ 5. ?
(a) 250
(b) 55
(c) 20
(d) 10
Answer:
(d) 10

Question 16.
How many coins of ₹ 2 are there in ₹ 26 ?
Answer:
₹ 26 ÷ ₹ 2 = 13.

Number of ₹ 2 coins in ₹ 26 = 13.

Question 17.
Check the rate list and answer the questions :

(i) Cost of 2 kilogram sugar
(ii) Cost of 500 g of tea leaves.
Answer:
(i) Cost of 2 kilograms sugar
= ₹ 60 × 2 = ₹ 120

(ii) 500 g = \(\frac{500}{1000}\) kg = \(\frac{1}{2}\) kg
Cost of 500 g = \(\frac{1}{2}\) kg tea leaves
= ₹ 320 × \(\frac{1}{2}\) = ₹ 160

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 4 Fractions MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 4 Fractions MCQ Questions

Which decimal number is being represented by the shaded part of figure :

(a) 0.3
(b) 0.03
(c) 0.7
(d) 0.07
Answer:
(a) 0.3

Tick (✓) the right answer :

Quetion 1.
Mark the (✓) sign against correct statement and that of (✗) sign against the wrong statement.

(i) The decimal form of \(\frac{3}{10}\) is 0.3 ( )
(ii) \(\frac{1}{5}\), \(\frac{1}{9}\), \(\frac{1}{7}\) are unit fraction. ( )
(iii) \(\frac{9}{4}\) is a proper fraction. ( )
(iv) The denominator of fraction \(\frac{3}{5}\) is 3. ( )
Answer:
(i) ✓
(ii) ✓
(iiii) ✗
(iv) ✗.

Quetion 2.
There was a birthday party of Raju. His father brought 24 caps for the children who have come to attend the party. If \(\frac{1}{3}\) rd of caps are red, \(\frac{1}{2}\) are of green colour and \(\frac{1}{6}\) are of yellow colour then answer the following questions :
(a) How many children can wear red colour caps ?
(b) How many children can wear yellow colour caps ?
Solution:
(a) Number of children who can wear ted colour cap = 24 × \(\frac{1}{3}\) = 8
(b) Number of children who can wear yellow coloured caps
= 24 × \(\frac{1}{6}\) = 4

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 \(\frac{12}{10}\)