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# Albert and Bob are painting rooms at constant, but different

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Intern
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Albert and Bob are painting rooms at constant, but different  [#permalink]

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05 Feb 2012, 19:24
1
11
00:00

Difficulty:

75% (hard)

Question Stats:

64% (03:15) correct 36% (03:53) wrong based on 191 sessions

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Albert and Bob are painting rooms at constant, but different rates. Albert takes 1 hour longer than Bob to paint n rooms. Working side by side, they can paint a total of 3n/5 rooms in 4/3 hours. How many hours would it take Albert to paint 3n rooms by himself?

A. 7
B. 9
C. 11
D.13
E. 15
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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06 Feb 2012, 02:31
9
7
sourabhsoni wrote:
Albert and Bob are painting rooms at constant, but different rates. Albert takes 1 hour longer than Bob to paint n rooms. Working side by side, they can paint a total of 3n/5 rooms in 4/3 hours. How many hours would it take Albert to paint 3n rooms by himself?

a) 7
b) 9
c) 11
d)13
e) 15

First of all, I don't like unnecessary variables. The data given is in terms of n rooms but the answer is independent of n. This means, no matter what value I put for n, the answer would be the same. So I will just put n = 1 and proceed.

Albert takes an hour longer than Bob to paint one room.
Time taken by Albert to paint 1 room = A hrs, Time taken by Bob = A-1 hrs

"they can paint a total of 3/5 rooms in 4/3 hours"
W = R*T
3/5 = R*(4/3)
R = 9/20 which is the combined rate of the two.

We obtained this by adding their rates together: 9/20 = 1/A + 1/(A-1)
Over here, please remember that numbers fall in place in GMAT questions. You should be able to figure that since the denominator is 20, A = 5 gives 5*4 in the denominator so it should work. Check before proceeding though.

So Albert paints a room in 5 hrs. For 3 rooms, he takes 3*5 = 15 hrs.
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Joined: 31 Jan 2012
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Re: Work and Rate tough question !!!  [#permalink]

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05 Feb 2012, 22:51
3
rate of bob to paint n rooms is n/T(bob)
rate of Albert to paint n rooms is n/T(albert).
albert paints the room 1 hour slower than bob, so T (albert) = T(bob) -1
Together they paint the 3n/5 rooms in 4/3 hours. Rate is equal to work over time

Therefore
n/x + n/x-1 = (3n/5)/(4/3) =
n/x + n/x-1 = 9n/20. Fastest way for me is to think how would make the denominator 20. 4*5 = 20 and it fits x and x-1 or you can solve the quadratic
4n/20 + 5n/20 = 9n/20. Therefore you know it takes Albert 5 hours to paint n room, since Albert's rate is n/5.
5*3 = 3n
15 = 3n.

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Joined: 27 Dec 2011
Posts: 53
Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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14 Sep 2012, 17:07
3
1
Hi,

This question can be done very quickly if you understand what they asking for.

I will present a very general way of solving ANY work,rate and time related question:

Question will always ask for something out of rate or time or work. (obviously coz its a work time rate related question )
ex. in this question they want "time" for "3n" work done by Albert, right??
This means ------> rate * TIME = 3n =====> All you have to do it find out the RATE for albert and then put it here to find the TIME!!!!! (easy right??)

Let see what is given here:
some relation between Albert and bob's individual time
and their combined time.

In combined time: some work and time is given, we don't care what it is, all we will do is find their rate by writing WORK/TIME
i.e. (3n/5) / 4/3 = 9n/20 = 9/20

From individual relationships just find Albert and Bob's respective rates:
A: takes 'a' time to complete n rooms===> rate becomes : n/a = 1/a
B: takes 1 hr less than A to complete n rooms =====> rate becomes : n/(a-1) = 1/a-1

now add individual rates and equate 'em to the combined rate:

1/a + 1/(a-1) = 9/20
From my experience I feel that from thus point its faster and much easier to put values than solve the quadratic:
if u see that answer I feel 9 and 15 stand out as they are multiples of 3 and they represent 3n work, we can do 3n=15 => n = 5 and put "5" in place of "a" in the equation. it will be equal to 9/20 thats ur answer.

BOTTOM LINE: first thing always look for what is need to in final answer and how can I get their, if yo start by what's given then u will start doing actions not even necessary to reach the solutions.

I hope this helps,

thanks,

-Kartik
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Joined: 06 Sep 2013
Posts: 1546
Concentration: Finance
Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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14 Jan 2014, 11:29
Sourabh2012 wrote:
Albert and Bob are painting rooms at constant, but different rates. Albert takes 1 hour longer than Bob to paint n rooms. Working side by side, they can paint a total of 3n/5 rooms in 4/3 hours. How many hours would it take Albert to paint 3n rooms by himself?

a) 7
b) 9
c) 11
d)13
e) 15

Did a similar approach but somewhat shorter

A = B+1

Then 4/3 (1/B+1/(B+1)) = 3/5

Hence 1/B + 1/(B+1) = 9/20

So B = 4

Therefore A = 5

And so we will take 5*3 = 15 to paint three rooms

Just my 2c

Hope it helps
Cheers!
J
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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29 Mar 2014, 00:57
Rate * Time = Work

Let B be the hours that Albert takes to paint 'n' rooms then

For (Albert) => n/B * B = n
For (BOB) => n/(B-1) ( (B-1) = n

Combined rate = n/B + n/(B-1)

Again Rate * Time = Work

Then {n/B + n/(B-1)} * 4/3 = 3n/5

2B-1/{B(B-1)}=9/20

Substituting values B=5 then answer takes 3B since 3n rooms then (E)
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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08 May 2014, 15:25
Here's another way. May be shorter

A = B+1
B = A-1

Now then we have

1/A + 1/ (A-1) = (3/5) / (3/4) = 9n/20

We need to find 3n

Let's see

To get 9/20, we need something like 1/4 + 1/5 = 9/20

So A must be 5

If A takes 5 to paint 'n', it takes 15 to paint '3n'

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Joined: 08 Apr 2012
Posts: 323
Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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11 May 2014, 03:05
VeritasPrepKarishma wrote:
Over here, please remember that numbers fall in place in GMAT questions. You should be able to figure that since the denominator is 20, A = 5 gives 5*4 in the denominator so it should work. Check before proceeding though.

So Albert paints a room in 5 hrs. For 3 rooms, he takes 3*5 = 15 hrs.

Hi Karishma,

What do you mean "numbers fall in place in GMAT"?
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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12 May 2014, 20:14
ronr34 wrote:
VeritasPrepKarishma wrote:
Over here, please remember that numbers fall in place in GMAT questions. You should be able to figure that since the denominator is 20, A = 5 gives 5*4 in the denominator so it should work. Check before proceeding though.

So Albert paints a room in 5 hrs. For 3 rooms, he takes 3*5 = 15 hrs.

Hi Karishma,

What do you mean "numbers fall in place in GMAT"?

It means that GMAT doesn't expect you to do any funny decimal calculations. You won't find yourself multiplying 2.3 by 6.7. If the rate of work comes out to be 1/11, I would expect that the time for which you would have to work to complete it would be 22 hrs/33 hrs... etc

Since GMAT doesn't give you a calculator, it doesn't expect you to waste time doing these calculations in the test when you can easily use a calculator in real life. You don't need to have the skill to perform arduous calculations in a short span of time. You need to understand concepts and that's what they test. The numbers they use to test those concepts will be clean and easy to work with. Recognizing the concept will be the tough part.
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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23 Mar 2016, 14:10
let t=Albert's time to paint n rooms
Albert's rate=n/t
Bob's rate=n/(t-1)
Albert and Bob's rate=(3n/5)/(4/3)=9n/20
dividing out n, 1/t+1/(t-1)=9/20
t=5 hours
3t=15 hours
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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25 Aug 2017, 02:41
1
Since options are taking about hrs taken by Albert to paint 3n rooms, we can use options which are divisible by 3, thus working on option B and E,
Option B
If for 3n rooms, Albert takes 9 hrs, so for n rooms he takes 3 hrs. Thus Bob will take 2 hrs to paint n rooms
Now if n = 60
Rate of Albert = $$\frac{60}{3}$$ = 20 rooms per hour
Rate of Bob = $$\frac{60}{2}$$ = 30 rooms per hour
So combined rate = 20+30 = 50 rooms per hour
So together they complete 50* $$\frac{4}{3} = \frac{200}{3}$$ rooms which is not equal to $$\frac{3*60}{5}$$

Option E:
If for 3n rooms, Albert takes 15 hrs, so for n rooms he takes 5 hrs. Thus Bob will take 4 hrs to paint n rooms
Now if n = 60
Rate of Albert = $$\frac{60}{5}$$ = 12 rooms per hour
Rate of Bob = $$\frac{60}{4}$$ = 15 rooms per hour
So combined rate = 12+15 = 27 rooms per hour
So together they complete 27* $$\frac{4}{3}$$ = 36 rooms which is equal to $$\frac{3*60}{5}$$
Thus option E
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Albert and Bob are painting rooms at constant, but different  [#permalink]

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26 Aug 2017, 13:03
Sourabh2012 wrote:
Albert and Bob are painting rooms at constant, but different rates. Albert takes 1 hour longer than Bob to paint n rooms. Working side by side, they can paint a total of 3n/5 rooms in 4/3 hours. How many hours would it take Albert to paint 3n rooms by himself?

A. 7
B. 9
C. 11
D.13
E. 15

let n=1
A's rate=1/t+1
B's rate=1/t
A+B rate=(3/5)/(4/3)=9/20
because the ratio of A's rate to B's rate=t/t+1,
assume A's rate to B's rate=(4/20)/(5/20)
20/4=5 hours
3*5=15 hours
E
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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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29 Aug 2017, 17:20
Sourabh2012 wrote:
Albert and Bob are painting rooms at constant, but different rates. Albert takes 1 hour longer than Bob to paint n rooms. Working side by side, they can paint a total of 3n/5 rooms in 4/3 hours. How many hours would it take Albert to paint 3n rooms by himself?

A. 7
B. 9
C. 11
D.13
E. 15

We can let b = the number of hours Bob takes to paint n rooms; thus, Bob’s rate = n/b and Albert’s rate = n/(b + 1). Knowing that they complete 3n/5 rooms in 4/3 hours, we can create the following equation:

(n/b)(4/3) + [n/(b + 1)](4/3) = 3n/5

4n/3b + 4n/[3(b + 1)] = 3n/5

Multiplying the entire equation by 3b x 5 x (b + 1) = 15b(b + 1), we have:

4n[5(b + 1)] + 4n(5b) = 3n[3b(b + 1)]

Dividing by n, we have:

20(b + 1) + 20b = 9b(b + 1)

20b + 20 + 20b = 9b^2 + 9b

9b^2 - 31b - 20 = 0

(9b + 5)(b - 4) = 0

b = -5/9 or b = 4

Since b can’t be negative, b = 4. Thus, it takes Albert 4 + 1 = 5 hours to paint n rooms and 5 x 3 = 15 hours to paint 3n rooms.

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Re: Albert and Bob are painting rooms at constant, but different  [#permalink]

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Re: Albert and Bob are painting rooms at constant, but different   [#permalink] 25 Aug 2019, 07:00
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