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20 Feb 2026, 09:54
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
49% (02:19) correct
51%
(02:19)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
Sofa ___ Regular Price ___Sale Price
A ———— $900 —-—— $720
B ________ $840 ________ $800
C ________ $750 _________ $600
D _______ $1,200 ________ $750
E _______ $954 _________ $795
Torrance bought two of these unique , floor model sofas at their respective sale prices. Which ones did buy?
(1) The sum of their regular prices was over $2,000 and the sum of their sale prices was less than $1,500.
(2) The regular price of one of the sofas he bought was 20% higher than the sale price of the other.
A ———— $900 —-—— $720
B ________ $840 ________ $800
C ________ $750 _________ $600
D _______ $1,200 ________ $750
E _______ $954 _________ $795
Torrance bought two of these unique , floor model sofas at their respective sale prices. Which ones did buy?
(1) The sum of their regular prices was over $2,000 and the sum of their sale prices was less than $1,500.
(2) The regular price of one of the sofas he bought was 20% higher than the sale price of the other.
20 Feb 2026, 10:43
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kevincan
Which sofa pair did Torrance buy?
Statement 1:
The sum of their regular prices was over $2,000 and the sum of their sale prices was less than $1,500.
If the sum of regular prices was more than $2000.
possible pairs are :
DE ( 1200+954 = $2154). Their respective sales prices were : ($750+$795 = $1545)
DB ( 1200+ 840 = $2040). Their respective sales prices were : ( $750+$800 = $1550)
DA ( 1200 + 900 = $2100). Their respective sales price were : ( $750+ $720 = $1470)
Only DA qualifies as per the criteria mentioned. Hence, Sufficient.
Statement 2:
The regular price of one of the sofas he bought was 20% higher than the sale price of the other.
Regular price of one sofa = (6/5) * sales price of other.
If sales price = 720, then Regular price = 864
If sales price = 800, then Regular price = 960
If sales price = 600, then Regular price = 720.
If sales price = 750, then Regular price = 900. This combination of sales price (D) and Regular price = 900 (A).
If sales price = 795, then Regular price = 954.
DA is the possible combination.
Hence, Sufficient
Option D
20 Feb 2026, 11:16
Kudos
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Great question — this is a classic DS setup where the secret weapon is enumeration over algebra. The moment you see a small, finite set of options (5 sofas → 10 possible pairs), stop thinking in variables and start filtering systematically.
The setup:
Sofa | Regular | Sale
A | $900 | $720
B | $840 | $800
C | $750 | $600
D | $1,200 | $750
E | $954 | $795
Statement (1): Sum of regular prices > $2,000 AND sum of sale prices < $1,500
The only sofa with a high enough regular price to push any pair over $2,000 is D ($1,200). So I only need to check the four pairs involving D:
D+A: Regular = $2,100 ✓ → Sale = $720 + $750 = $1,470 < $1,500 ✓ → Qualifies
D+B: Regular = $2,040 ✓ → Sale = $800 + $750 = $1,550 > $1,500 ✗
D+C: Regular = $1,950 < $2,000 ✗
D+E: Regular = $2,154 ✓ → Sale = $795 + $750 = $1,545 > $1,500 ✗
Only {A, D} satisfies both conditions. Sufficient.
Statement (2): The regular price of one sofa was 20% higher than the sale price of the other
This means Regular(X) = 1.2 × Sale(Y). For each regular price, divide by 1.2 and check if that value appears as any sofa's sale price:
Regular(A) = $900 → 900/1.2 = $750 = Sale(D) ✓ → {A, D} qualifies
Regular(B) = $840 → 840/1.2 = $700 → no match ✗
Regular(C) = $750 → 750/1.2 = $625 → no match ✗
Regular(D) = $1,200 → 1,200/1.2 = $1,000 → no match ✗
Regular(E) = $954 → 954/1.2 = $795 = Sale(E) → same sofa, invalid ✗
Only {A, D} qualifies. Sufficient.
Answer: D — each statement alone is sufficient.
The common trap here: most students check S1, find it works, then set up an algebra equation for S2 — and because Statement 2 involves a ratio, they assume it's complicated. The fix is just dividing each regular price by 1.2 and scanning the sale column. Takes 20 seconds.
Takeaway: When a Data Sufficiency question gives you a finite table of discrete options, enumerate and filter — it beats algebra every time and removes the risk of missing edge cases.
The setup:
Sofa | Regular | Sale
A | $900 | $720
B | $840 | $800
C | $750 | $600
D | $1,200 | $750
E | $954 | $795
Statement (1): Sum of regular prices > $2,000 AND sum of sale prices < $1,500
The only sofa with a high enough regular price to push any pair over $2,000 is D ($1,200). So I only need to check the four pairs involving D:
D+A: Regular = $2,100 ✓ → Sale = $720 + $750 = $1,470 < $1,500 ✓ → Qualifies
D+B: Regular = $2,040 ✓ → Sale = $800 + $750 = $1,550 > $1,500 ✗
D+C: Regular = $1,950 < $2,000 ✗
D+E: Regular = $2,154 ✓ → Sale = $795 + $750 = $1,545 > $1,500 ✗
Only {A, D} satisfies both conditions. Sufficient.
Statement (2): The regular price of one sofa was 20% higher than the sale price of the other
This means Regular(X) = 1.2 × Sale(Y). For each regular price, divide by 1.2 and check if that value appears as any sofa's sale price:
Regular(A) = $900 → 900/1.2 = $750 = Sale(D) ✓ → {A, D} qualifies
Regular(B) = $840 → 840/1.2 = $700 → no match ✗
Regular(C) = $750 → 750/1.2 = $625 → no match ✗
Regular(D) = $1,200 → 1,200/1.2 = $1,000 → no match ✗
Regular(E) = $954 → 954/1.2 = $795 = Sale(E) → same sofa, invalid ✗
Only {A, D} qualifies. Sufficient.
Answer: D — each statement alone is sufficient.
The common trap here: most students check S1, find it works, then set up an algebra equation for S2 — and because Statement 2 involves a ratio, they assume it's complicated. The fix is just dividing each regular price by 1.2 and scanning the sale column. Takes 20 seconds.
Takeaway: When a Data Sufficiency question gives you a finite table of discrete options, enumerate and filter — it beats algebra every time and removes the risk of missing edge cases.
