Machine A takes 2 more hours than machine B to make 20 widgets. If wor : Problem Solving (PS)

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

09 Sep 2015, 03:12

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Machine B takes T hours, Machine A takes=T+2 hours
In 1 hours, they can make= (20/T+20/T+2)
Therefore, in 3 hours, they can make= 3* (20/T+20/T+2)
Solving for T, T=4, T+2=6
Therefore, if A can takes 6 hours to make 20 widgets, he will take 6*40/20=12 hours to make 40 widgets
Answer E

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

09 Sep 2015, 05:49

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Solution:

Let x and y be the no. of widgets made in one hour by A and B and n be the no. of hours taken by B to complete 20 widgets.
Then, ny=20 and (n+2)x=20 ==> y=20/n and x=20/(n+2) -->(1)
Given, 3(x+y) = 25 ==> 3*20(1/n + 1/(n+2)) = 25.
Solving, we get n=4

So, x=20/6.

Time taken for machine A to make 40 widgets = 2(Time taken for machine A to make 20 widgets) = 2(6) =12

Option, E.

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

09 Sep 2015, 10:39

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

CONCEPT: Calculate one hour work of everyone to do further calculation

Let,
Time taken by Machine A to make 20 widgets = t hours
i.e. Time taken by Machine B to make 20 widgets = (t-2) hours

Calculating One hour work of each of them

Widgets made by Machine A in t Hour = 20
Widgets made by Machine A in 1 Hour = 20/t

Similarly,
Widgets made by Machine B in 1 Hour = 20/(t-2)

i.e. Widgets made by Machine A and B together in 1 Hour = (20/t) + [20/(t-2)] = 20(2t-2)/t(t-2)

But (Given),
Widgets made by Machine A and B together in 3 Hour = 25
i.e. Widgets made by Machine A and B together in 1 Hour = 25/3

i.e. 20(2t-2)/t(t-2) = 25/3
i.e. 4(2t-2)/t(t-2) = 5/3
i.e. 12(2t-2)=5t(t-2)
i.e. 24t-24 = 5t^2 - 10t
i.e. 5t^2 - 34t + 24 = 0
i.e. 5t^2 - 30t - 4t + 24 = 0
i.e. (t-6)(5t-4) = 0
i.e. t = 6 or 4/5
but 4/5 is not possible as (t-2) will be Negative in that case
therefore, t=6

i.e. A takes 6 hours to make 20 widgets
i.e. A will take 12 hours to make 40 widgets

Answer: option E

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

09 Sep 2015, 15:05

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

took 4.5 mins but got it finally

Rate of B = 20/t
Rate of A = 20/(t+2)

for 25 widgets, time taken by B = 5t/4 = T1
for 25 widgets, time taken by A = 5(t+2)/4 =T2

using time for both working together as T1*T2/(T1+T2) = 3 and substituting T1 and T2 from above equations, we get a quadratic equation.
solving this: t = 4 hrs.

so, time for A to produce 20 widgets = 6 hrs
for 40 widgets it takes double the time so 12 hrs.

E

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

19 Mar 2016, 14:46

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

i approached this one by plugging in numbers...started with C.
if 40 are made in 8 hours, then 20 are made in 4 hours. so time of A is 4, and time of B is 2.
rate together: 20/4 + 20/2 = 5+10 = 15. so in 1 hour, together make 15 widgets. in 3 hours = 45. way too much. we can eliminate right away C, B, and A - because B and A reduces the time - the total # of widgets made will be even higher.
now between D and E -> try only one ..if it doesn't work, then the other one is the answer.
i picked E:
12h to make 40 widgets, and 6 hours to make 20. this is the time of A. time of B=4 hours.
20/6 + 20/4 = 10/3 + 20/4
find LCM of 3 and 4 = 12. multiply first by 4, and second by 3:
40+60/12 = 100/12
divide by 4:
25/3
so this is the rate given.

E is the correct answer

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

22 Nov 2016, 00:14

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Time actually worked / time it takes to do the complete job = 1
It takes 4.8 hrs together to complete 40 widgets.
A takes 4 hours more than B to complete 40 widgets
4.8/A + 4.8/(A-4) = 1
Substitute and find the answer.
A=12

B

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

14 Aug 2017, 11:02

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I did the same calculations with solving a quadratic equation and it took me a while
There is a kind of shortcut which can be better option taking into account a tight GMAT time
Machine B rate 20/t
Machine A rate 20/(t+2)

let's write an equation to produce 25 widgets in 3 hours
(20/t + 20(t+2))* 3 = 25
60/t+ 60/(t+2) = 25

from now we can play a plug and guess game, we know that B has better performance so what number can be t to solve the equation for 25.
My second try was 4 and 6. Took me about a minute or a bit more for backward checking

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

04 May 2018, 23:02

VeritasPrepKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets. So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Dear VeritasPrepKarishma,

If all the options were more than 8.8, how would we solve this problem using this method? Should we test all the options?

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

06 May 2018, 05:58

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Mehemmed wrote:

VeritasPrepKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets. So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Dear VeritasPrepKarishma,

If all the options were more than 8.8, how would we solve this problem using this method? Should we test all the options?

\(\frac{1}{A} + \frac{1}{A + 4} = \frac{1}{4.8}\)

\(\frac{1}{A} + \frac{1}{A + 4} = \frac{5}{24}\)

Now note that if A is 10, A + 4 = 14. LCM would be 70 which is not a multiple of 24 so let's try another value.
If A = 12, A + 4 = 16. LCM would be 48, a multiple of 24 which is quite possible. Check it out.
and so on...

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

22 Jun 2020, 14:31

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VeritasKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets.

So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Hi VeritasKarishma

I did what you do till the red.

Then my logic was :

--- Given, Combined A + B can make 40 widgets in 4.8 hours
--- Then individually, if A and B were working at the same rate, then A and B would be able to complete 40 widgets at 9.6 hours (4.8 * 2) and B would be able to complete the job in 9.6 hours (4.8 hours * 2)

But we know A takes 4 hours more than B in terms of time when making 40 widgets

Hence the range would be 4 hours

Hence B is then taking 7.6 hours whereas A is taking 11.6 hours (the range is 4)

11.6 hours is closest to E and hence i chose E

Thoughts ?

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

22 Jun 2020, 22:00

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jabhatta@umail.iu.edu wrote:

VeritasKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets.

So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Hi VeritasKarishma

I did what you do till the red.

Then my logic was :

--- Given, Combined A + B can make 40 widgets in 4.8 hours
--- Then individually, if A and B were working at the same rate, then A and B would be able to complete 40 widgets at 9.6 hours (4.8 * 2) and B would be able to complete the job in 9.6 hours (4.8 hours * 2)

But we know A takes 4 hours more than B in terms of time when making 40 widgets

Hence the range would be 4 hours

Hence B is then taking 7.6 hours whereas A is taking 11.6 hours (the range is 4)

11.6 hours is closest to E and hence i chose E

Thoughts ?

The logic is fine here when the options are not that close. Think about it - if the options had 11.5, would you choose 11.5 or 12?
Note that the average time taken would be harmonic mean of the two individual time taken and harmonic mean is less than arithmetic mean if the numbers are different. So when the numbers are 8 and 12, their arithmetic mean will be 10 but harmonic mean will be less than 10.
But if the numbers are say 7.5 and 11.5, the harmonic mean will be less than 9.5 hence 11.5 will not be possible.

Approximating is great but it becomes error free only when you know at which side your error lies. Here, it is perfectly reasonable to say that time taken by both individually, if equal would be 9.6 hrs and since there is a range of 4, the time taken by A will be more than 11.6 hrs.

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

23 Jun 2020, 07:13

VeritasKarishma wrote:

jabhatta@umail.iu.edu wrote:

VeritasKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets.

So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Hi VeritasKarishma

I did what you do till the red.

Then my logic was :

--- Given, Combined A + B can make 40 widgets in 4.8 hours
--- Then individually, if A and B were working at the same rate, then A and B would be able to complete 40 widgets at 9.6 hours (4.8 * 2) and B would be able to complete the job in 9.6 hours (4.8 hours * 2)

But we know A takes 4 hours more than B in terms of time when making 40 widgets

Hence the range would be 4 hours

Hence B is then taking 7.6 hours whereas A is taking 11.6 hours (the range is 4)

11.6 hours is closest to E and hence i chose E

Thoughts ?

The logic is fine here when the options are not that close. Think about it - if the options had 11.5, would you choose 11.5 or 12?
Note that the average time taken would be harmonic mean of the two individual time taken and harmonic mean is less than arithmetic mean if the numbers are different. So when the numbers are 8 and 12, their arithmetic mean will be 10 but harmonic mean will be less than 10.
But if the numbers are say 7.5 and 11.5, the harmonic mean will be less than 9.5 hence 11.5 will not be possible.

Approximating is great but it becomes error free only when you know at which side your error lies. Here, it is perfectly reasonable to say that time taken by both individually, if equal would be 9.6 hrs and since there is a range of 4, the time taken by A will be more than 11.6 hrs.

Hi VeritasKarishma - You are right per the highlighted in yellow. I don't know which side of 11.6 is the actual answer. In fact, i thought A was actually taking 11.6 hours exactly for 40 widgets !

I tried to understand the harmonic mean mentioned above but given i am a non math / business background, i didn't understand it.

Any logic as to why the answer for A would be MORE than 11.6 ? (without going into the concept of harmonic mean specifically)

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

24 Jun 2020, 02:15

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jabhatta@umail.iu.edu wrote:

VeritasKarishma wrote:

jabhatta@umail.iu.edu wrote:

VeritasKarishma wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

Here is how you can do it without any variables:

Together they make 25 widgets in 3 hrs
So to make 40 widgets, together they take 3*(40/25) = 4.8 hrs
So each individually will certainly take more than 4.8 hrs to make 40 widgets.

Machine A takes 2 extra hrs to make 20 widgets so it will take 4 extra hrs to make 40 widgets.

So machine A will certainly take more than 8.8 hrs to make 40 widgets. So we need to pick out of (D) and (E) only.
Try (D) - If A takes 10 hrs, B takes 6 hrs and together they 1/(1/10 + 1/6) = 60/16 (less than 4 hrs)

Hence answer must be (E).

Hi VeritasKarishma

I did what you do till the red.

Then my logic was :

--- Given, Combined A + B can make 40 widgets in 4.8 hours
--- Then individually, if A and B were working at the same rate, then A and B would be able to complete 40 widgets at 9.6 hours (4.8 * 2) and B would be able to complete the job in 9.6 hours (4.8 hours * 2)

But we know A takes 4 hours more than B in terms of time when making 40 widgets

Hence the range would be 4 hours

Hence B is then taking 7.6 hours whereas A is taking 11.6 hours (the range is 4)

11.6 hours is closest to E and hence i chose E

Thoughts ?

The logic is fine here when the options are not that close. Think about it - if the options had 11.5, would you choose 11.5 or 12?
Note that the average time taken would be harmonic mean of the two individual time taken and harmonic mean is less than arithmetic mean if the numbers are different. So when the numbers are 8 and 12, their arithmetic mean will be 10 but harmonic mean will be less than 10.
But if the numbers are say 7.5 and 11.5, the harmonic mean will be less than 9.5 hence 11.5 will not be possible.

Approximating is great but it becomes error free only when you know at which side your error lies. Here, it is perfectly reasonable to say that time taken by both individually, if equal would be 9.6 hrs and since there is a range of 4, the time taken by A will be more than 11.6 hrs.

Hi VeritasKarishma - You are right per the highlighted in yellow. I don't know which side of 11.6 is the actual answer. In fact, i thought A was actually taking 11.6 hours exactly for 40 widgets !

I tried to understand the harmonic mean mentioned above but given i am a non math / business background, i didn't understand it.

Any logic as to why the answer for A would be MORE than 11.6 ? (without going into the concept of harmonic mean specifically)

Calculate the two cases:

1. A makes 8 widgets in 1 hour and B makes 12 widgets in 1 hr. If they work together, how many widgets will they make in 1 hr?

They will make 20 widgets in 1 hr together. This is equivalent to each having a rate of work of 10 widgets/hr if their rates were same.

This calculation average rate is arithmetic mean of the two rates.

2. A takes 8 hrs to complete a work and B takes 12 hrs to complete a work. If they both work together, how long will they take to complete the work?

Rate of A = 1/8 work/hrs
Rate of B = 1/12 work/hrs
Combined Rate = 1/8 + 1/12

Time taken = Work/Rate = 1/(1/8 + 1/12) = 24/5 = 4.8 hrs

This means that if each took 9.6 hrs to complete the work, they would finish it in 4.8 hrs.

This calculation of average time taken is the calculation of harmonic mean.

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

05 Jul 2020, 14:30

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy.

Machine A takes x hrs to make 40 widgets.

Rate of work done by machine A = Work done/Time taken = \(\frac{40}{x}\)

Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together.

Time taken by machine B to make 40 widgets = x – 4

Rate of work done by machine B = Work done/Time taken = \(\frac{40}{(x - 4)}\).

We know the combined rate of the machines is 25/3

So here is the equation:

\(\frac{40}{x} + \frac{40}{(x - 4)} = \frac{25}{3}\)

The steps till here are not complicated. Getting the value of x poses a bit of a problem.

Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options). To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too.

x cannot be 3 but it can be 6. If x = 6, \(\frac{40}{(6 - 4)} = 20\) i.e. the sum will certainly not be 20 or more since we have \(\frac{25}{3} = 8.33\) on the right hand side.

The only other option that makes sense is x = 12 since it has 3 in it.

\(\frac{40}{12} + \frac{40}{(12 - 4)} = \frac{10}{3} + 5 = \frac{25}{3}\)

Answer (E)

If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming.

Another method could be to bring 3 to the left hand side to get the following equation:
\(\frac{120}{x} + \frac{120}{(x - 4)} = 25\)

This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot of time.

Hi VeritasKarishma -- Was reviewing the veritas prep official solution posted above.

From the yellow highlight specifically -- where can i learn more about this strategy /theory where because i see a "3" on the Right hand side, X on the left hand side has to be a multiple of 3 as well.

I wasn't aware that x on the Left hand side has to be multiple of three necessarily because i thought perhaps the numerator and denominator share a 3 and the three could cancel each other out (referring to the left hand side)

Any blog post on the veritas prep website can perhaps give me some more additional theory on this ?

i need some additional theory on this to inculcate on d-day !

thank you for everything you do !!!

L

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

08 Jul 2020, 23:14

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jabhatta@umail.iu.edu wrote:

Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

We need to find the time taken by machine A to make 40 widgets. It will be best to take the time taken by machine A to make 40 widgets as the variable x. Then, when we get the value of x, we will not need to perform any other calculations on it and hence the scope of making an error will reduce. Also, value of x will be one of the options and hence plugging in to check will be easy.

Machine A takes x hrs to make 40 widgets.

Rate of work done by machine A = Work done/Time taken = \(\frac{40}{x}\)

Machine B take 2 hrs less than machine A to make 20 widgets hence it will take 4 hrs less than machine B to make 40 widgets. Think of it this way: Break down the 40 widgets job into two 20 widget jobs. For each job, machine B will take 2 hrs less than machine A so it will take 4 hrs less than machine A for both the jobs together.

Time taken by machine B to make 40 widgets = x – 4

Rate of work done by machine B = Work done/Time taken = \(\frac{40}{(x - 4)}\).

We know the combined rate of the machines is 25/3

So here is the equation:

\(\frac{40}{x} + \frac{40}{(x - 4)} = \frac{25}{3}\)

The steps till here are not complicated. Getting the value of x poses a bit of a problem.

Notice here that that the right hand side is not an integer. This will make the question a little harder for us, right? Wrong! Everything has its pros and cons. The 3 of the denominator gives us ideas for the values of x (as do the options). To get a 3 in the denominator, we need a 3 in the denominator on the left hand side too.

x cannot be 3 but it can be 6. If x = 6, \(\frac{40}{(6 - 4)} = 20\) i.e. the sum will certainly not be 20 or more since we have \(\frac{25}{3} = 8.33\) on the right hand side.

The only other option that makes sense is x = 12 since it has 3 in it.

\(\frac{40}{12} + \frac{40}{(12 - 4)} = \frac{10}{3} + 5 = \frac{25}{3}\)

Answer (E)

If we did not have the options, we might have tried x = 9 too before landing on x = 12. Nevertheless, these calculations are not time consuming at all since you can get rid of the incorrect numbers orally. Making a quadratic and solving it is certainly much more time consuming.

Another method could be to bring 3 to the left hand side to get the following equation:
\(\frac{120}{x} + \frac{120}{(x - 4)} = 25\)

This step doesn’t change anything but it helps if you face a mental block while working with fractions. Try to practice such questions using these techniques – they will save you a lot of time.

Hi VeritasKarishma -- Was reviewing the veritas prep official solution posted above.

From the yellow highlight specifically -- where can i learn more about this strategy /theory where because i see a "3" on the Right hand side, X on the left hand side has to be a multiple of 3 as well.

I wasn't aware that x on the Left hand side has to be multiple of three necessarily because i thought perhaps the numerator and denominator share a 3 and the three could cancel each other out (referring to the left hand side)

Any blog post on the veritas prep website can perhaps give me some more additional theory on this ?

i need some additional theory on this to inculcate on d-day !

thank you for everything you do !!!

Try this:

a + b = 25/3

Which two numbers without a denominator of 3 can you add to give 25/3? There needs to be a 3 in the denominator in at least one of a or b to get the sum of 25/3.

Similarly try a + b = 11/2
11/2 is 5.5. So the 0.5 has to come from somewhere to add up to 5.5

D

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

22 Jul 2020, 14:00

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Bunuel wrote:

Machine A takes 2 more hours than machine B to make 20 widgets. If working together, the machines can make 25 widgets in 3 hours, how long will it take machine A to make 40 widgets?

(A) 5
(B) 6
(C) 8
(D) 10
(E) 12

We can PLUG IN THE ANSWERS, which represent A's time to produce 40 widgets.
When the correct answer is plugged in, A and B will produce 25 widgets in 3 hours.

D: 10
Here, A can produce 40 widgets in 10 hours, implying that the time for A to produce 20 widgets = 5 hours.
Since A tales 2 hours longer than B, B's time to produce 20 widgets = 3 hours.

Since A takes 5 hours to produce 20 widgets, A's rate \( = \frac{work}{time} = \frac{20}{5} = 4\) widgets per hour.
Since B takes 3 hours to produce 20 widgets, B's rate \( = \frac{work}{time} = \frac{20}{3}\) widgets per hour.

Combined rate for A and B \(= 4+\frac{20}{3} = \frac{32}{3}\) widgets per hour.
Work produced by A and B in 3 hours \(= (rate)(time) = \frac{32}{3}*3 = 32\) widgets.

Here, A and B produce too many widgets and thus are working TOO FAST.
Implication:
For the rate to be reduced, A must take LONGER to produce 40 widgets, with the result that A and B will work more SLOWLY.

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Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

10 Nov 2020, 14:38

VeritasKarishma , thanks for the insights!

I came with an approach that I could not follow throw, can any one point me in the wright direction?

I tried to make sure that I was working with just one full work, and I thought that 100 widgets was a good number.

So A would take 10 more hours than machine B to make 100 widgets
AND
Working together, the machines can make 100 widgets in 12 hours

QUESTION
how long will it take machine A to make 40 widgets

Was this a good approach? How can one solve it?

[edit] I could solve in ~10 min, but this was not a good idea, right?

gmatclubot

Re: Machine A takes 2 more hours than machine B to make 20 widgets. If wor [#permalink]

10 Nov 2020, 14:38

GMAT Problem Solving (PS) Questions

Machine A takes 2 more hours than machine B to make 20 widgets. If wor