SOLUTION

On Monday morning a certain machine ran continuously at a uniform rate to fill a production order. At what time did it completely fill the order that morning?

(1) The machine began filling the order at 9:30 a.m. Clearly insufficient.

(2) The machine had filled 1/2 of the order by 10:30 a.m. and 5/6 to the order by 11:10 a.m.

In 40 minutes from 10:30 a.m. to 11:10 a.m., 5/6 - 1/2 = 1/3 of the job was done. Thus to do the entire job 3*40 minutes = 120 minutes = 2 hours are needed. Since by 10:30 a.m. half of the job was done, therefore for another half 1 more hour is needed. Therefore the entire job was done by 11:30 a.m. Sufficient.

Answer: B.
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On Monday morning a certain machine ran continuously at a uniform rate to fill a production order. At what time did it completely fill the order that
morning?


(1) The machine began filling the order at 9:30 a.m.
(2) The machine had filled 1/2 of the order by 10:30 a.m. and 5/6 to the order by 11:10a.m.

Sol: We need to find by what time the order was completely filled

St 1: Just tells starting time of order feeling. Not sufficient A and D ruled out
St 2: Let the order volume be V

So 1/2 V was completed by 10:30 am and 5/6V by 11: 10am

So in 40 mintues Volume filled (5/6-1/2)V ----> 1/3V so one third volume filled in 40 minutes so full volume will be filled in 120 minutes so remaining volume of 1/6V order will be finished in next 20 mins ie by 11:30am

Ans B

Difficulty level 600 is okay
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On Monday morning a certain machine ran continuously at a uniform rate to fill a production order. At what time did it completely fill the order that morning?

(1) The machine began filling the order at 9:30 a.m.
(2) The machine had filled 1/2 of the order by 10:30 a.m. and 5/6 to the order by 11:10 a.m.


Statement 1

In order to find the time when machine would completely fill the order, we need the following data:
a. Start time or any other time as reference
b. Rate to fill the order

Since, statement (1) does not provide Rate, the statement 1 alone is not sufficient.........(A) (D)

Statement 2

Rate of filling by machine \(= (\frac{5}{6} -\frac{1}{2})\)/Time interval\(= 1/3\) per\(40\) minutes

Given that, 5/6 of the order was filled by 11:10 a.m., the remaining 1/6 of the order would take 20 minutes more. So, the complete order was filled by 11:30 a.m.....Sufficient....(C) (E)

Answer: (B)

There is a typo in the above question, it should be 11:10 AM and not 11:00 AM

1) Insufficient as we do not know the rate

2) We can see that 1/3 of the work was done in 40 minutes, therefore whole whole work was done in 120 minutes or 2 hours ,Since half of the work would be done in 1 hour , so by 10:30, half of the work was done, another half work would take another hour, so by 11:30 the work was completed.
Sufficient
Answer B
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Why cant we solve in this fashion?
If the half of job is done by 10.30, another half will be 1+ hour . So 10.30 + 1 = 11.30 ,hm?

Solution:

We are given that a certain machine ran continuously at uniform rate to fill a production order. We need to determine at what time it filled the order.

Statement One Alone:

The machine began filling the order at 9:30 a.m.

Only knowing what time the machine began filling the order is not enough information to determine at what time the order was completely filled. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

The machine had filled 1/2 of the order by 10:30 a.m. and 5/6 of the order by 11:10 a.m.

Using the information in statement two, we can determine that the machine completed 5/6 - 1/2 = 5/6 - 3/6 = 2/6 = 1/3 of the job in 40 minutes. Since rate = work/time, the rate of the machine is (1/3)/40 = 1/120. We can now determine how many minutes it would take the machine to complete the remaining 1/6 of the job.

time = work/rate

time = (1/6)/(1/120) = 120/6 = 20 minutes.

Thus, the machine needs 20 more minutes to complete the remaining 1/6 of the job. That is, the machine completes the entire job at 11:30 a.m.

Answer: B
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Although a careful reading guarantees choosing the right answer without ANY need of writing,
let´s use this easy problem to practice UNITS CONTROL, one of our method´s most powerful tools!

\(?\,\,:\,\,{\text{time}}\,\,{\text{it}}\,\,{\text{finished}}\,\,{\text{the}}\,\,{\text{order}}\)

\(\left( 1 \right)\,\,{\text{If}}\,\,{\text{it}}\,\,\left\{ \begin{gathered}\\
{\text{takes}}\,\,30\min \,\,\, \Rightarrow \,\,\,? = 10\,\,{\text{a}}{\text{.m}}{\text{.}} \hfill \\\\
{\text{takes}}\,\,90\min \,\,\, \Rightarrow \,\,\,? = 11\,\,{\text{a}}{\text{.m}}{\text{.}} \hfill \\ \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\, \Rightarrow \,\,\frac{{\,\,\left( {\frac{5}{6} - \frac{1}{2} = } \right)\,\,\,\frac{1}{3}\,\,{\text{order}}\,\,}}{{40\,\,\min }}\)

\(\frac{1}{2}\,\,{\text{order}}\,\,\left( {\frac{{40\,\,\min }}{{\frac{1}{3}\,\,{\text{order}}}}} \right) = 60\,\,\min \,\,\,\left( {{\text{for}}\,\,{\text{half - order}}} \right)\,\,\)

\(?\,\, = \,\,10{\text{h}}30\min + 60\min \,\,\,\, \Rightarrow \,\,\,\,{\text{SUFF}}.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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