Question

if in a two digit number the digit at tens place is 7and the sum of the digit is is 8 times the digit at the unit place. then the number is​

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If in a two digit number the digit at tens place is 7 and the sum of the digit is 8 times the digit at the unit place.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the ten's place be 7

Let the one's place be r

$\boxed{\bf{The\:original\:number=10(7)+r}}}}$

$\mapsto\sf{The\:number=70+r...............(1)}$

A/q

$\longrightarrow\sf{7+r=8\times r}\\\\\longrightarrow\sf{7+r=8r}\\\\\longrightarrow\sf{r-8r=-7}\\\\\longrightarrow\sf{-7r=-7}\\\\\longrightarrow\sf{r=\cancel{\dfrac{-7}{-7} }}\\\\\longrightarrow\sf{\pink{r=1}}$

Putting the value of r in equation (1),we get;

$\longrightarrow\sf{The\:number=70+1}\\\\\longrightarrow\sf{\pink{The\:number=71}}$

Thus;

$\underbrace{\sf{The\:original\:number=71}}}$

Answer 2

Question  -

If in a two digit number the digit at tens place is 7 and the sum of the digit is is 8 times the digit at the unit place .

Then the number is​ ?

Solution -

In the above question, the following information is given -

In a two digit number the digit at tens place is 7 and the sum of

the digit is is 8 times the digit at the unit place .

So, let us assume that the units digit of the number is x.

Tens digit is 7.

Sum of the digits of the required number = x + 7

This is equal to 8 times the digit at the units place .

So,

8x = x + 7

7x = 7

x = 1

The required number - 7 1.

Hence the required number is 71.

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