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

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

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

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

Contact Rich at: [email protected]
www.empowergmat.com

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 ?) .

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

Don;t think the answers are right just based on the wording of the question. Least amount of time = both machines work together. That means their rate would be 1/3 + 1/7. Time would be 77/(1/3 +1/7).
­

­This is an official question, so the answer provided is correct, and the wording is as good as it gets. I suggest reviewing the question and solutions provided more carefully.

4t + 7(14 - t) = 77

t = 7

so A

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. ­

­
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
 ­

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.

bb @bunnel I got this question today in my Mock 5 - Official. Is there a easy way to approcah the question to solve in less than 2 mins?
­

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

Posted from my mobile device

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

the key is understanding what the question is asking

Loved your solution, bb!

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