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# PS. Rate with variables.

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PS: Machine rates [#permalink]  22 Mar 2009, 21:53
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Question Stats:

50% (08:15) correct 50% (00:00) wrong based on 1 sessions
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
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Re: PS: Machine rates [#permalink]  23 Mar 2009, 00:03
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.
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PS. Rate with variables. [#permalink]  26 May 2009, 16:57
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?

A. 4
B. 6
C. 8
D. 10
E. 12
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Re: PS. Rate with variables. [#permalink]  26 May 2009, 18:44
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.
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Re: PS. Rate with variables. [#permalink]  26 May 2009, 18:46
Y takes a days = w
X take a+2 days = w

X + Y takes 3 days = 5/4w
X+Y = 5 / 12 w

w/a + w/a+2 = 5/12w

1/a + 1/a+2 = 5/12

solve u will get 5a^2 - 14a -24 = 0

solving the eqn a = 4 and a = -6/5

a has to be +ve so a = 4

X will take 6 days = w
X will take 12 days = 2w

Ans E

X wi
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Re: PS. Rate with variables. [#permalink]  27 May 2009, 15:55
Thank you very much for your help and good luck with your study, too!

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Re: PS. Rate with variables. [#permalink]  28 May 2009, 16:48
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.
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Re: PS. Rate with variables. [#permalink]  29 May 2009, 17:18
Did you solve the quadratic equation by taking square roots ? or by substituting ?

tkarthi4u wrote:
Y takes a days = w
X take a+2 days = w

X + Y takes 3 days = 5/4w
X+Y = 5 / 12 w

w/a + w/a+2 = 5/12w

1/a + 1/a+2 = 5/12

solve u will get 5a^2 - 14a -24 = 0

solving the eqn a = 4 and a = -6/5

a has to be +ve so a = 4

X will take 6 days = w
X will take 12 days = 2w

Ans E

X wi
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Re: PS. Rate with variables. [#permalink]  29 May 2009, 18:34
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

Although the explanation seems long, the calculations will be faster as they are just simple calculations.
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Re: PS: Machine rates [#permalink]  20 Jul 2009, 16:07
Hi guys!

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.

Thank you very much!
Steve
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Re: PS: Machine rates [#permalink]  23 Jul 2009, 13:28
No replies yet
Maybe I made myself not clear enough. Let me know if you have trouble understanding my thoughts! I will try to be more clear then!

Thanks!
Steve
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Re: PS: Machine rates [#permalink]  17 Aug 2009, 08:24
Here is how I solved it:

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

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Re: PS: Machine rates [#permalink]  17 Aug 2009, 09:23
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h2polo wrote:
Here is how I solved it:

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

I went the following route

let work done by y in a day = w/y
work done by x in a day = w/y+2

now (w/y+w/y+2 )*3 = 5/4w
writing y in terms of x, then y+2 = x and y = x-2

(1/x+1/x-2)*3w = 5/4w

12(2x-2) = 5(x*x - 2x)

solving the quantratic equation , you get x = 6 ; x = 5/4

so X can make w widgets in 6 days , hence 2w in 12 days
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Re: PS. Rate with variables. [#permalink]  17 Aug 2009, 12:43
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

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Re: PS. Rate with variables. [#permalink]  17 Aug 2009, 23:34
traveller10 wrote:
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.

Ya i had done it the similar way. but is there any other simple way to go without forming equation.
Re: PS. Rate with variables.   [#permalink] 17 Aug 2009, 23:34
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