Question

A number cosists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits be reversed. Find the number.

• A. 253
• B. 325
• C. 523
• D. 311

Answer 1

The correct option is A 253

Let the numbers in the 100's place be x, 10's place be y and1's place be z.

H T O
x y z

As, sum of the digits is 10, we have,
$x+y+z=10.......\left(1\right)$

As, The middle term is equal to the sum of the other two, we have
$y=x+z$
$x-y+z=0.....\left(2\right)$

As, The number will reverse if 99 is added to it, we have,
$99+100x+10y+z=100z+10y+x$

$99x-99z=-9$
$x-z=-1.....\left(3\right)$

$\Rightarrow eqn\left(1\right)+eqn\left(2\right)$

$x+y+z=10$
$x-y+z=0$
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$2x+2z=10$
$x+z=\mathrm{5.....................}\left(4\right)$

$\Rightarrow eqn\left(3\right)+eqn\left(4\right)$

$x-z=-1$
$x+z=5$
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$2x=4$
$x=\mathrm{2.....................}\left(5\right)$


Now, substituting eqn(5) in eqn(3)
$x-z=-1$
$2-z=-1$
$z=\mathrm{3.....................}\left(6\right)$

Again, substituting eqn(5) and eqn(6) in eqn(2), we get,
$x-y+z=0$
$2-y+3=0$
$y=\mathrm{5.....................}\left(7\right)$

Hence the number is 253.

Check: 253 + 99 = 352 (digits are reversed.)