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Running at their respective constant rate, machine X takes 2 [#permalink]
22 Mar 2009, 20:53

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Difficulty:

45% (medium)

Question Stats:

61% (03:04) correct
39% (03:35) wrong based on 42 sessions

Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.

Re: PS: Machine rates [#permalink]
22 Mar 2009, 23:03

1

This post received KUDOS

What is the OA?

I got E. 12

Explanation: Suppose Y takes t days to deliver w work, then X takes t+2 days for w work. We can also say that, X works w/(t+2) work in 1 day and Y works w/t work in 1 day. So, if X & Y work together they deliver [w/(t+2)+w/t] work in 1 day. So, X+Y work 3*[w/(t+2)+w/t] in 3 days, which is equal to 5w/4 work. so, 3*[w/(t+2)+w/t] = 5w/4 or, 3*[1/(t+2)+1/t] = 5/4 or, (2t+2)/t(t+2) = 5/12 or, 24t+24 = 5t^2+10t or, 5t^2+-14t-24=0 or, 5t^2-20t+6t-24=0 or, 5t(t-4)+6(t-4) = 0 or,(t-4)(5t+6)=0 so, t=4 or t=-6/5 time can't be a negative value. hence, t=4 days

So, X does w work in t+2 days or in 4+2=6 days hence X does 2w work in 12 days.

PS. Rate with variables. [#permalink]
26 May 2009, 15:57

Could you guys provide your logic? Thank you, ---- (Test 1) Q26: Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

Re: PS. Rate with variables. [#permalink]
26 May 2009, 17:44

let machine y takes n days to produce w ==> x will take n+2 days. one day produce of y = (w)/(n) one day produce of x = (w)/(n+2) in 3 days both working together will produce = (3w)/(n) + (3w)/(n+2) =5w/4 (given) ==> w will get cancel in this equation as common in RHS and LHS ==> 5n^2-14n-24 = 0 ==> n = 4 (ignore the -ve as days cannot be -ve) ==> x one day produce = w/(4+2) ==> x will take 12 days to produce 2w. Ans = 12.

Re: PS. Rate with variables. [#permalink]
29 May 2009, 17:34

TriColor wrote:

BTW, have any of you thought of using substitution method instead?

I would like to see if there is a faster and smarter method of solving this problem. Thank you.

The faster way would save time on solving the quadratic equation.

To get to the point of quadratic equation it would take about 30 seconds. Assuming, (a + 2) is the number of days it takes X to produce w widgets, we need to find 2 (a + 2) = 2a + 4. For every answer choice subtract 4 and divide by two.

Choice a : Eliminate this as answer choice because 2a + 4 will be more than 4

Choice b : \frac{6-4}{2} = 1

Choice c : \frac{8-4}{2} = 2

Choice d : \frac{10-4}{2} = 3

Choice e : \frac{12-4}{2} = 4

Substitute 1,2,3 and 4 into the quadratic. Since 4 satisfies the quadratic, that is the answer. Although the explanation seems long, the calculations will be faster as they are just simple calculations.

sorry to bring this post up again but I need some guidance to beat this problem!

So here what I have trouble understanding:

(1) The text states that the 2 machines combined can produce 5/4w units in 3 hours. So 5/4w is the work output and 3 hours is the time, giving us 5/4w=5/12w*3 (pls correct me if my reasoning is wrong)

(2) Now, we need to set-up the equation for the combined rate:

5/12w=w/t + w/(t+2) (THAT'S WHAT EVERYONE ELSE GOT; I get something different; see below)

5/12w=(5/4w)/t + (5/4w)/(t+2) (See what I did? Since the rate 5/12w is related to the work output 5/4w, we need to take this into consideration when setting up the equation, right?)

I hope someone can trace my wrong reasoning and give me a hint how to solve this kind of problem in the future.

Re: PS. Rate with variables. [#permalink]
17 Aug 2009, 11:43

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?

A. 4 B. 6 C. 8 D. 10 E. 12

let Y = number of days for machine Y to produce w widgets

Rate of Y = w widgets / Y days Rate of X = w widgets / (Y+2) days

Formula: Rate (together) = Rate of Machine X + Rate of Machine Y

(5/4)*w/3 = w/Y+w/(Y+2) or (5/4)/3 = 1/Y+1/(Y+2)

simplify:

5*Y^2-14Y-24=0

then I used the quadratic formula to get Y = 4

therefore,

2w widgets * 1/Rate of X 2w widgets * (Y+2) days / w widgets 2w widgets * 6 days / w widgets = 12 days

Re: PS. Rate with variables. [#permalink]
28 Sep 2013, 09:58

Directly to the equation:

w/t + w/(t+2) = (w)5/12

5t^2 - 14t -24 =0

t=4

For w widgets machine x takes t+2 = 6

For 2w it would take 12

_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: PS. Rate with variables. [#permalink]
29 Sep 2013, 02:18

crunchboss wrote:

There are so many questions here i have only 5 weeks left, can some one advice me how to use this fiorum.

Do questions in the topics in which you are weak.

For example, this weekend I dedicated my time to Rate/Work problems as I was weak in this section and now I am not

_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: Running at their respective constant rate, machine X takes 2 [#permalink]
29 Sep 2013, 09:52

Expert's post

Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets. A. 4 B. 6 C. 8 D. 10 E. 12

For work problems one of the most important thin to know is rate*time=job \ done.

Let the time needed for machine X to produce w widgets be t days, so the rate of X would be rate=\frac{job \ done}{time}=\frac{w}{t};

As "machine X takes 2 days longer to produce w widgets than machines Y" then time needed for machine Y to produce w widgets would be t-2 days, so the rate of Y would be rate=\frac{job \ done}{time}=\frac{w}{t-2};

Combined rate of machines X and Y in 1 day would be \frac{w}{t}+\frac{w}{t-2} (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: 3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4} --> \frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}.

\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12} --> reduce by w --> \frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}.

At this point we can either solve quadratic equation: 5t^2-34t+24=0 --> (t-6)(5t-4)=0 --> t=6 or t=\frac{4}{5} (which is not a valid solution as in this case t-2=-\frac{6}{5}, the time needed for machine Y to ptoduce w widgets will be negatrive value and it's not possible). So t=6 days is needed for machine X to produce w widgets, hence time needed for machine X to produce 2w widgets will be 2t=12 days.

OR try to substitute the values from the answer choices. Remember as we are asked to find the time needed for machine X alone to produce 2w widgets then the answer should be 2t among answer choices: E work - 2t=12 --> t=6 --> \frac{1}{6}+\frac{1}{6-2}=\frac{5}{12}.