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Tom reads at an average speed of 30 pages per hour while Jas

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by imane81 » Sun Feb 07, 2010 1:15 pm
Your input is much appreciated in this one...thank you

Tom reads at an average speed of 30 pages per hour while Jason reads at an average speed of 40 pages per hour. If Tom starts at 4:30 and Jason at 5:20, at what time will they be reading the same page?
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Source: — Problem Solving |

by shashank.ism » Sun Feb 07, 2010 1:16 pm
imane81 wrote:Your input is much appreciated in this one...thank you

Tom reads at an average speed of 30 pages per hour while Jason reads at an average speed of 40 pages per hour. If Tom starts at 4:30 and Jason at 5:20, at what time will they be reading the same page?
you can see here time lag between Tom and Jason is 30+20 = 50 minutes = 50/60hrs. =5/6 hrs.

so let after X hrs from 5: 20 they are reading same page.
so Jason read 40X pages and Tom read 30(X+5/6) pages
both are equal i.e. 40X = 30(X+5/6) --> 10X =25 --> X = 2.5 hrs = 2hrs 30 min
So the time at which they are reading same page = 5:20 + 2:30 =7:50 Ans.
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by harsh.champ » Sun Feb 07, 2010 1:41 pm
imane81 wrote:Your input is much appreciated in this one...thank you

Tom reads at an average speed of 30 pages per hour while Jason reads at an average speed of 40 pages per hour. If Tom starts at 4:30 and Jason at 5:20, at what time will they be reading the same page?

Hey Imane 81 ,
Can you plz also post the answer choices while posting the question as it helps to apply and practice the elimination strategy which can come handy on the exam day.


As for the question,


Tom starts at 4:30 whereas Jason at 5:20 .
Let the time taken by Jason to reach the page be x minutes.
Then the time taken by Tom will be x+50 minutes.
So,40x = 30(x + 50)
Hence,x=[spoiler]150minutes or 2:30hrs[/spoiler] and the clock time would be [spoiler]5:20+ x = 07:50 hours[/spoiler].
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by shashank.ism » Sun Feb 07, 2010 10:40 pm
harsh.champ wrote: Hey Imane 81 ,
Can you plz also post the answer choices while posting the question as it helps to apply and practice the elimination strategy which can come handy on the exam day.


As for the question,


Tom starts at 4:30 whereas Jason at 5:20 .
Let the time taken by Jason to reach the page be x minutes.
Then the time taken by Tom will be x+50 minutes.
So,40x = 30(x + 50)
Hence,x=[spoiler]150minutes or 2:30hrs[/spoiler] and the clock time would be [spoiler]5:20+ x = 07:50 hours[/spoiler].
Yeah Harsh you are saying very right. Options should be posted so that we can try elimination process if required. This way we can have track on our time management.
Well if there is some conceptual problem you can post question without any options.
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by Stuart@KaplanGMAT » Sun Feb 07, 2010 11:49 pm
imane81 wrote:Your input is much appreciated in this one...thank you

Tom reads at an average speed of 30 pages per hour while Jason reads at an average speed of 40 pages per hour. If Tom starts at 4:30 and Jason at 5:20, at what time will they be reading the same page?
Here's a key rule for this type of question:

if two objects are travelling in the same direction, to find the relative rate, SUBTRACT the rates;

if two objects are travelling in opposite directions, to find the total rate, ADD the rates.

In this question, Tom and Jason are travelling in the same direction (i.e. both reading the book starting on the same page and moving toward the end of the book), so we SUBTRACT to get the relative rates.

40-30 = 10,

so Jason catches up 10 pages every hour.

Now that we have the rate, we can calculate the distance Jason needs to cover to catch up:

Tom has been reading for 50 minutes; 30 p/h is .5 p/m, so .5(50) = 25 pages.

time = distance/rate, so:

time to catch up = distance to catch up / rate of catching up

t = 25/10 = 2.5 hours

Tom starts at 5:20, so 5:20 + 2:30 = 7:50 for our catch up moment.

The other approaches posted certainly work (and if you see them may even be faster), but you can use the approach here to reason your way through any similar question.
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