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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

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time MCQ Questions and Answers.

PSEB 4th Class Maths Chapter 6 Time MCQ Questions

Question 1.
The number of hours in a day are:
(a) 24
(b) 12
(c) 18
(d) 16.
Answer:
(a) 24

Question 2.
How many days are there in a week?
(a) 6
(b) 8
(c) 7
(d) 31.
Answer:
(c) 7

Question 3.
Which of the following is a leap year?
(a) 2100
(b) 2000
(c) 2200
(d) 1900.
Answer:
(b) 2000

Question 4.
Which of the following is a leap year?
(a) 2013
(b) 2014
(c) 2015
(d) 2016.
Answer:
(d) 2016.

Question 5.
How many days are in a leap year?
(a) 365 days
(b) 361 days
(c) 366 days
(d) 360 days
Answer:
(c) 366 days

Question 6.
Which is 6th and 8th month of the Year?
(a) May and July
(b) June and September
(c) June and August
(d) August and May
Answer:
(c) June and August

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 3 Fractional Numbers Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 3 Fractional Numbers Ex 3.1

Question 1.
Match the fraction according to coloured portion:

Solution:
(a) \(\frac{1}{4}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{3}\)

Question 2.
Write the fraction of coloured as well as blank portion in the given space :


Solution:

3. Colour the figure according to the given fraction :

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

4. Mark (✓) on the correct fraction of the coloured portion of the figure:

Question 1.

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

Question 2.

Solution:
(d) \(\frac{3}{8}\)

Question 3.

Solution:
(c) \(\frac{1}{4}\)

5. Write the given fraction in words:

Question 1.
\(\frac{1}{2}\)
Solution:
Half

Question 2.
\(\frac{1}{4}\)
Solution:
One Fourth

Question 3.
\(\frac{1}{3}\)
Solution:
One third

Question 4.
\(\frac{2}{3}\)
Solution:
Two third

Question 5.
\(\frac{3}{4}\)
Solution:
Three fourth

Question 6.
\(\frac{1}{10}\)
Solution:
One tenth.

6. Write numerator and denominator of the given fractions :

Question 1.
\(\frac{2}{3}\)
Solution:
Numerator = 2 and Denominator = 3

Question 2.
\(\frac{1}{2}\)
Solution:
Numerator = 1 and Denominator = 2

Question 3.
\(\frac{1}{4}\)
Solution:
Numerator = 1 and Denominator = 4

Question 4.
\(\frac{3}{4}\)
Solution:
Numerator = 3 and Denominator = 4.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.5

Answer the following questions from Calendar of the years 2016 and 2018.

Question 1.
How many Sundays are there in January 2016 and January 2018 ?
Solution:
Number of Sundays in January
2016 = 5 (3, 10, 17, 24, 31)
Number of Sundays in January 2018 = 4 (7, 14, 21, 28)

Question 2.
On which day the Independence day falls in the Year 2018 ?
Solution:
In the year 2018 the Independence day falls on Wednesday.

Question 3.
What is date on first Monday in April 2018?
Solution:
The date on first Monday in April 2018 = 2nd

Question 4.
How many days are there in February 2016 and February 2018 ? What difference did you notice ?
Solution:
Number of days in February 2016 = 29
Number of days in February 2018 = 28
2016 is a leap year whereas 2018 is a non leap year.

Question 5.
What is the date on last Friday of the Year ?
Solution:
The date on last ffiday of this year i. e. 2018 = 28th December

Question 6.
Which is the day on 1 January 2018 and 31 December 2018 ?
Solution:
The day on 1st January 2018 = Monday
The day on 31 st December 2018 = Monday

Question 7.
Write the name 6f months which have 31 days.
Solution:
7. The name of months which have 31 days
January, March, May, July, August, October and December.

Question 8.
From the calendar find the date, month and day of your birthday.
Solution:
The date, month and day of my birthday :
Date =
Month =
Day =
Note : Write the date, months and day of your birthday in the blank.