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19 Feb 2026, 09:38
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:50) correct
67%
(03:24)
wrong
based on 60
sessions
History
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Time
Result
Not Attempted Yet
Students in a data science course complete three timed assignments and receive integer scores from 0 to 10 on each assignment. If any student’s scores are x, y, and z, where x is less than or equal to y and x is less than or equal to z, his or her course score is equal to 0.2x + 0.4(y + z ). Niloufar received a course score of 8.4. Which of the following must be true?
I. She received a score below 8 on at least one assignment.
II. She received a score of 9 on at least one assignment.
III. She received a score of at least 6 on every assignment.
A. I only
B. II only
C. III only
D. I and III only
E. None of the three
I. She received a score below 8 on at least one assignment.
II. She received a score of 9 on at least one assignment.
III. She received a score of at least 6 on every assignment.
A. I only
B. II only
C. III only
D. I and III only
E. None of the three
19 Feb 2026, 11:56
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Key concept: "Must be True" reasoning with a weighted formula — you're not solving for one answer, you're stress-testing each statement against every valid score combination.
The common trap: Most students find one set of (x, y, z) that gives 8.4 and check the statements against just that. But "must be true" means the statement has to hold for every valid combination. One counterexample kills a statement.
Step 1 — Simplify the formula.
Course score = 0.2x + 0.4(y + z) = 8.4
Multiply through by 5: x + 2(y + z) = 42
All scores are integers from 0–10, with x ≤ y and x ≤ z.
Step 2 — Find multiple valid (x, y, z) combinations.
- x = 2, y = 10, z = 10 → 2 + 40 = 42 ✓
- x = 4, y = 9, z = 10 → 4 + 38 = 42 ✓
- x = 6, y = 9, z = 9 → 6 + 36 = 42 ✓
- x = 8, y = 8, z = 9 → 8 + 34 = 42 ✓
Step 3 — Test each statement.
Statement I: She received a score below 8 on at least one assignment.
Counterexample: x = 8, y = 8, z = 9. All scores ≥ 8. Statement I fails here.
→ Does NOT have to be true.
Statement II: She received a score of 9 on at least one assignment.
Counterexample: x = 2, y = 10, z = 10. No score of 9 anywhere.
→ Does NOT have to be true.
Statement III: She received a score of at least 6 on every assignment.
Counterexample: x = 4, y = 9, z = 10. x = 4 is below 6, yet 4 + 38 = 42 ✓.
→ Does NOT have to be true.
Answer: E — None of the three.
Takeaway: On "Must be True" problems, your goal is counterexamples. The moment you find one valid scenario where a statement fails, cross it off. Never assume the first combination you find represents all possibilities.
The common trap: Most students find one set of (x, y, z) that gives 8.4 and check the statements against just that. But "must be true" means the statement has to hold for every valid combination. One counterexample kills a statement.
Step 1 — Simplify the formula.
Course score = 0.2x + 0.4(y + z) = 8.4
Multiply through by 5: x + 2(y + z) = 42
All scores are integers from 0–10, with x ≤ y and x ≤ z.
Step 2 — Find multiple valid (x, y, z) combinations.
- x = 2, y = 10, z = 10 → 2 + 40 = 42 ✓
- x = 4, y = 9, z = 10 → 4 + 38 = 42 ✓
- x = 6, y = 9, z = 9 → 6 + 36 = 42 ✓
- x = 8, y = 8, z = 9 → 8 + 34 = 42 ✓
Step 3 — Test each statement.
Statement I: She received a score below 8 on at least one assignment.
Counterexample: x = 8, y = 8, z = 9. All scores ≥ 8. Statement I fails here.
→ Does NOT have to be true.
Statement II: She received a score of 9 on at least one assignment.
Counterexample: x = 2, y = 10, z = 10. No score of 9 anywhere.
→ Does NOT have to be true.
Statement III: She received a score of at least 6 on every assignment.
Counterexample: x = 4, y = 9, z = 10. x = 4 is below 6, yet 4 + 38 = 42 ✓.
→ Does NOT have to be true.
Answer: E — None of the three.
Takeaway: On "Must be True" problems, your goal is counterexamples. The moment you find one valid scenario where a statement fails, cross it off. Never assume the first combination you find represents all possibilities.
06 Apr 2026, 13:04
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[*]If $x = 10$:
$10 + 2(y + z) = 42 \Rightarrow 2(y + z) = 32 \Rightarrow y + z = 16$.
Since $x=10$ and $x$ is the minimum, $y$ and $z$ would also have to be 10. But $10+10 = 20$, not 16. (Impossible)
[*]If $x = 2$:
$2 + 2(y + z) = 42 \Rightarrow 2(y + z) = 40 \Rightarrow y + z = 20$.
The only way for two scores to sum to 20 is if $y=10$ and $z=10$.
Set: $\{2, 10, 10\}$. This works.
[*]If $x = 4$:
$4 + 2(y + z) = 42 \Rightarrow 2(y + z) = 38 \Rightarrow y + z = 19$.
The only way to sum to 19 is $9 + 10$.
Set: $\{4, 9, 10\}$. This works.
[*]If $x = 6$:
$6 + 2(y + z) = 42 \Rightarrow 2(y + z) = 36 \Rightarrow y + z = 18$.
Possible pairs $(y, z)$ that are $\geq 6$: $(8, 10)$ or $(9, 9)$.
Sets: $\{6, 8, 10\}$ or $\{6, 9, 9\}$. Both work.
[*]If $x = 8$:
$8 + 2(y + z) = 42 \Rightarrow 2(y + z) = 34 \Rightarrow y + z = 17$.
Possible pair: $(8, 9)$.
Set: $\{8, 8, 9\}$. This works.
$10 + 2(y + z) = 42 \Rightarrow 2(y + z) = 32 \Rightarrow y + z = 16$.
Since $x=10$ and $x$ is the minimum, $y$ and $z$ would also have to be 10. But $10+10 = 20$, not 16. (Impossible)
[*]If $x = 2$:
$2 + 2(y + z) = 42 \Rightarrow 2(y + z) = 40 \Rightarrow y + z = 20$.
The only way for two scores to sum to 20 is if $y=10$ and $z=10$.
Set: $\{2, 10, 10\}$. This works.
[*]If $x = 4$:
$4 + 2(y + z) = 42 \Rightarrow 2(y + z) = 38 \Rightarrow y + z = 19$.
The only way to sum to 19 is $9 + 10$.
Set: $\{4, 9, 10\}$. This works.
[*]If $x = 6$:
$6 + 2(y + z) = 42 \Rightarrow 2(y + z) = 36 \Rightarrow y + z = 18$.
Possible pairs $(y, z)$ that are $\geq 6$: $(8, 10)$ or $(9, 9)$.
Sets: $\{6, 8, 10\}$ or $\{6, 9, 9\}$. Both work.
[*]If $x = 8$:
$8 + 2(y + z) = 42 \Rightarrow 2(y + z) = 34 \Rightarrow y + z = 17$.
Possible pair: $(8, 9)$.
Set: $\{8, 8, 9\}$. This works.
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