Question

the sum of two digit number and the number obtain by reversing the digit is 66 .if the digit of the number differ by 2, find the number?

Answer 1

Let unit's place = x

Ten,s place = x - 2

No. = 10(x-2) + x = 10x - 20 + x = 11x - 20

When digit is reversed then ,

unit's place = x-2, Ten's place = x

New no. = 10x + x - 2 = 11x - 2

Now, 11x - 20 + 11x - 2 = 66

⇒ 22x - 22 = 66

⇒ 22x = 88

x = 4

No. = 11x - 20 = 44 - 20 = 24

Answer 2

Hello student,

Let the unit digit be X and ten 's digit be Y. So, original digit be ( X+10.Y) , And by reversing the digit we get (Y+10.x)

SO ACCORDINGTO THE QUESTION

(X+10Y)+(Y+10X) = 66

11X+11Y = 66

11(X+Y) = 66

X+Y = 66/11

X+Y = 6 ------------------- 1

And, since digits differ by 2

So, X-Y = 2 --------------- 2

Adding 1 and 2 equations we get

2X = 8

X = 8/2

so, X = 4

And, puttingvalue of Xin equation 1

X + Y = 6

4 + Y = 6

Y = 6 - 4

Y = 2

Therefore, number is given by putting value of X & Y

Hence, number is ( 4+10*2) = 24...



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