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It takes printer A 4 more minutes than printer B to print 40 [#permalink]

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03 Aug 2010, 12:13

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It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages?

It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages? a. 12 b. 18 c. 20 d. 24 e. 30

Let the time needed to print 40 pages for printer A be \(a\) minutes, so for printer B it would be \(a-4\) minutes.

The rate of A would be \(rate=\frac{job}{time}=\frac{40}{a}\) pages per minute and the rate of B \(rate=\frac{job}{time}=\frac{40}{a-4}\) pages per minute.

Their combined rate would be \(\frac{40}{a}+\frac{40}{a-4}\) pages per minute. Also as "two printers can print 50 pages in 6 minutes" then their combined rate is \(rate=\frac{job}{time}=\frac{50}{6}\), so \(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\).

\(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\) --> \(\frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}\). At this point we can either try to substitute the values from answer choices or solve quadratic equation. Remember as we are asked to find time needed for printer A to print \(80\) pages, then the answer would be \(2a\) (as \(a\) is the time needed to print \(40\) pages). Answer D works: \(2a=24\) --> \(a=12\) --> \(\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\).

At your solution you write 1/a + 1/(a - 8) = 5/48, with only the number 1 at the denominator, very different from the other solutions that usually use the number 40 at the denominator.

Can you please explain?

Regards.

VeritasPrepKarishma wrote:

siddharthmuzumdar wrote:

Got the quadratic soon but took ages after that to solve the rest of the problem. The substitution, I guess, would take up a lot of time. How to solve within 2 mins?

Do not go up to the quadratic to solve it. The substitution will take a long time. Get your basic equation and then substitute. Let me show you what I mean.

Let me make the work same. We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins) We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages. Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins

Now, make your sum of rates equation:

1/a + 1/(a - 8) = 5/48

Now look at the options and substitute here. First check out the straight forward options. Say, a = 12 1/12 + 1/4 = 4/12 Nope

I will not try 18 and 20 because (18, 10) and (20, 12) doesn't give me 48, the denominator on right hand side.

I will try 24 instead. 1/24 + 1/16 = 5/48 Yes. Answer is 24.

In my solution, variable 'a' represents a different quantity. On top of that, I make the work same so that I can represent it as 1 complete work. This means that I take an extra step before forming the equation. Other solutions in which 40 is in the numerator simplify the equation later. Finally, you can use either method. Both are the same.

Compare my solution with Bunuel's solution to understand this:

Let the time needed to print 40 pages for printer A be a minutes, so for printer B it would be a−4 minutes. For printer A, Work done = 40 pages, time taken = a mins For printer B, work done = 40 pages, time taken = a - 4 mins For both printers together, work done is 50 pages and time taken is 6 mins.

\(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\)

Compare this with:

We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins) We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages. Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins. Now, I can say that printing 80 pages is 1 complete work.

or in other words, if you ignore the 1 complete work concept: \(\frac{80}{a} + \frac{80}{(a - 8)} = \frac{80}{48/5}\) 80 gets cancelled and you are left with the previous equation. _________________

B = x minutes for 40 pages A = x + 4 minutes for 40 pages A does in one in minute of 40 pages = 40/x+4 B does in one in minute of 40 pages = 40/x so, 6[(40/x+4)+40/x]= 50 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 x = 8 B = 8 minutes for 40 pages A = 12 minutes for 40 pages so, A needs 24 minutes for 80 pages. _________________

Got the quadratic soon but took ages after that to solve the rest of the problem. The substitution, I guess, would take up a lot of time. How to solve within 2 mins?

Do not go up to the quadratic to solve it. The substitution will take a long time. Get your basic equation and then substitute. Let me show you what I mean.

Let me make the work same. We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins) We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages. Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins

Now, make your sum of rates equation:

1/a + 1/(a - 8) = 5/48

Now look at the options and substitute here. First check out the straight forward options. Say, a = 12 1/12 + 1/4 = 4/12 Nope

I will not try 18 and 20 because (18, 10) and (20, 12) doesn't give me 48, the denominator on right hand side.

I will try 24 instead. 1/24 + 1/16 = 5/48 Yes. Answer is 24. _________________

Re: Solving work-rate equation (GMAT Club Test M17#20) [#permalink]

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15 Jul 2013, 04:35

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AndreG wrote:

It takes printer A 4 more minutes than printer B to print 40 pages. If working together, the two printers can print 50 pages in 6 minutes, how long will it take printer A to print 80 pages?

(C) 2008 GMAT Club - m17#20

* 12 * 18 * 20 * 24 * 30

I can get up to the part where: \(\frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\)

How do I solve this equation?? I wind up with some pretty complicated squared equation...

Thanks!

Even though it has been discussed before, and there is nothing to add to the solution, there is this one thing, which might just help.Apologies for the lack of brevity.

Imagine a quadratic which reads as : \(\frac {1}{x} + \frac{1}{x+1} =\frac{5}{6}\)

Try calculating this quadratic, it would take atleast say 1 minute. Now, take another look at the quadratic. We see that the RHS has 6 in the denominator. Thus, the LCM of both x and (x+1) should be 6. I say should be, because the RHS denotes a reduced fraction.

Now, think of the first two numbers which have an LCM of 6,and to top it up, they are consecutive \(\to\) 2 and 3[x and (x+1)]

Also, as because there is only ONE correct positive answer for such problems, you can afford the luxury to neglect the second root, which in most cases would turn out to be a negative entity.

Now back to the current problem :

\(\frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\)

Doesn't take long to figure out that the RHS is again a reduced fraction. Now, juggle between numbers which might give an LCM of 24 : (24,1),(6,8),(12,8),etc. But, notice that along with the LCM, the pair also subscribes to another restraint: they are seperated by 4 units. One can quickly recognize that (12,8) is one such pair, and this indeed gives us \(\frac{5}{24}\), when added.

Another way to look at it is by redistributing the numerator (5,in this case) into 2 parts, so that each part, is a factor of the common denominator(24), i.e. 5 \(\to\) (1,4) or (2,3). The first pair will lead to this \(\to \frac{1+4}{24}\) = \(\frac {1}{24} + \frac{1}{6}\) and we know that this is not a valid solution. Trying the other pair, we have something like this \(\to \frac{2+3}{24}\)= \(\frac {2}{24} + \frac{3}{24}\) = \(\frac {1}{8} + \frac{1}{12}\) and this is the correct solution.

Just to get the hang of it, let's try doing another problem :

Say \(\frac{1}{A} - \frac{1}{A+3} = \frac{1}{36}\).

LCM of 36\(\to\) (1,36),(4,9),(9,12),etc. But one pair which is seperated by 3 units \(\to\) (12,9) and it is the answer.

Another way, by redifining the numerator\(\to\) As the numertor in this case is anyways unity, we can think of it as the difference of 2 integers, both being factors of 36 \(\to\) (3,4).

Note: This method is not fail-proof, but knowing that the guys at GMAC design excellent problems,I am sure this method might just come in handy in some way or the other. _________________

Nothing's wrong here. Your equation: [{1/(T+4)} + {1/T}] = 5/24 is a simplified version of the actual equation: \(6*[{40/(T+4)}+{40/T}]=50\).

Which essentially has come from (what has been explained in above posts); i.e. In 1 minute, A can print 40/T+4 pages and B can print 40/T pages, so together in 1 min they can print 40/(T+4) + 40/T pages. Therefore, in 6 minutes, they can print 6*[{40/(T+4)}+{40/T}], which is given to be 50. When you simplify, you get the equation that you have written.

Pardon the repetitions or my lack of brevity

prasannajeet wrote:

Hi Bunuel

I may be crazy but let me understand what is wrong in the following equation and why I am not getting the right answer... Lets rate of B=T so A=T+4

Today I just happened to come across a work and time problem, which had a 95% difficulty level. I immediately thought of a way to do it and realized two ways were already discussed but not this one. All three would fall into category of standard methods.This prompted me to see if a fourth method could be found, a POE(Process Of Elimination) method. And I realized it was faster and brought down the difficulty level by a bit of thinking. The point is a bit of thinking can actually give us clues to try to solve the Qs in lesser time and different ways. But its important for that to try out different methods when we are practicing.

Here I will touch upon all 4 methods that can be used to solve the Q. I would prefer POE if the choices have values spread apart, otherwise any of the remaining three.

Question is:- It takes printer A 4 more minutes than printer B to print 40 pages. Working together, the two printers can print 50 pages in 6 minutes. How long will it take printer A to print 80 pages?

A. 12 B. 18 C. 20 D. 24 E. 30

1)

POE..

yes, you can use the method of elimination to get to the answer... what do we do?... we try and find the MINIMUM and MAXIMUM value it can take and then check for the choices..

first step:-

combined both do 50 pages in 6 minutes. 80 pages can be done in 6*80/50=48/5= 9 minutes 36 seconds..

second step:-

If there is a difference of 4 minutes while doing 40 pages, the difference will increase to 8 minutes for 80 pages.. We have one slower,A and one faster, B... A and B do 80 pages in 9min 36 secs.. Both will do 2times 80 pages in 19 min 12 secs... 2 of As will do 2 times 80 pages in > 19 min 12 secs,A works with a faster machine to clock a time of 19 min 36 secs... or A will do 1 time 80 pages >19 min 12 secs..

so 19 min 12 secs becomes our min time if the two A and B operate at the same speed.., but ofcourse it should be much more, since there is a 8 minutes diff between the two....

this eliminates all except 24 and 30..

third step:-

lets find the max it can be .. the difference between A and B is 8 min.. B is faster than A, so should take less their combined time 19 min 12 secs... Although A will be lesser than but, in no way, it can go beyond 19 min 12 secs. + 8 min, if we take B as 19 min 12 secs, the max possible (we are taking this,although in reality it will be lesser to check max value of A). so max time = 19 min 12 secs. + 8 min = 27 min 12 secs.. this eliminates 30 too.. ans 24 min...

2)

Second method

We can work on 50 pages and 6 min as there are two values avail.. there is a difference of 4 mins in 40 pages.. so, there will be a difference of 5 mins in 50 pages.. let the time taken by A be x min, then B will take x-5 min.. their combined one minute work= 50/x + 50/(x-5)... combined they do 50 pages in 6 min, so they will do 50/6 in one minute.. so \(\frac{50}{x} + \frac{50}{{x-5}} = \frac{50}{6}\).. removing 50 from both sides \(6x-30 + 6x= x^2-5x\)... x^2-17x+30 =0..... x=15 or -2... it cannot be negative, therefore x=15.. now A does 50 pages in 15 min, so it will do 80 pages in 15*80/50 = 24 min..

3)

Another Standard method

(I am copying from the earlier post, so thanks @Bunuel)

Quote:

Let the time needed to print 40 pages for printer A be \(a\) minutes, so for printer B it would be \(a-4\) minutes.

The rate of A would be \(rate=\frac{job}{time}=\frac{40}{a}\) pages per minute and the rate of B \(rate=\frac{job}{time}=\frac{40}{a-4}\) pages per minute.

Their combined rate would be \(\frac{40}{a}+\frac{40}{a-4}\) pages per minute. Also as "two printers can print 50 pages in 6 minutes" then their combined rate is \(rate=\frac{job}{time}=\frac{50}{6}\), so \(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\).

\(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\) --> \(\frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}\). At this point we can either try to substitute the values from answer choices or solve quadratic equation. Remember as we are asked to find time needed for printer A to print \(80\) pages, then the answer would be \(2a\) (as \(a\) is the time needed to print \(40\) pages). Answer D works: \(2a=24\) --> \(a=12\) --> \(\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\).

4)

another way in the same thread

..

We need time taken by printer A to print 80 pages. (Let's say this is 'a' mins) We know A takes 4 more mins for 40 pages so it will take 8 more minutes for 80 pages. Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins

Now, make your sum of rates equation:

1/a + 1/(a - 8) = 5/48

Now look at the options and substitute here. First check out the straight forward options. Say, a = 12 1/12 + 1/4 = 4/12 Nope

I will not try 18 and 20 because (18, 10) and (20, 12) doesn't give me 48, the denominator on right hand side.

I will try 24 instead. 1/24 + 1/16 = 5/48 Yes. Answer is 24...

Finer points

1. there can be various methods to solve the Qs, by substitution, by standard algebra, or by POE.. 2. we have to see how each fits in, but these have to be practiced in abundance to get a feel of each method. _________________

Solving work-rate equation (GMAT Club Test M17#20) [#permalink]

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19 Oct 2010, 15:10

It takes printer A 4 more minutes than printer B to print 40 pages. If working together, the two printers can print 50 pages in 6 minutes, how long will it take printer A to print 80 pages?

(C) 2008 GMAT Club - m17#20

* 12 * 18 * 20 * 24 * 30

I can get up to the part where: \(\frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\)

How do I solve this equation?? I wind up with some pretty complicated squared equation...

Denote \(A\) as the time it takes printer A to print 40 pages. Because working together the two printers can print 50 pages in 6 minutes, they can print 40 pages in \(\frac{40}{50}*6 = \frac{24}{5}\) minutes. Now, it follows from the stem that \(\frac{1}{A} + \frac{1}{B} = \frac{1}{A} + \frac{1}{A - 4} = \frac{5}{24}\) . The only acceptable solution to this equation is 12. So, printer A will print 40 pages in 12 minutes; therefore, it will print 80 pages in 24 minutes.

As for your question: after \(\frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}\) I recommend substitution from answer choices rather then solving quadratic equation. _________________

Re: Solving work-rate equation (GMAT Club Test M17#20) [#permalink]

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19 Oct 2010, 15:16

Bunuel wrote:

Merging similar topics.

As for your question from this point I recommend substitution from answer choices rather then solving quadratic equation.

Ok, thanks. So I would be trying answer choices divided by 2, right? (since answers give A work for 80 pages, and equation is set-up based on 40 pages)

As for your question from this point I recommend substitution from answer choices rather then solving quadratic equation.

Ok, thanks. So I would be trying answer choices divided by 2, right? (since answers give A work for 80 pages, and equation is set-up based on 40 pages)

Got the quadratic soon but took ages after that to solve the rest of the problem. The substitution, I guess, would take up a lot of time. How to solve within 2 mins? _________________

B = x minutes for 40 pages A = x + 4 minutes for 40 pages A does in one in minute of 40 pages = 40/x+4 B does in one in minute of 40 pages = 40/x so, 6[(40/x+4)+40/x]= 50 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 x = 8 B = 8 minutes for 40 pages A = 12 minutes for 40 pages so, A needs 24 minutes for 80 pages.

Can you explain this?

5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0

is there any faster way to solve such kind of quadratic expressions where "a" is different than 1??? Thanks

B = x minutes for 40 pages A = x + 4 minutes for 40 pages A does in one in minute of 40 pages = 40/x+4 B does in one in minute of 40 pages = 40/x so, 6[(40/x+4)+40/x]= 50 5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0 x = 8 B = 8 minutes for 40 pages A = 12 minutes for 40 pages so, A needs 24 minutes for 80 pages.

Can you explain this?

5x^2 - 28x - 96 = 0 (x - 8)(x+12) = 0

is there any faster way to solve such kind of quadratic expressions where "a" is different than 1??? Thanks

Re: It takes printer A 4 more minutes than printer B to print 40 [#permalink]

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12 Feb 2012, 02:08

Dear Surajm For this equation : 5x^2 - 28x - 96 = 0 The central term (-28) should be written is such a way which is combination of (5 * 96) [First term multiplied by last term] . 96*5 Can be factored as 8*2*6*5 (Prime factor will be 2*2*2*2*3*2*5)

Now -28 can be written as (-40 + 12)

-40 came from 8*5 from the bold factor above 12 came from 6*2 from the bold factor above

Hence 5x^2 - 28x - 96 = 0 5x^2 - 40x + 12x - 96 =0 5x(x-8) +12 (x-8) = 0 (5x +12)(x-8) = 0 ==> x = 8 (Other value of x will be negative, hence avoid it)

Re: It takes printer A 4 more minutes than printer B to print 40 [#permalink]

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14 Nov 2012, 00:46

Rate of Printer A + Rate of Printer B = Rate Together \(\frac{40}{a}+\frac{40}{a-4}=\frac{50}{6}\) Simplify further: \(\frac{1}{a}+\frac{1}{(a-4)}=\frac{5}{24}\)

The answer choices are equal to 2a = n.

Test all the answer choices and you will see D as the only one fulfilling the rate equation above.

D. 2a = 24 Thus a = 12 \(\frac{1}{12}+\frac{1}{(12-4)}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\)

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