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A newer machine, working alone at its constant rate

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A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
[Reveal] Spoiler: OA

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Hi Bunuel,
Could you please help on the below -

This is from Veritas practice test . I got the answer as "A", but as per veritas , the answer is E . Not sure where I went wrong .
Appreciate if experts can comment on this.

Let A and B , be old machines and C be the new machine .

----------- Rate -------------Tiime------------- Work
A -----------x --------------1/x------------------ 1

B ------------y --------------1/y--------------------1

A+B--------- x+y-----------1/(x+y)------------------ 1

C -----------2(x+y)---------1/2(x+y)------------ 1

combined rate of two old and one new machine is = (x+y)+2(x+y) = 3(x+y)
question : is time taken - i.e 1/3(x+y) >1 ? or (x+y) <1/3 ?

Statement 1 :- (1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

1/x <4 and 1/y <4
x>1/4 and y>1/4

hence x+y>1/2 , which answers our question as "No" x+y is not less than 1/3
Sufficient.

Statement 2 :- (2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
4(x+y) is the new rate . 1/4(x+y) <1 , hence (x+y) >1/4 . Hence (x+y)<1/3 may or may not be true.
Not sufficient.

I do not understand , where is it that I am going wrong on this one .


-Jyothi
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Last edited by gmacforjyoab on 27 Oct 2013, 08:35, edited 2 times in total.
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Re: A newer machine, working alone at its constant rate [#permalink]

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Let x, y be the time for 2 old machines to complete the job and z be the time for the newest machine to complete the same job.
When 2 old machine working together, they finish the job in \(\frac{1}{\frac{1}{x}+ \frac{1}{y}} = \frac{xy}{x+y}\). So \(z= \frac{xy}{2(x+y)}\).
(1) implies that x<8 and y<8. So z<2. If all three machines work together, the time \(\frac{1}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}\) is less than 4/3.
(2) implies that z< 2. Similarly, If all three machines work together, the time is less than 4/3.
Answer is E
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Re: A newer machine, working alone at its constant rate [#permalink]

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gmacforjyoab wrote:
Hi Bunuel,
Could you please help on the below -

This is from Veritas practice test . I got the answer as "A", but as per veritas , the answer is E . Not sure where I went wrong .
Appreciate if experts can comment on this.

Let A and B , be old machines and C be the new machine .

----------- Rate -------------Tiime------------- Work
A -----------x --------------1/x------------------ 1

B ------------y --------------1/y--------------------1

A+B--------- x+y-----------(x+y)------------------ 1

C -----------2(x+y)---------1/2(x+y)------------ 1

combined rate of two old and one new machine is = (x+y)=2(x+y) = 3(x+y)
question : is time taken - i.e 1/3(x+y) >1 ? or (x+y) <1/3 ?

Statement 1 :- (1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

1/x <4 and 1/y <4
x>1/4 and y>1/4

hence x+y>1/2 , which answers our question as "No" x+y is not less than 1/3
Sufficient.

Statement 2 :- (2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
4(x+y) is the new rate . 1/4(x+y) <1 , hence (x+y) >1/4 . Hence (x+y)<1/3 may or may not be true.
Not sufficient.

I do not understand , where is it that I am going wrong on this one .


-Jyothi




the analysis of the argument is wrongly done.

If Old m/c takes T time to complete the work, then the newer one will take T/2

so the equation for combined work will look like:

2/T(for new m/c)+ 2/T(for 2 old m/c)=4/T

Now the asked ques is-weather 4/T>1 or not?

which can be answered using both the given statements....
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 27 Oct 2013, 08:28
himang wrote:
gmacforjyoab wrote:
Hi Bunuel,
Could you please help on the below -

This is from Veritas practice test . I got the answer as "A", but as per veritas , the answer is E . Not sure where I went wrong .
Appreciate if experts can comment on this.

Let A and B , be old machines and C be the new machine .

----------- Rate -------------Tiime------------- Work
A -----------x --------------1/x------------------ 1

B ------------y --------------1/y--------------------1

A+B--------- x+y-----------(x+y)------------------ 1

C -----------2(x+y)---------1/2(x+y)------------ 1

combined rate of two old and one new machine is = (x+y)=2(x+y) = 3(x+y)
question : is time taken - i.e 1/3(x+y) >1 ? or (x+y) <1/3 ?

Statement 1 :- (1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

1/x <4 and 1/y <4
x>1/4 and y>1/4

hence x+y>1/2 , which answers our question as "No" x+y is not less than 1/3
Sufficient.

Statement 2 :- (2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
4(x+y) is the new rate . 1/4(x+y) <1 , hence (x+y) >1/4 . Hence (x+y)<1/3 may or may not be true.
Not sufficient.

I do not understand , where is it that I am going wrong on this one .


-Jyothi




the analysis of the argument is wrongly done.

If Old m/c takes T time to complete the work, then the newer one will take T/2

so the equation for combined work will look like:

2/T(for new m/c)+ 2/T(for 2 old m/c)=4/T

Now the asked ques is-weather 4/T>1 or not?

which can be answered using both the given statements....


Couple of issues in what you have explained above.

1. You cannot assume that time required for both the old machines is the same ( which you have assumed to be T). The problem just says , "work at THEIR CONSTANT RATE" .

2. Secondly , you can add the rates of three machines working together , which gives you a combined rate. But you cannot add the time taken by three machines and call that the combined time .
for instance - Lets say Machine L takes 2 hrs , M takes 2 hrs and N takes 2 hrs to complete a work independently. If they all work together , then as per your analysis , they take 2+2+2=6 hrs to complete . If they all work together , they would definitely complete it in less than 2 hrs.

3. Also, the answer to the question , from what veritas says , is E and not C/D as you mentioned.


-Jyothi
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 27 Oct 2013, 21:23
gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour



Hi jyothi,

I think picking up few nos for each statements is perhaps the fastest way to do this.Before doing this assume x+y is time taken by older machines and z by new machine

1)for 1st statement the older machines can fill the order in less than 4 hours , take for instance the time taken by 2 older machines be 6 hours,
so half-the production order will be in x+y/2=3 hours

time taken by new machine is 3 hours,

so in 6 hours 1.5 production order is filled
=> 1 production order in= 6/1.5= 4 hours (so answer to the statement is no)

but let us assume time taken by older machines x+y=1/4 hours

so half the prod order = 1/4*1/2=1/8

and z takes (1/4*1/2)=1/8

so in 1/4 hours 1.5 orders


1 prod order in 1/4*1/1.5 (which will be less than 1 hour, so answer is yes)

clearly therefore statment 1 is insufficient
so


2)for the answer to be yes (i,e statement sufficient to state it will take less than 1hour you can pick the same values as in 2 conditio of 1st statement i,e x+y=1/2 hours)

for the yes condition pick time taken by z = 10/6 ,, so if rate doubled it will be able to do in 5/6 hours,,
so x+y= 20/6
so in 30/6 2 orders are completed,

=> 1 order in 5/2 hours (so answer is no as order cannot be completed in less than in 1 hour),,
you can have the same value (z=10/6) and use it for statemet 1 as well.

hence both the values are insufficient.E is correct
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Re: A newer machine, working alone at its constant rate [#permalink]

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gmacforjyoab wrote:
Hi Bunuel,
Could you please help on the below -

This is from Veritas practice test . I got the answer as "A", but as per veritas , the answer is E . Not sure where I went wrong .
Appreciate if experts can comment on this.

Let A and B , be old machines and C be the new machine .

----------- Rate -------------Tiime------------- Work
A -----------x --------------1/x------------------ 1

B ------------y --------------1/y--------------------1

A+B--------- x+y-----------1/(x+y)------------------ 1

C -----------2(x+y)---------1/2(x+y)------------ 1

combined rate of two old and one new machine is = (x+y)+2(x+y) = 3(x+y)
question : is time taken - i.e 1/3(x+y) >1 ? or (x+y) <1/3 ?

Statement 1 :- (1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

1/x <4 and 1/y <4
x>1/4 and y>1/4

hence x+y>1/2 , which answers our question as "No" x+y is not less than 1/3
Sufficient.

Statement 2 :- (2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
4(x+y) is the new rate . 1/4(x+y) <1 , hence (x+y) >1/4 . Hence (x+y)<1/3 may or may not be true.
Not sufficient.

I do not understand , where is it that I am going wrong on this one .


-Jyothi


You misread statement (1).

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
Each of the older machines cannot fill a production order in less than 4 hrs. They can fill half the production order in less than 4 hrs. The actual time taken by them could be 1 hr, 2 hrs, 3 hrs, 3.9 hrs etc to fill HALF the order. Hence they can fill the complete order is less than 8 hrs.

1/x < 8
1/y < 8
x + y > 1/4
Not sufficient.
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 07 Nov 2013, 06:51
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
Hi Bunuel,
Could you please help on the below -

This is from Veritas practice test . I got the answer as "A", but as per veritas , the answer is E . Not sure where I went wrong .
Appreciate if experts can comment on this.

Let A and B , be old machines and C be the new machine .

----------- Rate -------------Tiime------------- Work
A -----------x --------------1/x------------------ 1

B ------------y --------------1/y--------------------1

A+B--------- x+y-----------1/(x+y)------------------ 1

C -----------2(x+y)---------1/2(x+y)------------ 1

combined rate of two old and one new machine is = (x+y)+2(x+y) = 3(x+y)
question : is time taken - i.e 1/3(x+y) >1 ? or (x+y) <1/3 ?

Statement 1 :- (1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

1/x <4 and 1/y <4
x>1/4 and y>1/4

hence x+y>1/2 , which answers our question as "No" x+y is not less than 1/3
Sufficient.

Statement 2 :- (2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
4(x+y) is the new rate . 1/4(x+y) <1 , hence (x+y) >1/4 . Hence (x+y)<1/3 may or may not be true.
Not sufficient.

I do not understand , where is it that I am going wrong on this one .


-Jyothi


You misread statement (1).

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
Each of the older machines cannot fill a production order in less than 4 hrs. They can fill half the production order in less than 4 hrs. The actual time taken by them could be 1 hr, 2 hrs, 3 hrs, 3.9 hrs etc to fill HALF the order. Hence they can fill the complete order is less than 8 hrs.

1/x < 8
1/y < 8
x + y > 1/4
Not sufficient.



Oh right !!
Thank you so much for clarifying !

-Jyothi
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Machines, work/rate problem [#permalink]

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New post 28 Jan 2015, 14:03
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
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New post 29 Jan 2015, 04:51
santorasantu wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


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Re: A newer machine, working alone at its constant rate [#permalink]

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Bunuel wrote:
santorasantu wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Please search before posting and name topics properly (check rule 3: rules-for-posting-please-read-this-before-posting-133935.html). Thank you.


Thanks Bunuel for redirecting and also on rules.

Could anybody please give an alternate quick solution for this question?
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gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Here is another solution:

New machine takes half the time taken by 2 old machines i.e. rate of new machine is twice the rate of 2 old machines. Say rate of old machine is R. Then rate of 2 old machines is 2R. Then rate of a new machine is 2*2R = 4R.
If all 3 machines work together, their combined rate is R + R + 4R = 6R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
This means the old machine takes less than 8 hrs to fill a complete order i.e. its rate is at least 1/8.
R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs. Is it actually less than 1 hr, we can't say. Not sufficient.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
If rate is 2*4R =8R, time taken is less than 1 hour to complete 1 full production lot. So rate is at least 1/8. This is exactly the same information as obtained in statement 1. This alone will also be insufficient.

Both statements give the same information and are insufficient. So together they will be insufficient.

Answer (E)
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 30 Jan 2015, 11:37
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Here is another solution:

New machine takes half the time taken by 2 old machines i.e. rate of new machine is twice the rate of 2 old machines. Say rate of old machine is R. Then rate of 2 old machines is 2R. Then rate of a new machine is 2*2R = 4R.
If all 3 machines work together, their combined rate is R + R + 4R = 6R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
This means the old machine takes less than 8 hrs to fill a complete order i.e. its rate is at least 1/8.
R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs. Is it actually less than 1 hr, we can't say. Not sufficient.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
If rate is 2*4R =8R, time taken is less than 1 hour to complete 1 full production lot. So rate is at least 1/8. This is exactly the same information as obtained in statement 1. This alone will also be insufficient.

Both statements give the same information and are insufficient. So together they will be insufficient.

Answer (E)



Thanks a lot, Karishma. I find this a bit more easier. +1
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A newer machine, working alone at its constant rate [#permalink]

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gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Let's call the two old machines A and B and new machine C.

Q: 1/A + 1/B + 1/C =< 1

(1)
0.5/A =< 0.5/4 or 1/A =< 1/8
0.5/B =< 0.5/4 or 1/B =< 1/8

Hence, 1/A + 1/B =< 1/8 + 1/8 =< 1/4

The Q tells us that C can fill the entire order is 1/2 the combined time of A and B:

1/C =< 2*(1/4) =< 1/2 (This means that C alone could fill the entire order in less than 2 hours)

All together, this information is insufficient because it doesn't tell us how less than 2 hours C alone can work or how less than 4 hours A and B can work combined. Therefore, we can't determine if all three together could fill the order in under 1 hour.

(2)

This tells us 1/2C =< 1... or 1/C =< 1/2 (the same information we had in 1). Still insufficient.

Answer: E

Hope this helps.
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A newer machine, working alone at its constant rate [#permalink]

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New post 22 Jul 2015, 12:11
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Here is another solution:

New machine takes half the time taken by 2 old machines i.e. rate of new machine is twice the rate of 2 old machines. Say rate of old machine is R. Then rate of 2 old machines is 2R. Then rate of a new machine is 2*2R = 4R.
If all 3 machines work together, their combined rate is R + R + 4R = 6R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
This means the old machine takes less than 8 hrs to fill a complete order i.e. its rate is at least 1/8.
R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs. Is it actually less than 1 hr, we can't say. Not sufficient.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
If rate is 2*4R =8R, time taken is less than 1 hour to complete 1 full production lot. So rate is at least 1/8. This is exactly the same information as obtained in statement 1. This alone will also be insufficient.

Both statements give the same information and are insufficient. So together they will be insufficient.

Answer (E)



Hi there Karishma,


Nice solution, I have a doubt since the statement 1: says each of the old machines completed half the production in less than 4 hours , I took a case where the machines finish their job in a matter of minutes thus the total time for all three will be less than 60 minutes or 1 hr here and an extreme case as you did. Hence I concluded insufficient. Is this correct ?? I took a similar approach for statement 2.

Answer :E

Thanks
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 23 Jul 2015, 02:12
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anurag356 wrote:
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Here is another solution:

New machine takes half the time taken by 2 old machines i.e. rate of new machine is twice the rate of 2 old machines. Say rate of old machine is R. Then rate of 2 old machines is 2R. Then rate of a new machine is 2*2R = 4R.
If all 3 machines work together, their combined rate is R + R + 4R = 6R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
This means the old machine takes less than 8 hrs to fill a complete order i.e. its rate is at least 1/8.
R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs. Is it actually less than 1 hr, we can't say. Not sufficient.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
If rate is 2*4R =8R, time taken is less than 1 hour to complete 1 full production lot. So rate is at least 1/8. This is exactly the same information as obtained in statement 1. This alone will also be insufficient.

Both statements give the same information and are insufficient. So together they will be insufficient.

Answer (E)



Hi there Karishma,


Nice solution, I have a doubt since the statement 1: says each of the old machines completed half the production in less than 4 hours , I took a case where the machines finish their job in a matter of minutes thus the total time for all three will be less than 60 minutes or 1 hr here and an extreme case as you did. Hence I concluded insufficient. Is this correct ?? I took a similar approach for statement 2.

Answer :E

Thanks


Yes, less than 4 hours could very well be a few seconds too. That is the extreme "minimum time" scenario. The approach is correct.
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 05 Dec 2015, 07:37
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gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour



-------------------------------------------

You may approach this question in efficiency method as well.

1) each can fill half a production in less than 4 hours, hence full production in less then 8 hours. Thus the efficiency of each may be greater than (1/8)*100 = 12.5 %.

Say i take the minimum as 12.5% of each machine, thus the newer machine will have a minimum efficiency of 25% , together the three will have minimum efficiency = 75% . Thus they can't complete the work in less than one hour ( note: for less than one hour the summation of those three should be greater than 100%)

Hence not sufficient.

2) The new machine if doubles it rates can complete work in less than one hour. Hence the efficiency of new machine should be >50% , and the sum of efficiency of the two older machine should be >25%

Taking minimum values of the three, i am getting an efficiency of 75% , which again shows that it may or may not take less than an hour.
Hence not sufficient.

And even if you combine both i can assume a value in such a way that the efficiency will be >75% but less than 100%. Thus even together they are not sufficient.

Ans -> E.
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A newer machine, working alone at its constant rate [#permalink]

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New post 21 Jan 2016, 04:56
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You cannot imagine how much I struggled with these rates pb.
So, every time I have a rate question, I proceed like this:

First, I draw a nice and lovely chart:

...................................................Rate.............Time.............Quantity
New Machine......................................................t/2..................1 Quant
2 old machines (counted as 1)....................................t..................1 Quant
The 2 things working for 1 Quant................................................. 1 Quant

From that, I start to plug the missing parts. The rate for the new machine is 2/t (2/t * t/2 =1 Quant). The rate for the old machines is 1/t. So from this I can guess that the 2 rates combined are 2/t + 1/t = 3/t. So now I have the combined rate. What would be the time, for the combined rate to make 1 Quant ? Only the reversal, T/3

Here is my new chart:


...................................................Rate.............Time.............Quantity
New Machine......................................2/t.............t/2.............1 Quant
2 old machines (counted as 1)..................1/t.............t.............1 Quant
The 2 things working for 1 Quant.............3/t.............t/3.............1 Quant

The 1st question says: Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
In other words, if each can do half in less than 4 hours, the 2 combined can achieve in less than 4 hours. So t has to be <4

So the question now is: if t<4, would t/3 (our combined time) be less than 1? The question is much easier ! if t is 3.99, its ok, the combined time is over 1 hour, but if t is 2, then the combined time would be 2/3, which is less than 1.

The 2st question says: If the newer machine were to double its rate, it could fill a production order working alone in less than one hour.
So here again, doubling its current rate means that 2/t (the rate of the new machine) x2, so 4/t. Therefore the time, which is still the reverse of the rate, is t/4. And the stem just says that t/4<1, which means t<4. So this is the same issue as the one stated in the first question. Still no way to know!

Combining both is useless.
So E
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Re: A newer machine, working alone at its constant rate [#permalink]

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New post 19 May 2016, 23:39
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
A newer machine, working alone at its constant rate, can fill a production order in half the time required by two older machines working together at their constant rates. If the three machines are used together, will they require more than 1 hour to fill a production order?

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour


Here is another solution:

New machine takes half the time taken by 2 old machines i.e. rate of new machine is twice the rate of 2 old machines. Say rate of old machine is R. Then rate of 2 old machines is 2R. Then rate of a new machine is 2*2R = 4R.
If all 3 machines work together, their combined rate is R + R + 4R = 6R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
This means the old machine takes less than 8 hrs to fill a complete order i.e. its rate is at least 1/8.
R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs. Is it actually less than 1 hr, we can't say. Not sufficient.

(2) If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
If rate is 2*4R =8R, time taken is less than 1 hour to complete 1 full production lot. So rate is at least 1/8. This is exactly the same information as obtained in statement 1. This alone will also be insufficient.

Both statements give the same information and are insufficient. So together they will be insufficient.

Answer (E)



Hi VeritasPrepKarishma / chetan2u,

I am not able to comprehend the following part of your explanation-

R is at least 1/8 so 6R is atleast 6/8 = 3/4. So time taken by all 3 machines together to complete 1 lot is maximum 4/3 hrs

If R is atleast 3/4, then why the complete 1 lot is ATMOST 4/3..?
I thought the complete time must also be ATLEAST 4/3..?
Can you please explain with some easy example..?

Thanks
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Re: A newer machine, working alone at its constant rate   [#permalink] 19 May 2016, 23:39

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