It is currently 18 Jan 2018, 23:29

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A newer machine, working alone at its constant rate

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
D
Joined: 02 Aug 2009
Posts: 5528

Kudos [?]: 6427 [0], given: 122

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

Show Tags

New post 19 May 2016, 22:50
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


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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


BANGALORE/-

Kudos [?]: 6427 [0], given: 122

Manager
Manager
User avatar
B
Joined: 16 Mar 2016
Posts: 134

Kudos [?]: 58 [0], given: 0

Location: France
GMAT 1: 660 Q47 V33
GPA: 3.25
GMAT ToolKit User
A newer machine, working alone at its constant rate [#permalink]

Show Tags

New post 20 May 2016, 14:26
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

Kudos [?]: 58 [0], given: 0

Manager
Manager
User avatar
Joined: 12 Jan 2015
Posts: 222

Kudos [?]: 84 [0], given: 79

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

Show Tags

New post 20 May 2016, 21:59
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
_________________

Thanks and Regards,
Prakhar

Kudos [?]: 84 [0], given: 79

Expert Post
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7866

Kudos [?]: 18473 [0], given: 237

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

Show Tags

New post 20 May 2016, 22:10
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 18473 [0], given: 237

Intern
Intern
avatar
Joined: 15 Dec 2014
Posts: 2

Kudos [?]: [0], given: 1

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

Show Tags

New post 21 Jun 2016, 13:09
What GMAT level question is this? I could not do it on the Veritas Prep test. :( Found it very very tough

Kudos [?]: [0], given: 1

Expert Post
Veritas Prep GMAT Instructor
User avatar
G
Joined: 16 Oct 2010
Posts: 7866

Kudos [?]: 18473 [0], given: 237

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

Show Tags

New post 21 Jun 2016, 21:03
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Kudos [?]: 18473 [0], given: 237

1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 18 Jan 2010
Posts: 257

Kudos [?]: 190 [1], given: 9

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

Show Tags

New post 22 Jun 2016, 01:22
1
This post received
KUDOS
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.

Kudos [?]: 190 [1], given: 9

Manager
Manager
avatar
G
Joined: 14 Oct 2012
Posts: 180

Kudos [?]: 63 [0], given: 966

Premium Member Reviews Badge
A newer machine, working alone at its constant rate [#permalink]

Show Tags

New post 31 Dec 2016, 21:15
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

Kudos [?]: 63 [0], given: 966

Manager
Manager
User avatar
B
Joined: 30 Apr 2017
Posts: 88

Kudos [?]: 2 [0], given: 74

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

Show Tags

New post 21 Oct 2017, 01:32
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

Kudos [?]: 2 [0], given: 74

A newer machine, working alone at its constant rate   [#permalink] 21 Oct 2017, 01:32

Go to page   Previous    1   2   [ 29 posts ] 

Display posts from previous: Sort by

A newer machine, working alone at its constant rate

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.