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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 [#permalink]  26 Oct 2013, 13:42
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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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Re: A newer machine, working alone at its constant rate [#permalink]  26 Oct 2013, 13:43
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Hi Bunuel,

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, 07:35, edited 2 times in total.
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Re: A newer machine, working alone at its constant rate [#permalink]  26 Oct 2013, 17:04
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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.
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Re: A newer machine, working alone at its constant rate [#permalink]  27 Oct 2013, 04:16
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gmacforjyoab wrote:
Hi Bunuel,

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]  27 Oct 2013, 07:28
himang wrote:
gmacforjyoab wrote:
Hi Bunuel,

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]  27 Oct 2013, 20:23
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]  06 Nov 2013, 19:55
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Expert's post
gmacforjyoab wrote:
Hi Bunuel,

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

(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]  07 Nov 2013, 05:51
VeritasPrepKarishma wrote:
gmacforjyoab wrote:
Hi Bunuel,

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

(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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Re: A newer machine, working alone at its constant rate [#permalink]  17 Nov 2014, 13:16
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Machines, work/rate problem [#permalink]  28 Jan 2015, 13:03
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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Re: A newer machine, working alone at its constant rate [#permalink]  29 Jan 2015, 03:51
Expert's post
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]  29 Jan 2015, 12:19
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

Thanks Bunuel for redirecting and also on rules.

Could anybody please give an alternate quick solution for this question?
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Re: A newer machine, working alone at its constant rate [#permalink]  29 Jan 2015, 21:46
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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.

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Re: A newer machine, working alone at its constant rate [#permalink]  30 Jan 2015, 10:37
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.

Thanks a lot, Karishma. I find this a bit more easier. +1
Re: A newer machine, working alone at its constant rate   [#permalink] 30 Jan 2015, 10:37
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