GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Jul 2020, 12:12

### 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

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

# Working simultaneously at their respective constant rates, M

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 03 Mar 2017
Posts: 394
Working simultaneously at their respective constant rates, M  [#permalink]

### Show Tags

17 Oct 2018, 18:27
We know that work done is 800 nails and we are asked to find that how many hours B take to produce same 800 nails.

Since work done is uniform , we take it as 1 instead of 800 to avoid heavy calculation.

Both A and B complete work in x hours.

1/A+1/B=1/x

Given A=y

So 1/B=1/x-1/y

Get B from here =(xy)/(y-x)

IMO E.
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
Intern
Joined: 20 Jan 2019
Posts: 2
Re: Working simultaneously at their respective constant rates, M  [#permalink]

### Show Tags

14 Mar 2019, 10:55
1
As the work is same (800 nails) let's not think about it.
If they can work together in x hours then their combined rate is 1/x
The rate of A is 1/y
so rate of b is = 1/x-1/y
= (y-x)/xy
so time will be the reverse of rate = xy/(y-x)
Manager
Joined: 20 Aug 2017
Posts: 122
Working simultaneously at their respective constant rates, M  [#permalink]

### Show Tags

06 Nov 2019, 19:03
since the work units are same i.e. 800 so we can take it as 1 unit as well.
Now work done by A + B = 1/x
work done by A alone = 1/y

$$\frac{1}{y}$$ + $$\frac{1}{B}$$ = $$\frac{1}{x}$$
On solving, we get B = $$\frac{xy}{(y-x)}$$

Director
Joined: 24 Oct 2016
Posts: 708
GMAT 1: 670 Q46 V36
GMAT 2: 690 Q47 V38
Re: Working simultaneously at their respective constant rates, M  [#permalink]

### Show Tags

17 Apr 2020, 03:25
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

Machine, R, T, T
A, 800/y, y, 800
B, 800/B, B, 800
A+B, 800/x, x, 800

Method 1: Using Time

Time = T1*T2/(T1+T2)
x = yB/(y+B) => xy + xB = yB => xy = yb - xB => y = xy/(y-x)

Method 2: Using Rate

800/B = 800/x - 800/y
1/B = 1/x - 1/y
1/B = (y - x)/xy
B = xy/(y-x)
Intern
Joined: 18 Jul 2017
Posts: 1
Re: Working simultaneously at their respective constant rates, M  [#permalink]

### Show Tags

07 Jul 2020, 04:01
Working simultaneously at their respective constant rates, Machines A and B produce 800 nails in x hours. Working alone at its constant rate, Machine A produces 800 nails in y hours. In terms of x and y, how many hours does it take Machine B, working alone at its constant rate, to produce 800 nails?

(A) x/(x+y)
(B) y/(x+y)
(C) xy/(x+y)
(D) xy/(x-y)
(E) xy/(y-x)

Clearly, x < y and since A and B together produces 800/x nails in 1 hr, (x+y) can never be an option. A,B,C goes off. and since x< y and time > 0, E is the option.
Re: Working simultaneously at their respective constant rates, M   [#permalink] 07 Jul 2020, 04:01

Go to page   Previous    1   2   3   [ 45 posts ]