Question

a number consists of two digits of which ten's digit exceeds the unit digit by 7. the number itself is equal to 10 times the sum of its digits. find the number.

Answer 1

Hi there!

Let the no. be 10x + y.

Then,
Ten's digit = x.
Unit's digit = y.

ATQ,

⇒x-7=y
⇒x-y=7                  ---(i)

And also,

⇒10x+y=10*(x+y)

⇒10x+y=10x+10y

⇒10x-10x=10y-y

⇒9y=0

⇒y=0

Putting value of y in equation (i),

⇒x-y=7

⇒x-0=7

⇒x=7

Therefore,
x=7 and y=0,

& the 2-digit number is 10x+y = 10 × 7 + 0 = 70

[ Thank you! for asking the question. ]
Hope it helps!