A number consists of two digits of which tens digits exceeds the unit digit by 7. The number itself is equal to 10 times the sum of its digits. Find the number.

Hi there!

Let the Original number be 10x + y.

Then,

• Ten's digit = x.
• Unit's digit = y.

According to th' Question :-

⇒x - 7 = y

⇒x - y = 7    ---(i)

And,

⇒10x + y = 10 (x + y)

⇒10x + y = 10x + 10y

⇒10x - 10x = 10y - y

⇒9y = 0

⇒y = 0

Substitute th' value of ' y ' in Eqn. (i)

⇒x - y = 7

⇒x - 0 = 7

⇒x = 7

∴ Original number = 10x + y = 10 × 7 + 0 = 70

Hence,
The required two digit number is 70

~ Hope it helps!
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