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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 2 Fundamental Operations on Numbers Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.6

Question 1.
The price of a cycle is ₹ 5699. What is price of 17 cycles ?
Solution:
Cost price of 1 cycle = ₹ 5699
Cost price of 17 cycles = ₹ 5699 × 17
= ₹ 96883

Question 2.
There are 12 tiles in a box. How many tiles are there in 4590 boxes ?
Solution:
Number of tiles in one box = 12
Number of tiles in 4590 in boxes = 4590 × 12 = 55080

Question 3.
Multiply 4 digit smallest number with 98.
Solution:
Smallest 4 digit number = 1000
Requires product = 1000 × 98 = 98000

Question 4.
Rate list of electrical equipment in electrical shop is as follows :

(i) Charan has ₹ 1 lakh with him. He buys 2 washing machines and one L.C.D. How much amount has he spent ?
(ii) Charan’s brother has ₹ 1 lakh. He buys one A.C., Two water Gysers and one Refrigerator. How much amount is left with him ?
Solution:
(i) Cost price of 1 washing machine
= ₹ 24999
Cost price of 2 washing machines

(ii) Cost price of 1 Water Gyser = ₹ 12999
Cost price of 2 Water Gyser

Question 5.
A factory manufactures 4990 toffees a day. How many toffees will be manufactured in 19 days ?
Solution:
Number of toffees manufactured in 1 day = 4990
Number of toffees manufactured in 19 days = 4990 × 19 = 94810

Question 6.
6798 bricks are loaded in a tractor in an hour. How many bricks will be loaded in 13 hours ?
Solution:
Number of bricks loaded in one hour = 6798
Number of bricks loaded in 13 hours = 6798 × 13 = 88374

Question 7.
A shopkeeper sells one mobile phone for t 5089. If he sells 18 such mobile phones in a day, how much amount would he collect in a day ?
Solution:
Selling price of 1 mobile phone = ₹ 5089
Selling price of 18 mobile phones = ₹ 5089 × 18
The amount he will collect in one day
= ₹ 91602

Question 8.
Multiply 3 digit largest number with 95.
Solution:
Greatest number of three digits
= 999
Required product = 999 × 95 = 94905

Question 9.
How many seconds are there in 24 hours ?
Solution:
Number of seconds in one hour
= 3600
Number of seconds in 24 hours = 3600 × 24 = 86400

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.5

Question 1.
Fill in the blanks:

(a) 451 × 1 = ____
Solution:
451 × 1 = 451

(b) 8135 × 10 = _____
Solution:
8135 × 10 = 81350

(c) 650 × 100 = _____
Solution:
650 × 100 = 65000

(d) 3090 × 0 = _____
Solution:
3090 × 0 = 0

(e) 129 × _____ = 12900
Solution:
129 × 100 = 12900

(f) _____ × 1000 = 13000
Solution:
13 × 1000 = 13000

(g) _____ × 791 = 0
Solution:
0 × 791 = 0

(h) _____ × 82 = 82 × 602
Solution:
602 × 82 = 82 × 602

(i) 8414 × 10 = _____
Solution:
8414 × 10 = 84140

(j) 67 × 100 = _____
Solution:
67 × 100 = 6700

(k) 91 × 1000 = _____
Solution:
91 × 1000 = 91000

(l) 100 × 1000 = _____
Solution:
100 × 1000 = 100000

(m) 545 × _____ = 5450
Solution:
545 × 10 = 5450

(n) _____ × 10 = 7060
Solution:
706 × 10 = 7060

(o) 798 × _____ = 798
Solution:
798 × 1 = 798

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 9 Volume Intext Questions and Answers.

PSEB 5th Class Maths Solutions Chapter 9 Volume Intext Questions

Page No. 195

(a) Tick (✓) the object which takes more space:

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

(b) Write in ascending order according to the space each object covers.

Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.4

Question 1.
Solve the following :
(a) 450 × 6
(b) 963 × 9
(c) 529 × 23
(d) 988 × 38
(e) 912 × 56
(f) 806 × 56
(g) 252 × 54
(h) 1888 × 19
(i) 2005 × 34
(j) 1560 × 64
(k) 10569 × 8
(l) 10210 × 9
(m) 230 × 150
(n) 400 × 225
Solution:

Question 2.
Find the product of the following:
(a) 4045 × 23
(b) 1609 × 30
(c) 363 × 134
(d) 455 × 208
(e) 105 × 120
(f) 1440 × 25
(g) 1530 × 61
(h) 3817 × 12
(i) 1908 × 35
(j) 1000 × 29
Solution:

Question 3.
Find the digits in place of * :

Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 9 Volume MCQ Questions and Answers.

PSEB 5th Class Maths Chapter 9 Volume MCQ Questions

Multiple Choice Questions

Tick (✓) the correct answer :

Question 1.
Volume of cube with side 9 cm is :
(a) 81 cubic cm
(b) 90 cubic cm
(c) 729 cubic cm
(d) 8 cubic cm.
Answer:
(c) 729 cubic cm

Question 2.
Find volume of cuboid with length 6 cm, breadth 4 cm and height 2 cm :
(a) 24 cubic cm
(b) 28 cubic cm
(c) 64 cubic cm
(d) 48 cubic cm.
Answer:
(d) 48 cubic cm.

Question 3.
Which is not the standard unit of volume?
(a) cubic cm
(b) sq. m
(c) cubic mm
(d) cubic metre.
Answer:
(b) sq. m

Question 4.
A cuboid with all sides equal is called ……………
(a) square
(b) cube
(c) cuboid
(d) rectangle.
Answer:
(b) cube

Question 5.
Count the number of cubes and find the volume of cuboid.

Solution:
Volume of cuboid = 8 cubic cm.

Question 6.
A trolly is full of bricks. Length of the trolly is 400 cm, breadth is 200 cm and height is 100 cm. Length, breadth and height of a brick are 20 cm, 10 cm and 6 cm respectively. How many bricks are there in the trolly?
Solution:
Length of trolly, L = 400 cm
Breadth of trolly B = 200 cm
Height of trolly, H = 100 cm
Volume of trolly = L × B × H
= 400 cm × 200 cm × 100 cm.
Length of a brick, l = 20 cm
Breadth of a brick; b = 10 cm
Height of a brick, h = 6 cm
Volume of a brick = 20 cm × 10 cm × 6 cm
Number of bricks in the trolly
= \(\frac{\text { Volume of trolly }}{\text { Volume of each brick }}\)
= \(\frac{400 \times 200 \times 100}{20 \times 10 \times 6}\)
= \(\frac{20 \times 20 \times 50}{3}\) = \(\frac{20000}{3}\)
= 6666 (Approx)

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 9 Volume Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 9 Volume Ex 9.1

1. Find the volume of the following by counting the number of cubes.

Question 1.

Volume = _________
Solution:
Volume = 4 cm³

Question 2.

Volume = _________
Solution:
Volume = 30 cm³

Question 3.

Volume = _________
Solution:
Volume = 36 cm³

2. Complete the table :

Solution:
(i) 6 cm³
(ii) 24 mm³
(iii) 6000 m³
(iv) 343 cm³
(v) 400 m³

3. Find the volume of cube whose side (edge) is :

Question 1.
6 cm
Solution:
Side (edge) of cube = 6 cm
Volume of cube = side × side × side
= 6 cm × 6 cm × 6 cm
= 216 cm³

Question 2.
8 m
Solution:
Side of cube = 8 m
Volume of cube = side × side × side
= 8m × 8m × 8m
= 512 m³

Question 3.
15 mm
Solution:
Side of cube = 15 mm
Volume of cube = side × side × side
= 15 mm × 15 mm × 15 mm
= 3375 mm³

Question 4.
21 m
Solution:
Side of cube = 21m
Volume of cube= side × side × side
= 21 m × 21 m × 21 m
= 9261 m³

4. Find the volume of cuboid whose length, breadth and height is as follows :

Question 1.
9 cm, 6 cm, 3 cm
Solution:
Length of cuboid = 9 cm
Breadth of cuboid = 6 cm
Height of cuboid = 3 cm
Volume of cuboid = Length × Breadth × Height
= 9 cm × 6 cm × 3 cm
= 162 cm³

Question 2.
12 mm, 9 mm, 4 mm
Solution:
Length of cuboid = 12 mm
Breadth of cuboid = 9 mm
Height of cuboid = 4 mm
Volume of cuboid = Length × Breadth × Height
= 12 mm × 9 mm × 4 mm
= 432 mm³

Question 3.
15 m, 13 m, 12 m
Solution:
Length of cuboid = 15 m
Breadth of cuboid = 13 m
Height of cuboid = 12 m
Volume of cuboid = Length × Breadth × Height
= 15 m × 13 m × 12 m
= 2340 m³

Question 4.
22 mm, 16 mm, 12 mm
Solution:
Length of cuboid = 22 mm
Breadth of cuboid = 16 mm
Height of cuboid = 12 mm
Volume of cuboid = Length × Breadth × Height
= 22 mm × 16 mm × 12 mm
= 4224 mm³

Question 5.
25 m, 23 m, 21 m.
Solution:
Length of cuboid = 25 m
Breadth of cuboid = 23 m
Height of cuboid = 21 m
Volume of cuboid = Length × Breadth × Height
= 25 m × 23 m × 21 m
= 12075 m³

Question 5.
A chalk box is 8 cm long, 6 cm broad and 10 cm high. Find the volume of the box.
Solution:
Length of the chalk box = 8 cm
Breadth of the chalk box = 6 cm
Height of the chalk box = 10 cm
Volume of the chalk box = Length × Breadth × Height
= 8 cm × 6 cm × 10 cm
= 480 cm³

Question 6.
A cardboard box is 50 cm long, 40 cm wide and 24 cm high. Find the volume of the box.
Solution:
Length of the cardboard box = 50 cm
Breadth of the cardboard box = 40 cm
Height of the cardboard box = 24 cm
Volume of the cardboard box = Length × Breadth × Height
= 50 cm × 40 cm × 24 cm
= 48000 cm³

Question 7.
Jashan’s tiffin box is 15 cm long, 10 cm wide and 8 cm high and Gurwinder’s tiffin box is 12 cm long, 10 cm wide and 10 cm high. Find the volume of both. Whose tiffin box has more volume ?
Solution:
Jashan :
Length of the tiffin box = 15 cm
Breadth of the tiffin box = 10 cm
Height of the tiffin box = 8 cm
Volume of the tiffin box = Length × Breadth × Height
= 15 cm × 10 cm × 8 cm
= 1200 cm³

Gurwinder :
Length of the tiffin box = 12 cm
Breadth of the tiffin box = 10 cm
Height of the tiffin box = 10 cm
Volume of the tiffin box = Length × Breadth × Height
= 12 cm × 10 cm × 10 cm
= 1200 cm³
Their tiffin boxes have equal volumes.

Question 8.
Find the volume of 25 cuboidal boxes with dimensions 12 cm long, 9 cm wide and 6 cm high each.
Solution:
Length of the cuboidal box =12 cm
Breadth of the cuboidal box = 9 cm
Height of the cuboidal box = 6 cm
Volume of the cuboidal box = Length × Breadth × Height
= 12 cm × 9 cm × 6 cm
= 648 cm³
Volume of 25 cuboidal boxes
= 648 × 25 cm³
= 16200 cm³

Question 9.
There are two types of powder boxes available in the market. One is of cubical shape with side 8 cm and other is of cuboidal shape with length 15 cm, breadth 8 cm and height 4 cm. Which box has more powder and how much? If both have same prices then which box will you prefer?
Solution:
Side of the cubical box = 8 cm
Volume of the cubical box = side x side x side
= 8 cm × 8 cm × 8 cm
= 512 cm³
Length of the cuboidal box = 15 cm
Breadth of the cuboidal box = 8 cm
Height of the cuboidal box = 4 cm
Volume of the cuboidal box = Length × Breadth × Height
= 15 cm × 8 cm × 4 cm
= 480 cm³
Cubical box has more powder
= 512 cm³ – 480 cm³
= 32 cm³
I will prefer cubical box.

Question 10.
How many bricks of size 24 cm long, 12 cm wide and 8 cm thick are required for* making a wall of 12 m long, 3 m high and 24 cm thick?
Solution:
Length of the wall = 12 m
= 12 × 100 cm
= 1200 cm
Height of the wall = 3 m
= 3 × 100 cm
= 300 cm
Breadth of the wall = 24 cm
Volume of the wall = Length × Breadth × Height
= 1200 cm × 24 cm × 300 cm
= 8640000 cm³
Length of each brick = 24 cm
Breadth of each brick = 12 cm
Height of each brick = 8 cm
Volume of each brick = Length × Breadth × Height
= 24 cm × 12 cm × 8 cm
= 2304 cm³
Number of bricks required
= \(\frac{8640000 \mathrm{~cm}^{3}}{2304 \mathrm{~cm}^{3}}\)
= 3750 bricks.

Question 11.
The length, breadth and height of a biscuit packet is 15 cm, 9 cm and 6 cm respectively. If a packet has 30 biscuits then Bind the volume of each biscuit.
Solution:
Length of the packet = 15 cm
Breadth of the packet = 9 cm
Height of the packet = 6 cm
Volume of the packet = Length × Breadth× Height
= 15 cm × 9 cm × 6 cm
= 810 cm³
Total number of biscuits in the packet = 30
Volume of each biscuit = \(\frac{810 \mathrm{~cm}^{3}}{30}\)
= 27 cm³

Question 12.
One trolley is full of bricks. It is 4 m long, 2 m wide and 60 cm deep. One brick is of size 20 cm x 10 cm x 6 cm. How many bricks are there in the trolley ?
Solution:
Length of the trolley
= 4m = 4 × 100 cm = 400 cm
Breadth of the trolley
= 2 m = 2 × 100 cm = 200 cm
Depth (Height) of the trolley = 60 cm
Volume of the trolley = Length × Breadth × Depth
= 400 cm × 200 cm × 60 cm
= 4800000 cm³
Length of the brick = 20 cm
Breadth of the brick = 10 cm
Height of the brick = 6 cm
Volume of the brick = Length × Breadth × Height
= 20 cm × 10 cm × 6 cm
= 1200 cm³
Number of bricks in the trolley
= \(\frac{4800000}{1200}\) cm³
= 4000.

13. Fill in the blanks :

Question 1.
Volume of cube = __________ × __________ × ________
Solution:
Side × Side × Side

Question 2.
Volume of cuboid = __________ × __________ × ________
Solution:
Length × Breadth × Height

Question 3.
The space occupied by a solid is called its ________
Solution:
Volume.

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.3

Think and Do :

Question 1.
(a) Find the sum of 60498,31292 and 7132.
(b) Find the difference of 70123 and 40268.
Solution:

Question 2.
27020 bricks are required for constructing a kitchen and 31275 bricks are required for constructing a room. How many bricks in total are required for construction of both ?
Solution:
No. of bricks required for the construction of a kitchen = 27020
No. of bricks required for the construction of a room = + 31275
Total no. of bricks required for the construction of the both = 58295.

Question 3.
Surjit had ₹ 20000 with him. He bought clothes costing ₹ 13750. How much amount was left with him ?
Solution:

Question 4.
In a library, there are 30155 Punjabi books, 28653 Maths books and 12376 English books. How many books are there in the library ?
Solution:

Question 5.
The sum of two numbers is 89000. If one number is 25450 then find the other number.
Solution:

Question 6.
What number must be added to 70429 to get 100000 ?
Solution:

Question 7.
Find the number which is :
(a) 7976 more than 36798
(b) 12967 less than 30067.
Solution:

Question 8.
If the price of a computer is ₹ 15560 and price of a laptop is ₹ 9050 more than price of the computer then find :
(a) Price of the laptop
(b) Total price of both the items.
Solution:

Question 9.
Find the greatest and smallest 5 digit numbers using digits 9, 3, 4, 0, 7. Also find their difference.
Solution:

Question 10.
Find the sum of greatest 2 digits, 3 digits and 4 digits numbers.
Solution:

Question 11.
Find the difference of place values of 6 and 7 in number 96074.
Solution:

Question 12.
Subtract 45555 from 6 digit smallest number.
Solution:

Question 13.
Satnam had ₹ 8765 with him. His uncle gave him ₹ 2500. Satnam gave ₹ 4770 to his sister out of his total money. How much money was left with him ?
Solution:

Question 14.
Mandeep had 110000. He bought a pair of shoes for ₹ 1050 and a suit for ₹ 3600. How much money was left with him ?
Solution:

Question 15.
Sandeep has t 78500 in his bank account. How much more amount should he deposit in the account so that he has ? 100000 in his account ?
Solution:

Question 16.
A person travels 165 km by car from Pathankot to Kashmir. Next day, he drives 138 km from Kashmir to Leh. How much distance did he cover ?
Solution:

Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.2

Question 1.
Fill the digit in place of *


Solution:

Question 2.
Evaluate the following :
(a) 1238 – 1025 + 5018
(b) 9386 – 2535 – 1002
(c) 6307 – 4052 + 2115
(d) 8107 + 2437 – 6089
(e) 18337 + 30947 – 34413
(f) 91206 – 70413 + 30824
(g) 10003 – 5911 – 3284
(h) 92319 – 65424 – 12105
Solution: