Working alone at its own constant rate, a machine seals k cartons in 8

Question

Working alone at its own constant rate, a machine seals k cartons in 8 hours, and working alone at its own constant rate, a second machine seals k cartons in 4 hours. If the two machines, each working at its own constant rate and for the same period of time, together sealed a certain number of cartons, what percent of the cartons were sealed by the machine working at the faster rate?

A. \(25%\)

B. \(33\frac{1}{3}%\)

C. \(50%\)

D. \(66\frac{2}{3}%\)

E. \(75%\)

A.  25\%

B.  33\frac{1}{3}\%

C.  50\%

D.  66\frac{2}{3}\%

E.  75\%

KarishmaB · Accepted Answer

or think of it this way:
First machine seals k cartons in 8 hours.
Second machine seals k cartons in 4 hours i.e. 2k cartons in 8 hours.

If they both are working for the same period of time i.e. 8 hours, they together seal 3k cartons. Faster machine (second one) seals 2k out of these 3k so it seals 2/3 = 66.66% of the total cartons.

jeeteshsingh · Answer

Answer is D...

1st Machine:
8 hrs --> K 
1 hr ---> K/8

2nd Machine:
4hrs --> K
1 hr ---> K/4

Both Machines together:
1 hr --> K/4 + K/8 = 3K/8

Faster Machine (2nd) %  = \frac{K/4 }{3K/8} * 100= \frac{K*8}{4*3K} * 100 = \frac{2}{3} * 100 = 66\frac{2}{3} %

bhushan252 · Answer

work will be done A(8hrs):B(4hrs) = 1:2
for first 2 hrs = 2 x (1/8+1/4) = 3/4 work is done
remaining work 1/4th will be done in 1:2 ratio
thus A will do 1/12 th work
B will do 1/6 work

Hence fatser m/c B will do = 1/4+1/4+1/6 = 2/3 =  66 2/3%

Bunuel · Answer

No need for lengthy calculations : since the second one works twice as fast as the first, working together second will do 2/3 of the job (their ratio 1/2).

Answer: D.

srini123 · Answer

if someone wants to double check and keep things simple, lets assume that they work for 8 hours and k=10, so the guy who takes 8 hours complets 10 and the guy who takes 4 hours complets 20 cartons

so faster guy % = 100* 20/30 = 66 2/3 %

R2I4D · Answer

If I am understanding this correctly, you could do the following:

Let's make  k  = 32 cartons. This means Machine A is doing this at a rate of 32 cartons/8 hours = 4 cartons/hour. Likewise, Machine B is working at a rate of 32 cartons/4 hours = 8 cartons/hour.  Machine B is working at a faster rate!

Let's multiply them by the same time to see their output; let's use 10 hours.

Machine A: (4 cartons per hour)*(10 hours) = 40 cartons
Machine B: (8 cartons per hour)*(10 hours) = 80 cartons
Total: 120 cartons

Now it's just a percentage of the total.

Machine A is is doing 40/120 = 1/3 of the work, and  Machine B is doing 2/3, or a little over 66% .

Answer: D

vinayaksatapute · Answer

Thanks for the update.

I have thought a general way.

Lets X machine seal K cartons in 8 hours : K/8

Lets Y machine seal K cartons in 4 Hours : K/4

Keeping in mind that they both sealed same period of time - Machine Y seals 2K cartons in 8 hours.

so K + 2K = 3K total carton sealed.
So machine Y sealed 2K/3K cartons that's 66%

mba4viplav · Answer

I solved it this way:

Step 1:

In one hour, the amount of work completed by Machine 1 is 1/8 and the amount of work completed by Machine 2 is ¼ i.e. Machine 2 works twice as fast as Machine 1.

(You can get the answer in Step 1 only. Anyhow, you will understand how to get the answer in the Step 1 by using a shortcut in Step 3)

Step 2:

In two hours, the amount of work completed by Machine 1 is 2/8 and the amount of work completed by Machine 2 is 2/4.

⇨ Total work completed in 2 hours by 2 machines = (2/8)+(2/4) => (1/4) + (1/2) => 0.25+0.50 = 0.75.
⇨ So the total amount of work completed by both the machines is 75 %. Out of the 75%, Machine 1 completes 50% of the work and Machine 2 completes 25% of the work.

Step 3:

Only 25% of the work is left. So the remaining work has to be divided between Machine 1 and Machine 2. As we know, Machine 2 can do the work twice as fast as Machine 1.

For example, if Machine 1 can do 10% of the work, Machine 2 can do 20% of the work in the same time. So, together they will complete 30% of the work. So from this example, we can understand that any work can be divided into 3 parts, and if Machine 1 completes 1 part of the work, and Machine 2 completes 2 parts of the work in the same time, irrespective of the quantum of work.

Now, the pending work is 25%. Let’s divide the work into 3 equal parts i.e. 8.33% per part. So, Machine 1 will complete 8.33% of the work whereas Machine 2 will complete 16.66% of the work, which is remaining.

So the total work completed by Machine 1 = 25%+8.33% = 33.33%
The total work completed by Machine 2 = 50% + 16.66% = 66.66%

So, the answer is (D)

We can apply Step 3 in Step 1 to get the solution faster, avoiding all the unnecessary calculations of Step 2 and Step 3.

amit2k9 · Answer

if each one has worked for 8 hrs then number of cartoons = k + 2k = 3k.
Mac B produces 2k of these.

hence % = 2k/3k = 66.66%

MHIKER · Answer

I solved the problem directly
Rate one machine = k/8
Rate of the other machine = k/4
Total k/8+ k/4 = 3k/8
Faster seal to total seal = k/4 x 8/k = 66.33
Ans. D

GyanOne · Answer

The second machine seals twice the number of cartons in the same time, so it is twice as fast. Therefore it will do twice the work that the first machine does. Divide 100% into two parts such that one is double the other to get (D) as the answer.

EMPOWERgmatRichC · Answer

Hi All,

Even if you don't see the immediate ratios involved, you can still answer this question rather easily by TESTing VALUES. Here's how:

We're told about 2 machines:

Machine A can produce K cartons in 8 hours
Machine B can produce K cartons in 4 hours

Let's TEST K = 2

So....
Machine A = 2 cartons every 8 hours
Machine B = 2 cartons every 4 hours

We're told that each machine works on its own for the SAME amount of time.

Let's say they both work for 8 hours. This means...

Machine A seals 2 cartons
Machine B seals 4 cartons
Total = 6 cartons

The question asks what ratio of the cartons the faster machine sealed. Machine B is the faster machine, and it sealed 4/6 of the cartons.

4/6 = 2/3 = 66 2/3%

Final Answer:  D

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

PareshGmat · Answer

Rate of Machine A  = \frac{k}{8}

Rate of Machine B  = \frac{k}{4}

Combined rate of A & B  = \frac{k}{8} + \frac{k}{4} = \frac{3k}{8}

Say they work for 1 hour

Work done by faster machine (Machine B)  = \frac{k}{4} * 1 = \frac{k}{4}

Combined work done in 1 hour  = \frac{3k}{8} * 1 = \frac{3k}{8}

Percentage work of faster machine  = \frac{\frac{k}{4}}{\frac{3k}{8}} * 100 = \frac{200}{3} = 66.66%

Answer = D

pacifist85 · Answer

I used the same aproach as Paresh. So, after finding the comdined rate as 3k/8, I assumed that they work for 1 hour. So, the individual work should be the same as the individual rates, i.e. k/8 for the slower one and k/4 for the faster one. It is then (k/4)/(3k/8).

What I didn't understand in the other approaches is why we used 8 hours as the amount of hours they worked for. I guess, we could also have chosen 4 hours or any other amount of hours?

EMPOWERgmatRichC · Answer

Hi pacifist85,

You are correct - we could use ANY number of hours when TESTing VALUES. However, since the prompt offers us two rates that are based on 8 hours of work and 4 hours of work, respectively, choosing 8 hours allows us to make the calculations as simple as possible. In many Quant questions on Test Day, the specific numbers that appear in the prompt (and in the Answer choices) can help you to make choices that will get you to the correct answer in the fastest/easiest way possible.

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

pacifist85 · Answer

Thank you Rich.

I was just wondering the reason why, since - as they were supposed to be working for  some  time - my intuition would be to use 1 hour, so I wouldn't have to calculate anything.

As I am not used to doing quick calculations and haven't achieved to immediately see the connection between numbers yet (except for finding the LCM which I have managed to do almost instantly!!!!!) I was wondering if there was a reasoning I should know about and can use during the actual test.

Thank god, I only need a 650, which is not really a low score for a psychologist to achieve... I am around the 600ts at the moment and can do most of the problems. However, calculations and time are the problem with me...

So, this is helpful! Thanx!

EMPOWERgmatRichC · Answer

Hi pacifist85,

Here's an interesting "drill" that you can do to help develop this skill. Go back and review all of the past Quant questions that you've done and look for situations in which TESTing VALUES is applicable. When you come across those questions, look for the various "clues" in the prompt that would help you make a smart choice for the value(s) that you'd pick. Look for what's listed in the answers, the wording/descriptions in the prompt, any fractions that you're given, etc. For example, when I see the % sign in the answers, my first thought is that I might be able to TEST the number 100 in this question... In this way, you'll be training to not only spot the potential uses for this tactic, but you'll also be training to test the values that will make solving the problem most efficient.

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

JeffTargetTestPrep · Answer

Since the second machine can seal k cartons in 4 hours, it can seal 2k cartons in 8 hours, whereas the first machine can seal only k cartons in 8 hours. So the second machine is twice as fast as the first machine and thus it will seal twice as many cartons as the the first machine. Therefore, when working together, the second machine (the faster machine) will finish ⅔ or 66⅔% of the work whereas the first machine (the slower machine) will finish ⅓ or 33⅓% of the work.

Alternate Solution:

In 8 hours, the slower machine will seal k cartons, and the faster machine will seal 2k cartons. In total, k + 2k = 3k cartons will be sealed, and, of these cartons, the faster machine will seal 2k/3k = 2/3 = 66 ⅔% of them.

Answer: D

Basshead · Answer

Since the question is asking for percentage, we can use ratios.

Machine A seals k cartons at a rate of 1/8
Machine B seals k cartons at a rate of 1/4

1/8 : 1/4
2 : 1
 66.6% : 33.3%

Answer is D.

Working alone at its own constant rate, a machine seals k cartons in 8

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Working alone at its own constant rate, a machine seals k cartons in 8

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