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23 Dec 2025, 21:58
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day 1 fine : 1$
day 2: 1.5$
day 3: 2$
day 4: 2.5$
day 5: 3$
for the sum to be integer (since options only have integral solutions)
either 4 day blocks or 5 day blocks starting from day 1 should be considered - reason in every 4 days blocks or 5 day blocks starting from day 1, there are two 0.5 decimals which would sum to 1
first 4 days sum: 7$
first 5 days sum: 10$
ans: B, E or 7,10
day 2: 1.5$
day 3: 2$
day 4: 2.5$
day 5: 3$
for the sum to be integer (since options only have integral solutions)
either 4 day blocks or 5 day blocks starting from day 1 should be considered - reason in every 4 days blocks or 5 day blocks starting from day 1, there are two 0.5 decimals which would sum to 1
first 4 days sum: 7$
first 5 days sum: 10$
ans: B, E or 7,10
23 Dec 2025, 23:10
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Bunuel
We have to combine the formulas for the sum of an arithmetic sequence with the formula for an arithmetic sequence for this. Let n be the number of days.
\(sum=\frac{n}{2}(1+a_n)\)
\(a_n=1+\frac{1}{2}(n-1)\)
Combined, we get:
\(sum=\frac{n}{2}(1+(1+\frac{1}{2}(n-1)))\)
This will take a bit of simplification, but we end up with:
\(sum=\frac{n(3+n)}{4}\)
So we're looking for an option from column 2, that when multiplied by 4, equals column 1 amount times column 1 amount plus 3.
Through trial and error, we find \(10=\frac{5(5+3)}{4}\) satisfies this. Therefore answers are 5 and 10.
23 Dec 2025, 23:45
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The fine is 1, 1.5, 2, 2.5, ....
Let n be the number of overdue days => The fine of last day = 1+(n-1)*0.5
Total fine = (1+1+(n-1)*0.5)*n/2 = (n/2+3/2)*n/2
Scan the options:
n = 5 => Total fine = 8/2*5/2 = 10
Answer: No. of overdue = 5, Total fine = 10
Let n be the number of overdue days => The fine of last day = 1+(n-1)*0.5
Total fine = (1+1+(n-1)*0.5)*n/2 = (n/2+3/2)*n/2
Scan the options:
n = 5 => Total fine = 8/2*5/2 = 10
Answer: No. of overdue = 5, Total fine = 10
Bunuel
