ML AGGARWAL CLASS 8 CHAPTER 5 Playing with Numbers Excercise 5.1

 Exercise 5.1

Question 1

Sol :
(i) 89

Let given number be ab i.e 10a+b
 ∴ 89=$10 \times 8+9$

(ii) $207=2 \times 100+0 \times 10+7$

(iii) $\quad 369=100 \times 3+10 \times 6+9$

Question 2

Sol:

Given number = 34 

Number obtained by reversing the digits = 43 

sum = 34 + 43 =77 = 7 x 11

(i) when sum is divided by 11, quotient is 7 

(ii) Whew sum is divided by 7 , quotient is 11


Question 3

Sol:

Given number =73
 
Number obtained by reversing the digits =37

Difference =73-37=36=$9 \times 4$

(i) Whew difference is divided by 9, quotient is 4

(ii) When difference is divided by 4 (difference of digits),

quotient is 9


Question 4


Sol:

given number =a b c

numbers obtained by reversing digits = b c a , c ab 

sum =a b c+b c a, c a b

(i) When sum divided by 111 , quotient is (a+b+c)

(ii) when sum divided by $(a+b+c),$ quotient is 111

(iii) whew sum divided by 37, quotient is  3(a+b+c)

(iv) when sum divided by 3, quotient is 37(a+b+c)$


Question 5

Sol :
Given number =843

number obtained by reversing digits =348

Difference $=843-348=495=99 \times 5$

(i) When difference divided by 99, quotient =5

(ii) when difference divided by 5, quotient =99


Question 6

Sol :

Let the given number is ab

Sum of digit=11 = a+b=11 $\rightarrow(1)$

From given condition

ba=ab-9

10 X b+a=10 X a+b-9

10(b-a)=(b-a)-9

$10(b-a) - (b-a)=-9

9(b-a) = - 9

b-a= -1....(2)

Solving (1) and (2)
b-a=-1
b+a=11
-------------
  2b = 10 

b= 5
b + a =11
5 + a = 11
a = 6

Question 7

Sol :

given : Let given number  is ab 

given condition 

ab - ba = 36

10 a+b-(10 b+a)=36

10(a-b)+(b-a)=36

10(a-b)-(a-b)=36

9(a-b)=36

$a-b=\frac{36}{9}$

a-b=4 

∴ Difference of two digits = 4 



Question 8

Sol :

Let given number be ab 

given condition 

sum of two digit number and the number 

obtained by reversing the digits is 55 

ab+b a=3655

10a+b+10 b+a=3655

10(a+b)+(b+a)=31655

11(a+b)=55

$(a+b)=\frac{55}{11}$

a+b=5

∴ Sum of digits is 5 


Question 9

Sol :

Let given digits number be abc 

Given   

unit's digits ten's digit and hundred's digit 

are in the ratio  1 : 2 : 3

 ∴ c : b : a = 1 : 2 : 3---------- ①

and

 $\quad a b c-c b a=594 \rightarrow(2)$

100 a+10 b+c-(100 c+10 b+a)=594

100(a-c)+10 b-10 b+(c-a)=594

100(a-c)+0-(a-c)=594

99(a-c)=594

$a-c=\frac{594}{99}$

$a-c=6 \rightarrow(3)$

given  c ; b: a=1: 2: 3

let 

$c=1 \times x$
$b=2 \times x$
$a=3 \times x$

Substitute a, c values in (3)

$\begin{aligned} 3 x-1 x &=6 \\ 2 x &=6 \\ x &=6 / 2 \\ x &=3 \end{aligned}$

$\begin{aligned} \therefore \quad a &=3 \times 3=9 \\ \ b &=2 \times 3=6 \\ c &=1 \times 3=3 \end{aligned}$

 ∴ given number is 963


Question 10

Sol :

Let the given number be = abc 

given 

       c = a + 1 --------1

       b = a - 1 ---------2

given sum of original number and numbers 

 obtained by reversing cyclically is 2664 

 abc + bca + cab = 2664 --------3

but   abc + bca + cab = 111 X ( a +  b + c )--------4


 ∴ from eq-(3) and eq-(4) 


$111 \times(a+b+c)=2664$

$a+b+C=\frac{2664}{111}$

a+b+c=24

a+a-1+a+1=24 (∵ From 1,2)

3a=24

$a=\frac{24}{3}$

a=8

c=a+1=8+1=9

b=a-1=8-1=7

 ∴ given number is 879

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