ML AGGARWAL CLASS 8 CHAPTER 5 Playing with Numbers Exercise 5.3

Exercise 5.3

Question 1

Sol :
(i) 87035

it is divisible by 5 


(ii) 75060

It is divisible by both 5 and 10 


(iii) 9685

It is divisible by 5 


iv. 10730

It is divisible by both 5 and 10 



Question 2

i. 67894

It is divisible by '2'


ii. 5673244

It is divisible by both 2 and 4


iii. 9685048

It is divisible by 2 , 4 and 8


iv. 75379

It neither divisible by 4 nor by 8 



Question 3

Sol :

(i) 45639

sum of digits  = 4 + 5 + 6 + 3 + 9 =27

27 is divisible by both 3 and 9

∴ 45639 is divisible by both 3 or 9 


(ii) 301248

Sum of digits = 3 + 0 + 1 + 2 + 4 + 8 = 18

18 is divisible by 9

∴ . 301248 is divisible by both 3 or 9 


(iii) 567081

Sum of digits =5+6+1+0+8+1=27

27 is divisible by 9

∴ 567081 is divisible by either 3 or by 9 


(iv)  345903

sum of digits =3+4+5+9+0+3=24

24 is divisible by only'3'

∴ 345903  is divisible by '3'only.


(v) 345046

sum of digits =3+4+5+0+4+6=22

22 is neither divisible by 3 nor by 9

∴ 345046 is neither divisible by 3 nor by 9 



Question 4

Sol :

(i) 10835

Sum of odd displace digits =5+8+1=14

Sum of even place digits =3+0=3

Difference =14-3=11

∴ 10835 is divisible by 11


(ii) 380237

Sum of odd place digits =1+2+8=17

Sum of even place digits =3+0+3=6

Difference =17-6=11

∴ 380237 is divisible by 11


(iii) 504670

Sum of odd place digits =0+6+0=6

Sum of even place digits =7+4+5=16

Difference = 16 - 6 = 10 

∴ 504670 is not divisible by 11 


(iv) 28248

Sum of odd place digits =8+2+2=12

Sum of even place digits =4+8=12

difference =12-12=0

 ∴ 28248 is divisible by 11 


Question 5

Sol :

(i)15414

15414 is divisible by '2' because units place contain '4'

1+5+4+1+4=15 divisible by 3

  ∴ 15414 in divisible by 3

  ∴ 15414 is divisible by 6


(ii) 213888

213888 is divisible by 2

Sum of digits =2+1+3+8+8+8=30 divisible by'3

  ∴ 213888 is  also divisible by 3

  ∴ 213888 is divisible by ' 6 '


(iii) 469876

469876 is divisible by 2

Sum of digits =4+6+9+8+7+6=40 not divisible by '3'

  ∴ 469876 in not divisible by 3

  ∴ 469876 in also not divisible by 6 '.



Question 6

Sol :
(i) 4618894875

Sum of digits of alternative blocks

875+618=1493  and 894+4=898

Difference $=595$ which is divisible by 7

   ∴ 4618894875 is divisible by 7 


(ii) 3794856

Sum of digits of alternative blocks

856+3=859 and 794

difference = 65 which is not divisible by 7 


(iii) 39823

sum of digits of alternative blocks

823 and 39

Difference 823-39=784 which is divisible by 7

 ∴ 3794856 is not divisible by 7 


(iii) 39823

sum of digits of alternative blocks

823 and 39

Difference 823-39=784 which is divisible by 7

 ∴  39823 is divisible by ' 7 '.




Question 7

Sol :

(i) Given number = 34x 

if 34x is multiple of 3 then (3+ 4 + x ) should be multiple of '3'

3+ 4+ x = 9+x

so to make sum of digits multiple of 3 
x should be 0 , 3 , 6 , 9 


(ii) 74 x 5284 is multiple of '3 'so

7+4+x+5+2+844=30+x

So to make sum of digits multiple of 3
x should be 0,3,6,9 .



Question 8

Sol :

Given number 43Z3 

If 42 z 3 in multiple of  then (4+2+z+3)
Should be multiple of 9

4+2+z+3=9+z

To make sum of digits multiple of a
should be 0,9 .


   

Question 9

Sol :


(i)  49*2207

If a number is divisible by 9 then its sum of digits 
should be divisible by 9 

4 + 9 + * + 2 + 2+ 0 + 7 = 24+*

The near by multiple of 9 is 27 so in place of the digit must be 27- 24 = 3


(ii) 5938*623

"If a number is divisible by 9 then it's sum of digits
Should be divisible by 9

5 + 9 + 5 + 8 + * + 6 + 2 + 3= 36 + *

36 is multiple of 9 so * should be either 0 or 9 


Question 10

Sol :

(i) 97*542

As unit's place is 2 , number in divisible by 2
irrespective of digit in * 'Place.

if a number divisible by 3 then its sum of digits should be divisible by 3

9+7-1 x+5+4+2=27+x

∴ 27 in a multiple of 3 so in place of '*' Should be

either 0,3,6, or 9

In place of * 0 , 3 , 6 , 9 number makes given 
number divisible by 6 (As it is divisible by both 2 and 3)


(ii)  709*94

As units digit in 4 given number divisible by
2  irrespective of number  in place of *

To make given number divisible by ,3 the sum of digits should be divisible by 3 

7+0+9+*+9+4=29+*

To make divisible $29, *$ should be either 1,4,7 

$\therefore$ In place ot *, 1,4 or 7 digits make the given number divisible by 6. (As it is divisible by 2 \  and 3


Question 11

Sol :
(i) given number $64 * 2456$

Sum of digits in odd places =6+4+*+6=16+*
Sum of digits in even places =5+2+4=11

Difference $=16+*-11=5+*$

* Should  Because to make given number
divisible by 11 , the difference in sum of digits in even
and odd place is either 0 or multiple of 11 

∴ should be 6


(ii) 86*6194

sum of digits in even places = 9 + 6+ 6= 21

sum of digits in odd places = 4+ 1 + * + 8 = 13+ *

so the difference should be either 0 or multiple 
of 11 to be divisible by 11

8 - * =0

* = 8

∴ to make 86*6194 divisible by 11 , in place of * digits must be place 

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