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Smithson sells two printer models, HL100 and HL200. Of the customers who purchased model HL200 last month,

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by BTGmoderatorLU » Tue Nov 03, 2026 10:22 am
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Smithson sells two printer models, HL100 and HL200. Of the customers who purchased model HL200 last month, \(\frac{3}{5}\) would have purchased model HL100 if HL200 had been unavailable, and of those who purchased model HL100, \(\frac{1}{20}\) would have purchased model HL200 if HL100 had been unavailable.

If \(\frac{7}{12}\) of the customers who purchased a Smithson printer last month would not have purchased either model if their chosen model had been unavailable, what fraction of the Smithson printers sold last month were model HL200?

A. \(\dfrac{2}{5}\)

B. \(\dfrac{7}{12}\)

C. \(\dfrac{3}{5}\)

D. \(\dfrac{2}{3}\)

E. \(\dfrac{4}{5}\)

The OA is D
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Source: — Problem Solving |

Answer: D

Turn each fraction into "would have bought nothing."

HL200 buyers: 3/5 would switch to HL100, so 2/5 would buy nothing.
HL100 buyers: 1/20 would switch to HL200, so 19/20 would buy nothing.
All buyers: 7/12 would buy nothing.

Let x = HL200 buyers and y = HL100 buyers.

(2/5)x + (19/20)y = (7/12)(x + y)

Multiply both sides by 60: 24x + 57y = 35x + 35y. So 22y = 11x, and x = 2y.

HL200 share = x/(x + y) = 2y/(3y) = 2/3.

Check with real numbers: take 40 HL200 buyers and 20 HL100 buyers. Then 16 + 19 = 35 of the 60 would buy nothing, and 35/60 = 7/12. The condition holds.

Faster route: treat 7/12 as a weighted average of 2/5 and 19/20. Over 60, those three numbers become 35, 24 and 57. The average sits 11 away from the HL200 rate and 22 away from the HL100 rate. The closer group weighs more, in the ratio 22 to 11. So HL200 buyers outnumber HL100 buyers 2 to 1.

Watch for A and B. A is the 2/5 from the first step, and B repeats the 7/12 from the question. Neither one answers what the question asks.
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