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29 Jun 2026, 19:00
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A stationery shop sells only notebooks and planners. Each notebook sells for $30 and costs the shop $25, while each planner sells for $20 and costs the shop $12. In June, the shop sold 140 notebooks and planners in total. Was the shop’s profit from these sales more than $900?
(1) The shop’s profit from planners was less than its profit from notebooks.
(2) The shop’s revenue from notebooks was greater than its revenue from planners.
(1) The shop’s profit from planners was less than its profit from notebooks.
(2) The shop’s revenue from notebooks was greater than its revenue from planners.
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M47-220
05 Jul 2026, 03:00
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GMAT Club Official Solution:
A stationery shop sells only notebooks and planners. Each notebook sells for $30 and costs the shop $25, while each planner sells for $20 and costs the shop $12. In June, the shop sold 140 notebooks and planners in total. Was the shop’s profit from these sales more than $900?
(1) The shop’s profit from planners was less than its profit from notebooks.
Each notebook gives a profit of $5, and each planner gives a profit of $8.
Let p be the number of planners sold. Then the number of notebooks sold is 140 - p.
Statement (1) gives:
8p < 5(140 - p)
13p < 700
p < 53.something
So p can be at most 53.
Since each planner gives more profit than each notebook, the total profit is greatest when p is as large as possible. Even at the highest possible value, p = 53, the total profit is:
53 * 8 + 87 * 5 = 424 + 435 = 859
So the profit is not more than $900.
Sufficient.
(2) The shop’s revenue from notebooks was greater than its revenue from planners.
This gives:
30(140 - p) > 20p
4200 - 30p > 20p
4200 > 50p
So p can be at most 83.
This is not enough. If p = 60, the profit is not more than $900, but if p = 80, the profit is more than $900.
Not sufficient.
Answer: A.
A stationery shop sells only notebooks and planners. Each notebook sells for $30 and costs the shop $25, while each planner sells for $20 and costs the shop $12. In June, the shop sold 140 notebooks and planners in total. Was the shop’s profit from these sales more than $900?
(1) The shop’s profit from planners was less than its profit from notebooks.
Each notebook gives a profit of $5, and each planner gives a profit of $8.
Let p be the number of planners sold. Then the number of notebooks sold is 140 - p.
Statement (1) gives:
8p < 5(140 - p)
13p < 700
p < 53.something
So p can be at most 53.
Since each planner gives more profit than each notebook, the total profit is greatest when p is as large as possible. Even at the highest possible value, p = 53, the total profit is:
53 * 8 + 87 * 5 = 424 + 435 = 859
So the profit is not more than $900.
Sufficient.
(2) The shop’s revenue from notebooks was greater than its revenue from planners.
This gives:
30(140 - p) > 20p
4200 - 30p > 20p
4200 > 50p
So p can be at most 83.
This is not enough. If p = 60, the profit is not more than $900, but if p = 80, the profit is more than $900.
Not sufficient.
Answer: A.
General Discussion
29 Jun 2026, 22:11
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Let notebooks = N, planners = P.
N + P = 140
Profit = 5N + 8P = 5(140 − P) + 8P = 700 + 3P
Need: Profit > 900?
(1) 8P < 5N
Substitute N = 140 − P:
8P < 5(140 − P)
13P < 700
P ≤ 53
Profit = 5(87) + 8(53) = 859
That's the max profit that can be earned with this condition
Therefore it's always less than 900
Sufficient.
(2) 30N > 20P 3N > 2P
Substitute N = 140 − P:
3(140 − P) > 2P
420 > 5P
P ≤ 83
If P = 67, profit = 901 (>900).
If P = 50, profit = 850 (<900).
Both possible.
Not sufficient.
Answer: A.
N + P = 140
Profit = 5N + 8P = 5(140 − P) + 8P = 700 + 3P
Need: Profit > 900?
(1) 8P < 5N
Substitute N = 140 − P:
8P < 5(140 − P)
13P < 700
P ≤ 53
Profit = 5(87) + 8(53) = 859
That's the max profit that can be earned with this condition
Therefore it's always less than 900
Sufficient.
(2) 30N > 20P 3N > 2P
Substitute N = 140 − P:
3(140 − P) > 2P
420 > 5P
P ≤ 83
If P = 67, profit = 901 (>900).
If P = 50, profit = 850 (<900).
Both possible.
Not sufficient.
Answer: A.
