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

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Two machines, X and Y, working together at their respective constant [#permalink]
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Vaishvii wrote:
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.

Check this video for Work Rate concepts: https://youtu.be/88NFTttkJmA
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Re: Two machines, X and Y, working together at their respective constant [#permalink]
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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
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Two machines, X and Y, working together at their respective constant [#permalink]
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nick13 wrote:
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?

a. 28
b. 30
c. 32
d. 34
e. 36

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 -

Attachment:

Screenshot 2023-10-23 081303.png [ 43.86 KiB | Viewed 13069 times ]

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 ?
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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If we let x = the rate of Machine X and y = the rate of Machine Y. Machine X works for 14 hrs total, while Machine Y works for 18 hours total. So according to the information given in the problem:

14(x + y) + 4 y = m

simplifying we get:

14x + 18y = m

We also are told that Machine X takes 8 hrs less to produce m items. So with that information, we can write the following equation:

(x)(t-8) = y(t) = m

Now check out the first equation. The time difference between the work done by Machine X and Machine Y is 4 hrs, so let's multiple everything by two.

28x + 36y = 2m

If (x)(t-8) = y(t) = m, then (x)(t-8) + y(t) = 2m

With this information, it's safe to say that E) 36 is indeed the correct answer.­

­This is so brilliant. I have no idea how I would recognize this in the actual exam though! Probably why it's an 800+
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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If we let x = the rate of Machine X and y = the rate of Machine Y. Machine X works for 14 hrs total, while Machine Y works for 18 hours total. So according to the information given in the problem:

14(x + y) + 4 y = m

simplifying we get:

14x + 18y = m

We also are told that Machine X takes 8 hrs less to produce m items. So with that information, we can write the following equation:

(x)(t-8) = y(t) = m

Now check out the first equation. The time difference between the work done by Machine X and Machine Y is 4 hrs, so let's multiple everything by two.

28x + 36y = 2m

If (x)(t-8) = y(t) = m, then (x)(t-8) + y(t) = 2m

With this information, it's safe to say that E) 36 is indeed the correct answer.­

­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 = but wrote X=Y-8 without factoring in the constant 't'
Love this solution
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Re: Two machines, X and Y, working together at their respective constant [#permalink]
my algebraic approach is too time consuming, any easier way around this? KarishmaB GMATNinja
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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.
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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?
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
Hi,

Don't think your first way is right. Half work will take (t+8)/2 less hours which = t/2 + 4.
chetan2u wrote:
nick13 wrote:
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?

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

A straight forward logical way
Y takes 8 hours more for entire work. So he will take 8/2 or 4 hrs extra to complete half work.
Here too, he is working for 4 hours extra = X works for 14 hrs and Y works for 18 hours 🧐
So, this must be the time for doing half the work by each. Thus X would take 14*2 and Y would take 18*2 or 36 hours.

Another convenient way is to look at total time they work for.
X works for 14 hrs while Y works for 14+4 or 17 hours.
Total work done by X and Y is $$\frac{14}{X}+\frac{18}{Y}$$.
As X=Y-8 => $$\frac{14}{Y-8}+\frac{18}{Y}=1$$
(I) Now use options => I would use the most friendly option, the one that would divide the numerator as finally we have to get 1.
36 would fit in perfectly as 36-8 or 28 would have common factors with 14 and 36 would have common factors with 18.
And 36 surely fits in properly. If it had not we would have gone for some other value.
OR
(II) you could solve the equation.
$$\frac{14}{Y-8}+\frac{18}{Y}=1$$
$$14Y+18Y-8*18=Y(Y-8)$$
$$Y^2-40Y+8*18=0$$
$$(Y-4)(Y-36)=0$$
Y cannot be less than 8, thus Y is 36

E

­
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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.
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
Thank you @chetan2u....Wonderful solution..
chetan2u wrote:
nick13 wrote:
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?

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

A straight forward logical way
Y takes 8 hours more for entire work. So he will take 8/2 or 4 hrs extra to complete half work.
Here too, he is working for 4 hours extra = X works for 14 hrs and Y works for 18 hours 🧐
So, this must be the time for doing half the work by each. Thus X would take 14*2 and Y would take 18*2 or 36 hours.

Another convenient way is to look at total time they work for.
X works for 14 hrs while Y works for 14+4 or 17 hours.
Total work done by X and Y is $$\frac{14}{X}+\frac{18}{Y}$$.
As X=Y-8 => $$\frac{14}{Y-8}+\frac{18}{Y}=1$$
(I) Now use options => I would use the most friendly option, the one that would divide the numerator as finally we have to get 1.
36 would fit in perfectly as 36-8 or 28 would have common factors with 14 and 36 would have common factors with 18.
And 36 surely fits in properly. If it had not we would have gone for some other value.
OR
(II) you could solve the equation.
$$\frac{14}{Y-8}+\frac{18}{Y}=1$$
$$14Y+18Y-8*18=Y(Y-8)$$
$$Y^2-40Y+8*18=0$$
$$(Y-4)(Y-36)=0$$
Y cannot be less than 8, thus Y is 36

E

­
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Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
chetan2u wrote:
nick13 wrote:
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?

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

A straight forward logical way
Y takes 8 hours more for entire work. So he will take 8/2 or 4 hrs extra to complete half work.
Here too, he is working for 4 hours extra = X works for 14 hrs and Y works for 18 hours 🧐
So, this must be the time for doing half the work by each. Thus X would take 14*2 and Y would take 18*2 or 36 hours.

Another convenient way is to look at total time they work for.
X works for 14 hrs while Y works for 14+4 or 17 hours.
Total work done by X and Y is $$\frac{14}{X}+\frac{18}{Y}$$.
As X=Y-8 => $$\frac{14}{Y-8}+\frac{18}{Y}=1$$
(I) Now use options => I would use the most friendly option, the one that would divide the numerator as finally we have to get 1.
36 would fit in perfectly as 36-8 or 28 would have common factors with 14 and 36 would have common factors with 18.
And 36 surely fits in properly. If it had not we would have gone for some other value.
OR
(II) you could solve the equation.
$$\frac{14}{Y-8}+\frac{18}{Y}=1$$
$$14Y+18Y-8*18=Y(Y-8)$$
$$Y^2-40Y+8*18=0$$
$$(Y-4)(Y-36)=0$$
Y cannot be less than 8, thus Y is 36

E

Hey why we equating the equation to 1? I'm not getting the equation frame efficiency * time = work done­
Re: Two machines, X and Y, working together at their repsective constant r [#permalink]
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