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21 Apr 2026, 05:48
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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:
95%
(hard)
Question Stats:
27% (02:29) correct
73%
(02:19)
wrong
based on 48
sessions
History
Date
Time
Result
Not Attempted Yet
Lamar and Mauricio arrived at a warehouse at noon , and each planned to unload all of the boxes therein by himself. However, they unloaded the boxes together, saving Lamar 5 minutes and Mauricio 45 minutes. What fraction of the boxes did Mauricio unload?
A)\(\frac{1}{10}\)
B)\( \frac{1}{9}\)
C) \(\frac{1}{8}\)
D)\( \frac{1}{4}\)
E)\( \frac{1}{3}\)
A)\(\frac{1}{10}\)
B)\( \frac{1}{9}\)
C) \(\frac{1}{8}\)
D)\( \frac{1}{4}\)
E)\( \frac{1}{3}\)
01 May 2026, 05:13
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Let the time taken by Lamar alone = L minutes
Let the time taken by Mauricio alone = M minutes
They start together at noon and finish together.
Since working together saved Lamar 5 minutes:
Together time = L - 5
Since working together saved Mauricio 45 minutes:
Together time = M - 45
So:
L - 5 = M - 45
Therefore:
L = M - 40
So Lamar is faster than Mauricio.
Now suppose they worked together for T minutes.
Then:
Lamar’s alone time = T + 5
Mauricio’s alone time = T + 45
So their rates are:
Lamar’s rate = 1/(T + 5)
Mauricio’s rate = 1/(T + 45)
Together, they finish the whole job in T minutes:
T/(T + 5) + T/(T + 45) = 1
Now solve:
T(T + 45) + T(T + 5) = (T + 5)(T + 45)
T^2 + 45T + T^2 + 5T = T^2 + 50T + 225
2T^2 + 50T = T^2 + 50T + 225
T^2 = 225
T = 15
So they worked together for 15 minutes.
Mauricio’s alone time would have been:
T + 45 = 60 minutes
Therefore, in 15 minutes, Mauricio unloaded:
15/60 = 1/4 of the boxes
The answer is D.
Let the time taken by Mauricio alone = M minutes
They start together at noon and finish together.
Since working together saved Lamar 5 minutes:
Together time = L - 5
Since working together saved Mauricio 45 minutes:
Together time = M - 45
So:
L - 5 = M - 45
Therefore:
L = M - 40
So Lamar is faster than Mauricio.
Now suppose they worked together for T minutes.
Then:
Lamar’s alone time = T + 5
Mauricio’s alone time = T + 45
So their rates are:
Lamar’s rate = 1/(T + 5)
Mauricio’s rate = 1/(T + 45)
Together, they finish the whole job in T minutes:
T/(T + 5) + T/(T + 45) = 1
Now solve:
T(T + 45) + T(T + 5) = (T + 5)(T + 45)
T^2 + 45T + T^2 + 5T = T^2 + 50T + 225
2T^2 + 50T = T^2 + 50T + 225
T^2 = 225
T = 15
So they worked together for 15 minutes.
Mauricio’s alone time would have been:
T + 45 = 60 minutes
Therefore, in 15 minutes, Mauricio unloaded:
15/60 = 1/4 of the boxes
The answer is D.
01 May 2026, 09:38
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- Lamar's rate of work = l.
- Mauricio's rate of work = m.
- Together, say they worked for time "t".
Work done = (l+m) * t
To do this work alone, what time would Lamar take?
Time taken by Lamar alone = Work/Rate = \(\frac{(l+m) * t }{ l}\)
To do this work alone, what time would Mauricio take?
Time taken by Mauricio alone = Work/Rate = \(\frac{(l+m) * t }{ m}\)
(1) Lamar saved 5 min by working with Mauricio
\(\frac{(l+m) * t }{ l}\) - 5 = t
\(\frac{(l+m) * t }{ l}\) - t = 5
(t) (\(\frac{(l+m)}{l}\) - 1) = 5
(t) (\(\frac{m}{l}\)) = 5
t = 5 (\(\frac{l}{m}\))
(2) Mauricio saved 45 min by working with Lamar
\(\frac{(l+m) * t }{ m}\) - 45 = t
\(\frac{(l+m) * t }{ m}\) - t = 45
(t) (\(\frac{(l+m)}{m}\) - 1) = 45
(t) (\(\frac{l}{m}\)) = 45
t = 45 (\(\frac{m}{l}\))
From (1) and (2)
t = 5 (\(\frac{l}{m}\)) = 45 (\(\frac{m}{l}\))
This boils down to ->
\(l^2\) = 9\(m^2\)
l = 3m.
This is all we need to solve the question.
Fraction of the work done by Mauricio = \(\frac{Mauricio's work }{ Total Work}\) = \(\frac{m(t)}{(l+m)(t)}\)
= \(\frac{m}{(l+m)}\) = \(\frac{m}{(m+3m)}\) = \(\frac{m}{4m}\) = \(\frac{1}{4}\). Choice D.
- Mauricio's rate of work = m.
- Together, say they worked for time "t".
Work done = (l+m) * t
To do this work alone, what time would Lamar take?
Time taken by Lamar alone = Work/Rate = \(\frac{(l+m) * t }{ l}\)
To do this work alone, what time would Mauricio take?
Time taken by Mauricio alone = Work/Rate = \(\frac{(l+m) * t }{ m}\)
(1) Lamar saved 5 min by working with Mauricio
\(\frac{(l+m) * t }{ l}\) - 5 = t
\(\frac{(l+m) * t }{ l}\) - t = 5
(t) (\(\frac{(l+m)}{l}\) - 1) = 5
(t) (\(\frac{m}{l}\)) = 5
t = 5 (\(\frac{l}{m}\))
(2) Mauricio saved 45 min by working with Lamar
\(\frac{(l+m) * t }{ m}\) - 45 = t
\(\frac{(l+m) * t }{ m}\) - t = 45
(t) (\(\frac{(l+m)}{m}\) - 1) = 45
(t) (\(\frac{l}{m}\)) = 45
t = 45 (\(\frac{m}{l}\))
From (1) and (2)
t = 5 (\(\frac{l}{m}\)) = 45 (\(\frac{m}{l}\))
This boils down to ->
\(l^2\) = 9\(m^2\)
l = 3m.
This is all we need to solve the question.
Fraction of the work done by Mauricio = \(\frac{Mauricio's work }{ Total Work}\) = \(\frac{m(t)}{(l+m)(t)}\)
= \(\frac{m}{(l+m)}\) = \(\frac{m}{(m+3m)}\) = \(\frac{m}{4m}\) = \(\frac{1}{4}\). Choice D.
