Question

The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the original number the new number is 4 more than 5 times the sum of its digits in the original number. Find the original two digit number

Answer 1

Answer:

55

Step-by-step explanation:

55+55 = 110

55-10 = 45

5+5=10×5+4=54

Answer 2

Answer:

64

Step-by-step explanation:

$\textbf{Let\: the\: digit\: at\: ones\:= \:M}$

and $\textbf{Digit \:at \:tens \:= \:N}$

☞ The sum of two digit number and the number formed by interchanging the digits is 110.

Let the original number be $\underline{\bold{10N\: +\: M}}$

Number formed by reversing the digit is $\underline{\bold{10M\:+\:N}}$

10N + M + 10M + N = 110

________ [AS SAID IN QUESTION]

11N + 11M = 110

11 (N + M) = 110

N + M = 10

N = 10 - M _______(eq 1)

☞ If 10 is subtracted from original number i.e. 10N + M

10N + M - 10

☞ The new number is (equal to) 4 more than 5 times the sum of it's digit in the original number.

5(M + N) + 4

• A.T.Q.

10N + M - 10 = 5(M + N) + 4

10N + M - 10 = 5M + 5N + 4

10N - 5N + M - 5M = 4 + 10

5N - 4M = 14 _______(eq 2)

5(10 - M) - 4M = 14 ___[From eq 1]

50 - 5M - 4M = 14

- 9M = 14 - 50

- 9M = - 36

9M = 36

M = 4

• Put value of M in eq 1

N = 10 - (4)

N = 6

☞ We have to find the original two digit numbers.

So, the original number is 10N + M

10(6) + 4

64

_______________________________

The original two digit number is 64.

___________________[ANSWER]