Question

Consider a three-digit number with the following properties.
I. If its digits in units place and tens place are interchanged, the number increases by $36$;
II. If its digits in units place and hundreds place are interchanged, the number decreases by $198$.
Now suppose that the digits in tens place and hundreds place are interchanged. Then the number.

• A. Increased by $180$
• B. Decreases by $270$
• C. Increases by $360$
• D. Decreases by $540$

Answer 1

The correct option is D Decreases by $540$

Let $a,b$ and $c$ are the hundreds, tens and units places digits respectively. then digit $=100a+10b+c$

Interchanging units and tens places digits, we get

$\Rightarrow 100a+10c+b=100a+10b+c+36\Rightarrow c-b=4\dots eqn\left(1\right)$

Now, Interchanging units and hundreds places digits, we get

$\Rightarrow 100c+10b+a=100a+10b+c-198\Rightarrow a-c=12\dots eqn\left(2\right)$

From equation $1$ and $2$,

$a-b=6\dots eqn\left(3\right)$

Suppose number is increased by $x$, If we change tens and hundreds digits,

$\Rightarrow 100b+10a+c=100a+10b+c+x\Rightarrow -x=90a-90b\Rightarrow x=-540$

Thus, number is decreased by $540.$