Question

The sum of the digits of a two digit number is 7. if the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.​

Answer 1

Answer:

Let the ones's digit =x.

So, sum of the digits =7

Tens's digit =7−x

Original no. =10(7−x)+x=70−10x+x=70−9x

Reversed no. =10x+7−x=9x+7

2(orig no) = reversed no. −2

2(70−9x)=9x+7−2140−18x=9x+527x=135x=5

Ten's digit =7−5=2

No. =25

Answer 2

Answer:

Let the digits be x and y... and so, the original number be 10x + y ( since x in tenth place and y in unit place)

so, given : sum of the digits = x + y = 7 ---------> (A)

reversing the digits means y in tenth place and x in unit place.. so, the reversed number be 10y + x

given: reversed number is decreased by 2 = twice original number

so, 10y + x - 2 = 2(10x + y)

simplifying : 19x - 8y = -2 ------------> (B)

solving 2 eqns (A) & (B) ,

we get x = 2 and y = 5

so the original number is 10x + y = 10(2) + 5 = 25

& the reversed number is 10y + x = 10(5) + 2 = 52

& the reversed number 52 when decreased by 2 is 50 which is twice the original number 25

Ans: the original number is 25

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