Question

If you add the digits in a two-digit number and multiply the sum by 7, you get the original number. If you reverse the digits in the two-digit number, the new number is 18 more than the sum of its two digits. What is the original number?

A.42
B.24
C.64
D.46
E.36

Answer 1

Let's assume the original two-digit number is represented by "10x + y," where x represents the tens digit and y represents the units digit.

According to the given information:
1) The sum of the digits multiplied by 7 is equal to the original number:
7(x + y) = 10x + y

2) Reversing the digits gives a new number that is 18 more than the sum of the digits:
10y + x = x + y + 18

We can solve this system of equations to find the values of x and y, which will give us the original two-digit number.

From the first equation:
7x + 7y = 10x + y
6x - 6y = 0
x = y

Substituting x = y into the second equation:
10y + y = y + y + 18
11y = 2y + 18
9y = 18
y = 2

Substituting y = 2 back into the first equation:
7x + 7(2) = 10x + 2
14 = 3x
x = 4

Therefore, the original two-digit number is 10x + y = 42.

So, the correct answer is A. 42.