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30 May 2026, 11:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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Question Stats:
90% (02:04) correct
10%
(02:34)
wrong
based on 25
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Elena and Riley each bought notebooks, pens, and folders from the same supplier. For each item type, the unit price was the same for Elena and Riley. If Riley’s total spending was twice Elena’s total spending, what fraction of Elena’s total spending was on folders?
(1) Elena bought 6 notebooks, 9 pens, and 10 folders.
(2) Riley bought 18 notebooks, 27 pens, and 15 folders.
The OA will be automatically revealed on Tuesday 2nd of June 2026 11:15:25 AM Pacific Time Zone
(1) Elena bought 6 notebooks, 9 pens, and 10 folders.
(2) Riley bought 18 notebooks, 27 pens, and 15 folders.
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The OA will be automatically revealed on Tuesday 2nd of June 2026 11:15:25 AM Pacific Time Zone
ankushsambare
Joined: 13 Jun 2022
Last visit: 01 Jun 2026
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Schools: ISB '27 Wharton '26 INSEAD Aug '26
GPA: 2.4
30 May 2026, 22:41
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Let notebook, pen, and folder prices be N, P, and F.
We want:
Elena's folder spending ÷ Elena's total spending.
(1) Elena bought 6 notebooks, 9 pens, and 10 folders.
Total spending = 6N + 9P + 10F
Folder spending = 10F
Required fraction = 10F / (6N + 9P + 10F)
Since prices are unknown, we cannot find a unique value.
(1) Insufficient.
(2) Riley bought 18 notebooks, 27 pens, and 15 folders.
We know Riley spent twice as much as Elena, but we don't know Elena's purchases.
(2) Insufficient.
Together:
Elena's spending = 6N + 9P + 10F
Riley's spending = 18N + 27P + 15F
Given Riley spent twice as much as Elena:
18N + 27P + 15F = 2(6N + 9P + 10F)
18N + 27P + 15F = 12N + 18P + 20F
6N + 9P = 5F
Substitute into Elena's spending:
E = (6N + 9P) + 10F = 5F + 10F = 15F
So the fraction spent on folders is:
10F / 15F = 2/3
A unique value is obtained.
Answer: C — Both statements together are sufficient, neither alone is sufficient.
We want:
Elena's folder spending ÷ Elena's total spending.
(1) Elena bought 6 notebooks, 9 pens, and 10 folders.
Total spending = 6N + 9P + 10F
Folder spending = 10F
Required fraction = 10F / (6N + 9P + 10F)
Since prices are unknown, we cannot find a unique value.
(1) Insufficient.
(2) Riley bought 18 notebooks, 27 pens, and 15 folders.
We know Riley spent twice as much as Elena, but we don't know Elena's purchases.
(2) Insufficient.
Together:
Elena's spending = 6N + 9P + 10F
Riley's spending = 18N + 27P + 15F
Given Riley spent twice as much as Elena:
18N + 27P + 15F = 2(6N + 9P + 10F)
18N + 27P + 15F = 12N + 18P + 20F
6N + 9P = 5F
Substitute into Elena's spending:
E = (6N + 9P) + 10F = 5F + 10F = 15F
So the fraction spent on folders is:
10F / 15F = 2/3
A unique value is obtained.
Answer: C — Both statements together are sufficient, neither alone is sufficient.
30 May 2026, 23:59
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Answer is C, and what makes this one click is realizing you never need to actually find the individual prices. The ratio is enough.
Let me set up the algebra. Let n, p, f = prices of notebook, pen, folder (same for both buyers).
The question asks: what fraction of Elena's total spending went to folders?
That's 10f / (6n + 9p + 10f).
Statement (1): Elena bought 6 notebooks, 9 pens, 10 folders.
Now we know Elena's quantities, but prices are still unknown. We can write her total as 6n + 9p + 10f, but we can't simplify the fraction without knowing the price ratios. Insufficient.
Statement (2): Riley bought 18 notebooks, 27 pens, 15 folders.
We know Riley's quantities, and we know Riley spent twice as much as Elena -- but we don't know Elena's purchase breakdown. Insufficient.
Combined:
Use the constraint: Riley's total = 2 x Elena's total.
18n + 27p + 15f = 2(6n + 9p + 10f)
18n + 27p + 15f = 12n + 18p + 20f
6n + 9p = 5f
Now substitute back into Elena's total:
6n + 9p + 10f = 5f + 10f = 15f
So the fraction on folders = 10f / 15f = 2/3.
The trap I fell for the first time I saw a problem like this: I spent 90 seconds trying to figure out whether the statements give us enough to solve for individual prices. They don't -- but that's not what you need. The "Riley spent twice as much" constraint links the two sets of quantities, which pins down the price relationship, which collapses the fraction.
This is a classic DS move: you don't need absolute values, just enough structure to fix a ratio.
Answer: C
Let me set up the algebra. Let n, p, f = prices of notebook, pen, folder (same for both buyers).
The question asks: what fraction of Elena's total spending went to folders?
That's 10f / (6n + 9p + 10f).
Statement (1): Elena bought 6 notebooks, 9 pens, 10 folders.
Now we know Elena's quantities, but prices are still unknown. We can write her total as 6n + 9p + 10f, but we can't simplify the fraction without knowing the price ratios. Insufficient.
Statement (2): Riley bought 18 notebooks, 27 pens, 15 folders.
We know Riley's quantities, and we know Riley spent twice as much as Elena -- but we don't know Elena's purchase breakdown. Insufficient.
Combined:
Use the constraint: Riley's total = 2 x Elena's total.
18n + 27p + 15f = 2(6n + 9p + 10f)
18n + 27p + 15f = 12n + 18p + 20f
6n + 9p = 5f
Now substitute back into Elena's total:
6n + 9p + 10f = 5f + 10f = 15f
So the fraction on folders = 10f / 15f = 2/3.
The trap I fell for the first time I saw a problem like this: I spent 90 seconds trying to figure out whether the statements give us enough to solve for individual prices. They don't -- but that's not what you need. The "Riley spent twice as much" constraint links the two sets of quantities, which pins down the price relationship, which collapses the fraction.
This is a classic DS move: you don't need absolute values, just enough structure to fix a ratio.
Answer: C
