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 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

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
­

Hi,

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

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. 

Hey why we equating the equation to 1? I'm not getting the equation frame efficiency * time = work done­

Harsha

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

All the given information and the question are about m items. So, we can consider m items 1 completed job.

Since X takes 8 fewer hours than Y takes, X is faster than Y.

Working together, they take 14 + 18 = 32 hours to complete the job. So, Y must take more than 32 hours since, working alone, Y would be slower than X and Y working together.

A. 28
B. 30
C. 32
D. 34
E. 36

So, of the choices, only 34 and 36 are possible values for the number of hours Y takes.

Test 34:

If Y takes 34 hours, X takes 34 - 8 = 26 hours to complete 1 job alone.

So, X's rate is 1/26 per hour and Y's rate is 1/34 per hour.

Using the given information on X and Y working together for 14 and 18 hours to complete 1 job, we get the following:

14/26 + 18/34 = more than 1 completed job since 14/26 and 18/34 are both greater than 1/2.

So, Y must go slower than 1/34 per hour and take more than 34 hours.

We're now done with this question after testing just one answer choice because 36 is the only remaining possible value. So, 36 must be the correct answer.

Just to see how 36 works, though, let's test it.

Test 36:

If Y takes 36 hours, X takes 36 - 8 = 28 hours to complete one job alone.

So, X's rate is 1/28 per hour and Y's rate is 1/36 per hour.

14/28 + 18/36 = 1

28 hours for X and 36 hours for Y work.

Correct answer: E

lets first consider second scenario.
Means Y taking 8 Hrs more than X.
So if they divide work evenly then Y will take 4 Hrs more.
Now look at this first scenario. It says exactly that.
Therefore, Total time taken by Machine Y = 2*(14 + 4) = 36

Answer is E.

KarishmaB do you have similar problems in your mind?

Ratios help you see the answer here quickly.

X works for 14 hrs and Y works for 18 hrs. Difference between the hours is 4. They produce m units.
So if X works for 28 hrs and Y works for 36 hrs, they will produce 2m units. Difference between the hours is 8.

We are given that X working alone to produce m items takes 8 hours less than does Y, working alone to produce m items. Since the difference we obtained between the hrs is 8 for total 2m units, it fits. X working for 28 hrs will produce m units and Y working for 36 hrs will produce m units.

Hence answer would be 36.

Answer (A)

If the values were a bit different, you might have needed to use algebra and equations etc. But knowing that GMAT would not expect too many calculations from us, I would always look at the given numbers and whether they fit to give the answer quickly.

Some other interesting work rate discussion:
https://anaprep.com/arithmetic-avoiding ... work-rate/
https://youtu.be/6CT3AXW-QzA
https://anaprep.com/arithmetic-work-rat ... s-and-all/

KarishmaB It looks like I'm not following this question at all. So we have:

(1/x + 1/y) * 14 h + (1/y) * 4 h = m

rate of X: m/(z-8)
rate of Y: m/z

and so on solving equation which gives:

(m/(z - 8) + m/z)×14 + m/z×4 = m


and z = 36 or z = 4

since 4 is not possible, then 36.

Equation will become super nasty. Any shortcuts?

Hi All, I ve resolved it in the following way that is maybe the best one:

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 --> Rx * 14 + Ry*14 + Ry*4 = m . So 14Rx + 18Ry =m. Thus, Rx:Ry = 9:7.
From this we have obtained that Rx=9/7 * Ry.

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 --> 9/7 *Ry * (t-8) = Ry * t --> t=36

This is probably a slower approach but this was the method that initially came to my mind.

14(x + y) + 4y = m → 14x + 18y = m

The time it takes to produce m units is 8 hours less for Machine X
m/x = m/y - 8 → 8 = m/y - mx → 8 = m(x - y)/xy → m = 8xy/(x-y)

Substituting for m
14x + 18y = 8xy/(x-y) → 14x2 - 4xy - 18y2 = 0 → 7x2 - 2xy - 9y2 = 0 → (7x - 9y)(x + y) = 0

Rate cannot be negative so the only solution is 7x = 9y or (7/9)x = y

Since we know X takes 8 hours less to produce m units, an equation can be formed
x(t - 8) = yt → x(t - 8) = (7/9)xt → t - 8 = (7/9)t → (2/9)t = 8 → t = 36