Question

a two digit number is such that the sum of its digit is 1/8 of the numbers when the digits of the number are reversed and the number is subtracted from the orignal number the result is 45 find orignal nummber

Answer 1

Answer:

$\mathfrak{72}$

Step-by-step explanation:

Let the tens and units digits of the original number be T, and U, respectively

Then number is: 10T + U, and when reversed, it becomes: 10U + T

Then:

$T +U=\frac{1}{8}(10T+U)$

$T +U=\frac{10T+U}{8}$

10T + U = 8(T + U) ------ Cross-multiplying

10T + U = 8T + 8U

10T - 8T = 8U - U

2T = 7U ------- eq (i)

Also, 10T + U - (10U + T) = 45

10T + U - 10U - T = 45 =====> 9T - 9U = 45 ======> 9(T - U) = 9(5) ======> T - U = 5 ======> T = 5 + U ------ eq (ii)

2(5 + U)  = 7U ------ Substituting 5 + U for T in eq (i)

10 + 2U = 7U

10 = 7U - 2U

10 = 5U

U, or units digit of original number = 10%2F5 = 2

T = 5 + 2 ------- Substituting 2 for U in eq (ii)

T, or tens digit of original number = 7

Original number:72