| (i) If sum of 3 digit numbers xyz, yzx and zxy is divided by (x + y + z), then quotient is \[\underline{\text{ P }}\]. |
| (ii) The difference between 2 digit numbers ab and ba, (where a > b) is divided by 3. The quotient is \[\underline{\text{ Q }}\]. |
| (iii) Sum of a 2 digit number and the number obtained by reversing its digits is always divisible by\[\underline{\text{ R }}\]. |
A)
P Q R 111 \[3(a+b)\] 11
B)
P Q R 99 \[(a+b)\] 7
C)
P Q R 111 \[3(a-b)\] 11
D)
P Q R 99 \[(a-b)\] 3
Correct Answer: C
Solution :
(i) As, \[xyz=100x+10y+z~~~~~...(i)\] \[yzx=100y+10z+x~~~~~~~~~~~~...(ii)\] \[zxy=-100z+10x+y~~~~~~~~~\text{ }...(iii)\] Adding (i), (ii) and (iii), we get \[xyz+yzx+zxy=100x+10y+z+100y\] \[+10z+x+100z+10x+y\] \[=111x+111y+111z\] \[=111(x+y+z)\] On dividing by \[(x+y+z)\], we get Quotient = 111. (ii) ab = 10 a + b and ba = 10 b+a ab - ba = 10 a + b - (10 b+a) \[=\text{9}a-\text{9}b=9(a-\text{b})\] On dividing by 3, we get Quotient \[=3(a-b)\] (iii) Let two digit number be\[10x+y\]. On reversing the digits, number becomes\[10y+x\]. \[\text{Sum=1}0x+y+10y+x=11x+11y=11(x+y)\]Which is always divisible by 11
You need to login to perform this action.
You will be redirected in
3 sec
