A newer machine, working alone at its constant rate

Question

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

KarishmaB · Accepted Answer

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)

tuanle · Answer

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

gmacforjyoab · Answer

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

himang · Answer

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....

gmacforjyoab · Answer

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

KarishmaB · Answer

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.

anurag356 · Answer

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

KarishmaB · Answer

Yes, less than 4 hours could very well be a few seconds too. That is the extreme "minimum time" scenario. The approach is correct.

sdsdssdds · Answer

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

PrakharGMAT · Answer

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

chetan2u · Answer

Hi  PrakharGMAT ,
speed and time are inversely related..
If time taken is more, speed will be less..
If time taken is less, speed will be more...

example..
let the distance be 100 and speed be s, so TIME = 100/s...
say the least speed =   \frac{1}{2} *s  .....then MAX TIME  = 100/\frac{1}{2} *s = \frac{200}{s},  which is  100*\frac{2}{s}.. ...
as can be seen here...
If speed is s/2... time becomes s/2

Alex75PAris · Answer

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

PrakharGMAT · Answer

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

KarishmaB · Answer

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.

adiagr · Answer

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.

GMATinsight · Answer

Answer: Option E Check the explanation in attachment Screenshot 2020-08-05 at 8.07.34 AM.png

rvgmat12 · Answer

OE:

This is a challenging question that requires quite a bit of scratch work. In general, yes/no data sufficiency questions involving inequalities and rates can be time-consuming.

We could start with some algebra, even before going to the statements. If the two older machines combined can do 1 production order in t hours, the newer machine can do the same 1 production order in half the time, or t/2 hours. Thus, the two older machines work at a combined rate of 1/t orders per hour, and the newer machine works at the rate of 2/t orders per hour. When the three machines work together, their rates add to a combined rate of 1/t + 2/t = 3/t orders per hour. The time required by all three machines working together to fill 1 production order at the rate of 3/t orders per hour would be 1/(3/t) = t/3 hours. The yes/no question asks whether this t/3 hours is greater than 1 hour, i.e. whether t is greater than 3.

Statement 1 provides enough information to find that t<4 hours. This is not sufficient to answer whether or not t is greater than 3. If each of the older machines can fill ½ production order in less than 4 hours, then both older machines working together could fill 1 entire order in less than 4 hours. Using our algebra from the question stem, t represents this time, so t is less than 4 hours.

Statement 2 provides enough information to find that t<4 hours. This is not sufficient to answer whether or not t is greater than 3. If the newer machine doubles its rate from 2/t to 4/t, it can fill 1 order in 1/(4/t) = t/4 hours. If this t/4 hours is less than 1 hour, then t is less than 4 hours.

Statements 1 and 2 are redundant, so combining them provides no new information. If each statement alone is insufficient, and they are redundant, then both statements combined are still insufficient. Answer E is correct.

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A newer machine, working alone at its constant rate

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