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Manager  Joined: 04 Apr 2010
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It takes printer A 4 more minutes than printer B to print 40  [#permalink]

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66 00:00

Difficulty:   85% (hard)

Question Stats: 61% (03:15) correct 39% (03:17) wrong based on 475 sessions

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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?

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

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bhandariavi wrote:
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: $$2a=24$$ --> $$a=12$$ --> $$\frac{1}{12}+\frac{1}{8}=\frac{5}{24}$$.

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where did you get the source of this problem? I got something complete different.

So ok.

A => B+4min = 40 pages
B=> Bmin = 40 pages

(B+4 + B)6 = 50 pages

(2B + 4)6 = 50

12B + 24 = 50

12B = 26

B = 26/12 = 13/6

so....

A = B + 4

A = (13/6) + 4

(13/6 + 6/6 + 6/6 + 6/6 + 6/6)t = 80

(37/6)t = 80

t = 12.97....

someone care to explain this?
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I got this question - unanswered somewhere around page 19-20 in PS Discussion section.
Is this a real GMAT type question?
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Solving work-rate equation (GMAT Club Test M17#20)  [#permalink]

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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!

Solution:
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.
Math Expert V
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Posts: 55732
Re: Solving work-rate equation (GMAT Club Test M17#20)  [#permalink]

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1
Merging similar topics.

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

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

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AndreG wrote:
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)

Hope it helps.
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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.
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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?
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Baten80 wrote:
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
Math Expert V
Joined: 02 Sep 2009
Posts: 55732

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Saurajm wrote:
Baten80 wrote:
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

Check this:

Though it's better to substitute the values rather than get the quadratics and then factor it (refer to my podt above to see how it can be done).

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

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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)
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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.

1/24 + 1/16 = 5/48 Yes.
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Re: It takes printer A 4 more minutes than printer B to print 40  [#permalink]

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

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1
1
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)]

$$\frac {1}{x} + \frac{1}{x+1}$$ = $$\frac {1}{2} + \frac{1}{3} =\frac{2+3}{6}$$ = $$\frac{5}{6}$$
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.
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Re: It takes printer A 4 more minutes than printer B to print 40  [#permalink]

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Hi Bunuel, could you please advice where am I going wrong here?

40/A - 40/B = 4

50/A + 50/B = 6

Is the algebraic translation of the givens correct? I'm getting some weird numbers when I try to solve for 1/A after factorizing the 40 and 50 in each respective equation

Something like 40 (1/A - 1/B) = 4 ---> 1/A - 1/B = 1/40

Then same here 50 (1/A + 1/B) = 6---> 1/A + 1/B = 6/50

2/A = 1/40 + 6/50 = 11/50

Then I get this funny value of 100/11 for A which I believe is incorrect

Thanks a lot
Cheers
J
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Re: It takes printer A 4 more minutes than printer B to print 40  [#permalink]

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jlgdr wrote:
Hi Bunuel, could you please advice where am I going wrong here?

40/A - 40/B = 4

50/A + 50/B = 6

Is the algebraic translation of the givens correct? I'm getting some weird numbers when I try to solve for 1/A after factorizing the 40 and 50 in each respective equation

Something like 40 (1/A - 1/B) = 4 ---> 1/A - 1/B = 1/40

Then same here 50 (1/A + 1/B) = 6---> 1/A + 1/B = 6/50

2/A = 1/40 + 6/50 = 11/50

Then I get this funny value of 100/11 for A which I believe is incorrect

Thanks a lot
Cheers
J

When writing an equation please make sure it's clear what each variable represents there.

If A and B are the rates in the first equation, then the second equation should be (rate of A) + (rate of B) = A + B = 50/6 = (combined rate).
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Bunuel wrote:
Saurajm wrote:
Baten80 wrote:
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

Check this:

Though it's better to substitute the values rather than get the quadratics and then factor it (refer to my podt above to see how it can be done).

Hope it helps.

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

(1/T+4)+(1/T)=5/24....????? Rgds
prasannajeet
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Joined: 02 Sep 2009
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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

(1/T+4)+(1/T)=5/24....????? Rgds
prasannajeet

If T is the rate how is T + (4 days) = (rate) + (time) = (rate) ???
If T is rate, then 1/T = 1/(rate) = (time), while 5/24 = (rate). How is (1/T+4)+(1/T) = (time) + (time) + (time) = 5/24 = (rate)???
_________________ Re: a good one   [#permalink] 19 Feb 2014, 00:34

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