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

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more