Two machines, X and Y, working together at their repsective constant r

IMO this a tough one if one tries to solve using an algebraic approach.

• Assume the rate of Machine X = \(x\)
  
• Assume the rate of Machine Y = \(y\)

Two machines, X and Y, working together at their respective constant rates for 14 hours, and then machine Y, working alone at its constant rate for an additional 4 hours, can produce a total of m items.

Inference → To produce m items ⇒ Machine X works for 14 hours, and Machine Y works for 18 hours

14x + 18y = Efforts required to produce m items

The time that it takes machine X, working alone at its constant rate, to produce m items is 8 hours less than the time that it takes machine Y, working alone at its constant rate, to produce m items.

Inference → X is faster than Y

Time taken by machine X to produce m items =\( \frac{\text{Efforts required to produce m items} }{\text{Rate of machine X} }\)

Time taken by machine X to produce m items = \(\frac{14x + 18y}{x} = 14 + 18 \frac{y}{x}\)

Time taken by machine Y to produce m items =\( \frac{\text{Efforts required to produce m items} }{\text{Rate of machine Y} }\)

Time taken by machine Y to produce m items = \(\frac{14x + 18y}{y} = 14\frac{x}{y} + 18 \)

Given

\((14\frac{x}{y} + 18) - (14 + 18 \frac{y}{x}) = 8\)

\(14\frac{x}{y} + 4 - 18 \frac{y}{x} = 8\)

\(14\frac{x}{y} - 18 \frac{y}{x} = 4\)

Dividing by 2, we get

\(7\frac{x}{y} - 9\frac{y}{x} = 2\)

If, time taken by machine X = \( X \); Rate of machine X ⇒ \(x = \frac{1}{X}\)

If, time taken by machine Y = \( Y \); Rate of machine Y ⇒ \(y = \frac{1}{Y}\)

Transforming our equation in terms of \(X\) and \(Y\)

\(7\frac{\frac{1}{X} }{\frac{1}{Y} } - 9\frac{\frac{1}{Y} }{\frac{1}{X} } = 2\)

\(7\frac{Y}{X} - 9\frac{X}{Y} = 2\)

Now we can use the options to our advantage. The options provide us the value of \(Y\); we can calculate the value of \(X\) as shown below -



My initial hunch is to try Option E first because in that option \(X\) is a multiple of 7, and \(Y\) is a multiple of 9. Hence, the numerator and denominator cancel out when we substitute the values in our equation. No other option provides that advantage (if I were running on a time crunch, I would choose this as my final answer).

LHS = \(7\frac{36}{28} - 9\frac{28}{36}\)

LHS = \(\frac{36}{4} - \frac{28}{4}\)

LHS = \(9 - 7 = 2\)

RHS = \(2\)

Hence, option E matches.

Option E

avigutman - I would love to see your approach to this question. Can this be solved in a shorter way ?

my algebraic approach is too time consuming, any easier way around this? KarishmaB GMATNinja

I decided to check on the options and thought of the following:

For the most part (14 hrs), both were working together. Then Y worked for another 4 hrs and hence the work got done in 18 hrs. Perhaps if X had continued working with Y for the whole time, the work may have gotten done in approximately 16 hrs. So if either one of them works alone (assuming equal rate of work for them) they would take 32 hrs alone each. But X is faster and takes 8 hrs less. So I could split the time taken by them into 28 hrs and 36 hrs for X and Y respectively.

Mind you, this is all an approximation but I see the figures in the question stem and the options so I would try 36.
If Y takes 36 hrs and X takes 28 alone, in 18 hrs of Y, he will do half the work and in 14 hrs of X, he will do half the work. Hence it works.

Answer (E)

Check this video for Work Rate concepts: https://youtu.be/88NFTttkJmA

GIven: Two machines, X and Y, working together at their respective constant rates for 14 hours, and then machine Y, working alone at its constant rate for an additional 4 hours, can produce a total of m items. The time that it takes machine X, working alone at its constant rate, to produce m items is 8 hours less than the time that it takes machine Y, working alone at its constant rate, to produce m items.

Asked: How many hours does it take machine Y, working alone at its constant rate, to produce m items?

Let the time taken by Machine X & Machine Y to produce m items be x & y hours respectively.

14/x + 18/y = 1

y - x = 8
x = y - 8

y = ?

14/(y-8) + 18/y = 1

A. 28
B. 30
C. 32
D. 34
E. 36: 14/28 + 18/36 = 1

IMO E

­This is so brilliant. I have no idea how I would recognize this in the actual exam though! Probably why it's an 800+

This is the hardest problem I have attempted to date. I did get it correct via the algebraic approach but it took me far far far far too long. Kudos to people who got this question in a reasonable amount of time, I take my hat off to you.

­This is the best response. After going through the other responses it's even clearer. Initially, I also started with the equations 14(x + y) + 4 y = m but wrote X=Y-8 without factoring in the constant 't'
Love this solution

Picking a fake number for work units (m) seems to be a bad strategy here, but I can't understand why. It seems like some work problems benefit from choosing fake numbers but others don't. How can I tell when reading a problem?

nick13
Given: Two machines, X and Y, working together at their respective constant rates for 14 hours, and then machine Y, working alone at its constant rate for an additional 4 hours, can produce a total of m items. The time that it takes machine X, working alone at its constant rate, to produce m items is 8 hours less than the time that it takes machine Y, working alone at its constant rate, to produce m items.

Asked: How many hours does it take machine Y, working alone at its constant rate, to produce m items?
Let the time taken by machine X & machine Y to produce m items be x & y hours respectively

14/x + 14/y + 4/y = 1

x = y - 8 

y = ?

14/x + 18/y = 1
14/(y+8) + 18/y = 1
14y + 18(y+8) = y(y+8)
42y + 18*8 = yˆ2 + 8y
yˆ2 - 34y - 2ˆ4*3ˆ2 = 0
yˆ2 - 36y - 4y - 2ˆ4*3ˆ2 = 0
(y-36) - 4(y-36) = 0
(y-4)(y-36) = 0
y = 4 or 36
Since x = y-4 > 0 ; y = 4 is NOT POSSIBLE

y = 36

IMO E
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Hi,

Don't think your first way is right. Half work will take (t+8)/2 less hours which = t/2 + 4. 

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unicornilove
It is perfectly fine. 
Full work: X= t and Y=t+8
hakf work: X=t/2 and Y=(t+8)/2 or t/2 + 4. You are making a mistake by taking t/2 + 4 as the less time. 
X takes (t/2 + 4) - t/2 = 4 hours less and that is exactly what has been mentioned. 

Thank you @chetan2u....Wonderful solution..

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