Question

Conclude from the given information:

$x,y,s$, and $t$ are positive integers.
$x<y<10$
$s<t<10$

A: The maximum possible value of $y-x$.

B: The minimum possible value of $s+t$.

• A. The quantity A is greater than B.
• B. The quantity B is greater than A.
• C. The two quantities are equal
• D. The relationship cannot be determined from the information given.

Answer 1

Answer: (A)

The quantity A is greater than B.

Given $x<y<10$ and $x,y$ are positive integers.

So the maximum value of $y-x$ occurs when $y$ is maximum and $x$ is minimum.

Therefore maximum value of $y-x$ is $9-1=8$.

Given $s<t<10$ and $s,t$ are positive integers.

So the minimum value of $s+t$ occurs when $s$ is minimum and $t$ is minimum.

Therefore minimum value of $s+t$ is $1+1=2$.

Therefore the quantity in $1$st column is greater.

So, option $A$ is correct.