Question

A number of two digits exceeds four times the sum of the digits by 3 and the number
obtained by interchanging the digits exceeds six times their sum by 5. Find the number.
(Ans.: 35)​

Answer 1

Step-by-step explanation:

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

• A number of two digits exceeds four times the sum of the digits by 3.

• The number obtained by interchanging the digits exceeds six times their sum by 5.

$\bf\underline{\underline{\blue{To\:Find:-}}}$

• The original number.

$\bf\underline{\underline{\green{Solution:-}}}$

Let the tens digit of the number be x

And ones digit of the number be y

The original number = 10x + y

The number obtained by interchanging the digits = 10y + x

Case 1:-

The number of exceeds four times the sum of the digits by 3.

$\rm \implies The \: number = 4(sum \: of \: digits) + 3$

$\rm \implies 10x + y= 4(x + y) + 3$

$\rm \implies 10x + y= 4x + 4y + 3$

$\rm \implies 10x + y - 4x - 4y = 3$

$\rm \implies 6x - 3y = 3$

Dividing the whole equation by 3

$\rm \implies 2x - y = 1.......(i)$

Case 2:-

The number obtained by interchanging the digits exceeds six times their sum by 5.

$\rm Revered\: number = 6(sum\: of \: digits) + 5$

$\rm \implies 10y + x = 6(x + y ) + 5$

$\rm \implies 10y + x = 6x + 6y + 5$

$\rm \implies 10y + x - 6x - 6y = 5$

$\rm \implies 4y - 5x = 5......(ii)$

Multiplying equation (i) with 4

$\rm \implies 4(2x - y = 1)$

$\rm \implies 8x - 4y = 4.....(iii)$

Adding equations (ii) and (iii)

$\rm \implies 4y - 5x + 8x - 4y= 5 + 4$

$\rm \implies 3x = 9$

$\rm \implies x = 3$

Substituting x = 3 in equation (i)

$\rm \implies 2x - y = 1$

$\rm \implies 2(3) - y = 1$

$\rm \implies 6 - y = 1$

$\rm \implies - y = 1-6$

$\rm \implies -y = -5$

$\rm \implies y = 5$

We have :-

• x = 3
• y = 5

Now:-

The original number = 10x + y

= 10(3) + 5

= 30 + 5

= 35

$\underline{\boxed{\purple{\rm \therefore The \: number = 35}}}$