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

its A.42

Answer 2

but the procedure?

Answer 3

let the required number be,

\(A_1A_0 = A_0 + 10A_1\)

Answer 4

just see the first condition add the digits 4+2=6
now 6*7=42 done only this option satisfies the condition so its the answer

Answer 5

Thanks!

Answer 6

If you add the digits in a two-digit number and multiply the sum by 7, you get the original number.

\((A_1 + A_0)\times 7 = A_0 + 10A_1\) ----------(1)

Answer 7

If you reverse the digits in the two-digit number, the new number is 18 more than the sum of its two digits.

\(A_0A_1 = A_1 + 10A_0 = A_1 + A_0 + 18\) -------(2)

Answer 8

donnt wright formula's like that it confuses
dont feel bad but i get confused

Answer 9

two linearly independent equations, two unknowns, you can solve

Answer 10

you're free to confuse

Answer 11

take direct examples and do

Answer 12

what would you do when you don't have options ?