Class 4

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.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.4

Question 1.
Write the names of the months which starts with “J”.
Solution:
January, June, July

Question 2.
Write the name of the months having 31 days.
Solution:
The name of the months having 31 days : January, March, May, July, August, October, December.

Question 3.
Write the names of the months which have less than 30 days.
Solution:
The names of the months which have less than 31 days : February.

Question 4.
In which month of year do you celebrate your birthday ?
Solution:
I celebrate my birthday in the month of ……….
Note : In the blank write the month of your birth.

Question 5.
In which months you have Summer vacations and winter vacations ?
Solution:
Summer vacations = In June
Winter vacations = In December

Question 6.
Shivansh went to visit historical places with his uncle from 28 May to 15 August. How many days he spent on vacation ? (28 May and 15 August both the days included).
Solution:
Shivansh starts his journey on = 28th may
Number of days of May
(28, 29, 30, 31) = 4
Number of days of June = 30
Number of days of July = 31
Number of days of August = 15
Number of days he spent on vacation = 80

Question 7.
26th January to 15th August = No. of days of January (26, 27, 28, 29, 30, 31) = 6
No. of days of February = 28
No. of days of March = 31
No. of days of April = 30
No. of days of May = 31
No. of days of June = 30
No. of days of July = 31
No. of days of August = 15

The total number of days from 26th January 2018 to 15th August, 2018

Question 8.
(a) 6 June to 22nd November Number of days of June = 26 (31 – 5 = 26)
Number of days of July = 31
Number of days of August = 31
Number of days of September = 30
Number of days of October = 31
Number of days of November = 22
Total number of days from 6th June to 22nd November = 171

(b) Number of winter holidays = 24th December to 31st December
(24,25,26,27,28,29,30,31) = 8

(c) 3rd June to 4th July
Number of days of June = 28
(30 – 2 = 28)
Number of days of July = 4
Total number of holidays = 32

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.3

Question 1.
What is the time after 2 hours ?
(a) 9:20 AM
(b) 12:00 noon
(c) 11:15 PM
(d) 5:10 PM
(e) 3:30 PM
(f) 7:35 AM.
Solution:
(a) 11:20 AM
(b) 2:00 PM
(c) 01:15 AM
(d) 7:10 PM
(e) 5:30 PM
(f) 9:35 AM.

Question 2.
What is the time 1 hour before the given time ?
(a) 12:00 mid night
(b) 3:30 afternoon
(c) 11:00 before noon
(d) 4:00 before noon
(e) 9:00 afternoon
(f) 8:50 before noon.
Solution:
(a) 11:00 PM
(b) 2:30 PM
(c) 10:00 AM
(d) 3:00 AM
(e) 8:00 PM
(f) 7:50 AM.

Question 3.
Add :
(a) 2 hours 15 minutes in 3 hours 28 minutes
(b) 15 hours 28 minutes in 4 hours 12 minutes
(c) 8 hours 48 minutes in 3 hours 22 minutes
(d) 4 hours 32 minutes in 3 hours 48 minutes
Solution:

Question 4.
Subtract :
(a) 3 hours 27 minutes from 1 hours 13 minutes
Solution:

(b) 15 hours 14 minutes from 3 hours 5 minutes
Solution:

(c) 12 hours 17 minutes from 4 hours 27 minutes
Solution:

(d) 9 hours 28 minutes from 3 hours 38 minutes
Solution:

Question 5.
A train starts at 7:40 A.M. and reaches its destination at 2:15 p.m. How much time it will takes to complete its journey. Find out the time interval of journey.
Solution:
The time when the train starts the journey = 7:40 AM
The time gap from 7:40 AM to 8:00 AM = 20 mins
The time gap from 8:00 AM to 12:00 Noon = 4 hours
The time gap from 12:00 Noon to 2:00 PM = 2 hours
The time gap from 2:00 PM to 2:15 PM = 15 mins
Total time = 6 hours 35 mins
The time interval of journey = 6 hours 35 mins

Second Method :
The time when the train starts the journey = 7:40AM = 7 : 40 hours
The time when the train reaches its destination = 2 : 15 PM
= 2 : 15 + 12 : 00
= 14 : 15 hours

Therefore, the time interval of journey = 6 h 35 mins

Question 6.
Shikha starts her journey by car at 6:40 A.M. and completes her journey at 3:50 p.m. How long did she drive the car ?
Solution:
The time when Shikha starts the journey = 6:40 AM
The time gap from 6:40 AM to 7:00 AM = 20 mins
The time gap from 7: 00 AM to 12:00 Noon = 5 h
The time gap from 12:00 Noon to 3:00 PM = 3 h
The time gap from 3:00 PM to 3:50 PM = 50 mins
The time for which she drives the car = 8 hours 70 mins
= 8 hours + 70 mins
= 8 hours + 60 mins + 10 mins
= 8 hours + 1 hour + 10 mins
= 9 hours 10 mins
Second Method:
The time when Shikha starts the journey = 6 : 40 AM 6 : 40 hours
The time when Shikha reaches her destination = 3 : 50 PM
= 3:50+ 12:00
= 15:50 hours

Question 7.
A cricket match starts at 9:30 p.m. and ends at 1:25 a.m. For how long did the match continue?
Solution:
The time when the cricket match starts = 9:30 PM
The time gap from 9:30 PM to 10:00 PM = 30 minutes
The time gap from 10:00 PM to 12:00 mid night = 2 hours
The time gap from 12:00 mid night to 11:00 AM = 1 hour
The time gap from 1:00 AM to 1:25 AM = 25 minutes
The time for which match continues = 3 hours 55 minutes

Question 8.
Sunny starts his bhangra practice at 4:15 p.m. For how long did he practice for bhangra?
Solution:
The time when sunny starts his bhangra practice = 4:15 PM
The time gap from 4:15 PM to 5.00 PM = 45 minutes
The time gap from 5:00 PM to 6:00 PM = 1 hour
The time. gap from 6:00 PM to 6:10 PM = 10 minutes
The time spent for practice = 1 hour 55 minutes
Second method:
Because starting time and finishing time both are in PM
∴ Therefore, the the spent by Sunny for practice = 1 hour 55 minutes

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.2

Question 1.
Fill in the blanks :

(a) 15 minutes to 9 = …. minutes past 8.
(b) Quarter to 6 = …. minutes past 5.
(c) Half to 9 = minutes past 8.
(d) 20 minutes to 8 = minutes past 7.
Solution:
(a) 45
(b) 45
(c) 30
(d) 40.

Question 2.
Write afternoon times in figures :
(a) 15 minutes to 5
(b) 15 minutes past 4
(c) 35 minutes to 9
Solution:
(a) 4:45 PM.
(b) 4:15 PM
(c) 8:25 PM.

Question 3.
Write the time in a.m. or p.m.
(a) 5:20 in the morning
(b) 6:40 in the evening
(c) 9:35 at night
(d) 11:10 in the morning
(e) 8:40 in the morning.
Solution:
(a) 5:20 AM.
(b) 6:40 PM
(c) 9:35 PM
(d) 11:10 AM
(e) 8:40 AM.

Question 4.
Change the following in 24 hours notation :
(a) 9:45 in the morning
(b) 9:45 at night
(c) 10:15 in the morning
(d) 10:15 at night
(e) 3:20 in the morning
(f) 3:20 afternoon.
Solution:
(a) 09:45 hours
(b) 21:45 hours
(c) 10:15 hours
(d) 22:15 hours
(e) 03:20 hours
(f) 15:20 hours

Question 5.
Change 24 hours notation into 12 hours with use of a.m. and p.m.
(a) 08:48
(b) 20:48
(c) 13:13
(d) 07:20
(e) 06:00
(f) 19:30
Solution:
(a) 8:48 AM
(b) 8:48 PM
(c) 1:13 PM
(d) 7:20 AM
(e) 6:00 AM
(f) 7:30 PM.

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers MCQ Questions and Answers.

PSEB 4th Class Maths Chapter 2 Fundamental Operations on Numbers MCQ Questions

Question 1.
573 + 227 = ____________
(a) 798
(b) 799
(c) 800
(d) 801.
Answer:
(c) 800

Question 2.
__________ + 336 = 868
(a) 632
(b) 528
(c) 532
(d) 1204.
Answer:
(c) 532

Question 3.
700 – 125 = ____________
(a) 475
(b) 575
(c) 675
(d) 825.
Answer:
(b) 575

Question 4.
801 – __________ = 602
(a) 201
(b) 1403
(c) 100
(d) 199.
Answer:
(d) 199

Question 5.
53 × 8 = 8 × __________
(a) 3
(b) 53
(c) 40
(d) 159
Answer:
(b) 53

Question 6.
716 × ________ = 716
(a) 0
(b) 1
(c) 716
(d) 2
Answer:
(b) 1

Question 7.
573 × 0 = _________
(a) 573
(b) 1
(c) 0
(d) 57
Answer:
(c) 0

Question 8.
_________ × 1 = 600
(a) 1
(b) 200
(c) 600
(d) 300
Answer:
(c) 600

Question 9.
7 × 1000 = ___________
(a) 7
(b) 1000
(c) 7000
(d) 700
Answer:
(c) 7000

Question 10.
53 × 30 = __________
(a) 159
(b) 1590
(c) 83
(d) 1690
Answer:
(b) 1590

Question 11.
128 ÷ 16 = __________
(a) 9
(b) 10
(c) 12
(d) 8
Answer:
(d) 8

Question 12.
126 ÷ 14 = 9, which is divisor?
(a) 14
(b) 9
(c) 126
(d) 0
Answer:
(a) 14

Question 13.
15 × 12 + 8 = __________
(a) 168
(b) 198
(c) 178
(d) 188
Answer:
(d) 188

Question 14.
1509 ÷ 1 = ___________
(a) 1
(b) 1509
(c) 3
(d) 0
Answer:
(b) 1509

Question 15.
In a school there are 22 students in 1st, 25 students in 2nd, 23 students in 3rd, 27 students in 4th and 23 students In 5th class. Find out total number of students in the school.
(a) 120
(b) 130
(c) 145
(d) 160
Answer:
(a) 120

Question 16.
Which number must be added in 779 to get the smallest number of 4 digit ?
(a) 231
(b) 220
(c) 321
(d) 221
Answer:
(d) 221

Question 17.
How many hours are there in the month of May ?
(a) 31
(b) 744
(c) 24
(d) 720
Answer:
(b) 744

Question 18.
Find the difference between 4 digits greatest number and the smallest number by using digits 2,0, 4 and 6.
(a) 3747
(b) 6174
(c) 2046
(d) 4374.
Answer:
(d) 4374

Question 19.
Find the product of 3 digits smallest number and 2 digits greatest number.
(a) 9900
(b) 10,000
(c) 290
(d) 9700.
Answer:
(a) 9900

Question 20.
After distributing 178 toffees among 15 children equally, find the number of toffees left.
(a) 13
(b) 14
(c) 12
(d) 11
Answer:
(a) 13

Question 21.
19 × 300 = _________
(a) 57000
(b) 5700
(c) 2200
(d) 319
Answer:
(b) 5700

Question 22.
225 × ___________ = 2250
(a) 1
(b) 10
(c) 100
(d) 0
Answer:
(b) 10

Brain Teaser

Question 1.
Fill in the blank boxes replacing sign of ?

Solution:

Question 2.

Solution:

Question 3.

Solution: