Bunuel
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.