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M13-28

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M13-28 [#permalink]

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Working together, John and Jack can type 20 pages in one hour. If they would be able to type 22 pages in one hour provided Jack increased his typing speed by 25%, what is the ratio of Jack's normal typing speed to that of John?

A. \(\frac{1}{3}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{5}\)
[Reveal] Spoiler: OA

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Re M13-28 [#permalink]

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New post 15 Sep 2014, 23:50
Official Solution:

Working together, John and Jack can type 20 pages in one hour. If they would be able to type 22 pages in one hour provided Jack increased his typing speed by 25%, what is the ratio of Jack's normal typing speed to that of John?

A. \(\frac{1}{3}\)
B. \(\frac{2}{5}\)
C. \(\frac{1}{2}\)
D. \(\frac{2}{3}\)
E. \(\frac{3}{5}\)


Let the rate of John be \(x\) pages per hour and the rate of Jack \(y\) pages per hour. Then as we can sum the rates and \(rate*time=job\) then \((x+y)*1=20\). So, we have that \(x+y=20\);

"They would be able to type 22 pages in one hour provided Jack increased his typing speed by 25%": \((x+1.25y)*1=22\). So, we also have that \(x+1.25y=22\);

Question: \(\frac{y}{x}=?\)

Subtract (1) from (2): \(x+1.25y-(x+y)=22-20\), which gives \(y=8\), so \(x=12\). Therefore \(\frac{y}{x}= \frac{8}{12}=\frac{2}{3}\).

OR: if by increasing the rate of Jack by 25% 2 more pages can be typed in one hour, then we can directly write: \(0.25y=2\), which gives \(y=8\) and \(x=12\). Therefore \(\frac{y}{x}=\frac{8}{12}=\frac{2}{3}\).


Answer: D
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Re: M13-28 [#permalink]

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New post 02 Nov 2014, 07:16
we know that 25% of jack's speed is 2 then its total speed must be 8 and remaining 12 is speed of john.. 8/12 =2/3
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Re: M13-28 [#permalink]

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New post 20 May 2016, 12:37
awal_786@hotmail.com wrote:
we know that 25% of jack's speed is 2 then its total speed must be 8 and remaining 12 is speed of john.. 8/12 =2/3


Is this an accurate approach? I feel like you can't simply say 1/20 to 1/22 is 2 pages though I'm not sure why exactly that would be wrong.
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Re: M13-28 [#permalink]

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New post 21 May 2016, 20:43
redfield wrote:
awal_786@hotmail.com wrote:
we know that 25% of jack's speed is 2 then its total speed must be 8 and remaining 12 is speed of john.. 8/12 =2/3


Is this an accurate approach? I feel like you can't simply say 1/20 to 1/22 is 2 pages though I'm not sure why exactly that would be wrong.


Hi,

Ofcourse getting 2 from 1/20 and 1/22 is not correct..
But here they initially do 20 pages in one hour so their combinrd speed = 20pages per hour ..

and after an increase of 25% in jack's speed, 2 pages more are created in 1 hour...
from above sentence we can deduce that 25% of jack's speed can create 2 pages.... OFCOURSE speed has to be in pages per hour..

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Re: M13-28 [#permalink]

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New post 17 Apr 2017, 02:15
How did we sum up the rates? From what I understand, we can invert the two rates and then add them up to make combined rate. Im sure Im missing some basic concept here. Please help me out.
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Re: M13-28 [#permalink]

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New post 18 Apr 2017, 06:53
TheMastermind wrote:
How did we sum up the rates? From what I understand, we can invert the two rates and then add them up to make combined rate. Im sure Im missing some basic concept here. Please help me out.


THEORY
There are several important things you should know to solve work problems:

1. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance in rate problems.

\(time*speed=distance\) <--> \(time*rate=job \ done\). For example when we are told that a man can do a certain job in 3 hours we can write: \(3*rate=1\) --> \(rate=\frac{1}{3}\) job/hour. Or when we are told that 2 printers need 5 hours to complete a certain job then \(5*(2*rate)=1\) --> so rate of 1 printer is \(rate=\frac{1}{10}\) job/hour. Another example: if we are told that 2 printers need 3 hours to print 12 pages then \(3*(2*rate)=12\) --> so rate of 1 printer is \(rate=2\) pages per hour;

So, time to complete one job = reciprocal of rate. For example if 6 hours (time) are needed to complete one job --> 1/6 of the job will be done in 1 hour (rate).

2. We can sum the rates.

If we are told that A can complete one job in 2 hours and B can complete the same job in 3 hours, then A's rate is \(rate_a=\frac{job}{time}=\frac{1}{2}\) job/hour and B's rate is \(rate_b=\frac{job}{time}=\frac{1}{3}\) job/hour. Combined rate of A and B working simultaneously would be \(rate_{a+b}=rate_a+rate_b=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\) job/hour, which means that they will complete \(\frac{5}{6}\) job in one hour working together.

3. For multiple entities: \(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}\), where \(T\) is time needed for these entities to complete a given job working simultaneously.

For example if:
Time needed for A to complete the job is A hours;
Time needed for B to complete the job is B hours;
Time needed for C to complete the job is C hours;
...
Time needed for N to complete the job is N hours;

Then: \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+...+\frac{1}{N}=\frac{1}{T}\), where T is the time needed for A, B, C, ..., and N to complete the job working simultaneously.

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that \(t_1\) and \(t_2\) are the respective individual times needed for \(A\) and \(B\) workers (pumps, ...) to complete the job, then time needed for \(A\) and \(B\) working simultaneously to complete the job equals to \(T_{(A&B)}=\frac{t_1*t_2}{t_1+t_2}\) hours, which is reciprocal of the sum of their respective rates (\(\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}\)).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

\(T_{(A&B&C)}=\frac{t_1*t_2*t_3}{t_1*t_2+t_1*t_3+t_2*t_3}\) hours.

Theory on work/rate problems: http://gmatclub.com/forum/work-word-pro ... 87357.html

All DS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=46
All PS work/rate problems to practice: http://gmatclub.com/forum/search.php?se ... &tag_id=66


Hope this helps
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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.


What are GMAT Club Tests?
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Re: M13-28   [#permalink] 18 Apr 2017, 06:53
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