Free GMAT Problem Solving Practice Question

Translate each archive into a power of the same base, then set the two expressions equal.

Method (common base): Write both starting amounts and both growth factors as powers of 2. Archive M starts at 2, which is 2¹, and quadruples every 4 days. Because quadrupling is the same as two doublings, growing by a factor of 4 every 4 days means multiplying by 2 twice every 4 days, so after t days Archive M has 2⁽¹ ⁺ ᵗ/²⁾ records. Archive N starts at 16, which is 2⁴, and doubles every 6 days, so after t days it has 2⁽⁴ ⁺ ᵗ/⁶⁾ records.

Setting the two equal and matching exponents on the common base: 1 + t/2 = 4 + t/6. Multiply through by 6: 6 + 3t = 24 + t, so 2t = 18 and t = 9.

Check: at t = 9, Archive M has 2⁽¹ ⁺ ⁴·⁵⁾ = 2⁵·⁵, and Archive N has 2⁽⁴ ⁺ ¹·⁵⁾ = 2⁵·⁵. They match, so the answer is (B).

Method (doubling-gap reasoning): Archive N starts 8 times as large as Archive M, and 8 = 2³, a gap of three doublings. Archive M doubles every 2 days while Archive N doubles every 6 days, so Archive M gains ground at a rate of ½ − ⅙ = ⅓ of a doubling per day. To close three doublings takes 3 ÷ ⅓ = 9 days, confirming (B).

Each wrong choice is a single specific slip, either in translating the word problem into exponents or in converting one archive's growth to the common base, while (B) applies every step correctly.