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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]

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New post 20 May 2016, 15:26
I don't understand where my reasoning is false :

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

From the question we know three things :


(1) Ra * (1/2) t1 = 1
(2) 2 Rb * t1 = 1
(3) (Ra+2Rb) * t2 = 1

Where,
Ra and Rb, the speed rates of respectively newer and older machines ;
t1, the time for 2 older machines to do the production ;
t2, the time for one newer and 2 older to do the production, when the three machines are used together.

We want to know if t2 > 1h.

So,
From (1), Ra = 2/t1
From (2), Rb = 1/2t1
From (3), Ra*t2 + 2Rb*t2 = 1

So (2*t2)/t1 +t2/t1 = 1 => (3*t2)/t1=1 =>3*t2 = t1 (4)

(1) from statement 1

Rb * t3 = 1, where t3 < 4h
So, 1/Rb < 4h
But, From (2), Rb = 1/2*t1, then 2t1 < 4h
So, From (4) 6*t2 <4h => t2 <4/6 h => t2<2/3h => t2 <40min, sufficient
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 20 May 2016, 22:59
Hi Alex75PAris,

Its just the rate which is different for older machines and New machine. However, as I see from your equations you have changed the time as well.

Further, you have taken 2 variable for 2 machines Ra and Rb, if you are already multiplying the rate of new machine by 2 then you don't need to take 2 variables.
The more you take variables in a question, more difficult it would be to crack

Simple eauation will be

Work = Rate x Time

For old machines-
1= R x T
T= 1/R (This is the combined rate of two old machines)

For new machine (Single machine)-
1= 2R x T
T= 1/2R

(1) Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
If 1 old machine take less than 4 hour to complete HALF a PRUDUCTION
the to complete a FULL production it will take less than 8 hours
If 1 machine take less than 8 hours to complete FULL production in less than 8 hours
than 2 machine will take less than 4 hours to complete
So, the new machine will take less than 2 hour to complete FUll production (because its rate is double from two pld machines)

For, easy understanding lets remove less than and treat this question as 4 hours and 2 hours
Means- Two old machine take = 4 hrs to complete a work
New machine take = 2 hours to complete a work
So, working together all three machines can complete a work in
(1/4) + (1/2) = 1/R
R= 4/3

SO, Time taken to complete a work will be 3/4= 0.75 hours

Remember we replace LESS THAN with equal time
We can cay that to complete a work both will take ATLEAST 0.75 hours weather it will be actually more than 1 hr, WE ARE NOT SURE.

So, statement 1 is inufficient
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 20 May 2016, 23:10
PrakharGMAT 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 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


Say, your friend lives 120 miles away from your home. You are trying to figure out how long it will take him to reach you. You know that his speed is ATLEAST 60 mph. He could drive at 60 mph or 65 mph or 70 mph etc. So how long will it take him to reach you?
If he drives at his slowest speed of 60 mph, he will take 120/60 = 2 hrs to reach you.
What if he drives at a faster speed? Say 70 mph? Will he take more time or less? Less, right?
If his speed is more, time taken will be less.

Similarly, in this question, the old machine takes less than 8 hrs. So its speed is more than 1/8th lot per hour. So the speed of all three machines together is more than 3/4th lot per hour. Hence the time they will take is at most 4/3 hours.
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 21 Jun 2016, 14:09
What GMAT level question is this? I could not do it on the Veritas Prep test. :( Found it very very tough
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 21 Jun 2016, 22:03
189anish wrote:
What GMAT level question is this? I could not do it on the Veritas Prep test. :( Found it very very tough


It is a 700 level question, so certainly on the higher side of the difficulty range. I hope that after reading the explanations, you understand it now.
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 22 Jun 2016, 02:22
2
1
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



Suppose old machine completes work in x hours.

2 old machines will complete work in (x/2) hours

New machine completes work in (x/4) hours

In one hour Old machine does (1/x) work
In one hour New machine does (4/x) work

When all 3 work together then in one hour they do

\(\frac{1}{x}+\frac{1}{x}+\frac{4}{x} = \frac{6}{x}\)

Full work is done in \(\frac{x}{6}\) hours

Question asked is \(\frac{x}{6}\) > 1; or x > 6.


Statement 1

Old machine completes (1/2) order in less than 4 hours

so Old machine completes full order in less than 8 hours

x < 8; so x = 7 or x =1. I cant say x is necessarily more than 6.

Not sufficient.

Statement 2

New machine doubles its rate.

It completes order in (x/8) hours.

Given (x/8) < 1

or x < 8.

This is the same condition that we got from statement (1). So again not sufficient.

Even if we combine statements (1) and (2), we dont get anything new. we only get x<8

E is the answer.
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A newer machine, working alone at its constant rate  [#permalink]

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New post 31 Dec 2016, 22:15
Hello Friends, my 2 cents...

m/c --------------- work load ---------------------time
O1 --------------------- f/x --------------------- 1hr
O2 --------------------- f/y --------------------- 1hr
O1 + O2 ------------- f*[(x+y)/xy] ------------- 1hr
N1 ------------------- 2f*[(x+y)/xy] ------------- 1hr
O1+O2+N1 --------- 3f*[(x+y)/xy] ------------- 1hr
--------------------- f --------------------- (1/3)*[xy/(x+y)] hrs
--------------------- --------------------- (1/3)*[xy/(x+y)] < 1
--------------------- --------------------- xy/(x+y) < 3hr ?
1) Each older machine can fill (1/2) prod order in < 4 hrs
O1 --------------------- f/2x ---------------------- < 4hrs
O2 --------------------- f/2y ---------------------- < 4hrs
O1+O2 ---------------- f/2*[1/x+1/y] ---------- < 4hrs
--------------------- f/2*[x+y/xy] ------------ < 4hrs
N1 -------------------- f*[x+y/xy] --------------- < 4hrs
O1+O2+N1 ----------- f*[x+y/xy]*3/2 ---------- < 4hrs
--------------------- f --------------------- < (2/3)*[xy/(x+y)] hrs
--------------------- --------------------- (2/3)*[xy/(x+y)] < 4 hrs
--------------------- --------------------- xy/(x+y) < 6 hrs NOT SUFFICIENT
2) If new m/c doubles its rate, it can fill prod order in < 1 hr
N1 --------------------- 2f*[(x+y)/xy] --------------------- 1hr
N1 --------------------- 4f*[(x+y)/xy] --------------------- < 1hr
O1 + O2 -------------- 2f*[(x+y)/xy] --------------------- < 1hr
O1+O2+N1 ----------- 6f*[(x+y)/xy] --------------------- < 1hr
--------------------- f --------------------- (1/6)*xy/(x+y)
--------------------- --------------------- (1/6)*xy/(x+y) < 1
--------------------- --------------------- xy/(x+y) < 6hr NOT SUFFICIENT
1) + 2)
Both statments give us the same information. Thus answer E
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A newer machine, working alone at its constant rate  [#permalink]

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New post 21 Oct 2017, 02:32
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)



Dear VeritasPrepKarishma
the question says: rate of a new machine is twice the rate of 2 old machines.

statement 2 says: If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
we know from the question newer is twice two old, so two old can do half of part
for example newer:4R two old together:4R
if newer were 8R, work would be done in less than 1 hour
so, if we add 4R (two old) we can get that

it's logical, I cant see why it is wrong
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 12 Dec 2018, 00:25
VeritasKarishma 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 Karishma,

Why the combined rate as been assumed as equal R? and so 2R

Can we take it as simple R for both and for the new machine as 2R.

Now total rate sum =3R

Now from stat 1 : older machines can fill the prod order in 4 hours

So R can complete the task in less than 4 hours
hence 3R will complete the work in less than 4/3 hours .
So statement one insufficient.

Is my approach right ?
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 09 Feb 2019, 09:35
VeritasKarishma, how do we assume that the rates of the two older machines are the same?
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 09 Feb 2019, 21:57
tuanle wrote:
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



I got explanation of your first statement well.. perhaps easiest way to solve this question..
can please ellab. on 2-statement for my better understanding.
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Re: A newer machine, working alone at its constant rate  [#permalink]

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New post 14 Jun 2019, 15:13
There is a very simple method to this question, the trick is to glance at what the statements are giving, and re-write the question stem in terms of time.

Draw a basic RTW table as the picture below, and assign t as the time taken for the newer machine to complete its order, associate the older machines too with the question stem:
Attachment:
Machines.PNG
Machines.PNG [ 2.78 KiB | Viewed 94 times ]


The combined rates of newer and 2 older is as below:
\(\frac{1}{t}+\frac{1}{2t}=\frac{3}{2t}\)
Therefore it takes \(\frac{2t}{3}\) hours for the machines to complete the job.
The question stem is then asking if \(\frac{2t}{3}>1\)?
Rewriting, the question then becomes, is \(t>\frac{3}{2}\)?

Statement 1: 2 old machines finish a job in 2t hours, therefore 1 machine finishes half a job in the same 2t hours.
It takes less than 4 hours to complete the job so 2t < 4 and t < 2.
Since the question asks if t > 1.5, the answer could be yes (t = 1.75) or no (t = 1) NOT SUFFICIENT

Statement 2: Newer machine rate = \(\frac{2}{t}\), therefore time taken to complete a job =\(\frac{t}{2}\)
It takes less than an hour to complete the job so \(\frac{t}{2}<1\)
And t<2, so NOT SUFFICIENT as is it the same as Statement 1

Statements 1 & 2: Both statements give the same values, and both are inefficient, ANSWER E
Attachments

Machines.PNG
Machines.PNG [ 3.48 KiB | Viewed 94 times ]

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Re: A newer machine, working alone at its constant rate   [#permalink] 14 Jun 2019, 15:13

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