Nedas coffee shop sells small, medium, and large coffees for $2, $3

Question

Neda’s coffee shop sells small, medium, and large coffees for $2, $3, and $4, respectively. On a particular day, her shop earned $404 from coffee sales. The number of medium coffees sold was greater than the number of small coffees and also greater than the number of large coffees, and at least one coffee of each size was sold. What is the difference between the greatest and smallest possible total numbers of coffees Neda’s shop sold that day?

A. 32
B. 38
C. 44
D. 50
E. 56

Show Answer

Official Answer

C

A. 32
B. 38
C. 44 
D. 50
E. 56

officiainventore · Answer

@kevincan is it possible to provide an explanation for this one? thanks!

kevincan · Answer

corrected ! I will post a solution tomorrow

dec2023 · Answer

Would like to understand what is wrong with my calculations?

404=2S+3M+4L

maximise M, minimise S/L, while keeping each size >1:
404=2(8)+3(128)+4(1)

Difference:
128-1=127

rutikaoqw · Answer

For greatest, you shall min L, let L be 1.
2S + 3M = 400
Now, since M is greater than S, convert everything to S. 2S + 3(S+x) = 400
Instead of x, you will need something that 5 can divide, hence 3(s+5). This gives S as 77 and M as 82.
Total 160
Solve similarly for smallest,
Min S. Hence S will be 1.
3M + 4L= 402
3(L+x)+ 4(L) = 402. (since M>L)
X has to be defined in a way that the result is divided by 7. X=1
This gives L=57, M =58
Total 116
and find difference.
You get 44

Edskore · Answer

The key concept here is Min-Max optimisation under integer constraints — and this question has two separate sub-problems: maximise the total coffees sold, and minimise it.

Setup: 2S + 3M + 4L = 404, with M > S >= 1, M > L >= 1, all integers.

--- PART 1: Maximise total (S + M + L) ---

To maximise count, we want as many cheap coffees as possible. Small ($2) is cheapest, but M must be the largest group. Strategy: set L = 1 (minimum), then maximise S while keeping M > S.

With L = 1: 2S + 3M = 400, and M > S.
From 3M = 400 - 2S → M = (400 - 2S)/3.
For M to be an integer: 400 - 2S must be divisible by 3. Since 400 ≡ 1 (mod 3), we need 2S ≡ 1 (mod 3), so S ≡ 2 (mod 3).
For M > S: (400 - 2S)/3 > S → 400 > 5S → S < 80.

Largest valid S: S = 77 (since 77 ≡ 2 mod 3 ✓ and 77 < 80 ✓).
→ M = (400 - 154)/3 = 246/3 = 82.
Check: M = 82 > S = 77 ✓, M = 82 > L = 1 ✓.
Max total = 77 + 82 + 1 = 160.

--- PART 2: Minimise total (S + M + L) ---

To minimise count, use the most expensive coffees. Large ($4) is priciest, but M must still be the largest group. Strategy: set S = 1 (minimum), then maximise L while keeping M > L.

With S = 1: 3M + 4L = 402, and M > L.
From 3M = 402 - 4L → M = (402 - 4L)/3.
For M integer: 402 - 4L divisible by 3. Since 402 ≡ 0 (mod 3) and 4L ≡ L (mod 3), need L ≡ 0 (mod 3).
For M > L: (402 - 4L)/3 > L → 402 > 7L → L < 57.4.

Largest valid L: L = 57 (since 57 ≡ 0 mod 3 ✓ and 57 < 57.4 ✓).
→ M = (402 - 228)/3 = 174/3 = 58.
Check: M = 58 > L = 57 ✓, M = 58 > S = 1 ✓.
Min total = 1 + 58 + 57 = 116.

--- ANSWER ---

Difference = 160 - 116 = 44. Answer: C.

---

Common trap (dec2023's approach above): Setting M = 128 violates the problem constraints — you also need M > S, so S can't be as small as 1 when M = 128 unless S < 128. But that approach also doesn't correctly maximise the total number of coffees. The goal is to maximise S + M + L, not just maximise M.

Takeaway: On Min-Max word problems with three variables, always anchor one variable at its extreme (min or max), then use the divisibility constraint to find the binding limit on the other.

(Kavya | 725 on GMAT Focus Edition)

Nedas coffee shop sells small, medium, and large coffees for $2, $3

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