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Half an hour after Car A started traveling from Newtown to

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Half an hour after Car A started traveling from Newtown to [#permalink]

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New post 17 May 2013, 08:48
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Half an hour after Car A started traveling from Newtown to Oldtown, a distance of 62 miles, Car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after Car B started it's trip. If Car A traveled at a constant rate that was 8 miles per hour greater than Car B's constant rate, how many miles had Car B driven when they met?

A. 14 miles
B. 12 miles
C. 10 miles
D. 9 miles
E. 8 miles

You're suposed to convert the hours into fractions (i.e. Car A traveled for .75 hours and Car B traveled for .25 hours) but why can't I solve just using minutes (i.e. 45 and 15 minutes respectively)?


Thanks!
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 May 2013, 16:26, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Distance/Rate Problem [#permalink]

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\(space = speed * time\)

A travels 45 minutes => \(\frac{3}{4}\) hours
B travels 15 minutes => \(\frac{1}{4}\) hours

speedA=speedB+8

In this time A travels \((Sb+8)*\frac{3}{4}\)
In this time B travels \((Sb)*\frac{1}{4}\)

The sum must equal 62 km, so \((Sb+8)*\frac{3}{4}+(Sb)*\frac{1}{4}=62\) B=56 km/h

\(space=56*\frac{1}{4}=14\)

WholeLottaLove wrote:
You're suposed to convert the hours into fractions (i.e. Car A traveled for .75 hours and Car B traveled for .25 hours) but why can't I solve just using minutes (i.e. 45 and 15 minutes respectively)?


Here the speed is in km pour HOUR, if you want to work with minutes you have to convert everything into MINUTES

You must work with similar "quantities" => all in hours or all in minutes, cannot mix the two.
Hope it's clear
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Re: Distance/Rate Problem [#permalink]

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WholeLottaLove wrote:
Half an hour after car A started traveling from Newtown to Oldtown, a distance of 62 miles, car B started traveling along the same road from Oldtown to Newtown. The cars each met each other on the road 15 minutes after car B started it's trip. If Car A traveled at a constant rate that was 8 MPH greater than car B's constant rate, how many miles had car B driven when they met?

A) 14
B) 12
C) 10
D) 9
E) 8

You're suposed to convert the hours into fractions (i.e. Car A traveled for .75 hours and Car B traveled for .25 hours) but why can't I solve just using minutes (i.e. 45 and 15 minutes respectively)?


Thanks!


Using Fractions
First case:-

Car A started first and travelled for half an hour.

Distance travelled by car A = 0.5A (A and B are the speeds of the respective cars)

Second case:-
Next They travel for 15 min and meet each other on the road.

During this time

Distance travelled by car A = 0.25 A
Distance travelled by car B = 0.25B


Now summing all the distances wee should get 62 miles

So 0.25A + 0.25B + 0.5A = 62

But A = B+8

So we will get A as 64 and B as 56. So distance travelled by car B = 56 *0.25 = 14

In Minutes.

First case:- Distance travelled by car A = 30 * A (A is in miles per minute)

Second Case

Distance travelled by A = 15*A
Distance travelled by B = 15*B

So total distance travelled = 62 = 15*A + 15*B + 30*A
=> 62 = 45A + 15B
= 15(3A + B)

But A = B + 8/60(miles/minute)

So 62 = 15(4B + 2/5)

=> 62 = 6 + 4(15B)

=> 15B = 14.

Now distance travelled by Car B = 15 * B

So answer is 14
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Re: Distance/Rate Problem [#permalink]

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New post 17 May 2013, 09:33
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In questions where you have the speeds given in miles/hour, its convenient to convert the time from minutes to hours.
The one who started half an hour (or 30 mins.) earlier must have traveled for 30+15=45 mins when he meets the other person driving from oldtown.
Now it is always easy for problems like these(where you have figures like 15 mins. or 45 mins) to convert time expressed in mins to hours.
45 mins=3/4 of an hr
15 mins=1/4 of an hr.
Now, distance traveled by the one who started earlier when he meets the other on his way=\(\frac{3.(x+8)}{4}\) miles
where x-speed of the person( in miles/hr) who started later.
When the meeting happens, the one who started late would have traveled \(\frac{x}{4}\) miles
Hence, \(\frac{3.(x+8)}{4} + \frac{x}{4}=62\)
Solving, we get x=56.
Therefore, the answer would be \(\frac{x}{4}=\frac{56}{4}=14\)
I hope this answers your question. :)
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Re: Half an hour after Car A started traveling from Newtown to [#permalink]

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New post 08 May 2014, 13:17
Here's how I did this one

\((2B+8)(1/4) = (62- ((B+8)/2)\)

Solving we get \(B=56\)

Therefore, \(56/4= 14\)

Answer: A
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Re: Half an hour after Car A started traveling from Newtown to [#permalink]

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Re: Half an hour after Car A started traveling from Newtown to [#permalink]

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New post 03 Sep 2016, 02:23
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Re: Half an hour after Car A started traveling from Newtown to [#permalink]

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New post 03 Sep 2016, 14:41
Half an hour after car A started traveling from Newtown to Oldtown, a distance of 62 miles, car B started traveling along the same road from Oldtown to Newtown. The cars each met each other on the road 15 minutes after car B started it's trip. If Car A traveled at a constant rate that was 8 MPH greater than car B's constant rate, how many miles had car B driven when they met?

let b/4=B's miles
b/4=62-(b+8)(3/4)
b/4=14 miles
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Re: Half an hour after Car A started traveling from Newtown to [#permalink]

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New post 12 Nov 2016, 18:58
My way of solving by substitution


speed of car B (Y) = x/0.25
speed of car A(Y+8) = 62-x/0.75

x/0.25 + 8 = 62-x/0.75

x=14 miles

Hope its clear!!
Re: Half an hour after Car A started traveling from Newtown to   [#permalink] 12 Nov 2016, 18:58
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