Question

sum of the digit of two digit number is 9 when we interchange the digit it is found that the resulting new number is greater than the original number by 27 what is the original number

Answer 1

Heya mate,

here is your answer,

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Let the digits at tens place and ones place be xand 9 − x respectively.

Therefore, original number = 10x + (9 − x) = 9x+ 9

On interchanging the digits, the digits at ones place and tens place will be x and 9 − xrespectively.

Therefore, new number after interchanging the digits = 10(9 − x) + x

= 90 − 10x + x

= 90 − 9x

According to the given question,

New number = Original number + 27

90 − 9x = 9x + 9 + 27

90 − 9x = 9x + 36

Transposing 9x to R.H.S and 36 to L.H.S, we obtain

90 − 36 = 18x

54 = 18x

Dividing both sides by 18, we obtain

3 = x and 9 − x = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 9x + 9 = 9 × 3 + 9 = 36


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Answer 2

hola!

→ Let the digit in ones place as x and the digit in tens place is y
→ The original number is (10y + x)
→ Number obtained by reversing the digits = (10x + y

ATQ,

♦ First condition is :-
Sum of digits of the number is 9:-

x + y = 9   ---(i)

♦ Second condition is:-
Number obtained by interchanging the digits is greater than the original number by 27

(10x + y) = (10y + x) 27

⇒ 10x + y – 10y – x = 27

⇒ 9x – 9y  = 27

∴ x – y = 3 -----(ii)

Adding (i) and (ii)
we get :-

2x = 12

x = 6

Put x = 6 in eqn. (i), we get

6 + y = 9

y = 9 – 6 = 3

Hence, The required answer is :-
The original number = 10y + x = 10(3) + 6 = 36

hope it helps! :)