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19 Feb 2026, 08:07
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Irene and Tina are two of the five members of a swimming team, who each weigh at least 100 pounds. If on average, the team members weigh 117 pounds, does Irene weigh less than 150 pounds?
(1) The median weight of the 5 team members is 120.
(2) Irene weighs 5 pounds more than Tina.
(1) The median weight of the 5 team members is 120.
(2) Irene weighs 5 pounds more than Tina.
19 Feb 2026, 10:07
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kevincan
Let the five members be I, T, x,y,z respectively.
Each weigh at least 100 pounds.
Average weight of all 5 is 117. Total = 117*5 = 585.
We need to find: Is Irene weigh less than 150 pounds ?
Statement 1:
The median weight of the 5 team members is 120.
x+y+z+T+I = 585
if z is 120, we need maximise I.
so, x=y= 100. And T can be equal to median =120.
100+100+120+120+I = 585.
I max can be 145. Which is less than 150.
Hence, Sufficient
Statement 2:
Irene weighs 5 pounds more than Tina.
I = 5+ T
x+y+z+T+T+5 = 585
x+y+z+2T = 580
To maximize T let’s put x=y=z =100.
2T = 280
T=140
I = 145.
I max can be 145. Which is less than 150.
Hence, Sufficient
Option D
19 Feb 2026, 11:59
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Key concept: Data Sufficiency "maximize/minimize" — with a bounded inequality, you need to show whether every possible scenario gives the same yes/no answer.
The common trap: Students see "median = 120" in Statement 1 and assume Irene is somewhere near 120. But the median only pins down the middle value in the sorted list — Irene could be the heaviest member, far above 120.
Setup: 5 members, each ≥ 100 lbs, average = 117 → total = 585 lbs. Question: Is Irene < 150?
Step 1 — Statement 1: Median = 120.
Sort the five weights: w1 ≤ w2 ≤ w3 ≤ w4 ≤ w5. So w3 = 120.
To find the maximum Irene could weigh, minimize everyone else's contribution:
Set w1 = w2 = 100 (minimum allowed), w3 = 120 (fixed), w4 = 120 (minimum ≥ median).
Then: 100 + 100 + 120 + 120 + Irene = 585 → Irene = 145.
Even at the absolute maximum, Irene = 145 < 150. The answer is always YES.
→ Statement 1 alone is Sufficient.
Step 2 — Statement 2: Irene = Tina + 5.
Again, minimize the other three: set them all to 100.
100 + 100 + 100 + T + (T + 5) = 585 → 2T = 280 → T = 140, Irene = 145 < 150.
Can the other three weigh more, pushing Irene down? Yes — that would only make Irene lighter. And the floor is Irene ≥ 105 (since Tina ≥ 100). So Irene is always between 105 and 145 — always less than 150. Answer is always YES.
→ Statement 2 alone is Sufficient.
Answer: D — each statement alone is sufficient.
Takeaway: In yes/no DS problems with an inequality, the key question is always "what's the maximum (or minimum) the unknown can be?" If even the worst case still answers YES, the statement is sufficient.
The common trap: Students see "median = 120" in Statement 1 and assume Irene is somewhere near 120. But the median only pins down the middle value in the sorted list — Irene could be the heaviest member, far above 120.
Setup: 5 members, each ≥ 100 lbs, average = 117 → total = 585 lbs. Question: Is Irene < 150?
Step 1 — Statement 1: Median = 120.
Sort the five weights: w1 ≤ w2 ≤ w3 ≤ w4 ≤ w5. So w3 = 120.
To find the maximum Irene could weigh, minimize everyone else's contribution:
Set w1 = w2 = 100 (minimum allowed), w3 = 120 (fixed), w4 = 120 (minimum ≥ median).
Then: 100 + 100 + 120 + 120 + Irene = 585 → Irene = 145.
Even at the absolute maximum, Irene = 145 < 150. The answer is always YES.
→ Statement 1 alone is Sufficient.
Step 2 — Statement 2: Irene = Tina + 5.
Again, minimize the other three: set them all to 100.
100 + 100 + 100 + T + (T + 5) = 585 → 2T = 280 → T = 140, Irene = 145 < 150.
Can the other three weigh more, pushing Irene down? Yes — that would only make Irene lighter. And the floor is Irene ≥ 105 (since Tina ≥ 100). So Irene is always between 105 and 145 — always less than 150. Answer is always YES.
→ Statement 2 alone is Sufficient.
Answer: D — each statement alone is sufficient.
Takeaway: In yes/no DS problems with an inequality, the key question is always "what's the maximum (or minimum) the unknown can be?" If even the worst case still answers YES, the statement is sufficient.
