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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
increases by $180$
decreases by $270$
increases by $360$
decreases by $540$
Solution
(d)
Let three digits number be
According to problem,
$100 x+10 y+z=100 x+10 z+y-36$
According to problem,
$100 x+10 y+z=100 x+10 z+y-36$
$\begin{aligned} \Rightarrow & & 9 y-9 z+36 &=0 \\ \Rightarrow & y-z+4 &=0 \end{aligned}$
$\Rightarrow \quad y-z+4=0$
$\Rightarrow \quad 100 x+10 y+z=100 z+10 y+x+198$
$\Rightarrow \quad x-z-2=0$
Now, $(100 x+10 y+z)-(100 y+10 x+z)$
$=90(x-y)$
$=90(6) \quad[\because$ from Eqs. (i) and (ii) $]$
$=540$
$\therefore$ So, on interchanging for digit at tens place and hundred place, the value of number is decreased by $540$.
