Question

The sum of a two digit number is 7. if the number formed by reversing the digits is less than the original number by 27,find the original number.

Answer 1

AnswEr:-

❀ Your Answer Is 52.

ExplanaTion:-

Given:-

• The sum of a digits of a two digit number is 7.

• If the number formed by reversing the digits is less than the the original number by 27.

To Find:-

• The Original Number.

So Now,

• Let the digit at ten's place be x.

• Let the digit at one's place be y.

So the original number will be 10x+y.

Hence the number formed after reversing the digits would be 10y+x.

Therefore ATQ:-

$\\ \tt\longrightarrow \fbox{x + y = 7... \: eq(1).}$

Also the number formed by reversing the digits is 27 less than Original number.

$\\ \tt\longrightarrow(10y + x)= (10x + y )- 27.\\ \\ \tt\longrightarrow(10x + y) - (10y + x) = 27. \\ \\ \tt\longrightarrow10x + y - 10y - x = 27. \\ \\ \rm(b y\: \: ope ning \: \: brackets). \\ \\ \tt\longrightarrow9x - 9y = 27. \\ \\ \tt\longrightarrow9(x - y) = 27. \\ \\ \rm(taking \: \: 9 \: \: as \: \: common). \\ \\ \tt\longrightarrow x - y = \cancel\dfrac{27}{9} . \\ \\ \tt\longrightarrow \fbox{ x - y = 3... \: eq(2).}$

Now,

By adding eq(1) And eq(2) we get:

$\\ \tt\leadsto(x + y) + (x - y) = 7 + 3. \\ \\ \tt\leadsto x \: \cancel{+ y} \: + x \: \cancel{ - y} = 10. \\ \\ \tt\leadsto2x = 10. \\ \\ \tt\leadsto x = \cancel \frac{10}{2} . \\ \\ \tt\leadsto{\underline {\underline{ \: \: x = 5. \: \: }}}$

By putting the value of y in eq(1) we get:-

$\\ \tt\leadsto x + y = 7. \\ \\ \tt\leadsto 5 + y = 7. \\ \\ \tt\leadsto y = 7 - 5. \\ \\ \tt\leadsto {\underline {\underline{ \: \: y = 2. \: \: }}}$

Therefore,

$\\ \rm {\underline{ \underline {\bigstar \: \: T he \: \: original \: \: number \: \: is: - }}} \\ \\ \tt\mapsto10x + y. \\ \\ \tt = 10(5) + 2. \\ \\ \tt = 50 + 2. \\ \\ \tt \fbox {\underline{ \underline{ = 52.}}}$

$\tt\therefore$ The Original Number Is 52.