Bhagya

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:

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 6 Time Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 6 Time Ex 6.1

Question 1.
Write the time shown in each clock given below:
(a)

Solution:
1:55

(b)

Solution:
7:10

(c)

Solution:
9:05

(d)

Solution:
2:40

(e)

Solution:
10:40

(f)

Solution:
11:45

Question 2.
Draw clocks in your note book and show the time as given below :
(a) 4:20
(b) 7:35
(c) 4:45
(d) 3:15
(e) 11:40
(f) 9:15.
Solution:

Question 3.
How many minutes a minute hand will take to reach the time shown between the first clock and the second clock ?
(a)

Solution:
15 minutes

(b)

Solution:
25 minutes

Question 4.
Tell the time shown in the given clock and write.

Solution:
4:18

Question 5.
Tell the time shown in the given clock and write.

Solution:
5:58

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.9

Question 1.
In morning assembly 161 students stand in 7 rows equally. How many students stand in each row ?
Solution:
Total number of student = 161
Number of rows = 7
Number of students in each row = 161 ÷ 7 = 23

Question 2.
I have 72 apples I have to put these in 3 buckets equally. How many apples each basket will contain ?
Solution:
The total number of apples = 72
Number of baskets = 3
Number of apples in each basket = 72 ÷ 3 = 24

Question 3.
A farmer has produced 4250 kg wheat in his Held. One bag is required to fill for 50 kg of wheat. How many such bags are required to fill 4250 kg wheat ?
Solution:
Total quantity of wheat produced = 4250 kg
Quantity of wheat in one bag = 50 kg
Number of bags required = 4250 ÷ 50 = 85

Question 4.
By which number to multiply 25 so that product becomes 625 ?
Solution:
Product of the two numbers = 625
One of the number = 25 Second number = 625 ÷ 25 = 25

Question 5.
A gardener has 120 flowers. He has to make garland of 24 flowers. How many such garlands are made from 120 flowers ?
Solution:
The total number of flowers = 120
Number of flowers in 1 garland = 24
Number of garlands that can be made = 120 ÷ 24 = 5

Question 6.
How many ₹ 50 notes are required to make ₹ 2000 ?
Solution:
Total amount = ₹ 2000
Value of 1 note = ₹ 50
Number of notes required = ₹ 2000 ÷ ₹ 50

Question 7.
I want to exchange my ₹ 500 note. How many notes of following denomination will I get ?
(a) If all are ₹ 100 notes …………………….
(b) If all are ₹ 50 notes ……………………
(c) If all are ₹ 10 notes ……………………..
Solution:
(a) If all are ₹ 100 notes then ₹ 500 ÷ ₹ 100

= 5 notes are required to exchange ₹ 500 note.

(b) If all are ₹ 50 notes then ₹ 500 ÷ ₹ 50

= 10 notes are required to exchange
= ₹ 500 note

(c) If all are ₹ 10 notes then ₹ 500 ÷ ₹ 10

= 50 notes are required to exchange a ₹ 500

Question 8.
A labourer picks up 20 bricks in 1 round. How many number of rounds are required to pick up 1000 bricks ?
Solution:
Total number of bricks = 1000
Number of bricks picked in 1 round = 20
Number of rounds = 1000 ÷ 20 = 50

Question 9.
A railway ticket costs ₹ 24. palak gave ₹ 576 to the Station Master. How many mumber of tickets did she get ?
Solution:
The cost of 1 ticket = ₹ 24
Palak gave the money = ₹ 576
No of tickets = ₹ 576 ÷ 24 = 24

Question 10.
Kashvi brought a toffees packet on her birthday. There are 175 toffees in this packet and there are 35 students in her class. How many toffees did each of them get.
Solution:
Total ho. of toffees = 175
No. of children = 35
No. of toffees that each child will get = 175 ÷ 35 = 5

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.8

1. Fill in the boxes of Q. No. 1 and Q.no. 2 as directed :

Question 1.

Question 2.
9 × 4 = 36 __________ ____________
Solution:
36 ÷ 9 = 4, 36 ÷ 4 = 6

Question 3.
6 × 8 = 48 __________ ____________
Solution:
48 ÷ 6 = 8, 48 ÷ 8 = 6

Question 4.
10 × 4 = 40 __________ ____________
Solution:
40 ÷ 10 = 4, 40 ÷ 4 = 10

2.

Question 1.

Question 2.
35 ÷ 7 = 5 _____________ _____________
Solution:
5 × 7 = 35, 7 × 5 = 35

Question 3.
56 ÷ 8 = 7 _____________ _____________
Solution:
7 × 8 = 56, 8 × 7 = 56

Question 4.
150 ÷ 10 = 15 _____________ _____________
Solution:
10 × 15 = 150, 15 × 10 = 150

Question 5.
120 ÷ 10 = 10 _____________ _____________
Solution:
10 × 12 = 120, 12 × 10 = 120

3. Divide and verify the solution :

Question 1.
66 ÷ 6
Solution:

Quotient = 11
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
66 = 11 × 6 + 0
66 = 66

Question 2.
431 ÷ 7
Solution:

Quotient = 61
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
431 = 61 × 7 + 4
431 = 427 + 4
431 = 431

Question 3.
728 ÷ 8
Solution:

Quotient = 91
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
728 = 91 × 8
728 = 728

Question 4.
648 ÷ 9
Solution:

Quotient = 72
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
648 = 72 × 9 + 0
648 = 648

Question 5.
960 ÷ 5
Solution:

Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
960 = 192 × 5 + 0
960 = 960

4. Solve the following :

Question 1.
666 ÷ 6
Solution:

Quotient = 111

Question 2.
655 ÷ 5
Solution:

Quotient = 131

Question 3.
787 ÷ 7
Solution:

Quotient = 112
Remainder = 3

Question 4.
877 ÷ 7
Solution:

Quotient = 125
Remainder = 2

Question 5.
598 ÷ 6
Solution:

Quotient = 99
Remainder = 4

Question 6.
566 ÷ 8
Solution:

Quotient = 70
Remainder = 6

Question 7.
707 ÷ 7
Solution:

Quotient = 101
Remainder = 0

5. Solve the following :

Question 1.
2150 ÷ 2
Solution:

Quotient = 1075

Question 2.
4050 ÷ 3
Solution:

Quotient = 1350

Question 3.
8048 ÷ 8
Solution:

Quotient = 1006

Question 4.
5106 ÷ 6
Solution:

Quotient = 851

Question 5.
3043 ÷ 3
Solution:

Quotient = 1014
Remainder = 1

Question 6.
7890 ÷ 7
Solution:

Quotient = 1127
Remainder = 1

Question 7.
4050 ÷ 5
Solution:

Quotient = 810

6. Divide and verify the following:

Question 1.
96 ÷ 12
Solution:

Quotient = 8
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
96 = 8 × 12 + 0
96 = 96

Question 2.
98 ÷ 14
Solution:

Quotient = 7
Verification : Dividend = Quotient × Divisor + Remainder
98 = 14 × 7 + 0
98 = 98

Question 3.
78 ÷ 16
Solution:

Quotient = 4
Remainder = 14
Verification : Dividend = Quotient × Divisor + Remainder
78 = 4 × 16 + 14
78 = 64 + 14
78 = 78

Question 4.
760 ÷ 19
Solution:

Quotient = 40
Verification : Dividend = Quotient × Divisor + Remainder
760 = 40 × 19 + 0
760 = 760

Question 5.
550 ÷ 13
Solution:

Quotient = 42
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
550 = 42 × 13 + 4
550 = 446 + 4
550 = 550

Question 6.
894 ÷ 24
Solution:

Quotient = 37
Remainder = 6
Verification : Dividend = Quotient × Divisor + Remainder
894 = 37 × 24 + 6
894 = 888 + 6
894 = 894

Question 7.
913 ÷ 66
Solution:

Quotient = 13
Remainder = 55
Verification : Dividend = Quotient × Divisor + Remainder
913 = 13 × 66 + 55
913 = 858 + 55
913 = 913

Question 8.
826 ÷ 34
Solution:

Quotient = 24
Remainder = 10
Verification : Dividend = Quotient × Divisor + Remainder
826 = 24 × 34 + 10
826 = 816 + 10
826 = 826

Question 9.
7645 ÷ 12
Solution:

Quotient = 637
Remainder = 1
Verification : Dividend = Quotient × Divisor + Remainder
7645 = 637 × 12 + 1
7645 = 7644 + 1
7645 = 7645

Question 10.
7813 ÷ 13
Solution:

Quotient = 601
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
7813 = 601 × 13 + 0
7813 = 7813

Question 11.
5375 ÷ 25
Solution:

Quotient = 215
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
5375 = 215 × 25 + 0
5375 = 5375

Question 12.
6767 ÷ 33
Solution:

Quotient = 205
Remainder = 2
Verification : Dividend = Quotient × Divisor + Remainder
6767 = 205 × 33 + 2
6767 = 6765 + 2
6767 = 6767

Question 13.
9600 ÷ 50
Solution:

Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9600 = 192 × 50 + 0
9600 = 9600

Question 14.
9999 ÷ 33
Solution:

Quotient = 303
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9999 = 303 × 33 + 0
9999 = 9999

Question 15.
9660 ÷ 60
Solution:

Quotient = 161
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9660 = 161 × 60 + 0
9660 = 9660

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.7 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.7

1. Fill in the blank:

Question 1.
18 ÷ 9 = ____________
Solution:
2

Question 2.
77 ÷ 7 = ___________
Solution:
11

Question 3.
48 ÷ 8 = __________
Solution:
6

Question 4.
78 ÷ ________ = 6
Solution:
13

Question 5.
42 ÷ 7 = __________
Solution:
6

Question 6.
84 ÷ 14 = ____________
Solution:
6

Question 7.
28 ÷ __________ = 7
Solution:
4

Question 8.
0 ÷ 8 = __________
Solution:
0

Question 9.
50 ÷ 5 = _________
Solution:
10

Question 10.
12 ÷ 1 = __________
Solution:
12

Question 11.
54 ÷ __________ = 9
Solution:
6

Question 12.
________ ÷ 15 = 1
Solution:
15

Question 13.
70 ÷ 5 = ___________
Solution:
14

Question 14.
100 ÷ 10 = _________
Solution:
10

Question 15.
81 ÷ 9 = __________
Solution:
9

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.6

Question 1.
Cost of 1 notebook is ₹ 15. Find the cost of 9 such notebooks ?
Solution:
Cost of 1 notebook = ₹ 15
Cost of 9 notebooks = ₹ 15 × 9 = ₹ 135.

Question 2.
There are 75 pencils in a box. How many pencils are there in 19 such boxes ?
Solution:
Number of pencils in a packet = 75
Number of pencils in 19 packets = 75 × 19
= 1425.

Question 3.
There are 79 beads in a chain. How many beads are there in 68 such chains ?
Solution:
Number of beads in one chain = 79
Number of beads in 68 chains = 79 × 68
= 5372.

Question 4.
The cost of a toycycle is ? 1560. Find total cost of 6 such toycycles ?
Solution:
The cost of 1 toycycle = ₹ 1560
The cost of 6 toycycles = ₹ 1560 × 6
= ₹ 9360.

Question 5.
There are 11 players in a cricket team. How many players are there in 12 such teams ?
Solution:
Number of players in 1 cricket team = 11
Number of players in 12 cricket teams = 11 × 12 = 132.

Question 6.
A box contains 1440 soaps. Find the numbers of soaps in 6 such boxes ?
Solution:
Number of soaps in 1 box = 1440
Number of soaps in 6 boxes = 1440 × 6
= 8640

Question 7.

Your mom went to market—
(a) She bought 2 kg apples, 2 kg. guava. How much amount she will pay to the seller ?
Solution:
Cost price of 2 kg apples
= ₹ 120 × 2 = ₹ 240
Cost price of 2 kg guava
= ₹ 35 × 2 = ₹ 70
Amount she will pay to the seller = ₹ 310.

(b) If she bought 3 kg oranges and 2 kg pomegranate. What amount she will pay to seller ?
Solution:
Cost price of 3 kg oranges
= ₹ 45 × 3 = ₹ 135
Cost price 2 kg pomegranate
= ₹ 40 × 2 = ₹ 80
Amount she will pay to seller = ₹ 215

Question 8.
These all notes and coins Karan received on his birthday. How much total amount did Karan receive ?

Solution:
= ₹ 500 × 5 + ₹ 50 × 3 + ₹ 10 × 7 + ₹ 2 × 3
= ₹ 2500 + ₹ 150 + ₹ 70 + ₹ 6
= ₹ 2726.

Question 9.
A car covers a distance of 16 km in a litre. How much distance it will cover in 28 litres ?
Solution:
Distance covered in 1 litre = 16 km
Distance covered in 28 litres = 16 km × 28

= 448 km.

Question 10.
A factory produces 125 soap bars in an hour. How many such soap bars will be produced in 8 hours ?
Solution:
Number of soaps produced in 1 hour = 125
Number of soaps produced in 8 hours = 8 × 125
= 1000

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 7 Shapes Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 7 Shapes Ex 7.2

Question 1.
Which shapes can we get from the following net ?

Solution:

Question 2.
How does a brick look from the top view ?

Solution:

Question 3.
Complete the pattern by filling colours :

Solution:
Filling colours:

Question 4.
Which tile would complete the following desings?

Solution:
I.

II.