Half an hour after Car A started traveling from Newtown to Oldtown, a

Question

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

Thanks![/spoiler]

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!

Zarrolou · Accepted Answer

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

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

SrinathVangala · Answer

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

GMATtracted · Answer

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.

jlgdr · Answer

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

gracie · Answer

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

let b/4=B's miles
b/4=62-(b+8)(3/4)
b/4=14 miles

Nitu27 · Answer

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!!

ScottTargetTestPrep · Answer

Let’s let the rate of Car B be v. Then, the rate of Car A is v + 8.

Since Car A had been traveling for 30 minutes when Car B started its trip and since the two cars met 15 minutes after, the total travel time for Car A is 30 + 15 = 45 minutes = 3/4 hour, and the total travel time for Car B is 15 minutes = 1/4 hour. Since the sum of the distances traveled by the two cars must equal the total distance between the two towns, which is 62, we can create the following equation:

(v + 8)(3/4) + v(1/4) = 62

Let’s multiply each side of this equation by 4:

(v + 8)(3) + v = 248

3v + 24 + v = 248

4v = 224

v = 56

Since the rate of Car B is v = 56 mph and since Car B traveled for 1/4 hour when the two cars met, Car B traveled 56 * 1/4 = 14 miles.

Answer: A

AcceptMePLZ · Answer

Why do we know we can add these two rate*times of the two cars together to get 62? The problem does not say that if we were to add both, they would reach the end of the road so how is this assumed?

Eternal · Answer

From question we know that both cars started from the opposite ends and before meeting Car A had traveled for 45 mins and Car B had traveled for 15 min.
So the total Distance 62 miles will be equal to the sum of their distances traveled.

RastogiSarthak99 · Answer

took me a while to figure out. But the main trick here is to:

add both the distances together

(8 +x) * 3/4 + x/4 = 62

x = 56

Distance covered by B is 56/4 which is 14.

A

Aboyhasnoname · Answer

Back Solving....Let the distance be = x ...

So Speed of Car is B ....4x miles/hr (As x travelled in 15mins)
Speed of Car A is 4x+8 miles/hr....

Car A travels for 3/4 hours and Car b Travels for 1/4 hours ...to cover a total of 62 miles... 
So (4x+8) x 3/4 + (4x) x 1/4 = 62 
3x + 6 + x = 62 
4x = 56 | x = 14......

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