JAMB 2015 · UTME · Q17

Find a two-digit number such that three times the tens digit is 2 less than twice the units digit, and twice the number is 20 greater than the number obtained by reversing the digits.

Worked solution (try it first)
  1. Let the tens digit be tt and the units digit uu, so the number is 10t+u10t + u.
  2. The first fact gives 3t=2u−23t = 2u - 2, or 3t−2u=−23t - 2u = -2.
  3. Reversed, the number is 10u+t10u + t.
  4. The second fact gives 2(10t+u)=10u+t+202(10t + u) = 10u + t + 20, which simplifies to 19t−8u=2019t - 8u = 20.
  5. Multiply the first equation by 4, giving 12t−8u=−812t - 8u = -8, and subtract it from the second: 7t=287t = 28, so t=4t = 4.
  6. Then 2u=3t+2=142u = 3t + 2 = 14, so u=7u = 7.
  7. The number is 47, option D.

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