Question

A number consists of two digits whose sum is 9. If 27 is subtracted from the number,its digit are reversed.find the number

Answer 1

no. on ten's place = x

no. on one's place = y

ATQ    x + y = 9    ...(1)

Also  10x + y - 27 = 10y + x

         9x - 9y = 27

          x - y = 3      ...(2)

Adding (1) and (2)

2x = 12

x = 6

(put value of x in (1)

6 + y = 9

y = 3

Hence, the number formed is 63

Answer 2

Answer:

$\bf given \: that$

$\bf let \: once \: place \: digit \: of \: a$

$\bf number \: be \: x$

$\bf ⇒sum \: of \: 2 \: digits = 9$

$\bf tens \: place \: digit \: of \: a$

$\bf number \: = 9 - x$

$\bf ⇒10(9 - x)x = 90 - 10x - x$

$\bf ⇒10(9 - x)x = 90 - 9x$

$\bf 27 \: is \: subtracted \: from \: the \:$

$\bf number \: we \: get \: the \: digits$

$\bf are \: reversed$

$\bf ⇒90 - 9x - 27 =10(x) + 9 - x$

$\bf ⇒90 - 9x - 27 = 10x + 9 - x$

$\bf ⇒- 9x + 63 = 9x + 9$

$\bf ⇒63 - 9 = 9x + 9x$

$\bf ⇒54 = 18x$

$\bf ⇒\frac{54}{18} = x$

$\bf ⇒x = 3$

$\therefore therefore$

$\bf ⇰once \: place \: = \: 3$

$\bf ⇰tens \: place \: = \: 9 - 3$

$\bf ⇰ten s\: place \: = \: 6$