It takes printer A 4 more minutes than printer B to print 40

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?

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


Let's denote the time taken by printer A to print 40 pages as \(a\) minutes. By the given information, printer B can accomplish the same task in \(a-4\) minutes.

The printing rate of A can be calculated as \(rate = \frac{job}{time} = \frac{40}{a}\) pages per minute, and for B it would be \(rate = \frac{job}{time} = \frac{40}{a-4}\) pages per minute.

The combined printing rate of A and B together is therefore \(\frac{40}{a}+\frac{40}{a-4}\) pages per minute. From the stem we also know that both printers together can print 50 pages in 6 minutes, which means that the combined printing rate of A and B together is \(rate = \frac{job}{time} = \frac{50}{6}\). Hence, \(\frac{40}{a}+\frac{40}{a-4} = \frac{50}{6}\).

Dividing this equation by 40, we obtain \(\frac{1}{a}+\frac{1}{a-4}=\frac{5}{24}\). At this point, we can either substitute values from answer choices or solve the resulting quadratic equation. It's crucial to remember that we're asked for the time printer A would take to print 80 pages. Therefore, our answer should be \(2a\) (since \(a\) is the time required to print 40 pages).

Answer D fits the equation:

If \(2a=24\), then \(a=12\).

Substituting \(a\) into the equation gives \(\frac{1}{12}+\frac{1}{8}=\frac{5}{24}\), which holds true. Hence, the solution is \(2a = 24\).


Answer: D
_________________

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.

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?

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)]

\(\frac {1}{x} + \frac{1}{x+1}\) = \(\frac {1}{2} + \frac{1}{3} =\frac{2+3}{6}\) = \(\frac{5}{6}\)
x = 2.

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).

Thus, \(\frac{1}{36} = \frac{4-3}{36} = \frac{4}{36} - \frac{3}{36} = \frac{1}{9} - \frac{1}{12}.\)

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.
_________________

Hello Everyone,

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)


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.
_________________

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*80/50 = 48/5 mins.
Now, I can say that printing 80 pages is 1 complete work.

\(\frac{1}{a} + \frac{1}{(a - 8)} = \frac{5}{48}\)

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.

*Edited a typo*

We have a combined worker problem for which we can use the following formula:

work ( machine 1) + work ( machine 2) = total work completed

We are given that it takes printer A 4 more minutes than printer B to print 40 pages. Recall that rate = work/time, and if we let the rate of printer B = 40/t, in which t is the time for printer B to print 40 pages, then the rate of printer A = 40/(t+4).

We are also given that working together the two printers can print 50 pages in 6 minutes.

Since work = rate x time, we can calculate the work done by each printer.

printer A work = 40/(t+4) x 6 = 240/(t+4)

printer B work = 40/t x 6 = 240/t

Since the two machines print 50 pages together, we can use that value in our combined work formula and determine t.

work (printer A) + work (printer B) = 50 pages

240/(t+4) + 240/t = 50

Multiplying the entire equation by t(t+4) gives us:

240t + 240(t+4) = 50t(t+4)

240t + 240t + 960 = 50t^2 + 200t

50t^2 - 280t - 960 = 0

5t^2 - 28t - 96 = 0

(5t + 12)(t - 8) = 0

t = -12/5 or t = 8

Since t can’t be negative, t = 8. That is, it takes printer B 8 minutes to print 40 pages. Thus it will take printer A 12 minutes to print 40 pages and 24 minutes to print 80 pages (since 80 pages are twice as many as 40 pages).

Answer: D
_________________

1. Let us calculate everything for 80 pages
2. So it takes A 8 more minutes than B to print 80 pages
3. Working together they can print 80 pages in (80/50)*6= 9.6 minutes
4. Let x be the time taken by A. So 1/(x+8) +1/x = 1/9.6.
x=16 and x+8=24. Therefore it takes A 24 minutes to print 80 pages.

It is easier to substitute values in the equation and find the answer..
_________________

chetan2u VeritasKarishma Gladiator59 Bunuel generis

My question is very different from others. The difficulty level of this problem is very high. One is not only supposed to come up with equations, but also somehow quickly understand that he / she needs to plug answer choices (and this also contains a trick with 2x). How does one develop a skill to bail on similar questions? I assume only <1% of all GMAT takers are actually capable of solving this kind of question in under 2.5 minutes...

I wouldn't suggest one to bail on this question! The only thing different about this problem is that "work done" is different in 3 different cases so you need to first bring them all on same platform (by bringing work done to 80 pages in each case). This can be done orally.

Thereafter, you have a simple equation in which you need to plug in.

If one is targeting a lower score, say 600, one can skip this question. There are a few indicators - work is different in each case, when you try to make work done same, time is in fractions. Printer A takes 4 MORE MINS than printer B (not twice as fast etc). The addition of time here will be complicated to handle. Multiplication are division are far easier - does so and so in half the time etc.

VeritasKarishma, is it not 6*80/50 here, "Together, they print 50 pages in 6 mins so they will print 80 pages in 6*50/80 = 48/5 mins", and pls explain me how u arrive at this...i have an explanation as follows: 6 min for 50 pages, so for 1 page min taken will be 6/50 and for 80 pages => 6*80/50...pls let me know ur thoughts on this...

It is and that is what I have written here: https://gmatclub.com/forum/it-takes-pri ... l#p1628806

Check the third highlight:
Together, they print 50 pages in 6 mins so they will print 80 pages in 6*80/50 = 48/5 mins.

Given: 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.

Asked: How long will it take printer A to print 80 pages?

Let the time required by printer A to print 80 pages be x minutes.
The time required by printer A to print 40 pages = x/2 minutes.
The time required by printer B to print 40 pages = x/2 - 4 minutes.

The time required by both printers to print 40 pages = 1/ (2/x + 2/(x-8)) = 1 / (2(2x-8)/x(x-8)) = x(x-8)/2(2x-8)
The time required by both printers to print 50 pages = 5x(x-8)/8(2x-8) = 6
5x(x-8) = 48(2x-8)
5x^2 - 40x = 96x - 384
5x^2 - 136x + 384 = 0
x = 24 minutes

IMO D