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Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 4 Money (Currency) Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 4 Money (Currency) Ex 4.1

1. Draw 50 paise coins for given notes/coins :

Question 1.

Solution:

= Take four coins of 50 paise

Question 2.

Solution:

= Take ten coins of 50 paise

Question 3.

Solution:

= Take twenty coins of 50 paise

Question 4.

Solution:

= Take fourteen coins of 50 paise

Question 5.

Solution:

= Take eight coins of 50 paise.

2. Draw notes/coins, converting 50 paise into rupees.

Question 1.

Question 2.

Solution:

Question 3.

Solution:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 5 Measurement Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 5 Measurement Ex 5.2

Question 1.
Measure the length of given items in cm and mm :
(a)

Solution:
7 cm 8 mm

(b)

Solution:
3 cm 4 mm

(c)

Solution:
3 cm 8 mm

(d)

Solution:
6 cm 5 mm.

Question 2.
Measure the length of line segments in cm and mm :
(a)

Solution:
3 cm 7 mm

(b)

Solution:
4 cm 6 mm

(c)

Solution:
5 cm 2 mm

(d)

Solution:
6 cm 8 mm

(e)

Solution:
8 cm 3 mm

(f)

Solution:
12 cm 5 mm

Question 3.
Measure the length and breadth of given currency notes :

(a) Length ……. = ……. cm …. mm
(b) Breadth …… = …… cm …… mm

(c) Length ……. = ……. cm …. mm
(d) Breadth …… = …… cm …… mm
Solution:
Take a ₹2000 and a ₹200 note and measure their length and breadth
(a) 16 cm 8 mm
(b) 6 cm 6 mm
(c) 14 cm 6 mm
(d) 6 cm 6 mm

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 5 Measurement Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 5 Measurement Ex 5.1

Question 1.
Complete the table

Solution:
Take the above mentioned items and measure each of them and complete the table.

Question 2.
Find the distance between given dots and give the following answers:

(a) Distance from point A to B
= ……… cm
Solution:
4 cm

(b) Distance from point B to D
= ……… cm
Solution:
6 cm

(c) Distance from point A to E
= ……… cm
Solution:
5 cm

(d) Distance from point C to D
= ……… cm
Solution:
4 cm

(e) Distance from point B to E
= …….. cm
Solution:
11 cm

(f) Distance from point A to D
= …….. cm
Solution:
6 cm.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 5 Measurement Revision Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 5 Measurement Revision Exercise

Question 1.
The length of pencil is 19 ………
(centimetre, kilogram, metre)
Solution:
Centimeter

Question 2.
Weight of a brick is 3 ……..
(litre, kilogram, metre)
Answer:
Kilogram

Question 3.
There is 2 …….. water in the jug.
(litre, kilogram, metre)
Solution:
Litre

Question 4.
Draw a picture on weighing scale showing less and more weight.

Solution:

Question 5.
Colour the given container having capacity upto 2 litre

Solution:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 8 Perimeter and Area MCQ Questions and Answers.

PSEB 4th Class Maths Chapter 8 Perimeter and Area MCQ Questions

Question 1.
Sum of the length of all sides of a closed figure is called ………
(a) Perimeter
(b) Area
(c) Shadow
(d) None of these
Answer:
(a) Perimeter

Question 2.
Find the perimeter of a triangle having sides 5 cm, 7 cm and 9 cm :
(a) 15 cm
(b) 20 cm
(c) 27 cm
(d) 21 cm.
Answer:
(d) 21 cm.

Question 3.
Find the perimeter of a shape if each side of square is 1 cm :
(a) 12 cm
(b) 7 cm
(c) 28 cm
(d) 14 cm

Answer:
(a) 12 cm

Question 4.
The perimeter of given figure is 22 m. The four sides of this figure are 4 m, 6 m, 6 m and 3 m. Find the fifth side.

(a) 4 m
(b) 3 m
(c) 5 m
(d) 2 m.
Answer:
(b) 3 m

Question 5.
Find the perimeter of a square whose side is 5 cm.
(a) 25 cm
(b) 15 cm
(c) 20 cm
(d) 16 cm.
Answer:
(c) 20 cm

Question 6.
Find the perimeter of a rectangle whose length is 4 cm and breadth is 5 cm ?
(a) 9 cm
(b) 12 cm
(c) 15 cm
(d) 18 cm.
Answer:
(d) 18 cm.

Question 7.
Find the area of the given figures :

Which figure has greater area ?
(a) (iv)
(b) (iii)
(c) (i)
(d) (ii).
Answer:
(a) (iv)

Question 8.
Find the area of a square whose side is 6 cm.
(a) 24 square cm
(b) 36 cm
(c) 36 square cm
(d) 12 square cm
Answer:
(b) 36 cm

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 8 Perimeter and Area Ex 8.3

Question 1.
Colour the area covered by each figure :

Solution:

Solution:

Solution:

Solution:

Note: You can fill any colour.

Question 2.
Find the area of each figure on the basis of number of squares if side of each square is 1 cm and area of each square is 1 square cm.

Solution:
(a) Number of squares = 13
Area of the figure = 13 × 1 cm² = 13 cm²

(b) Number of squares = 20
Area of the figure = 20 × 1 cm² = 20 cm²

(c) Number of squares = 5
Area of the figure = 5 × 1 cm² = 5 cm²

(d) Number of squares = 9
Area of the figure = 9 × 1 cm² = 9 cm²

(e) Number of squares = 12
Area of the figure = 12 × 1 cm² = 12 cm²

(f) Number of squares = 16
Area of the figure = 16 × 1 cm² = 16 cm²

Question 3.
How many square cm area is covered by each figure ?

Solution:
(a) 6
(b) 7
(c) 10
(d) 7
(e) 13
(f) 4

Question 4.
In a notebook with squares draw your favourite figure in which the number of square boxes are:
(a) 20
(b) 27
(c) 15.
Solution:
Try yourself

Question 5.
Look at this picture. Can you divide the figure into four equal parts. How many squares are there in each part?
Solution:
Number of squares = 12
Number of parts into which to be divided = 4
Number of squares in each part = 12 ÷ 4 = 3

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 8 Perimeter and Area Ex 8.2

Which of the following figures have covered more surface. Tick the figure which (✓) has greater area.

(a)

Solution:
First Figure

(b)

Solution:
Second Figure

(c)

Solution:
First Figure

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 4 Money (Currency) Revision Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 4 Money (Currency) Revision Exercise

1. Fill in the blanks :

Question 1.
There are ………………. paise in a rupee.
Solution:
100

Question 2.
There are ……………… 50 paise coins in a rupee.
Solution:
2

Question 3.
To write one rupee …………… sign is used.
Solution:
₹

Question 4.
We will get …………….. five rupee coins for a ₹ 10 note.
Solution:
2

Question 5.
A ₹ 20 note = …………….. ₹ 5 notes.
Solution:
4

Question 6.
A ₹ 50 note = …………………… ₹ 10 notes.
Solution:
5

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 3 Fractional Numbers Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 3 Fractional Numbers Ex 3.2

1. Colour the given figures in such a way that they show equal fraction and also write their fraction in the given box :

Question 1.

Solution:

Question 2.

Solution:

2. Write next five equal fractions of the given each fraction :

Question 1.
\(\frac{1}{2}\)
Solution:
\(\frac{2}{4}\), \(\frac{3}{6}\), \(\frac{4}{8}\), \(\frac{5}{10}\), \(\frac{6}{12}\)

Question 2.
\(\frac{3}{4}\)
Solution:
\(\frac{6}{8}\), \(\frac{9}{12}\), \(\frac{12}{16}\), \(\frac{15}{20}\), \(\frac{18}{24}\)

Question 3.
\(\frac{1}{3}\)
Solution:
\(\frac{2}{6}\), \(\frac{3}{9}\), \(\frac{4}{12}\), \(\frac{5}{15}\), \(\frac{6}{18}\)

Question 4.
\(\frac{2}{5}\)
Solution:
\(\frac{4}{10}\), \(\frac{6}{15}\), \(\frac{8}{20}\), \(\frac{10}{25}\), \(\frac{12}{30}\)

3. Fill in the blanks to make equal fractions to the given each fraction :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 8 Perimeter and Area Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 8 Perimeter and Area Ex 8.1

Question 1.
Find the perimeter of given figures:
(a)

Solution:
Perimeter of Figure = Sum of all sides of the figure
= 7 mm + 9 mm + 13 mm
= 29 mm

(b)

Solution:
Perimeter of Figure = Sum of all sides of the figure
= 9 m + 11 m + 15 m + 18 m = 53 m.

(c)

Solution:
Perimeter of Figure = Sum of all sides of the figure = 2 cm + 2 cm + 3 cm + 3 cm + 2 cm + 2 cm
= 14 cm

Question 2.
Find the perimeter of given figures :
(a)

Solution:
Perimeter of Figure
= Sum of all sides of the figure = 1 cm + 1 cm + 3 cm + 4 cm + 5 cm + 4 cm = 18 cm

(b)

Solution:
Perimeter of Figure
= Sum of all sides of the figure = 1 cm + 4 cm + 6 cm + 4 cm + 1 cm + 3 cm + 4 cm + 3 cm = 26 cm

(c)

Solution:
Perimeter of Figure
= Sum of all sides of the figure = 1 cm +’ 1 cm + 1 cm + 1 cm + 1 cm + 1 cm + 2 cm + 1 cm+ 3 cm + 4 cm = 16 cm

Question 3.
Find the perimeter of given figures :
(a)

Solution:
Perimeter of Figure
= Sum of all sides of the figure
= 6 cm + 7 cm + 15 cm + 6 cm 4 + 7 cm + 15 cm
= 56 cm

(b)

Solution:
Perimeter of Figure
= Sum of all sides of the figure
= 4m+8m + 14m + 4m + 10m + 4m = 44 m

(c)

Solution:
Perimeter of Figure
= Sum of all sides of the figure
= 35 cm + 35 cm + 55 cm + 35 cm + 60 cm + 80 cm + 150 cm + 80 cm
= 530 cm

(d)

Solution:
Perimeter of Figure
= Sum of all sides of the figure
= 15 cm + 25 cm + 5 cm + 7 cm + 3 cm + 3 cm + 7 cm + 5 cm + 25 cm
= 95 cm

Question 4.
In given figures, the perimeter of which figure is less and by how much ?
(a)

Solution:
Perimeter of Figure (a)
= 12 m + 16 m + 14 m + 18 m = 60 m

(b)

Solution:
Perimeter of Figure (b)
= 10 m + 12 m + 17 m + 20 m = 59 m
Perimeter of Figure (b) is less than figurea (a) by =
(60 m – 59 m) = 1 m
Perimeter of Figure (b) is less by 1 m

Question 5.
Find the length of side with (?) of the given figures :

Solution:
(a) Sides of the figure
= 30 m, 25 m and x m
Perimeter of the figure = 70 m
Length of the Side of the figure (x) = Perimeter – Sum of the two sides
x = 70 m – 55 m = 15 m.

(b) Sides of the figure = 34 cm, 43 cm
50 cm and x cm
Perimeter of the figure = 150 cm
Length of the Side of the figure (x) = Perimeter – Sum of the other three sides
x = 150 cm – (34 cm + 43 cm + 50 cm)
= 150 cm – 127 cm
= 23 cm

(c) Sides of the figure = 32 m, 68 m, 25 m,
37 m and x m.
Perimeter of the figure = 207 m
Length of the Side of the figure (x) = Perimeter – Sum of the other four sides
= 207 m – (32 m + 68 m + 25 m + 37 m)
= 207 m – 162 m = 45 m

Question 6.
(a) Four sides of a Geld are 40 m, 35 m, 25 m and 28 m. Find its perimeter.
Solution:
Four sides of the field = 40 m, 35 m 25 m and 28 m
Perimeter of the field
= Sum of all sides
= 40m+35m+25m+28m
= 128m

(b) Length and breadth of a tennis court are 25 m and 9 m respectively. A net is required on four sides of the tennis court so that players do not face any difficulty. What is the length of the net required to cover the 4 sides of tennis court ?
Solution:
Length of the tennis court = 25 m
Breadth of the tennis court = 9 m
Perimeter of the tennis court = Length + Length + Breadth + Breadth
= 25 m + 25 m + 9 m + 9 m
= 68 m
The length of the net required to cover the 4 sides of tennis court = 68 m