stne wrote:

Bunuel wrote:

If the tens digit x and the units digit y of a positive integer n are reversed, the resulting integer is 9 more than n. What is y in terms of x?

A. 10 - x

B. 9 - x

C. x + 9

D. x - 1

E. x + 1

\(n=10x+y\) and \(n'=10y+x\) --> \(n'-n=(10y+x)-(10x+y)=9\) --> \(y=x+1\).

Answer: E.

Can anybody clear this for me

suppose I take first number as 10y + x and reverse it to get 10x + y

Then according to the equation (10x + y) - (10y + x ) = 9

9x -9y= 9

x-y=1

y= x-1

so why are we getting two different answers.

if I take first number to be 10x + y and reverse it to get 10y +x

then (10y + x) - (10x +y)= 9y- 9x=9

y-x=1

y= x+1

why two different answers ?

You cannot arbitrary assign which will be the "first" number and which will be the "second", since the stem explicitly clears that.

Positive integer \(n\) has the tens digit x and the units digit y, so \(n=10x+y\);

Reversed integer, say \(n'\), has the tens digit y and the units digit x, so \(n'=10y+x\);

We are also told that " the resulting integer (so \(n'\)) is 9 more than \(n\)", which means \(n'-n=(10y+x)-(10x+y)=9\) --> \(y=x+1\).

Hope it's clear.

Its clear now , great. Thank you