Class 5

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.9

Question 1.
Find the estimated answers :
(a) 753 + 525
Solution:

(b) 11526 + 8748.
Solution:

(c) 980 – 489
Solution:

(d) 5897 – 2987
Solution:

(e) 440 × 28
Solution:

(f) 6198 × 13
Solution:

(g) 563 ÷ 34
Solution:
563 rounded off = 600
34 rounded off = 30
Estimated value = 600 ÷ 30 = 20

(h) 7541 ÷ 43
Answer:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.8

Question 1.
In a stadium, in the match of cricket there are 84000 people sitting in 24 rows. How many people are sitting in a row?
Solution:
Total number of people = 84000
Number of rows = 24
Number of people sitting in each row
= 84000 ÷ 24
= 3500

Question 2.
You have ₹ 99825 which is to be distributed equally among 33 friends. How much amount will each friend get ?
Solution:
Total amount = ₹ 99825
Number of friends = 33
Each friend will get = ₹ 99825 ÷ 33
= ₹ 3025

Question 3.
My grandfather divided ₹ 72000 equally among four brothers sisters. How much will each get?
Solution:
Total amount = ₹ 72000
Number of brothers and sisters = 4
Each will get
= 72000÷4
= 18000

Question 4.
What number must be multiplied with 26 to get 14508?
Solution:
Number to be obtained = 14508
The given number = 26
The required number = 14508 ÷ 26
= 558

Question 5.
The gardener has 23976 flowers to make garlands. One garland has 24 flowers in it. How many garland can be made from 23976 flowers ?
Solution:
Total number of flowers = 23976
Number of flowers in one garland = 24
Total number of garlands = 23976 ÷ 24
= 999

Question 6.
How many ₹ 2000 notes are there in forty thousand rupees ?
Solution:
Total amount = ₹ 40,000
Value of one note = ₹ 2000
Number of notes = ₹ 40,000 ÷ ₹ 2000
= 20

Question 7.
I need change of ₹ 25,000. How many following notes shall I get ?
(a) Number of notes of ₹ 1000 = ……….
(b) Number of notes of ₹ 500 = ………..
(c) Number of notes of ₹ 100 = ………..
Solution:
Total amount = ₹ 25,000
(a) Number of ₹ 1000 notes
= ₹ 25000 ÷ ₹ 1000

(b) Number of ₹ 500 notes
= ₹ 25000 ÷ ₹ 500

(c) Number of ₹ 100 notes
= ₹ 25000 ÷ ₹ 100

Question 8.
A J.C.B. machine picks 900 bricks in a round. How many rounds will it take to pick 99000 bricks ?
Solution:
Total number of bricks = 99,000
Number of bricks picked in one round = 900
Number of rounds
= 99,000 ÷ 900
= \(\frac{99000}{900}\) = 110

Question 9.
The cost of a railway ticket is ₹ 78. Palak gave ₹ 7722 for buying tickets. How many tickets will she get ?
Solution:
Cost of 1 ticket = ₹ 78
Total amount = ₹ 7722
Number of tickets she will get
= ₹ 7722 ÷ ₹ 78
= 99

Question 10.
A factory manufactures 45540 ice cream cones in the month of June. How many ice cream cones are manufactured in a day ?
Solution:
Total number of ice cream cones manufactured = 45540
Number of the days in the month of June = 30
Number of cones manufactured in one day = 45540 ÷ 30 = 1518

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 8 Perimeter and Area MCQ Questions

Multiple Choice Questions

Tick (✓) the right answer :

Question 1.
Which type of figure is notebook’s page ?
(a) Square
(b) Rectangle
(c) Triangle
(d) Pentagon.
Answer:
(b) Rectangle

Question 2.
What is the perimeter of the square if its side is 6 cm ?
(a) 36 cm
(b) 18 cm
(c) 24 cm
(d) 21 cm.
Answer:
(c) 24 cm

Question 3.
The four sides of square are
(a) different
(b) equal
(c) two equal pairs
(d) none.
Answer:
(b) equal

Question 4.
The length and breadth of rectangle is 6 m and 4 m. Find its perimeter.
(a) 36 m
(b) 16 m
(c) 20 m
(d) 10 m.
Answer:
(c) 20 m

Question 5.
A rectangular park is 65 m long and 35 m wide. Mukesh takes 4 rounds of it. How much distance is covered by him ?
(a) 100 m
(b) 200 m
(c) 400 m
(d) 800 m.
Answer:
(d) 800 m

Question 6.
What will be the area of a square whose side is 13 cm ?
(a) 169 cm
(b) 169 sq. cm
(c) 52 sq. cm
(c) 26 sq. cm.
Answer:
(a) 169 cm

Question 7.
A chart is 125 cm long and 8 cm wide. Its area = ………………..
(a) 100 sq. cm
(b) 1000 sq. cm
(c) 1250 sq. cm
(d) 1100 sq. cm.
Answer:
(b) 1000 sq. cm

Question 8.
If length and breadth of the rectangle is equal then it is called ………………
(a) Rectangle
(b) Length
(c) Square
(d) Perimeter.
Answer:
(c) Square

Question 9.
Side × Side is the area of a ………………
(a) Square
(b) Rectangle
(c) Breadth
(d) Circle.
Answer:
(a) Square

Question 10.
Area of a rectangle is 96 sq. cm. If its length is 12 cm then its breadth is :
(a) 8 cm
(b) 9 cm
(c) 10 cm
(d) 108 cm.
Ans.
(a) 8 cm

Question 11.
Find the area of given rectangle.

(a) 10 sq. cm
(b) 10 cm
(c) 8 sq. cm
(d) 12 sq. cm.
Answer:
(a) 10 sq. cm

Question 12.
Look at the following two figure carefully and select the correct option :

(a) Area of fig. 1 is more than area of fig. 2.
(b) Area of fig. 1 is less than area of fig. 2.
(c) Area of fig. 1 is equal to area of fig. 2.
(d) Area of both the figures is equal.
Answer:
(b) Area of fig. 1 is less than area of fig. 2.

Question 13.
Below is given a picture of a field. Find the area of field.

Answer:
Length of field = 96 m
Breadth of field = 64 m
Area of field = Length × Breadth
= 96 × 64 sq. m
= 6144 sq. m

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 8 Perimeter and Area Ex 8.2

1. Find the area of following rectangles whose length and breadth are as follows :

Question 1.
9 m and 7 m
Solution:
Length of rectangle = 9 m
Breadth of rectangle = 7 m
Area of rectangle = Length × Breadth
= 9 m × 7 m
= 63 m²

Question 2.
85 cm and 76 cm
Solution:
Length of rectangle = 85 cm
Breadth of rectangle = 76 cm
Area of rectangle = Length × Breadth
= 85 cm × 76 cm
= 6460 cm²

Question 3.
23 mm and 18 mm
Solution:
Length of rectangle = 23 mm
Breadth of rectangle = 18 mm
Area of rectangle = Length × Breadth
= 23 mm × 18 mm
= 414 mm²

Question 4.
5 m and 85 cm
Solution:
Length of rectangle
= 5 m
= 5 × 100 cm
= 500 cm
Breadth of rectangle = 85 cm
Area of rectangle = Length × Breadth
= 500 cm × 85 cm
= 42500 cm²

Question 5.
840 cm and 7 m
Solution:
Length of rectangle = 840 cm
Breadth of rectangle = 7 m
= 7 × 100 cm
= 700 cm
Area of rectangle = Length × Breadth
= 840 cm × 700 cm
= 588000 cm²

2. Find the area of a square whose side is :

Question 1.
25 cm
Sol.
Side of square = 25 cm
Area of square = side × side
= 25 cm × 25 cm
= 625 cm²

Question 2.
48 cm
Solution:
Side of the square = 48 cm
Area of the square = side × side
= 48 cm × 48 cm
= 2304 cm²

Question 3.
27 mm
Solution:
Side of the square = 27 mm
Area of the square = side × side
= 27 mm × 27 mm
= 729 mm²

Question 4.
87 m
Solution:
Side of the square = 87 m
Area of the square = side × side
= 87 m × 87 m
= 7569 m²

Question 3.
Find the area of rectangular park whose length is 62 m and breadth is 38 m.
Solution:
Length of the rectangular park = 62 m
Breadth of the rectangular park = 38 m
Area of the rectangular park = Length × Breadth
= 62 m × 38 m
= 2356 m²

Question 4.
The side of a carrom-board is 60 cm. Find its area.
Solution:
Side of the carrom-board = 60 cm
Area of the carrom-board = side × side
= 60 cm × 60 cm
= 3600 cm²

Question 5.
The length and breadth of a rectangular field is 100 m and 45 m. What is the cost of levelling its floor at the rate of ₹ 8 per sq. m?
Solution:
Length of the rectangular field = 100 m
Breadth of the rectangular field = 45 m
Area of the rectangular field = Length × Breadth
= 100 m × 45 m
= 4500 m²
The rate of levelling = ₹ 8 per m²
The cost of levelling = ₹ 8 × 4500
= ₹ 36000

Question 6.
A carpet has a length 8 m and breadth 5 m. In an auditorium, 125 such carpets are being set on the floor. Find the area of the floor of the auditorium.
Solution:
Length of the carpet = 8 m
Breadth of the carpet = 5 m
Area of each carpet = Length × Breadth
= 8 m × 5 m
= 40 m²
Area of 125 carpets = 125 × 40 m²
= 5000 m²
Therefore, area of the floor of the auditorium = 5000 m².

Question 7.
The verandah of Gurpreet’s home is 52 m long and 32 m wide and the verandah of Pankaj’s home is of square shape with side 41 m. which person’s home has a roof of verandah bigger and by how much ?
Solution:
Gurpreet :
Length of verandah = 52 m
Breadth of verandah = 32 m
Area of verandah = Length × Breadth
= 52 m × 32 m
=1664 m²

Pankaj :
Side of square verandah = 41 m
Area of square verandah = Side × side
= 41 m × 41 m
= 1681 m²
Area of verandah of Pankaj’s home is bigger by
= 1681 m² – 1664 m²
= 17 m²

Question 8.
Roof of Amarjeet’s home is of length 9 m and breadth 6 m. There is a leakage of water from the roof. He wants to fix tiles of size 30 cm long and 20 cm wide for plugging the leakage. How many tiles does he need ?
Solution:
Length of the roof = 9 m
= 9 ² 100 cm
= 900 cm
Breadth of the roof = 6 m
= 6 ² 100 cm
= 600 cm
Area of the roof = Length × Breadth
= 900 cm × 600 cm
= 540000 cm²
Length of each tile = 30 cm
Breadth of each tile = 20 cm
Area of each tile = Length × Breadth
= 30 cm × 20 cm
= 600 cm²
Area of the roof
Number of tiles = \(\frac{\text { Area of the roof }}{\text { Area of each tile }}\)
= \(\frac{540000}{600}\) = \(\frac{900 \times 600}{600}\)
= 900.

9. Fill in the blanks :

Question 1.
Area of rectangle = ……………… × ………………
Solution:
Length × Breadth

Question 2.
Area of square = ……………… × ………………
Solution:
side × side

Question 3.
1 sq. m. = ……………… sq. cm.
Solution:
10000

Question 4.
The space covered by a closed figure is called its ………………
Solution:
Area.

Question 10.
Complete the table :

Solution:
(a) 56 m²
(b) 2 cm
(c) 6 mm
(d) 700 cm²

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.7

Question 1.
Solve the following :
(a) 117 ÷ 13
(b) 135 ÷ 15
(c) 72 ÷ 12
(d) 108 ÷ 9
(e) 78 ÷ 13
(f) 121 ÷ 11
(g) 140 ÷ 20
(h) 144 ÷ 16
(i) 98 ÷ 14
(j) 119 ÷ 17.
Solution:

Question 2.
Divide the following and verify :
(a) 54598 ÷ 12 .
(b) 8975 ÷ 21
(c) 77552 ÷ 18
(d) 88001 ÷ 17
(e) 12896 ÷ 11.
Solution:

Quotient = 4549
Remainder = 10
Verification :
Dividend = Quotient × Divisor + Remainder
54598 = 4549 × 12 + 10
54598 = 54588 + 10
54598 = 54598

Quotient = 427
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
8975 = 427 × 21 + 8
8975 = 8967 + 8
8975 = 8975

Quotient = 4308
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
77552 = 4308 × 18 + 8
77552 = 77544 + 8
77552 = 77552

Quotient = 5176
Remainder = 9
Verification :
Dividend = Quotient × Divisor + Remainder
88001 = 5176 × 17 + 9
88001 = 87992 + 9
88001 = 88001

Quotient = 1172
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
12896 = 1172 × 11 + 4
12896 = 12892 + 4
12896 = 12896

Question 3.
Solve the following and verify :
(a) 760 ÷ 12
(b) 550 ÷ 14
(c) 894 ÷ 21
(d) 913 ÷ 19
(e) 826 ÷ 25
(f) 7645 ÷ 24
(g) 89781 ÷ 9
(h) 99999 ÷ 80
(i) 82525 ÷ 75
(j) 70008 ÷ 14
(k) 50205 ÷ 15
(l) 16258 ÷ 36
(m) 96000 ÷ 50
(n) 45457 ÷ 35
Solution:

Quotient = 63
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
760 = 63 × 12 + 4
760 = 756 + 4
760 = 760

Quotient = 39
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
550= 39 × 14 + 4
550 = 546 + 4
550 = 550

Quotient = 42
Remainder = 12
Verification :
Dividend = Quotient × Divisor + Remainder
894 = 42 × 21 + 12
894 = 882 + 12
894 = 894

Quotient = 48
Remainder = 1
Verification :
Dividend = Quotient × Divisor + Remainder
913 = 48 × 19 + 1
913 = 912 + 1
913 = 913

Quotient = 33
Remainder = 1
Verification :
Dividend = Quotient × Divisor + Remainder
826 = 33 × 25 + 1
826 = 825 + 1
826 = 826

Quotient = 318
Remainder = 13
Verification :
Dividend = Quotient × Divisor + Remainder
7645 = 318 × 24 + 13
7645 = 7632 + 13 .
7645 = 7645

Verification :
Dividend = Quotient × Divisor + Remainder
89781 = 9975 × 9 + 6
89781 = 89775 + 6 89781 = 89781

Quotient = 1249
Remainder = 79
Verification :
Dividend = Quotient × Divisor + Remainder
99999 = 1249 × 80 + 79
99999 = 99920 + 79
99999 = 99999

Quotient = 1100
Remainder = 28
Verification :
Dividend = Quotient × Divisor + Remainder
82525 = 1100 × 75 + 25
82525 = 82500 + 25
82525 = 82525

Quotient = 5000
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
70008 = 5000 × 14 + 08
70008 = 70000 + 08
70008 = 70008

Quotient = 3347
Remainder = 0
Verification :
Dividend = Quotient × Divisor + Remainder
50205 = 3347 × 15 + 0
50205 = 50205

Quotient = 451
Remainder = 22
Verification :
Dividend = Quotient × Divisor + Remainder
16258 = 451 × 36 + 22
16258 = 16258

Quotient = 1920
Remainder = 0
Verification :
Dividend = Quotient × Divisor + Remainder
96000 = 1920 × 50 + 0
96000 = 96000

Quotient = 1298
Remainder = 27
Verification :
Dividend = Quotient × Divisor + Remainder
45457= 1298 × 35 + 27
45457 = 45430 + 27
45457 = 45457

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 8 Perimeter and Area Ex 8.1

1. Find the perimeter :

Question 1.

Solution:
Perimeter of a rectangle = Length + Breadth + Length + Breadth
= 2 (Length + Breadth)
= 2 (8 m + 3 m)
= 2 × 11 m
= 22 m.

Question 2.

Solution:
Perimeter of a square = side + side + side + side
= 4 × side
= 4 × 5 cm
= 20 cm.

2. Find the perimeter of the rectangle whose length and breadth are as follows :

Question 1.
3 cm, 2 cm
Solution:
Length of the rectangle = 3 cm
Breadth of the rectangle = 2 cm
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (3 cm + 2 cm)
= 2 × 5 cm
= 10 cm.

Question 2.
12 m, 10 m
Solution:
Length of the rectangle = 12 m
Breadth of the rectangle = 10 m
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (12 m + 10 m)
= 2 × 22 m
= 44 m.

Question 3.
15 cm, 8 cm.
Solution:
Length of the rectangle = 15 cm
Breadth of the rectangle = 8 cm
Perimeter of the rectangle = 2 (Length + Breadth)
= 2 (15 cm +8 cm)
= 2 × 23 cm
= 46 cm.

3. Find the perimeter of the square, whose side is :

Question 1.
4 cm
Solution:
Side of the square = 4 cm
Perimeter of the square = 4 × side
= 4 × 4 cm
= 16 cm.

Question 2.
8 cm
Solution:
Side of the square = 8 cm
Perimeter of the square = 4 × side
= 4 × 8 cm
= 32 cm.

Question 3.
10 cm
Solution:
Side of the square = 10 cm
Perimeter of the square = 4 × side
= 4 × 10 cm
= 40 cm.

Question 4.
72 mm
Solution:
Side of the square = 72 mm
Perimeter of the square = 4 × side
= 4 × 72 mm
= 288 mm.

4. Find the side of the square whose perimeter is :

Question 1.
48 cm
Solution:
Perimeter of the square = 48 cm
Side of the square = \(\frac{\text { Perimeter }}{4}\)
= \(\frac{48}{4}\) cm
= 12 cm

Question 2.
80 m
Solution:
Side of the square = \(\frac{\text { Perimeter }}{4}\)
= \(\frac{80}{4}\) cm
= 20 cm

Question 3.
24 m
Solution:
Side of the square = \(\frac{\text { Perimeter }}{4}\)
= \(\frac{24}{4}\) cm
= 6 cm

Question 5.
The length and breadth of a rectangular park is 96 m and 64 m respectively. Find the length of wire which can fence it all around.
Solution:
Length of rectangular park = 96 m
Breadth of rectangular park = 64 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (96 m + 64 m)
= 2 × 160 m
= 320 m.
∴ Length of wire which can fence it all round = 320 m.

Question 6.
The perimeter of the rectangular park is 84 m. Find its breadth if length is 24 m.
Solution:
Perimeter of the rectangular park = 84 m
Length of the rectangular park = 24 m
Breadth of the rectangular park = \(\frac{\text { Perimeter }}{2}\) – length
= \(\frac{84 m}{2}\) – 24 m
= 42 m – 24 m
= 18 m

Question 7.
A player runs around a square track of side 50 m. How many rounds will he take to complete the race of 2000 m?
Solution:
Side of the square track = 50 m
Perimeter of the square track = 4 × side
= 4 × 50 m
= 200 m
Total distance of race = 2000 m
Number of rounds = \(\frac{2000}{200}\)
= 10

8. Fill in the blanks :

Question 1.
Perimeter of rectangle = 2 × (length + ………………)
Solution:
Breadth

Question 2.
Perimeter of square = …………… × side
Solution:

Question 3.
The perimeter of a closed figure, made of line segments, is ……………… of lengths of its all sides.
Solution:
sum.

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

PSEB 5th Class Maths Solutions Chapter 4 Fractions InText Questions

Try These : (Textbook Page No.86)

Question 1.
Write the fraction of coloured stars.

Solution:
\(\frac{1}{2}\)

Solution:
\(\frac{3}{4}\)

Solution:
\(\frac{5}{8}\)

Question 2.
Colour the diagram according to given fraction :

(a) \(\frac{2}{3}\)



Solution:

Question 3.
In fraction \(\frac{2}{3}\), numerator is and denominator is .
Solution:
In fraction \(\frac{2}{3}\), numerator is and denominator is .

Question 4.
In fraction \(\frac{1}{2}\), numerator is and denominator is .
Solution:
In fraction \(\frac{1}{2}\), numerator is and denominator is .

Question 5.
Write the fraction with numerator 4 and denominator 5:
Solution:
Write the fraction with numerator 4 and denominator 5:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 6 Measurement Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 6 Measurement Ex 6.5

Question 1.
The cost of 1 m cloth for pants is ₹ 265.50 and there is 24 m cloth in a roll. Find the cost of one bundle.
Solution:
Cost of 1 m cloth for pants = ₹ 265,50
Cost of 24 m cloth in a roll
= ₹ 265.50 × 24
= ₹ 6372.00
= ₹ 6372

Question 2.
The weight of a box of mangoes is 32.4 kg. A shopkeeper wants to make 6 packets from this. How many kilograms of mangoes will be there in each packet?
Solution:
The weight of the box of mangoes = 32.4 kg
Number of packets = 6
Weight of mangoes in each box = 32.4 kg ÷ 6

= 5.4 kg

Question 3.
A vessel contains 28.5 7 oil. It is poured into. 5 small containers. How much milk will be there in one small container?
Solution:
Quantity of milk in the given vessel = 28.5 l
Number of small containers = 5
Quantity of milk in each of small container = 28.5 ÷ 5

= 5.7 l

Question 4.
1 bundle of copies weighs 9.8 kgs. Find the weight of 14 such bundles.
Solution:
Weight of 1 bundle of copies = 9.8 kg
Weight of 14 bundles of copies
= 9.8 × 14 kg
= 137.2 kg.

Question 5.
The length of a stick is 12.7 cm. Find the length of 7 such sticks.
Solution:
The length of 1 stick = 12.7 cm
The length of 7 sticks = 12.7 × 7 cm
= 88.9 cm

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 3 HCF and LCM MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 3 HCF and LCM MCQ Questions

Tick (✓) the right answer :

Question 1.
Which number is the smallest even Prime number ? ………
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(c) 2

Question 2.
Which number is neither Prime nor Composite ?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 3.
Which numbers are Prime numbers between 70 and 80 ?
(a) 71, 72, 73
(b) 71, 75, 79
(c) 71, 80
(d) 71, 73, 79.
Answer:
(d) 71, 73, 79.

Question 4.
HCF of 75 and 90 = ……
(a) 5
(b) 10
(c) 15
(d) 20.
Answer:
(c) 15

Question 5.
LCM of 12, 18 and 24 = …….
(a) 72
(b) 36
(c) 48
(d) 24
Answer:
(a) 72

Question 6.
If HCF of any two numbers is 8 then out of following which can not be LCM of that numbers ?
(a) 48
(b) 60
(c) 24
(d) 56
Answer:
(b) 60

Question 7.
What is length of the largest tape which can measure the lengths of 24 m and 30 m ?
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m.
Answer:
(c) 6 m

Question 8.
What is the smallest number which is divisible by 8 and 12 ?
(a) 16
(b) 48
(c) 72
(d) 24
Answer:
(d) 24

Question 9.
LCM of 26 and 39 = ……
(a) 13
(b) 78
(c) 39
(d) 26
Answer:

Question 10.

(a) 5
(b) 65
(c) 12
(d) 13.
Answer:
(d) 13.

Question 11.
Which number is composite number in the following?
(a) 43
(b) 23
(c) 21
(d) 37.
Answer:
(c) 21

Question 12.
Out of following, which number is multiple of 19?
(a) 171
(b) 172
(c) 173
(d) 174.
Answer:
(a) 171

Question 13.
HCF of 15, 45 and 105 = ………
(a) 15
(b) 5
(c) 30
(d) 45.
Answer:
(a) 15

Question 14.
What is the HCF of two prime numbers ?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 15.
Three bells ring with the time gap of 10 min, 15 min and 20 min respectively in a school. If all bells are rung together at 9:00 am then after how long the bells would ring together again ?
(a) 11:00 o’clock
(b) 08:00 o’clock
(c) 10:00 o’clock
(d) 12:00 o’clock
Answer:
(c) 10:00 o’clock

Read this pattern carefully and answer questions (16 – 20)

Question 16.
Read the above pattern and find the sum of first 6 Odd numbers.
(a) 30
(b) 12
(c) 25
(d) 36.
Answer:
(d) 36.

Question 17.
Read the above pattern and find the sum of first 10 Odd numbers.
(a) 20
(b) 50
(c) 100
(d) 40.
Answer:
(c) 100

Question 18.
Read the above pattern and find the sum of first 8 Even numbers.
(a) 16
(b) 24
(c) 72
(d) 64.
Answer:
(c) 72

Question 19.
Read the above pattern and find sum of first 9 even numbers.
(a) 19
(b) 18
(c) 45
(d) 90.
Answer:
(d) 90.

Question 20.
On a given road, the poles are erected at a distance of 24 m each and piles of stones are lying at a distance of 30 m each. If the 1st pile of stones is lying adjacent to the first pole, at what distance will the pole and the pile of stones be together again ?
(a) 100 m
(b) 110 m
(c) 150 m
(d) 120 m.
Answer:
(d) 120 m.

Question 21.
What is the length of the largest inchitape which can measure the lengths of 24 m and 30 m completely:
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m.
Answer:
(c) 6 m

Question.
Write the smallest number which is divisible by 8 and 12.
Solution:
The smallest number which is divisible by 8 and 12 = LCM of 8 and 12.

= 2 × 2 × 2 × 3 = 24

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 3 HCF and LCM Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 3 HCF and LCM Ex 3.3

Question 1.
Find LCM of the following :
(a) 5, 10
(b) 6,18
(c) 25, 50
(d) 9, 24
Solution:
(a) Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, …….
Multiples of 10 = 10, 20, 30, 40, 50, ……..
Common multiples of 5 and 10 = 10, 20, 30, 40, 50, ……….
The lowest common multiple = 10
So, LCM of 5 and 10 = 10

(b) Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, …….
Multiples of 18 = 18, 36, 54, ……….
Common multiples of 6 and 18 = 18, 36, 54, ………
The lowest common multiple = 18
So, LCM of 6 and 18= 18

(c) Multiples of 25 = 25, 50, 75, 100, 125, 150, …….., …….
Multiples of 50 = 50, 100, 150, 200, ……, ……, ……..
Common multiples of 25 and 50= 50, 100, 150, ………..
The lowest common multiple = 50
So, LCM of 25 and 50 = 50

(d) Multiples of 9 = 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, ………
Multiples of 24 = 24, 48, 72, 96, …., ….., …….
Common multiples of 9 and 24 = 72,
So, LCM of 9 and 24 = 72

Question 2.
Find LCM of the follówing:
(a) 4, 8 and 12
(b) 6, 12 and 24
(e) 15, 18 and 27
(d) 24, 36 and 40
Solution:
(a) Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36,. 40, 44, 48,
Multiples of 8 = 8, 16, 24, 32, 40, 48, …….
Multiples of 12 = 12, 24, 36, 48, 60, ……..
Common multiples of 4, 8 and 12 = 24, 48
The lowest common multiple = 24
So, LCM of 4, 8, 12 = 24

(b) Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, ……
Multiples of 12 = 12, 24, 36, 48, 60, 72, …….
Multiples of 24 = 24, 48, 72, 96, ……..
Common multiples of 6, 12 and 24 = 24, 48, 72, …..
The lowest common multiple = 24
So, LCM of 6, 12, 24 = 24

LCM of 15, 18 and 27 = 3 × 3 × 5 × 2 × 3 = 270

LCM of 24, 36 and 40 = 2 × 2 × 2 × 3 × 3 × 5 = 360

Question 3.
Find LCM of following using Prime factorisation :
(a) 32,40
(b) 24, 36
(c) 15, 30 and 45
(d) 40, 44 and 48
Solution:
(a)

32 = 2 × 2 × 2 × 2 × 2
40 = 2 × 2 × 2 × 5
Common factors = 2 × 2 × 2
Remaining factors = 2 × 2 × 5
So, LCM = 2 × 2 × 2 × 2 × 2 × 5 = 160

(b) 24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3×3

Common factors = 2 × 2 × 3
Remaining factors = 2 × 3
So, LCM = 2 × 2 × 2 × 3 × 3 = 72

(c) 15 = 3 × 5
30 = 2 × 3 × 5
45 = 3 × 3 × 5

Common factors = 3 × 5
Remaining factors = 2 × 3
So, LCM = 3 × 5 × 2 × 3 = 90

(d) 40 = 2 × 2 × 2 × 5
44 = 2 × 2 × 11
48 = 2 × 2 × 2 × 2 × 3

Common factors = 2 × 2
Remaining factors = 5 × 11 × 2 × 2 × 3
So, LCM = 2 × 2 × 2 × 2 × 5 × 3 × 11
= 2640

Question 4.
Find LCM of following using Division method :
(a) 15, 20
(b) 12, 38
(c) 30, 45 and 50
(d) 40, 68 and 60
Solution:
(a)

LCM of 15 and 20 = 2 × 2 × 5 × 3 = 60
(b)

LCM of 12 and 38 = 2 × 2 × 3 × 19 = 228

(c)

LCM of 30, 45 and 50 = 2 × 3 × 3 × 5 × 5 = 450

(d)

LCM of 40, 68 and 60 = 2 × 2 × 2 × 3 × 5 × 17 = 2040

Question 5.
Find the smallest number which is divisible by 12, 15 and 20 completely ?
Solution:
LCM of 12, 15 and 20
LCM = 2 × 2 × 3 × 5 = 60
Smallest number which is divisible by 12, 15 and 20 = 60

Question 6.
One child jumps 3 feet high and another jumps 4 feet high. If both the children continue jumping together in same direction then after how many feet they will be together again ?
Solution:
We are to find the LCM of 3 and 4
LCM = 3 × 4 = 12
They will be together again after 12 feet.

Question 7.
How many minimum number of students are required from a class to make groups of 4 each and 5 each so that no student is left ?
Solution:
We are to find the LCM of 4 and 5
LCM = 4 × 5 = 20
Number of students required = 20

Question 8.
Three bells ring with a time gap of 10 min, 20 min and 30 min respectively in a school. If all bells are rung together at 8:00 am then after how long the beUs would ring together again ?
Solution:
We are to find the LCM of 10, 20 and 30
LCM = 2 × 5 × 2 × 3 = 60

The bells will ring together after 60 min. i.e. 1 hours.
Thus, the bells will ring together at 9:00 am