Machine H produces a certain product at a constant rate of 3 dozen

Question

Machine H produces a certain product at a constant rate of 3 dozen units per hour, and machine K produces the same product at a constant rate of 4 dozen units per hour. The two machines produced 77 dozen units during a 14-hour period, and at least one of the two machines was working at any time in that period. What was the least amount of time that the two machines could have worked simultaneously in that period to complete the production of 77 dozen units?

A. 7 hr
B. 7 hr 30 min
C. 8 hr 45 min
D. 10 hr 15 min
E. 11 hr

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Official Answer

A

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bb · Accepted Answer

Wow. What a convoluted question. I don’t think I would’ve read it the correct way if I hadn’t known the OA.

First, you have to calculate how many hours the machines had worked together during the 14 hours to produce 77 units, and it has to be the least so you have to assume the best case scenario. in this case, you would assume that the 4 unit machine is working the entire 14 hours and makes 56 units and then the other 3-unit machine would have to make 21 units, which would take 7 hrs.

So the four unit machine would work 14 hours straight without stopping and the three unit machine would work only half the time for seven hours to complete the 77 unit production.

Frankly, I feel the question could have been clearer and easier to understand. I don’t think GMAT plays tricks like this because I misunderstood the question after reading it the first time. Maybe it’s just me

Posted from my mobile device

sachi-in · Answer

4*h + 7*(14 - h) = 77 => h = 7
products done by K in h hours + products done by K-H togather ( 4 + 3 ) in the remaining hours

Logically  -  To minimize the work done by 2 machines togather, we would try to run the machine that produces the maximum product for maximum time. That would mean running K for as time time as possible to complete all products remaining time we will run both togather.  
Imagine the scenario if 1 machine alone can complete 77 products in 14hrs why use 2 machines ?

Hence run only machine K and K-H togather. Let's say we would run K for h hours then.

chetan2u · Answer

Taking it further from where  bb  left..
Since we are looking for the least time for which both work together, the slowest machine would never work alone.

(I) one hour work 
time taken by faster machine =  \frac{77}{4} , so one hour work of the faster machine =  \frac{4}{77} 
time taken by both machines together =  \frac{77}{4+3} , so one hour work of the faster machine =  \frac{1}{11} 
say both work together for x hours, so faster machine works alone for 14-x h
=>  \frac{x}{11}+\frac{4(14-x)}{77}=1......7x+56-4x=77....3x=21....x=7

(II) Weighted Average method 
 faster machine = 4 units per hour
 both machine = 4+3 or 7 units per hour
Average required = 77/14 or 5.5 units per hour
 hours both work together =  \frac{5.5-4}{7-4}*14=7 
Or we can see 5.5 is exactly half way from 4 and 7, so both will work for equal time or 7 hrs each.

(II) Logical or BB's method 
The faster has to work for entire 14 hrs, thereby making 4*14 or 56 units. 
The remaining have to be made by the slower machine, so time taken =  \frac{77-56}{3}  or 7 hours, meaning for these SEVEN hrs both are working together.

A

gmatophobia · Answer

This is how I approached this question

1) Ignored the dozen → because each of the terms has a dozen associated with it. Example " ...product at a constant rate of 3 dozen units per hour, and... ", " ...constant rate of 4 dozen units per hour. ... ", and " ...77 dozen units during a 14-hour period.. ". I assumed that - Machine H produces ⇒ 3 units/hour Machine K produced ⇒ 4 units/ hour Together they produced ( \frac{77}{14} = 5.5 ) ⇒ 5.5 units / hour 2) Had machines H and K been working simultaneously for one hour they would have produced 4 +3 ⇒ 7 units per hour. However, in actuality, they produced 5.5 units/ hour. This is 1.5 units less than the expected number. Surprisingly 1.5 is half of 3 units, which means that for each hour, machine H produced half the actual number. This can only happen when it works half the time. Hence, in every one hour, machine H worked for only half an hour. In the other half an hour it worked simultaneously with Machine K. In 14 hours, the overlap = \frac{1}{2} * 14 = 7 hours Option A

EMPOWERgmatRichC · Answer

Hi nick13,

While this question is oddly worded (for example, the "dozens" that are mentioned are essentially meaningless; since all of the measures are set in 'dozens' per hour', you can simply ignore that word and just refer to the numbers involved), there are a number of Tactical shortcuts that you can use to eliminate a bunch of the answers and ultimately solve this question with some basic Arithmetic.

First, with Machine rates of 3 units/hour and 4 units/hour - and a TOTAL of 77 units - we're not going to get to that integer result (re: the 77) if we're spending a fraction-of-an-hour (during which BOTH machines are working) that creates a non-integer number of products.

For example, if we spent 30 minutes (re: HALF an hour) at the two rates, then we'd have (1/2)(3) + (1/2)(4) = 3.5 units - and we can't get to 77 if we're just adding a bunch of 3s and 4s to that non-integer. With a little arithmetic, we can logically eliminate Answers B, C and D for that reason.

Even if you did not recognize that shortcut though, you can still TEST THE ANSWERS. Since the prompt asks for the LEAST amount of time that the two machines could have been working, let's start with Answer A: 7 hours.

IF... the two machines were working together for 7 hours, then (7)(3) + (7)(4) = 21+28 = 49 units would be produced.

This means that 77 - 49 = 28 additional units would need to produced over the remaining 14 - 7 = 7 hours of work-time. Can we get to 28 units by adding some combination of individual 3s and 4s? YES.... if all 7 of those remaining hours were producing 4 units/hour. Thus, we have our answer.

Final Answer:

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A

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]
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sachi-in · Answer

After wasting 3 mins I realised it just needs 10s to solve. I think a lot of gmat FE 750+ questions I have faced in mocks are using more tricks in phrasing the problem rather then making the problem difficult.
Concept looks easy but in maximum questions am having difficulty comprehending what they mean. But if we can find the loopholes the questions are very easy, I am unsure about what they will actually ask in the FE ( may be in FE this is going to be the norm ?) .

sayan640 · Answer

I think it would be "k" in place of "h".  bb   Bunuel   gmatphobia

Raman109 · Answer

To find the least amount of time that machines H and K could have worked simultaneously during a 14-hour period to produce 77 dozen units, we'll first determine their combined production rates and then adjust their individual working hours to minimize the simultaneous work time while still meeting the production goal.

Step 1: Determine combined production rate. 
 
 Machine H produces 3 dozen units per hour. 
 Machine K produces 4 dozen units per hour. 
 Together, they produce 3+4=7 dozen units per hour when working simultaneously.  
 Step 2: Calculate the minimum hours required if both machines work simultaneously all the time. 
 
 Total production needed is 77 dozen units. 
 Working together at 7 dozen units per hour, it would take 77/7 ​=11 hours for both machines working simultaneously to produce 77 dozen units.  
 Step 3: Consider the entire 14-hour period. 
 
 If both machines worked together for 11 hours straight, they would produce all 77 dozen, leaving 3 hours where neither machine needs to work. However, the problem states at least one machine must be working at all times.  
 Step 4: Maximize the working hours of one machine over the other to minimize simultaneous working hours. 
 
 Let's assume machine K, which is faster, works alone for some hours to increase total production without H.  
 Step 5: Minimize simultaneous work time. 
 
 Assume machine K works alone for x hours at 4 dozen per hour and then both work together for y hours at 7 dozen per hour. 
 We need 4x+7y = 77. and x + y = 14 (total working hours constraint).  
Solve the equations, y = 7 hours.

KarishmaB · Answer

Rate of H = 3 per hr (ignore dozens since all values are given in dozens. I did check for this the moment I read dozens because it could be a trap) Rate of K = 4 per hr Rate together = 7 per hr (Rates are additive) Required Rate = 77/14 = 5.5 per hr We need to make both work together to give a rate of 5.5 per hr but we need to minimize the together work. Hence we should make K work alone as much as possible (K is faster) 5.5 is right in the middle of 4 and 7 and hence the 14 hrs are split evenly between machine K alone and both machines together. Hence both together worked for at least 7 hrs. Answer (A) Here is a video discussion on work-rate: https://youtu.be/88NFTttkJmA

nguyenpham1309 · Answer

First impression 
As a Work rate question, my first step after reading (quite annoying phrase with "at least" - max min type problem) is to draw a table with Rate - Time - Work.

Because through my first reading, I see all the "dozens" so I just skip it and just thing about the number

1. Imagine the two machines work simultaneously, so H and K can produce 
 3 + 4 = 7 products in 1 hour

2. Under the time pressure, I might not be calm enough to understand the question, so I just think how about two machines work simultaneously for the whole 14 hours. 
 That would be 7 * 14 = 98 products in total. 
Therefore, it is easier to get the idea of what the question is asking now -> The time that two machines work simultaneously, surely fewer than 14 hours (or even 11 hours as the largest AC) to produce 77 products

3. Because of the word "at least", I test the answer, starting from the fewest amount of time - 7 hours
 7 * 7 = 49 products in total

4. The needed number of products left: 77 - 49 = 28 (products). The number of product is an integer number, therefore, the needed number left must be divisible by either 3 or 4. It is 4
 28 : 4 = 7 (hours). So in case H works all the time, K needs only to work 7 hours -> A

5. If I am more generous with time, I can test by assuming a scenario in the opposite way: K works all the time with 4*14 = 56 products, the needed products left: 77 - 56 = 21 -> 21 : 3 = 7(hours)

The hardest thing, I believe, is to understand the question and see what I can do with kind of "min-max" thing. But if I cannot recognize which method should I use, maybe turn back to the above method I am familiar with (or other better solutions that are familiar with you) is a proper choice to do.

Aryaa03 · Answer

I solved it orally in this manner. 
a) If machine A would have worked alone for 14 hours, it would have completed: 14*3 = 42 dozen.
b) If machine B would have worked alone for 14 hours, it would have completed: 14*4 = 56 dozen

Total required to be completed = 77 Dozens

Now, for minimum overlap time, B should work maximum and complete 56.

Remaining : 77 - 56 = 21

This A can complete in 21/3 = 7 hours, and this is the minimum overlap.

Answer-A

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A_Nishith · Answer

a) If machine A would have worked alone for 14 hours, it would have completed: 14*3 = 42 dozen.
b) If machine B would have worked alone for 14 hours, it would have completed: 14*4 = 56 dozen

Total required to be completed = 77 Dozens

Now, for minimum overlap time, B should work maximum and complete 56.

Remaining : 77 - 56 = 21

This A can complete in 21/3 = 7 hours, and this is the minimum overlap.

Answer-A

KarishmaB · Answer

Loved your solution,   bb  !

Dbrunik · Answer

This should not be labled 700-800 difficulty. this is a 500 level question.

H=3 per hour
k = 4 per hour
H+k=7 per hour.

7*14=98 if they work together. too high...

start with the middle answer choice 8 3/4

8*7=56. 77-56=21. 6 remaining hours * rate of 4 for only one machine working=24. This is too much, we can go lower!

try A

7*7=49. 77-49=28. 4*7=28. perfect match.

answer A

Aboyhasnoname · Answer

Bunuel   bb   KarishmaB  If you could review this approach...I'll be grateful..

Here's How I did it..But of course . after nearly spending 15 mins on it....

If at least 1 of the machines is working for 14 hours ..means at least 3 Units are being made at an given time.... Means at least 14 x 3 = 42 Units.... 
Now ...We are left with time duration of 1 unit and 4 units (1 unit when only Y is working and 4 Unit when both are working) to make 77-42 = 35 Units 
Let the time duration be of both working together be = x, So Y alone will be 14-x 
Now 1(14-x) + 4(x) = 35 
14 + 3x = 35 
3x = 21 
x = 7......

Thanks

Bunuel · Answer

That's a correct approach.

Aboyhasnoname · Answer

Thanks Bunuel. You are a star!

yogeshr2 · Answer

I used the answer choices to solve this question quickly -

Since all the units are in dozens, I disregarded the "dozen".
(rate)*(time) = units; (3 units/hr)*x + (4 units/hr)*7 + (7 units/hr)*z = 77 units
x + y + z = 14

The question asks for the least amount of time machines can work simultaneously, as in, the least value of "z". In other words, x and y cannot be negative. The answer choices seem confusing at first, with fractions of an hour, but you can take whole number hours for z and plug and chug. Answer choice A worked coincidentally, but lets assume "B" was the answer. Plugging in Z=7 would give you negative values for either x or y, and plugging in Z=8 would give you valid values for x and y. But Z=8 is not an answer choice, so the answer must be between 7 and 8, thus B in this hypothetical scenario.

Is my thought process flawed or would this be good to do in the actual test?

jaykayes · Answer

The wording of the question is very tricky!

They are NOT asking us to find time taken for both machines working simultaneously to produce 77 dozen units!

If that were the case, the question would have been too simple and the answer would have been far too obvious – 11 hours.
Too good to be true. Always be suspicious. Eliminate option E.

What they actually want us to find is time taken to produce 77 dozen units to minimize the number of hours in that period (i.e., within 14 hours) when both machines need to work simultaneously.

In other words, we want to run only one machine as far as possible, and use the second machine only for the minimum time required so as to complete the total job in 14 hours.

Clearly, we will first try to use the faster machine alone for the entire 14 hours.

In 14 hours, the faster machine will produce 14 × 4 = 56 dozen units.

Therefore, the remaining 77 - 56 = 21 dozen units will need to be produced by the slower machine alone
This would require 21 ÷ 3 = 7 hours.

Machine H produces a certain product at a constant rate of 3 dozen

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Machine H produces a certain product at a constant rate of 3 dozen

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