Question

a number consists of two digits whose sum is 9 if 27 is subtracted from the number its digits are reversed find the number​

Answer 1

Answer:

Let us assume, x and y are the two digits of the two-digit number

Therefore, the two-digit number = 10x + y  and reversed number = 10y + x

Given:

x + y = 9 -------------1

also given:

10x + y - 27 = 10y + x

9x - 9y = 27

x - y = 3 --------------2

Adding equation 1 and equation 2

2x = 12

x = 6

Therefore, y = 9 - x = 9 - 6 = 3

The two-digit number = 10x + y = 10*6 + 3 = 63

Answer 2

Answer:

Given that:‐

• let once place digit of a number be x.

• sum of 2 digits = 9

• tens place digit of a number = 9 - x

Then:-

➤10 ( 9 - x ) x = 90 - 10x - x

= 90 - 9x

Now:-

• 27 is subtracted from the number we get the digits are reversed.

Solution:-

➠ 90 - 9x - 27 = 10(x) + 9 - x

➠ 90 - 9x - 27 = 10x + 9 - x

➠ -9x + 63 = 9x + 9

➠ 63 - 9 = 9x - 9x

➠ 54 = 18x

➠ $\bf \frac{54}{18} = x$

➠ x = 3

$\therefore therefore$

⇒ once place = 3

⇒ Tens place = 9-3

⇒ Tens place = 6