Exercise 5.1
Question 1
(i) 89
Let given number be $a b$ i.e $10 a+b$
∴ 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 reverssing the digits = 43
sum = 34 + 43 =77 = 7 x 11
i) when sum is divited 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, quoptient is 4
(ii) When difference is divided by 4 (differen ot 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 som divided by $(a+b+c),$ quctient is 111
iii) whew sum divided by 37, quotient is 3(a+b+c)
iv) when sum dividect by 3, quotient is 37(a+b+c)$
Question 5
Given number =843
number obtained by reversing digits =348
Difference $=843-348=495=99 \times 5$
i) When difrerence divided by 99, quotient =5
ii) when difference divided by 5, quotient =99
Question 6
Let the given number is ab
Sumof digit=11 = a+b=11 $\rightarrow(1)$
From given condition
$b a=a b-9$
$10 X b+a=10 X a+b-9$
$10(b-a)=(b-a)-9$
$10(b-a) - (b-a)=-9$
$9(b-a) = - 9$
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
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
let given number be ab
given condition
sum of two digit number and the number
obtained by reversing the digits is 55
$a b+b a=3655$
$10 a+b+10 b+a=3655$
$10(a+b)+(b+a)=31655$
$11(a+b)=55$
$(a+b)=\frac{55}{11}$
$a+b=5$
∴ Sum of digits is 5
Question 9
Let given3 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 $\quad c ; b: a=1: 2: 3$
let
$c=1 \times x$
$b=2 \times x$
$a=3 \times x$
Substitube $a, c$ 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
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
$111 \times(a+b+c)=2664$
$a+b+C=\frac{2664}{111}$
$\quad a+b+c=24$
$a+a-1+a+1=24(\because$ From 1,2$)$♀
$3 a=24$
$a=\frac{24}{3}$
$a=8$
$c=a+1=8+1=9$
$b=a-1=8-1=7$
∴ given number is 879
Exercise 5.2
Question 1
4 A
3 5
--------
B 2
--------
A+5= we get a number whose unit's
digit in 2, A+5 should not exceed 14
$\begin{aligned} \therefore A+5 &=12 \\ A &=12-5 \\ A &=7 \end{aligned}$
4 7
3 5
-------
8 2
-------
B = 8
∴ A = 7 , B = 8
Question 2
5 A
7 9
---------
CB 3
---------
A+9 Should Not exceeD 18 . and for A+9= anumber
whose unit's place is 3 this:s possible when A=44
(daigram should be added)
∴ A = 4 , C = 1 , B = 3
Question 3
4 2 A
2 A 5
---------
A 0 2
---------
A+5 sum should not exceed 14 and it is a
number whose Unit's digit is 2 This is possible
when A=7
(daigram should be added)
A = 7
satisfies given condition
A = 7
Question 4
A A
+ A A
-----------
BA 8
case(i)
$\begin{aligned} A+A &=8 \\ 2 A &=8 \\ A &=4 \end{aligned}$
check
4 4
4 4
-------------
8 8
but given the sum
as there digit number
case (ii)
A + A 18
2A = 18
A = 9
check : 9 9
9 9
-----1----------
198
----------------
A = 9 , B = 1
A = 9 , B = 1
Question 5
1 8 A
+B A 7
-----------
C B 2
-----------
A + 7 sum should not exceed 16 and it is a
number whose unit, s digit is 2
∴ A + 7 = 12
A = 12 - 7
A = 5
$1+8+A \Rightarrow 1+8+5=13$
1+B=C
C=1+4
C=5
∴ A = 5 , B = 4 , C = 5
Question 6
A 2 1 B
+1 C A B
--------------
B 4 9 6
B + B sum should not exceed 18 and its units digits is 6
case (i)
B + B = 6
2B = 6
B = 3
Then 1 + A = 9
A = 8
A + 1 = B
8 + 1 = B
B = 9 but we got B = 3 earlier
so B ≠ 3
case (ii)
B+B=16
2 B=16
$B=\frac{16}{2}$
B=8
7 2 1 8
1 2 7 8
--------------
8 4 9 6
A + 1 = B
A + 1 = 8
A = 8 - 1
A = 7 A = 7
ஃ A = 7 B = 8 C = 2
Question 7
B 3 4 5
C 9 B A
--------------
8 B A 2
--------------
--------------
Sum 5 + A = 12
A = 12 - 5
A = 7
1 + 4 + B = A
1 + 4 + B = 7
B = 7 - 5
B = 2
$\begin{aligned} B+C &=8 \\ 2+C &=8 \\ C &=8-2 \end{aligned}$
c=6
$A=7, \quad B=2, C=6$
Question 8
$\begin{array}{r}A B \\(-) B 6 \\\hline 4 7 \\\hline\end{array}$
$\begin{aligned} B-6 &=7 \\ B &=7+6 \\ B &=13 \end{aligned}$
means B = 3 , because B should be single digit
$\begin{aligned} A-B-1 &=4 \\ A-3-1 &=4 \\ A-4 &=4 \\ A &=8 \\ \therefore \quad A=8, \quad B &=3 \end{aligned}$
Question 9
Sol :
$\begin{array}{r}2 A \\\times 3 A \\\hline B 7 A\end{array}$
A x A = A means A should be 1 or 5 or 0
case (i) A = 0
$\begin{array}{r}20 \\30 \\\hline 600 \\\hline\end{array}$
≠ B7A
∴ This is not given number so A ≠ 0
case (ii) A = 1
$\begin{array}{l}21 \\31 \\\hline 21 \\\hline 23 \\\hline 251 \\\hline\end{array}$
≠ B71
∴ B ≠ 1
case (iii)
$\begin{array}{l}25 \\35 \\\hline 125 \\75 \\\hline 875\end{array}$
∴ B = 8
$\therefore \quad A=5, \quad B=8$
Question 10
$\begin{array}{r}A B \\A B \\\hline 6 A B\end{array}$
B should either 0 , 1 or 5
case (i) B = 0
$\begin{array}{r}A 0 \\B 0 \\\hline 00 \\A B O \\\hline A B B \quad 0 \end{array}$
≠ 6AB
∴ A ≠ 0
case (ii) B = 1
(DIAGRAM SHOULD BE ADDED )
≠ 6AB
case (iii)
B = 5
A 5
A 5
------
X25 = 6AB (∵ may be any number)
A = 2 (∵from rules by squar number)
Question 11
$\begin{array}{l}A A \\4 A \\\hline 9 A 4\end{array}$
A = 2
A = 2 is correct
Question 12
(DIAGRAM SHOULD BE ADDED )
Question 13
(DIAGRAM SHOULD BE ADDED )
Question 14
(DIAGRAM SHOULD BE ADDED )
Exercise 5.3
Question 1
(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
i. 45639
sumot 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 divisibe by only'3'
∴ 345903 is divisibly 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
i. 10835
Sum of odd daplace digits =5+8+1=14
Sum of even place digits =3+0=3
Disfference =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 odol place digits =0+6+0=6
Sum of even place digits =7+4+5=16
Difference = 16 - 6 = 10
∴ z504670 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
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
Sumot digits =4+6+9+8+7+6=40 not divisible by '3'
∴ 469876 in not divisibie by 3
∴ 469876 in also not divisible by 6 '.
Question 6
i. 4618894875
Sum of digits of alternative blocks
875+618=1493 and 894+4=898
Dibterence $=595$ whicl 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 whichin divisible by 7
∴ 3794856 is not divisible by 7
iii. 39823
sum of digits of alternative blocks
823 and 39
Difference 823-39=784 whichin divisible by 7
∴ 39823 is divisible by ' 7 '.
Question 7
i. Given number = 34x
it 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
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 sumob digits multiple of a
should be 0,9 .
Question 9
(i) 49*2207
If a number is divisble by 9 then its sum of digits
should be divisble 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
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 divisble 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
(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
Dilterence $=16+*-11=5+*$
* Should bí Becouse to make given number
divisible by 11 , the diffrerence in sum of digits in even
and odd place is either 0 or mulliple 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
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