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25 Feb 2026, 20:00
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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:
55%
(hard)
Question Stats:
55% (01:58) correct
45%
(02:23)
wrong
based on 87
sessions
History
Date
Time
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Not Attempted Yet
One school building serves as a combined grade school, middle school, and high school for a small town. The number of high schoolers who study in the building is twice the number of grade schoolers and 50 more than the number of middle schoolers. In all, how many students study in this building?
(1) The number of grade schoolers and middle schoolers in the building is 130.
(2) The probability of a randomly selected student being a high schooler are 12/25.
(1) The number of grade schoolers and middle schoolers in the building is 130.
(2) The probability of a randomly selected student being a high schooler are 12/25.
26 Feb 2026, 13:52
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Let G=Grade schoolers, M=Middle schoolers, and H=High schoolers
Then as per the given,
Building = (G+M+H)
Also, H=2G and
H=(M+50) OR M=(H-50) OR M=(2G-50)
Plugging this in our original equation : Building = (G+2G+(2G-50)) OR (5G-50). This is what we have to find out.
(1) The number of grade schoolers and middle schoolers in the building is 130.
(G+M)=130
Since, we know that M=(2G-50)
The whole equation can be rewritten as (G+(2G-50))=130 OR 3G=180
OR G=60
Now, notice in the prompt that we targeted G all we needed to solve was for that only.
So, overall students that study in the building : (5(60)-50) OR 250. Sufficient
(2) The probability of a randomly selected student being a high schooler are 12/25.
Given = (H)/((G+M+H)=(12/25)
Again, let's plug in the value for G from prompt, it would become = (2G)/(G+(2G-50)+2G)= 12/25
OR 50G = 60G-600 OR G=60.
Again the total students would be (5(60)-50) OR 250. Sufficient
Option D
Then as per the given,
Building = (G+M+H)
Also, H=2G and
H=(M+50) OR M=(H-50) OR M=(2G-50)
Plugging this in our original equation : Building = (G+2G+(2G-50)) OR (5G-50). This is what we have to find out.
(1) The number of grade schoolers and middle schoolers in the building is 130.
(G+M)=130
Since, we know that M=(2G-50)
The whole equation can be rewritten as (G+(2G-50))=130 OR 3G=180
OR G=60
Now, notice in the prompt that we targeted G all we needed to solve was for that only.
So, overall students that study in the building : (5(60)-50) OR 250. Sufficient
(2) The probability of a randomly selected student being a high schooler are 12/25.
Given = (H)/((G+M+H)=(12/25)
Again, let's plug in the value for G from prompt, it would become = (2G)/(G+(2G-50)+2G)= 12/25
OR 50G = 60G-600 OR G=60.
Again the total students would be (5(60)-50) OR 250. Sufficient
Option D
ExpertsGlobal5
26 Feb 2026, 14:22
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The key concept being tested here is expressing multiple unknowns as a single variable before evaluating the statements — the classic DS move for "find the total" questions with linked quantities.
Step 1: Set up variables and use the stem's relationships.
Let G = grade schoolers, M = middle schoolers, H = high schoolers.
The stem tells us: H = 2G and H = M + 50, so M = 2G − 50.
Total = G + M + H = G + (2G − 50) + 2G = 5G − 50
The entire question now reduces to: can we determine G? This is the unlock — once you see this, both statements become straightforward to evaluate.
Step 2: Evaluate Statement 1 — "G + M = 130"
Substitute M = 2G − 50:
G + (2G − 50) = 130 → 3G = 180 → G = 60
Total = 5(60) − 50 = 250 students. → Sufficient.
Step 3: Evaluate Statement 2 — "Probability of selecting a high schooler = 12/25"
This means H / Total = 12/25 → 2G / (5G − 50) = 12/25
Cross-multiply: 50G = 60G − 600 → 10G = 600 → G = 60
Total = 250 students. → Sufficient.
Answer: D
Common trap: Most students who miss this either try to work with three separate unknowns without reducing to one, which makes both statements look insufficient, or they evaluate Statement 2 by plugging in numbers until something works — that approach eats up time and often leads to errors. The moment you collapse the total to 5G − 50, both statements hand you the answer cleanly.
Takeaway: In DS questions where the stem links multiple unknowns together, always reduce to a single variable first — it tells you exactly what information each statement needs to deliver.
(Kavya | 725 on GMAT Focus Edition)
Step 1: Set up variables and use the stem's relationships.
Let G = grade schoolers, M = middle schoolers, H = high schoolers.
The stem tells us: H = 2G and H = M + 50, so M = 2G − 50.
Total = G + M + H = G + (2G − 50) + 2G = 5G − 50
The entire question now reduces to: can we determine G? This is the unlock — once you see this, both statements become straightforward to evaluate.
Step 2: Evaluate Statement 1 — "G + M = 130"
Substitute M = 2G − 50:
G + (2G − 50) = 130 → 3G = 180 → G = 60
Total = 5(60) − 50 = 250 students. → Sufficient.
Step 3: Evaluate Statement 2 — "Probability of selecting a high schooler = 12/25"
This means H / Total = 12/25 → 2G / (5G − 50) = 12/25
Cross-multiply: 50G = 60G − 600 → 10G = 600 → G = 60
Total = 250 students. → Sufficient.
Answer: D
Common trap: Most students who miss this either try to work with three separate unknowns without reducing to one, which makes both statements look insufficient, or they evaluate Statement 2 by plugging in numbers until something works — that approach eats up time and often leads to errors. The moment you collapse the total to 5G − 50, both statements hand you the answer cleanly.
Takeaway: In DS questions where the stem links multiple unknowns together, always reduce to a single variable first — it tells you exactly what information each statement needs to deliver.
(Kavya | 725 on GMAT Focus Edition)
