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01 Apr 2026, 03:31
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Difficulty:
95%
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Question Stats:
25% (02:54) correct
75%
(02:36)
wrong
based on 56
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The average bill at Veracruz Restaurant yesterday was $48 per customer. Each customer chose to leave either a 15% tip or a 20% tip. Did customers yesterday leave on average a tip of more than $8 per customer?
(1) Most of yesterday’s customers chose to leave a 20% tip.
(2) The tips collected yesterday accounted for more than 16% of the total amount paid by yesterday’s customers.
(1) Most of yesterday’s customers chose to leave a 20% tip.
(2) The tips collected yesterday accounted for more than 16% of the total amount paid by yesterday’s customers.
02 Apr 2026, 01:41
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Nice problem. Let me work through both statements.
First, anchor the numbers. Average bill = $48. 15% tip = $7.20 per customer. 20% tip = $9.60 per customer. The question asks if the average tip exceeded $8. Since $8 / $48 = 16.67%, we're really asking: did the effective tip rate exceed 16.67%?
Statement (1): Most customers chose the 20% tip.
"Most" means strictly more than half. So if even 50.01% of customers chose 20%, the weighted average tip is just above (0.5 x 9.60 + 0.5 x 7.20) = $8.40. And since any majority at 20% pushes the average higher than the 50/50 baseline, the average tip is always above $8.40 whenever statement 1 holds. Comfortably above $8. SUFFICIENT.
I fell for a version of this trap once where I thought "most" could sneak you close to 50% and make it borderline - but here, even the absolute minimum case (barely more than half choosing 20%) gives $8.40+. There's no ambiguity.
Statement (2): Tips accounted for more than 16% of the total amount paid by yesterday's customers.
"Total amount paid" = bill + tip. So if T is total tips and B is total bills:
T / (B + T) > 0.16
Solving: T > 0.16B + 0.16T, so 0.84T > 0.16B, which means T/B > 0.16/0.84 = 4/21, roughly 19.05%.
Average tip per customer > 19.05% x $48 = $9.14. That's well above $8. SUFFICIENT.
The trap here is misreading "total amount paid" as just the bill total (not bill + tip). If you do that, you get tips > 16% of bills = $7.68 average, which doesn't answer the question. But if you read it correctly as the full amount paid (which includes the tip), then statement 2 is also sufficient.
Answer: D
The key conceptual move: convert the tip ratio question into a tip-rate question first, then evaluate each statement against that threshold.
First, anchor the numbers. Average bill = $48. 15% tip = $7.20 per customer. 20% tip = $9.60 per customer. The question asks if the average tip exceeded $8. Since $8 / $48 = 16.67%, we're really asking: did the effective tip rate exceed 16.67%?
Statement (1): Most customers chose the 20% tip.
"Most" means strictly more than half. So if even 50.01% of customers chose 20%, the weighted average tip is just above (0.5 x 9.60 + 0.5 x 7.20) = $8.40. And since any majority at 20% pushes the average higher than the 50/50 baseline, the average tip is always above $8.40 whenever statement 1 holds. Comfortably above $8. SUFFICIENT.
I fell for a version of this trap once where I thought "most" could sneak you close to 50% and make it borderline - but here, even the absolute minimum case (barely more than half choosing 20%) gives $8.40+. There's no ambiguity.
Statement (2): Tips accounted for more than 16% of the total amount paid by yesterday's customers.
"Total amount paid" = bill + tip. So if T is total tips and B is total bills:
T / (B + T) > 0.16
Solving: T > 0.16B + 0.16T, so 0.84T > 0.16B, which means T/B > 0.16/0.84 = 4/21, roughly 19.05%.
Average tip per customer > 19.05% x $48 = $9.14. That's well above $8. SUFFICIENT.
The trap here is misreading "total amount paid" as just the bill total (not bill + tip). If you do that, you get tips > 16% of bills = $7.68 average, which doesn't answer the question. But if you read it correctly as the full amount paid (which includes the tip), then statement 2 is also sufficient.
Answer: D
The key conceptual move: convert the tip ratio question into a tip-rate question first, then evaluate each statement against that threshold.
03 Apr 2026, 10:24
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Careful ! We know that the average bill was $48, not that every bill was $48
