Question

Ajay attempted to add ten two-digit numbers. One ofthem, a, was the reverse of one of the others. If a wasreplaced by another two-digit number, b and thereverse of a was replaced by the reverse of b and theaverage was found, it would be 2.2 more. The sum ofthe digits in b exceeds the sum of the digits in a by
(A) 1
(B) 2
(C) 3
(D) 4​

Answer 1

Answer:

2

Step-by-step explanation:

kkrjfnntifjdjjdjddjejejekeejejejeje

Answer 2

Answer:

The sum of the digits in b exceeds the sum of the digits in a by 2 The Correct answer is option B

Step-by-step explanation:

Let avg of initial 10 digital be P

Avg of 10 digits after replacement be Q

let initial number and it's reverse be A= 10x+y and 10y+x

Replaced Number and it's reverse be B= 10a+b and 10b+a

The difference of their avg is 2.2

so the difference in total sum is 2.2*10 i.e 22

considering all other numbers to be unchanged.

the difference Q-P = 22

therefore, 10a+b + 10b+a - (10x+y + 10y+x) = 22

or, 11(a+b) - 11(x+y) = 22

or, (a+b) - (x+y) = 2 , i.e. the sum of the digits in B that exceeds the sum of digits in A.

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