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Working together, printer A and printer B would finish the

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Working together, printer A and printer B would finish the  [#permalink]

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New post Updated on: 06 Sep 2012, 09:01
6
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A
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D
E

Difficulty:

  55% (hard)

Question Stats:

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Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500

I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !

Originally posted by gauravnagpal on 01 Sep 2010, 10:12.
Last edited by Bunuel on 06 Sep 2012, 09:01, edited 2 times in total.
Edited the question.
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Re: solution required  [#permalink]

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New post 01 Sep 2010, 11:56
16
13
gauravnagpal wrote:
.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600
b. 800
c. 1000
d. 1200
e. 1500

answer: A
I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\);
Printer A alone would finish the task in 60 minutes --> \(60a=x\);
Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.
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Re: solution required  [#permalink]

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New post 03 Sep 2010, 10:26
16
9
B in a minute=x/40
A in a minute=x/60
then,
x/40-x/60=5

Solving x=600
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Re: solution required  [#permalink]

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New post 06 Sep 2012, 07:09
1
Bunuel wrote:
gauravnagpal wrote:
.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600
b. 800
c. 1000
d. 1200
e. 1500

answer: A
I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\);
Printer A alone would finish the task in 60 minutes --> \(60a=x\);
Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.


Hi Bunnel,

I have always one confusion in rate and work problems can you please clarify this?,

When do we add rates i.e what you did above...... \(24(a+b)=x\);

and when do we divide by rates i.e something . rate =(Job done/ time)

Regards

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Re: solution required  [#permalink]

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New post 06 Sep 2012, 09:39
kotela wrote:
Bunuel wrote:
gauravnagpal wrote:
.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600
b. 800
c. 1000
d. 1200
e. 1500

answer: A
I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\);
Printer A alone would finish the task in 60 minutes --> \(60a=x\);
Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.


Hi Bunnel,

I have always one confusion in rate and work problems can you please clarify this?,

When do we add rates i.e what you did above...... \(24(a+b)=x\);

and when do we divide by rates i.e something . rate =(Job done/ time)

Regards

Srinath


You can denote rate directly by some variable (a in the solution) or express rate as a reciprocal of time. For example, say printer A needs t minutes to print 1 page and printer B needs m minutes to print 1 page, then the rate of printer A would be job/time=1/t pages per minute and the rate of printer B would be 1/m pages per minute (rate is a reciprocal of time, so 1/t=a and 1/m=b). In this case the equation would be 24(1/t+1/m)=x.

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 08 Sep 2012, 04:51
3
First thing i did was i found out time that is required to B to complete the task alone: 1/24-1/60=3/120=1/40. Then i looked at the information which states that the rate of B is 5+ page than that of A so, lets say x is the number of pages printed by A per minute, so the task consists of 60*x or 40*(x+5) pages. I can make an equation: 60x=40(x+5), 20x=200, x=10, total number of pages is 60*10=600 or 40*15=600

Answer is A.

It is clear but it took me about 3 min to do it, does it because i am doing it slow or i am using longer route?
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 15 Nov 2012, 04:37
2
4
\(\frac{1}{A}=\frac{1}{60}\)
\(\frac{1}{B}+\frac{1}{A}=\frac{1}{24}\)

Get: \(\frac{1}{B}\)

\(\frac{1}{B}=\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\)

Let p be the number of pages produced by A.
Let p+5 be the number of pages produced by B.

\(24(p + p+5) = 60(p)==> p=10pages\)

Answer: 60(p)=600pages
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 03 Mar 2013, 23:40
2
gauravnagpal wrote:
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500

I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


total time taken by B = 24 * 60 / (60 -24) = 40 min.

A take 60 min. B takes 40 min to complete a task.

Now, divide the values given in option (in Ans) to get the rate per min.

option A: 600 / 10 = 60 & 600/40 = 15...> this satisfies the condition given in question stem that printer B prints 5 pages a minute more than printer A ?
. therefore A
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 21 Sep 2013, 10:31
1
2
Ta = 60min
Ra = 1/Ta = 1/60
Rb = 1/Tb

Combined task completion time 24min.
=Ra + Rb
=1/60 + 1/Tb = 1/24
Tb = 40 min.

Ra = X/Ta Rb = X/Tb

Ra + 5 = Rb
X/Ta + 5 = X/Tb
X/60 + 5 = X/40
X=600 Ans.
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 25 Sep 2013, 03:09
mbaiseasy wrote:
Let p be the number of pages produced by A.
Let p+5 be the number of pages produced by B.

\(24(p + p+5) = 60(p)==> p=10pages\)

Answer: 60(p)=600pages



How do you come up with these??
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Re: solution required  [#permalink]

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New post 20 Nov 2013, 15:01
Bunuel wrote:
gauravnagpal wrote:
.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600
b. 800
c. 1000
d. 1200
e. 1500

answer: A
I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\);
Printer A alone would finish the task in 60 minutes --> \(60a=x\);
Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.



I'm curious about the 1/a+1/b=1/24 solution as well. I started out trying that way and got stuck at the same point as the other fellow. Is there any way to solve it once you have 1/60+1/40 for their combined rates? Or is that just a dead end?


edit: Also, question #2:

Shouldn't the equation be 24(1/a+1/b)=x; ? Since you have 24 minutes in which the machines are working at their individual rates, doing 1/a and 1/b of the job per minute? I don't get how one can know when to arbitrarily use "a" instead of '1/a" or "b" instead of "1/b"
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Re: solution required  [#permalink]

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New post 21 Nov 2013, 01:40
1
AccipiterQ wrote:
Bunuel wrote:
gauravnagpal wrote:
.Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

a. 600
b. 800
c. 1000
d. 1200
e. 1500

answer: A
I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Let the rate of printer A be \(a\) pages per minute, the rate of printer B be \(b\) pages per minute and whole task be \(x\) pages.

\(time*rate=job \ done\):

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(a+b)=x\);
Printer A alone would finish the task in 60 minutes --> \(60a=x\);
Printer B prints 5 pages a minute more than printer A --> \(b=a+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.

Some work problems with solutions:
time-n-work-problem-82718.html?hilit=reciprocal%20rate
facing-problem-with-this-question-91187.html?highlight=rate+reciprocal
what-am-i-doing-wrong-to-bunuel-91124.html?highlight=rate+reciprocal
gmat-prep-ps-93365.html?hilit=reciprocal%20rate
questions-from-gmat-prep-practice-exam-please-help-93632.html?hilit=reciprocal%20rate
a-good-one-98479.html?hilit=rate

Hope it helps.



I'm curious about the 1/a+1/b=1/24 solution as well. I started out trying that way and got stuck at the same point as the other fellow. Is there any way to solve it once you have 1/60+1/40 for their combined rates? Or is that just a dead end?


edit: Also, question #2:

Shouldn't the equation be 24(1/a+1/b)=x; ? Since you have 24 minutes in which the machines are working at their individual rates, doing 1/a and 1/b of the job per minute? I don't get how one can know when to arbitrarily use "a" instead of '1/a" or "b" instead of "1/b"


In this case we'd have:
The rate of printer A = 1/a pages per minute, where a is the time to print 1 page.
The rate of printer B = 1/b pages per minute, where b is the time to print 1 page.

Working together, printer A and printer B would finish the task in 24 minutes" --> \(24(\frac{1}{a}+\frac{1}{b})=x\);
Printer A alone would finish the task in 60 minutes --> \(60*\frac{1}{a}=x\);
Printer B prints 5 pages a minute more than printer A --> \(\frac{1}{b}=\frac{1}{a}+5\).

Solving for \(x\) --> \(x=600\).

Answer: A.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 18 Mar 2014, 06:00
2
1
Rate A= X
Rate B= X+5

Work(A)=> X * 60 = 60X

Rate(A+B) * 24 = Work

(2X+5) * 24 = 60X

X=10
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 22 Apr 2014, 20:56
A does in one minute =x pages
Therefore, in 60 minutes=60x pages

B does in one minute= x+5
In 24 minutes both do 60x pages
24x+24(x+5)=60x
X=10 total work=60*10=600 pages
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 15 May 2014, 10:02
2
2
Easiest way to do this: Machine A and B can do the task in 24 minutes thus Rate of A and B = 1/24. Now given A can do the task in 60 minutes therefore Rate of A= 1/60. We know that Rate of A and B = Rate of A + Rate of B therefore Rate of B= Rate of A and B - Rate of A = 1/24-1/60= 1/40. Now we know that Rate of B = 1/40 thus B can do the work in 40 minutes.

Let pages printed per minute by A = x, given that pages printed by B per minute is 5 more than that of A
Pages printed by B per minute = x+5
Now Complete task is done by A in 60 minutes therefore total number of pages printed by A = x * 60
Also Complete task is done by B in 40 minutes therefore total number of pages printed by B = (x+5) * 40
therefore x * 60 = (x+5) * 40
therefore x=10
thus the total number of pages in task = x*60 = 10*60 = 600 :-D
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 18 Jul 2016, 04:29
1
gauravnagpal wrote:
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500

I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !



Okay..this is how I did it..
Let the task(number of pages) be 120x(LCM of all numbers given in the problem)..

A and B take 24 minutes to complete it..thus, pages per min = 5x
A's pages per minute = 2x
B's pages per minute = 3x

Difference
3x - 2x = 5
=> x = 5
Thus, 120x = 600..(A)

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New post 25 Feb 2017, 19:30
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gauravnagpal wrote:
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500

I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !


Another approach after getting rate for B
as rate of B = 1/40
let A prints x pages a minute , then B will print x+5pages a minute
A work for 60 minutes and B work for 40 minutes alone and so ,they are able to print same no. of pages
thus,
40(x+5) = 60x
40x+200 =60x
20x=200
x=10
thus total no of pages = 60x or 40(x+5) =600 pages
Ans A
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New post 03 Jul 2017, 03:48
gauravnagpal wrote:
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500

I know this question is relatively symol if make an equation in one vaibale ...
I tried to do it by applying the fundamental of A = Jobs per min * time ( the way we typically solve the work problems ) and i was stuck

I did jobs per minute A , 1/60
combined rate = 1/24

so rate of b = 1/24 - 1/60 = 1/40

but could not arrive at the solution ... i tried to form the equation by assuming x as the total numbe of pages So x/60+ x+5/40 = cld nt take ot forward from here
kindly see where am I losing the track !

Given that, A can finish in 60 minutes, combined 24 minutes.
so, the rate of B = 1/24- 1/60 equals 1/40 thus B takes 40 minutes to accomplish the job alone.
As B can finish 5 more pages than A in a minute. Let, A can print x pages per minutes then, B = x+5
60x = 40( x+5)
x= 10
Thus total number of pages together can print = 60 *10

A. 600
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New post 06 Aug 2018, 07:21
gauravnagpal wrote:
Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500


Let the total work be 120 units

So, Efficiency of A and B is 5 units/min & Efficiency of A is 2 units/min

Thus , the efficiency of B is 3 units/min

So, 1unit/min = 5 Pages

Hence, 120 Units = 120*5 => 600 Units, Answer must be (A)
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Re: Working together, printer A and printer B would finish the  [#permalink]

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New post 10 Dec 2018, 12:23
[quote="gauravnagpal"]Working together, printer A and printer B would finish the task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A ?

A. 600
B. 800
C. 1000
D. 1200
E. 1500


Total task = 60A
B = A+5
24(A+B) = 60A
24(A+A+5) = 60A
24(2A+5)=60A
48A+120= 60A
120=12A
A=10

Total tasks =60A = 600. Check the answers to confirm

Hence option A is the answer.
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Re: Working together, printer A and printer B would finish the &nbs [#permalink] 10 Dec 2018, 12:23

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