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Working simultaneously at their respective constant rates, M

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Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 08:58
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Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)
[Reveal] Spoiler: OA
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 11:28
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Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = 800/x ..... 1
a = 800/y ..... 2


Use 2 in 1... we get

b = 800 (y-x) / xy

Finally

Rate of B * time = Work done by B (we want time)

800 (y-x) / xy * t = 800

t = xy / (y-x)
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 06 Dec 2012, 09:06
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Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


Pick some smart numbers for x and y.

Say x=1 hour and y=2 hours (notice that y must be greater than x, since the time for machine A to do the job, which is y hours, must be more than the time for machines A and B working together to do the same job, which is x hours).

In this case, the time needed for machine B to do the job must also be 2 hours: 1/2+1/2=1.

Now, plug x=1 and y=2 in the options to see which one yields 2. Only option E fits.

Answer: E.
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 18 Sep 2013, 05:49
RateA + RateB = 800/x

RateA = 800/y
RateB = 800/z

So --> 800/y + 800/z = 800/x

1/y + 1/z = 1/x --> 1/z = 1/x - 1/y
z = xy/(y-x)
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 04 Oct 2013, 07:31
Jp27 wrote:
Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = 800/x ..... 1
a = 800/y ..... 2


Use 2 in 1... we get

b = 800 (y-x) / xy

Finally

Rate of B * time = Work done by B (we want time)

800 (y-x) / xy * t = 800

t = xy / (y-x)


Yeah, R*T=W is a lengthy way to solve these problems but, I have seen that it is almost a sure shot way to solve most of the problems on this concept. Picking up the smart numbers may be a neat way to solve these questions but it highly depends on the mental state when you are taking the exam.
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 20 Oct 2013, 15:48
A and B : 1/x = 1/800
A alone: 1/y = 1/800
B alone: A and B - A: 1/x - 1/y = 0

solving it: (y - x)/xy => total time it takes is the reciprocal therefore B alone = xy/(y-x)

Answer: E
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 02 Nov 2013, 07:20
Jp27 wrote:
Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = 800/x ..... 1
a = 800/y ..... 2


Use 2 in 1... we get

b = 800 (y-x) / xy

Finally

Rate of B * time = Work done by B (we want time)

800 (y-x) / xy * t = 800

t = xy / (y-x)


I had a problem with this.
(1/A + 1/B) X = 800

(1/A)Y = 800

and when comparing both, I have too many unknowns....
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 03 Nov 2013, 10:27
Expert's post
ronr34 wrote:
Jp27 wrote:
Walkabout wrote:
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)


The above sol is awesome.... but i did it the longer way, algebraically...

rate of A be a and B be b

a + b = 800/x ..... 1
a = 800/y ..... 2


Use 2 in 1... we get

b = 800 (y-x) / xy

Finally

Rate of B * time = Work done by B (we want time)

800 (y-x) / xy * t = 800

t = xy / (y-x)


I had a problem with this.
(1/A + 1/B) X = 800

(1/A)Y = 800

and when comparing both, I have too many unknowns....


That's because you stop on a halfway. Try to continue as suggested by Jp27 in the post you are quoting.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 17 Nov 2013, 01:11
Simplest solution Here-

RA + RB = 1/X

RA = 1/Y

RB = (1/X - 1/Y) = (Y-X)/XY

Time = 1/RB = XY/(X+Y)


I have treated 800 as equivalent to unity(= 1), as it's presence in final answer was trivial, as it will eventually cancel out, taking it unity has make the solution quite Un Complex..
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 29 Mar 2014, 02:50
Another approach:

Rate(A) = 800/y
Rate(A+B) = 800/x

Rate A + Rate B = Rate(A+B)

=> Rate(B) = Rate(A+B) - Rate(A)
= 800(y-x)/xy

Then the GODFATHER equation Rate * Time = Work

800(y-x)/xy * Time = 800

Time = xy/(y-x)

Rgds,
TGC!
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 06 Apr 2014, 15:49
I managed to solve this via plugging in numbers but I went over 3 minutes! With these plug-in-numbers type of problems, I strive to pick simple numbers but there always seems to be one, usually the one i'm solving for, that ends up being a rather complicated number. My question is:

1) I plugged in Rate A and Rate A+B and then solved for time. Should I have plugged in numbers for time directly. Is there a general rule as to what number I should be plugging in?
2) In this case, to keep things super simple, I could have plugged the total Rate to be 8 and Ra and Rb to both be 4. Is it bad practice to choose the same numbers for the individual rates/times?
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Re: Working simultaneously at their respective constant rates, M [#permalink] New post 07 Apr 2014, 00:04
russ9 wrote:
I managed to solve this via plugging in numbers but I went over 3 minutes! With these plug-in-numbers type of problems, I strive to pick simple numbers but there always seems to be one, usually the one i'm solving for, that ends up being a rather complicated number. My question is:

1) I plugged in Rate A and Rate A+B and then solved for time. Should I have plugged in numbers for time directly. Is there a general rule as to what number I should be plugging in?
2) In this case, to keep things super simple, I could have plugged the total Rate to be 8 and Ra and Rb to both be 4. Is it bad practice to choose the same numbers for the individual rates/times?



Just refer to method of Bunuel; he did using plug-ins.

I used same variables available & got correct answer (Had taken 800 = 1 as done by honchos)
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Re: Working simultaneously at their respective constant rates, M   [#permalink] 07 Apr 2014, 00:04
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