AP ICET 2016 Paper

Question : 80 of 200

Marks: +1, -0

A three digit number 4a3 is added to another three digit number 984 to give the four digit number 13b7 which is divisible by 11. Then

(a+b)^2=

Solution:

1 3 b 7 is divisible by 11.

⇒(1+b)-(3+7)=0⇒b-9=0

⇒b=9

Now,
4 a 3
9 8 4
13 b 7

⇒a+8=b

⇒a=9-8=1

∴(a+b)^2=(1+9)^2=100

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