Question

Sum of the digit of a two digit no. is 9. When we interchange the digits, it's found that the resulting new number is greater than the original number by 27. What is the two-digit number​

Answer 1

Answer:

Question:-

Sum of the digits of a two digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. Find the original number.

Solution:-

Let the digits of the number be:-

x and y

x + y = 9 — (i)

Original number = xy

= 10x + y [acc to place value]

New number = yx

= 10y + x [acc to place value]

yx > xy by 27

yx = xy + 27

10y + x = 10x + y + 27

10y + x - 10x - y = 27

9y - 9x = 27

9(y - x) = 27

y - x = $\frac{27}{9}$

y - x = 3 — (ii)

Now we compare both the equations using subtraction method:-

$x + y = 9\\y - x = 3\\2y = 12\\y = \frac{12}{2} \\y = 6$

Now to find the value of x:-

$x + y = 9\\x + 6 = 9\\x = 9 - 6\\x = 3$

ORIGINAL NUMBER = xy

= 36

REVERSED NUMBER = yx

= 63

Verify:-

yx = xy + 27

63 = 36 + 27

63 = 63

LHS = RHS

Hence, verified

Hope it helps

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Answer 2

Answer:

$\huge\green{\boxed{\sf Answer}}$

Given

The sum of the two digits = 9

On interchanging the digits, the resulting new number is greater than the original number by 27.

Let us assume the digit of units place = x

Then the digit of tens place will be = (9 – x)

Thus the two-digit number is 10(9 – x) + x

Let us reverse the digit

the number becomes 10x + (9 – x)

As per the given condition

10x + (9 – x) = 10(9 – x) + x + 27

⇒ 9x + 9 = 90 – 10x + x + 27

⇒ 9x + 9 = 117 – 9x

On rearranging the terms we get,

⇒ 18x = 108

⇒ x = 6

So the digit in units place = 6

Digit in tens place is

⇒ 9 – x

⇒ 9 – 6

⇒ 3

Hence the number is 36