Question

in a two digit number digit at the tens place is twice the digit at unit place if the number obtained by interchanging the digit is added to the original number the sum is 66 find the number

Answer 1

let's see the a answer

Lets take the the digit on the tens place as x and the digit on the units place be y

So the number will be : 10x + y

Now According to given Conditions

x = 2 × y ...(1)

Now if the digits are interchanged the number will be : 10y + x

So According to given Conditions

10y + x + 10x + y = 66. Solving it...

11x + 11y = 66. Dividing both the sides by 11

We get : x + y = 6

Now According to (1)

(2 × y) + y = 6
2y + y = 6
3y = 6
y = 6/3 = 2

Hence y = 2

Now let's find the value of x by replacing y = 2

x = 2 × y
x = 2 × 2
x = 4

So the original number will be

10x + y = 10(4) + 2

40 + 2

42

Hence the number is 42

Hope it helps