S.chand publication New Learning Composite mathematics solution of class 8 Chapter 4 Playing with Numbers Exercise 4A

 Exercise 4A


Q1  |Ex-4A |Class 8 | Playing with Numbers |S.Chand | New Learning Composite | myhelper


Question 1 

The sum of the digits of a 2-digit number is 12. The number is 6 times the units digit. Find the number.

Sol :

2 digit number=10a+b

a+b=12⇒b=12-a

b×6=10a+b

6b=10a+b

10a+(12-a)=6(12-a)

10a+12-a=72-6a

9a+6a=72-12=60

15a=60

$a=\frac{460}{15}=4$

b=12-4=8


The number=10a+b

=(10×4)+8=48



Q2  |Ex-4A |Class 8 | Playing with Numbers |S.Chand | New Learning Composite | myhelper

Question 2

The sum of the digits of a 2 digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the original number.

Sol :

The original number=10a+b

a+b=7⇒b=7-a...(i)

Reversed number=10b+a

according to question

4(10a+b)-(10b+a)=3

40a+b-10b-a=3

40a-a+b-10b=3

39a-6b=3

39a-6(7-a)=3  [From (i)]

39a-42+6a=3

39a+6a=3+42

45a=45

a=1


∴b=7-1=6

∴original number=10a+b=10×1+6

=16



Q3  | Ex-4A |Class 8 | Playing with Numbers |S.Chand | New Learning Composite | myhelper

Question 3

The sum of a number of two digits  and of the number formed by reversing the digits is 110 and the difference of the digits is 6. Find the number.

Sol :

b-a=6⇒b=6+a

Original number=10a+b

Reversed number=10b+a

(10a+b)+(10b+a)=110

10a+b+10b+a=110

11a+11b=110

a+b=10

a+6+a=10

a=2


∴b=6+2=8

∴original number=10a+b=10×2+8

=28



Q4 |Ex-4A |Class 8 | Playing with Numbers |S.Chand | New Learning Composite | myhelper

Question 4

A certain number between 10 and 100 is 8 times the sum of its digits. If 45 is subtracted from it the digits will be reversed. Find the number.

Sol :

The number between 10 and 100

The number will be 2 digit number=10a+b

According to the question

(10a+b)=8(a+b)

10a+b=8a+8b

10a-8a=8b-8b

2a=7b

2a-7b=0


According to question

(10a+b)-45=10b+a

10a+b-(10b+a)=45

10a+b-10b-a=45

9a-9b=45

a-b=5

a=5+b


Now, 20-7b=0

2(5+b)-7b=0

10+2b-7b=0

-5b=-10

b=2

∴a=5+b=5+2=7

∴The number=10×7+2=72



Q5 |Ex-4A |Class 8 | Playing with Numbers |S.Chand | New Learning Composite | myhelper

Question 5

A number consists of three digits, the right hand being zero. if the left hand and the middle digits are interchanged the number is diminished by 180. If the left hand digit is halved and the middle and right hand digits are interchanged the number is diminished by 454. Find the number.

Sol :

Original number=100a+10b+0

=100a+10b

According to the question

(100a+10b)-(100b+10a)=180

100a+10b-100b-10a=180

90a-90b=180

a-b=2

According to question

$100\times \frac{a}{2}+0+b=50a+b$

(10a+10b)-(50a+b)=454

50a+9b=454

50(2+b)+9b=454

100+50b+90=454

59b=454-100

59b=354

$b=\frac{354}{59}$

b=6


∴a=2+b=2+6=8


The number=100a+10b+0

=(100×8)+10×6+0

=800+60

=860

2 comments:

  1. Hmmm.. there is a mistake you should solve it in question number 2.

    I was also confused that why it is 6b
    BTW thanks for the solution and thanks for helping me for my preparation

    ReplyDelete
    Replies
    1. thanks for letting us know . we fixed it :) keep using .

      Delete

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