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25 Apr 2026, 08:40
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Currently, \(\frac{1}{3}\) of the loans administered by Maple Bank are in arrears. If \(\frac{1}{5}\) of the loans in arrears are made current, at least 30 new loans would have to be granted so that less than \(\frac{1}{4}\) of the loans administered by the bank are in arrears.
What could be the number of Maple Bank loans in arrears?
I. 105
II. 120
III. 135
(A) III only
(B) II and III
(C) I and III
(D) I, II and III
(E) None
What could be the number of Maple Bank loans in arrears?
I. 105
II. 120
III. 135
(A) III only
(B) II and III
(C) I and III
(D) I, II and III
(E) None
25 Apr 2026, 11:56
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The trap here is subtle and I fell for it the first time I saw this type of question. The phrase "at least 30 new loans would have to be granted" isn't just saying 30 loans get added. It's telling you that 30 is the MINIMUM needed. That means 29 loans don't cut it, but 30 does.
Let N = total current loans. Loans in arrears = N/3.
After 1/5 of arrears are made current, remaining arrears = (4/5)(N/3) = 4N/15.
Now here's where the "at least 30" constraint gives us two inequalities:
1. With 29 new loans, the condition fails:
4N/15 >= (N + 29)/4
Multiply through by 60: 16N >= 15N + 435
So N >= 435
2. With 30 new loans, the condition holds:
4N/15 < (N + 30)/4
Multiply through by 60: 16N < 15N + 450
So N < 450
Combining: 435 <= N < 450.
Loans in arrears = N/3, so: 145 <= N/3 < 150.
Check the options:
- I. 105 arrears means N = 315. That's way below 435. Doesn't fit.
- II. 120 arrears means N = 360. Also below 435. Doesn't fit.
- III. 135 arrears means N = 405. Still below 435. Doesn't fit.
Answer: E (None).
Most people set up just one inequality (N < 450) and get I, II, and III all working, which leads them to D. The whole point of "at least 30" is that it pins down a range, not just a ceiling. The lower bound (29 not being enough) is the constraint that eliminates all three options.
Classic Problem Solving inequality question where reading the constraint precisely is the whole game.
Let N = total current loans. Loans in arrears = N/3.
After 1/5 of arrears are made current, remaining arrears = (4/5)(N/3) = 4N/15.
Now here's where the "at least 30" constraint gives us two inequalities:
1. With 29 new loans, the condition fails:
4N/15 >= (N + 29)/4
Multiply through by 60: 16N >= 15N + 435
So N >= 435
2. With 30 new loans, the condition holds:
4N/15 < (N + 30)/4
Multiply through by 60: 16N < 15N + 450
So N < 450
Combining: 435 <= N < 450.
Loans in arrears = N/3, so: 145 <= N/3 < 150.
Check the options:
- I. 105 arrears means N = 315. That's way below 435. Doesn't fit.
- II. 120 arrears means N = 360. Also below 435. Doesn't fit.
- III. 135 arrears means N = 405. Still below 435. Doesn't fit.
Answer: E (None).
Most people set up just one inequality (N < 450) and get I, II, and III all working, which leads them to D. The whole point of "at least 30" is that it pins down a range, not just a ceiling. The lower bound (29 not being enough) is the constraint that eliminates all three options.
Classic Problem Solving inequality question where reading the constraint precisely is the whole game.
