Question

A number consists of two digits whose sum is 9 if 27 is subtracted from the number it digits are reversed find the numbers

Answer 1

Step-by-step explanation:

Let us assume, x and y are the two digits of the two-digit number

Therefore, the two-digit number = 10x + y and reversed number = 10y + x

Given:

x + y = 9 -------------1

also given:

10x + y - 27 = 10y + x

9x - 9y = 27

x - y = 3 --------------2

Adding equation 1 and equation 2

2x = 12

x = 6

Therefore, y = 9 - x = 9 - 6 = 3

The two-digit number = 10x + y = 10*6 + 3 = 63

Answer 2

Answer:

$\bf given \: that$

$\bf lets \: once \: place \: digit \: of \: a \: number$

$\bf be \: x$

$\bf sum \: of \: two \: digits \: = 9$

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

$\bf be \:9 - x$

$\bf ⇒10(9 - x)x \: = \: 90 - 10x - x$

$\bf ⇒10(9 - x)x \: = \:90 - 9x$

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

$\bf we \: get \: the$

$\bf digit s\: are \: reversed$

$\bf ⇒90 - 9x - 27 = 10(x) + 9 - x$

$\bf ⇒90 - 9x - 27 = 10x + 9 - x$

$\bf ⇒- 9x + 63 = 9x + 9$

$\bf ⇒63 - 9 = 9x + 9x$

$\bf ⇒54 = 18x$

$\bf ⇒\frac{54}{18} = x$

$\bf ⇒x = 3$

$\therefore therefore$

$\bf ⇒ones \: place \: = 3$

$\bf ⇒tens \: place \: = 9 - 3$

$\bf ⇒tens \: place \: = 6$

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