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15 Apr 2026, 11:23
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
42% (02:22) correct
58%
(03:13)
wrong
based on 19
sessions
History
Date
Time
Result
Not Attempted Yet
Belinda, Carlota, and Darcy each shelled the same number of kilograms of shrimp at constant rates. When Belinda finished, Carlota had 20 kilograms left to shell and Darcy had 22 kilograms left to shell, and when Carlota finished, Darcy had 10 kilograms left to shell. How many kilograms of shrimp did the three workers shell in total?
(A) 75
(B) 81
(C) 90
(D) 105
(E) 120
(A) 75
(B) 81
(C) 90
(D) 105
(E) 120
15 Apr 2026, 14:23
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Answer is (A) 75.
This is a Work and Rate problem, but the real trap is treating it like a simple ratio problem without properly tracking what happens across two separate time intervals.
Let T = the total amount each person shells (all three shell the same amount).
1. When Belinda finishes, figure out the rate ratios.
Time Belinda takes = T / r_B. In that same time, Carlota shells (T - 20) kg and Darcy shells (T - 22) kg.
So: r_C / r_B = (T - 20) / T, and r_D / r_B = (T - 22) / T
2. Use the second time interval.
After Belinda finishes, Carlota has 20 left. When Carlota finishes those 20, Darcy (who had 22 left) now has 10 left. So Darcy shelled 12 kg while Carlota shelled 20 kg.
That gives us: r_D / r_C = 12 / 20 = 3 / 5
3. Solve for T.
r_D / r_C = (T - 22) / (T - 20) = 3 / 5
5(T - 22) = 3(T - 20)
5T - 110 = 3T - 60
2T = 50
T = 25
4. Total = 3 x 25 = 75 kg.
Quick sanity check: set r_B = 1. Then r_C = 1/5, r_D = 3/25. Belinda takes 25 time units. In that time: Carlota shells 5 kg (25 left at start minus 5 = 20 left - check), Darcy shells 3 kg (25 left minus 3 = 22 left - check). Then Carlota takes 100 more time units to finish her 20 kg. In that time, Darcy shells 12 more, leaving exactly 10. Everything checks out.
The trap most people fall into: they see "20 left" and "10 left" and set up a 2:1 ratio directly without accounting for the time difference between the two intervals. Classic Work and Rate misdirection - the question is testing whether you can set up two separate time windows correctly.
This is a Work and Rate problem, but the real trap is treating it like a simple ratio problem without properly tracking what happens across two separate time intervals.
Let T = the total amount each person shells (all three shell the same amount).
1. When Belinda finishes, figure out the rate ratios.
Time Belinda takes = T / r_B. In that same time, Carlota shells (T - 20) kg and Darcy shells (T - 22) kg.
So: r_C / r_B = (T - 20) / T, and r_D / r_B = (T - 22) / T
2. Use the second time interval.
After Belinda finishes, Carlota has 20 left. When Carlota finishes those 20, Darcy (who had 22 left) now has 10 left. So Darcy shelled 12 kg while Carlota shelled 20 kg.
That gives us: r_D / r_C = 12 / 20 = 3 / 5
3. Solve for T.
r_D / r_C = (T - 22) / (T - 20) = 3 / 5
5(T - 22) = 3(T - 20)
5T - 110 = 3T - 60
2T = 50
T = 25
4. Total = 3 x 25 = 75 kg.
Quick sanity check: set r_B = 1. Then r_C = 1/5, r_D = 3/25. Belinda takes 25 time units. In that time: Carlota shells 5 kg (25 left at start minus 5 = 20 left - check), Darcy shells 3 kg (25 left minus 3 = 22 left - check). Then Carlota takes 100 more time units to finish her 20 kg. In that time, Darcy shells 12 more, leaving exactly 10. Everything checks out.
The trap most people fall into: they see "20 left" and "10 left" and set up a 2:1 ratio directly without accounting for the time difference between the two intervals. Classic Work and Rate misdirection - the question is testing whether you can set up two separate time windows correctly.
