Question

The sum of digits 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:

Step-by-step explanation:

Answer 2

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☞ Your Answer Is 52

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$\huge\sf\blue{Given}$

✭ The sum of a digits of a two digit number is 7

✭ The new number formed by reversing the digits is less than the the original number by 27

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◈ Original Number?

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Assume that,

◕ 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

$\sf\twoheadrightarrow {x + y = 7\qquad -eq(1)}$

$\bullet\underline{\textsf{As Per the Question}}$

$\sf\twoheadrightarrow(10y + x)= (10x + y )- 27\\ \\ \sf\twoheadrightarrow (10x + y) - (10y + x) = 27 \\ \\ \sf\twoheadrightarrow 10x + y - 10y - x = 27 \\ \\ \sf\twoheadrightarrow 9x - 9y = 27\\ \\ \sf\twoheadrightarrow 9(x - y) = 27 \\ \\ \sf\twoheadrightarrow x - y = \dfrac{27}{9} \\ \\ \sf\twoheadrightarrow { x - y = 3\qquad -eq(2)}$

Adding eq(1) And eq(2) we get,

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

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

$\sf\leadsto x + y = 7 \\ \\ \sf\leadsto 5 + y = 7 \\ \\ \sf\leadsto y = 7 - 5 \\ \\ \sf\leadsto {\red{ y = 2}}$

Therefore,

$\sf\dashrightarrow 10x + y \\ \\ \sf \dashrightarrow 10(5) + 2 \\ \\ \sf \dashrightarrow 50 + 2\\ \\\sf\orange{\dashrightarrow{{ 52}}}$

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