Question

in a two digit number digit at the ten's place is twice the digit at unit's place if the number obtained by interchanging the digits is added to the original number the sum is 66 find the number
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Answer 1

Answer:

Let the digit at the unit's place be x. ∴ The unit's digit = 2. Tens Digit = 2 × 2 = 4. Hence, the number is 42.

Answer 2

Step-by-step explanation:

Given :-

In a two digit number digit at the ten's place is twice the digit at unit's place if the number obtained by interchanging the digits is added to the original number the sum is 66.

To find :-

Find the number ?

Solution :-

Let the digit at 10's place be X

Then the value of X = 10×X = 10X

Let the digit at 1's place be Y

Then the value of Y = 1×Y = Y

So ,The Original number = 10X+Y

The number obtained by reversing the digits

= 10Y++1

Given that

The digit at the ten's place is twice the digit at unit's place

=> X = 2Y ------------(1)

and

The number obtained by interchanging the digits is added to the original number the sum is 66.

=> (10X+Y)+(10Y+X) = 66

=> 10X+X +10Y+Y = 66

=> 11X+11Y = 66

=> 11(X+Y) = 66

=> X+Y = 66/11

=> X+Y = 6-----------(2)

On Substituting the value of X from (1) in (2)

=> 2Y+Y = 6

=> 3Y = 6

=> Y = 6/3

=> Y = 2

On Substituting the value of Y in (1) the

=> X = 2(2)

=> X = 4

The digit at 10's place = 4

The digit at 1's place = 2

The number = 42

Answer:-

The required number for the given problem is 42

Check :-

The number = 42

=> Digit at 10's place = 4

=> 2×2

=> 2 × Digit at 1's place

The number obtained by reversing the digits = 24

Their sum = 42+24 = 66

Verified the given relations in the given problem.