A newer machine, working alone at its constant rate

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
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I don't understand where my reasoning is false :

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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

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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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.

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

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

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 ?

VeritasKarishma, how do we assume that the rates of the two older machines are the same?

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.

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:


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

I did not get this question right in the test yesterday. I hate that fact.

However, this is my solution.

Let's say units of work 32 units.

Old machines does 4 units an hour. Two old combined will do 8 units, that means they will take 32/8=4 hours. That means new one is taking 2 hours and that's 16 units an hour.

Combined all 3, 32/(4+4+16)=32/24. More than an hour.

Old machines does 8 units an hour. Two old combined will do 16 units, that means they will take 32/16=2 hours. That means new one is taking 1 hours and that's 32 units an hour.

Well, clearly all 3 will take less than 1 hour.

Not Sufficient.

Statement B is just rephrasing of statement 1. I did realize that in the exam but it did not click me the answer can be E. Confidently I marked D.

Right Answer. (E)

P.S: Do you generally find VP tests to be on the higher side of difficulty ?

Lets assume the total work to be 12 toys == 1 production order

New machine makes x toys per hour; Each of the older machines be y and z toys per hour

We are given x = 2 (y+z) toys in 1 hour
Questions asks, is x + y + z > 12

Statement 1: we get y=z >=(12/2)/4 toys per hour; so y+z >= 3 toys per hour
or x >=6 toys per hour
so x+y+z >= 9 toys per hour; so x+y+ z can 9, 10, 11, 12, 13....hence insufficient.

Statement 2: we get 2x > 12 or x >=6 toys per hour
=> y+z >= 3 toys per hour
so, again, x+y+z >= 9 toys per hour; Hence insufficient. since statement 1 and 2 essentially are the same both together are insufficient.

Answer: Option E

Check the explanation in attachment

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given
rate of A, B & C be ; a,b,c
we know that a= 1/2*(b+c)
or say 2a = b+c
target find whether If the three machines are used together, will they require more than 1 hour to fill a production order
i.e a+b+c>1 or not ; or say 3a>1 ; a>1/3

#1
Each of the older machines, working alone at its constant rate, can fill half a production order in less than 4 hours.
i.e b+c<4
so value of b+c can be 3.9 , 3,2, 1
therefore value of a can be (3.9/2) ; 1.95 ; (3/2) ; 1,5 ; (2/2) ; 1 ; (1/2) ; .5
so 3a = 5.85,4.5,3,1.5 yes & no insufficient
#2
If the newer machine were to double its rate, it could fill a production order working alone in less than one hour
we know
a= 1/2*(b+c)
and now given double rate ; so a = b+c
target becomes is 2a>1
with double rate a<1
so a = .95,.9,0.5,0.1
so
2a ; 1.9,1.8,1,.2 ; again yes & no
insufficient
from 1 &2
we get we get both yes & no for values a as given under #2
insufficient
OPTION E

Let the time needed for each old machines be O

Time needed together = O/2

New machine needs time = O/4

Time for 2 old and one new = O/6

Question asks if O/6 > 1 or O > 6

Statement 1:
O < 8

Not sufficient

Statement 2:

O/8 < 1

O <8
Not Sufficient

1&2: Not sufficient

E

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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