Question

The result of dividing a no. of 2 digits by the no obtained by reversing the digits is 7/4. If the sum of digits is 20 find the no.​

Answer 1

Answer:-

140

Step by Step Solution:-

Let:-

The two digit number be in the form

10x + y.

And the number obtained by reversing the digits is 10y + x.

As given:-

Original number /reversed number = 7/4

=> 10x + y/10y + x = 7/4

=> 4(10x+y) = 7(10y+x)

[by cross multiplication]

=> 40x + 4y = 70y + 7x

=> 40x - 7x = 70y - 4y

=> 33x = 66y

=> x = 2y

Now-:

Given x + y = 20 so

=> 2y + y = 20

=> 3y = 20

=> y = 20/3

As x + y = 20

=> x + 20/3 = 20

=> x = 20 - 20/3

=> x = 40/3

As the number is 10x + y

$= > 10( \frac{40}{3} ) + \frac{20}{3} \\ = > \frac{400}{3} + \frac{20}{3} \\ = > \frac{400 + 20}{3} \\ = > \frac{ \cancel420}{ \cancel3} \\ = > 140$

Therefore:-

The required number is 140.

Answer 2

Answer:

Suppose the two digit number be in the form of 10x + y .

and the number obtained by reserving the digit

= 10y + x.

Original number/reserved number = 7/4.

10x + y /10y + x = 7/4.

4(10x + y) = 7(10y + x) .

40x + 4y = 70y + 7x .

40x - 7x = 70y + 4y .

33x = 66y.

x = 2y.

Given x + y = 20.

Now ,

2y + y = 20.

3y = 20.

y = 20/3.

Therefore,as x + y = 20.

x + 20/3 = 20.

x = 20 - 20/3.

x = 40/3.

Therefore, number is ;

10x + y = 10 × 40/3 +20/3.

= 140.