PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4

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.

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4 1

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4 2
The total number of days from 26th January 2018 to 15th August, 2018

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4 3

(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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4 4
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.4 5

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3

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.

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3

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:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 1
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 2

Question 4.
Subtract :
(a) 3 hours 27 minutes from 1 hours 13 minutes
Solution:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 3

(b) 15 hours 14 minutes from 3 hours 5 minutes
Solution:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 4

(c) 12 hours 17 minutes from 4 hours 27 minutes
Solution:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 5

(d) 9 hours 28 minutes from 3 hours 38 minutes
Solution:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 6

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 7
Therefore, the time interval of journey = 6 h 35 mins

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 8

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

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3

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
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.3 9

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.2

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.

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.2

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

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.2

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.

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1

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

PSEB Solutions for Class 4 Maths Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1

ਪ੍ਰਸ਼ਨ 1.
ਨੋਟਾਂ/ਸਿੱਕਿਆਂ ਬਦਲੇ 50 ਪੈਸੇ ਦੇ ਸਿੱਕੇ ਬਣਾਓ ।
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 1
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 6

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 2
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 7

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 3
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 8

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 4
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 9

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 5
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 10

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1

ਪ੍ਰਸ਼ਨ 2.
50 ਪੈਸੇ ਦੇ ਸਿੱਕਿਆਂ ਨੂੰ ਰੁਪਇਆਂ ਵਿੱਚ ਬਦਲਦੇ ਹੋਏ ਨੋਟ/ਸਿੱਕੇ ਬਣਾਓ ।

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 11
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 14

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 12
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 15

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 13
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Ex 4.1 16

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

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

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

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

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

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

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

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

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

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

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

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

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

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 ?
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 1
Solution:
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 2

Question 2.
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 3
Solution:
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 4

Question 3.
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 5
Solution:
PSEB 4th Class Maths MCQ Chapter 2 Fundamental Operations on Numbers 6

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

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1

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)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 1
Solution:
1:55

(b)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 2
Solution:
7:10

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1

(c)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 3
Solution:
9:05

(d)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 4
Solution:
2:40

(e)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 5
Solution:
10:40

(f)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 6
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:
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 7
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 8

PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1

Question 3.
How many minutes a minute hand will take to reach the time shown between the first clock and the second clock ?
(a)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 9
Solution:
15 minutes

(b)
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 10
Solution:
25 minutes

Question 4.
Tell the time shown in the given clock and write.
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 11
Solution:
4:18

Question 5.
Tell the time shown in the given clock and write.
PSEB 4th Class Maths Solutions Chapter 6 Time Ex 6.1 12
Solution:
5:58

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

Punjab State Board PSEB 4th Class Maths Book Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise Questions and Answers.

PSEB Solutions for Class 4 Maths Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

ਦੁਹਰਾਈ

ਪ੍ਰਸ਼ਨ 1.
ਸ਼ਬਦਾਂ ਵਿੱਚ ਲਿਖੋ :
(a) 598
ਹੱਲ:
ਪੰਜ ਸੌ ਅਠਾਨਵੇਂ
(b) 608
ਹੱਲ:
ਛੇ ਸੌ ਅੱਠ

(c) 328
ਹੱਲ:
ਤਿੰਨ ਸੌ ਅਠਾਈ

(d) 999.
ਹੱਲ:
ਨੌ ਸੌ ਨੜਿਨਵੇਂ ।

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

ਪ੍ਰਸ਼ਨ 2.
ਅੰਕਾਂ ਵਿੱਚ ਲਿਖੋ :
(a) ਦੋ ਸੌ ਅਠੱਤਰ
ਹੱਲ:
278

(b) ਸੱਤ ਸੌ ਦਸ
ਹੱਲ:
710

(c) ਚਾਰ ਸੌ ਛੇ
ਹੱਲ:
406

(d) ਅੱਠ ਸੌ ਛਿਆਸੀ ।
ਹੱਲ:
886.

ਪ੍ਰਸ਼ਨ 3.
ਵਿਸਤ੍ਰਿਤ ਰੂਪ ਵਿੱਚ ਲਿਖੋ :
(a) 298
ਹੱਲ:
298 = 200 + 90 + 8

(b) 183
ਹੱਲ:
183 = 100 + 80 + 3

(c) 709
ਹੱਲ:
709 = 700 +9

(d) 840.
ਹੱਲ:
840 = 800 + 40 + 0.

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

ਪ੍ਰਸ਼ਨ 4.
ਹੇਠ ਲਿਖੀਆਂ ਸੰਖਿਆਵਾਂ ਨੂੰ ਗਿਣਤਰੇ ਉੱਪਰ ਰੰਗ ਭਰ ਕੇ ਦਿਖਾਓ :
(a) 803
PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise 1
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise 3

(b) 999
PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise 2
ਹੱਲ:
PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise 4

ਪ੍ਰਸ਼ਨ 5.
ਹੇਠ ਲਿਖੇ ਅੰਕਾਂ ਨੂੰ ਵਰਤਦੇ ਹੋਏ ਬਣਨ ਵਾਲੀਆਂ ਸਾਰੀਆਂ ਸੰਖਿਆਵਾਂ ਲਿਖੋ :
(a) 2, 5, 4
ਹੱਲ:
254, 524, 452, 425, 245, 542

(b) 5, 3, 2.
ਹੱਲ:
532, 253, 235, 523, 352, 325

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

(c) 3, 1, 9
ਹੱਲ:
319, 193, 391, 139, 931, 913

(d) 4, 0, 8
ਹੱਲ:
408, 804, 480, 840

ਪ੍ਰਸ਼ਨ 6.
ਸਮਝੇ ਅਤੇ ਖਾਲੀ ਖਾਨੇ ਭਰੋ :
(i)
(a) 761 762
ਹੱਲ:

(b) 400 ____
ਹੱਲ:
400 401

(c) 678 ____
ਹੱਲ:
678 679

(d) 962 ____
ਹੱਲ:
962 963

(e) 348 ____
ਹੱਲ:
348 349

(f) 824 ____
ਹੱਲ:
824 825

(ii)
(a) 863 864

(b) ____ 112
ਹੱਲ:
111 112

(c) ___ 456
ਹੱਲ:
455 456

(d) ___ 562
ਹੱਲ:
561 562

(e) ___ 715
ਹੱਲ:
714 715

(f) ___ 950
ਹੱਲ:
949 950

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

(iii)
(a) 387 ___ 389
ਹੱਲ:
387 388 389

(b) 680 ___ 682
ਹੱਲ:
680 681 682

(c) 996 ___ 998
ਹੱਲ:
996 997 998

(d) 514 ___ 516
ਹੱਲ:
514 515 516

(e) 788 ___ 790
ਹੱਲ:
788 789 790

(f) 200 ___ 202
ਹੱਲ:
200 201 202

ਪ੍ਰਸ਼ਨ 7.
ਖ਼ਾਲੀ ਖਾਨੇ ਵਿੱਚ >, <, ਜਾਂ = ਚਿੰਨ੍ਹ ਭਰੋ :

(a) 761 ___ 671
ਹੱਲ:
761 > 671

(b) 137 ___ 106
ਹੱਲ:
137 > 106

(c) 115 ___ 162
ਹੱਲ:
115 < 162

(d) 492 ___ 492
ਹੱਲ:
492 = 492

(e) 987 ___ 989
ਹੱਲ:
987 < 989

(f) 134 __ 431
ਹੱਲ:
134 < 431

(g) 768 __ 876
ਹੱਲ:
768 < 876

(h) 617 __ 617
ਹੱਲ:
617 =

(i) 146 __ 416
ਹੱਲ:
146 < 416

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

(j) 768 ___ 98
ਹੱਲ:
768 > 98

ਹੁਣ ਤੁਸੀਂ ਹੇਠਾਂ ਲਿਖੇ ਸਵਾਲ ਹੱਲ ਕਰੋ :

ਪ੍ਰਸ਼ਨ 8.
ਹੇਠ ਲਿਖੀਆਂ ਸੰਖਿਆਵਾਂ ਵਿੱਚੋਂ ਸਭ ਤੋਂ ਵੱਡੀ ਸੰਖਿਆ ਲਿਖੋ :

(a) 96, 279, 961, 899, 99
ਹੱਲ:
961

(b) 163, 894, 534, 106, 119
ਹੱਲ:
894

(c) 764, 895, 564, 381, 678
ਹੱਲ:
895

(d) 161, 37, 153, 275, 891
ਹੱਲ:
891

(e) 800, 190, 700, 861, 199
ਹੱਲ:
861

(f) 221, 448, 868, 88, 992
ਹੱਲ:
992.

ਪ੍ਰਸ਼ਨ 9.
ਹੇਠ ਲਿਖੀਆਂ ਸੰਖਿਆਵਾਂ ਵਿੱਚੋਂ ਸਭ ਤੋਂ ਛੋਟੀ ਸੰਖਿਆ ਲਿਖੋ :
(a) 99,638, 125, 369, 581
ਹੱਲ:
99

(b) 163, 894, 534, 106, 119
ਹੱਲ:
106

(c) 764, 895, 564, 381, 678
ਹੱਲ:
381

(d) 161, 37, 153, 275, 891
ਹੱਲ:
37

(e) 800, 190, 700, 861, 199
ਹੱਲ:
190

(f) 221, 448, 686, 88, 992
ਹੱਲ:
88

ਪ੍ਰਸ਼ਨ 10.
ਵੱਧਦੇ ਕ੍ਰਮ ਵਿੱਚ ਲਿਖੋ :
(a) 269, 781, 683, 453, 239
ਹੱਲ:
239, 269, 453, 683, 781

(b) 196, 638, 700, 699, 824
ਹੱਲ:
196, 638, 699, 700, 824

(c) 910, 800, 816, 72, 16
ਹੱਲ:
16, 72, 800, 816, 910

(d) 361, 482, 469, 756, 29
ਹੱਲ:
29, 361, 469, 482, 756.

PSEB 4th Class Maths Solutions Chapter 1 ਸੰਖਿਆਵਾਂ Revision Exercise

(e) 235, 568, 567, 245, 961
ਹੱਲ:
235, 245, 567. 568, 961

ਪ੍ਰਸ਼ਨ 11.
ਘੱਟਦੇ ਕ੍ਰਮ ਵਿੱਚ ਲਿਖੋ :
(a) 619, 564, 72, 12, 169
ਹੱਲ:
619, 564, 169, 72, 12

(b) 781, 890, 967, 961, 119
ਹੱਲ:
967, 961, 890, 781, 119

(c) 543, 650, 790, 798, 260
ਹੱਲ:
798, 790, 650, 543, 260

(d) 806, 818, 76, 82, 9
ਹੱਲ:
818, 806, 82, 76, 9

(e) 582, 254, 184, 784, 591
ਹੱਲ:
784, 591, 582, 254, 184

ਪ੍ਰਸ਼ਨ 12.
ਸਮਝੋ ਅਤੇ ਲਿਖੋ :
ਹੱਲ:
(a) 11, 22, 33, 44, 55, 66, 77, 88
(b) 10, 20, 30, 40, 50, 60, 70, 80
(c) 44, 48, 52, 56, 60, 64, 68, 72
(d) 52, 54, 56, 58, 60, 62, 64, 66
(e) 81, 83, 85, 87, 89, 91, 93, 95

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

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 1

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

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

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 3

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

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 6

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

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 7
= 5 notes are required to exchange ₹ 500 note.

(b) If all are ₹ 50 notes then ₹ 500 ÷ ₹ 50
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 8
= 10 notes are required to exchange
= ₹ 500 note

(c) If all are ₹ 10 notes then ₹ 500 ÷ ₹ 10
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 9
= 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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 10

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

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 11

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
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.9 12

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

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.
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 1

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

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

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

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

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

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

3. Divide and verify the solution :

Question 1.
66 ÷ 6
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 3
Quotient = 11
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
66 = 11 × 6 + 0
66 = 66

Question 2.
431 ÷ 7
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 4
Quotient = 61
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
431 = 61 × 7 + 4
431 = 427 + 4
431 = 431

Question 3.
728 ÷ 8
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 5
Quotient = 91
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
728 = 91 × 8
728 = 728

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

Question 4.
648 ÷ 9
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 6
Quotient = 72
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
648 = 72 × 9 + 0
648 = 648

Question 5.
960 ÷ 5
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 7
Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
960 = 192 × 5 + 0
960 = 960

4. Solve the following :

Question 1.
666 ÷ 6
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 8
Quotient = 111

Question 2.
655 ÷ 5
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 9
Quotient = 131

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

Question 3.
787 ÷ 7
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 10
Quotient = 112
Remainder = 3

Question 4.
877 ÷ 7
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 11
Quotient = 125
Remainder = 2

Question 5.
598 ÷ 6
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 12
Quotient = 99
Remainder = 4

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

Question 6.
566 ÷ 8
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 13
Quotient = 70
Remainder = 6

Question 7.
707 ÷ 7
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 14
Quotient = 101
Remainder = 0

5. Solve the following :

Question 1.
2150 ÷ 2
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 15
Quotient = 1075

Question 2.
4050 ÷ 3
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 16
Quotient = 1350

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

Question 3.
8048 ÷ 8
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 17
Quotient = 1006

Question 4.
5106 ÷ 6
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 18
Quotient = 851

Question 5.
3043 ÷ 3
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 19
Quotient = 1014
Remainder = 1

Question 6.
7890 ÷ 7
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 20
Quotient = 1127
Remainder = 1

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

Question 7.
4050 ÷ 5
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 21
Quotient = 810

6. Divide and verify the following:

Question 1.
96 ÷ 12
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 22
Quotient = 8
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
96 = 8 × 12 + 0
96 = 96

Question 2.
98 ÷ 14
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 23
Quotient = 7
Verification : Dividend = Quotient × Divisor + Remainder
98 = 14 × 7 + 0
98 = 98

Question 3.
78 ÷ 16
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 24
Quotient = 4
Remainder = 14
Verification : Dividend = Quotient × Divisor + Remainder
78 = 4 × 16 + 14
78 = 64 + 14
78 = 78

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

Question 4.
760 ÷ 19
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 25
Quotient = 40
Verification : Dividend = Quotient × Divisor + Remainder
760 = 40 × 19 + 0
760 = 760

Question 5.
550 ÷ 13
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 26
Quotient = 42
Remainder = 4
Verification : Dividend = Quotient × Divisor + Remainder
550 = 42 × 13 + 4
550 = 446 + 4
550 = 550

Question 6.
894 ÷ 24
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 27
Quotient = 37
Remainder = 6
Verification : Dividend = Quotient × Divisor + Remainder
894 = 37 × 24 + 6
894 = 888 + 6
894 = 894

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

Question 7.
913 ÷ 66
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 28
Quotient = 13
Remainder = 55
Verification : Dividend = Quotient × Divisor + Remainder
913 = 13 × 66 + 55
913 = 858 + 55
913 = 913

Question 8.
826 ÷ 34
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 29
Quotient = 24
Remainder = 10
Verification : Dividend = Quotient × Divisor + Remainder
826 = 24 × 34 + 10
826 = 816 + 10
826 = 826

Question 9.
7645 ÷ 12
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 30
Quotient = 637
Remainder = 1
Verification : Dividend = Quotient × Divisor + Remainder
7645 = 637 × 12 + 1
7645 = 7644 + 1
7645 = 7645

Question 10.
7813 ÷ 13
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 31
Quotient = 601
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
7813 = 601 × 13 + 0
7813 = 7813

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

Question 11.
5375 ÷ 25
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 32
Quotient = 215
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
5375 = 215 × 25 + 0
5375 = 5375

Question 12.
6767 ÷ 33
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 33
Quotient = 205
Remainder = 2
Verification : Dividend = Quotient × Divisor + Remainder
6767 = 205 × 33 + 2
6767 = 6765 + 2
6767 = 6767

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

Question 13.
9600 ÷ 50
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 34
Quotient = 192
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9600 = 192 × 50 + 0
9600 = 9600

Question 14.
9999 ÷ 33
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 35
Quotient = 303
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9999 = 303 × 33 + 0
9999 = 9999

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

Question 15.
9660 ÷ 60
Solution:
PSEB 4th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.8 36
Quotient = 161
Remainder = 0
Verification : Dividend = Quotient × Divisor + Remainder
9660 = 161 × 60 + 0
9660 = 9660

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Revision Exercise

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

PSEB Solutions for Class 4 Maths Chapter 4 ਧਨ (ਕਰੰਸੀ) Revision Exercise

ਦੁਹਰਾਈ

ਪ੍ਰਸ਼ਨ 1.
ਖ਼ਾਲੀ ਥਾਂਵਾਂ ਭਰੋ :
(i) ₹ ਵਿੱਚ …….. ਪੈਸੇ ਹੁੰਦੇ ਹਨ ।
ਹੱਲ:
100

(ii) ₹ 1 ਵਿੱਚ 50 ਪੈਸੇ ਦੇ …….. …… ਸਿੱਕੇ ਹੁੰਦੇ ਹਨ ।
ਹੱਲ:
ਦੋ

(iii) 1 ਰੁਪਏ ਨੂੰ ਲਿਖਣ ਲਈ …………….. ਚਿੰਨ੍ਹ ਦੀ ਵਰਤੋਂ ਕੀਤੀ ਜਾਂਦੀ ਹੈ ।
ਹੱਲ:

PSEB 4th Class Maths Solutions Chapter 4 ਧਨ (ਕਰੰਸੀ) Revision Exercise

(iv) ਤੋਂ 10 ਦੇ ਨੋਟ ਬਦਲੇ ਤੋਂ 5 ਦੇ ………….. ਸਿੱਕੇ ਮਿਲਣਗੇ ।
ਹੱਲ:
ਦੋ

(v) ₹ 20 ਦੇ ਇੱਕ ਨੋਟ ਦਾ ਮੁੱਲ, ₹ 5 ਦੇ ………. ਨੋਟਾਂ ਦੇ ਮੁੱਲ ਦੇ ਬਰਾਬਰ ਹੁੰਦਾ ਹੈ ।
ਹੱਲ:
ਚਾਰ

(vi) ₹ 50 ਦੇ ਇੱਕ ਨੋਟ ਦਾ ਮੁੱਲ, ਤੋਂ 10 ਦੇ ………… ਨੋਟਾਂ ਦੇ ਮੁੱਲ ਦੇ ਬਰਾਬਰ ਹੁੰਦਾ ਹੈ ।
ਹੱਲ:
ਪੰਜ ॥