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At the same time Fredrick started walking toward...

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At the same time Fredrick started walking toward...

by AAPL » Wed Jan 24, 2018 6:36 am
At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Bernard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes

The OA is D.

Is there a strategic approach to this question? Can any experts help, please?

I can solve it of the following way,

I know the total distance, it is 70 blocks.

And can I say that the total speed will be the sum of the Fredrick and Bernard's speed? Then, the total speed is 8 + 6 = 14 blocks every ten minutes or 1.4 blocks per minute.

We know the formula to determine the speed,
$$Speed=\frac{Dist}{Time}$$
Then, the time,in minutes, that did it take them to meet will be,
$$Time=\frac{Dist}{Speed}$$ $$Time=\frac{Dist}{Speed}=\frac{70}{1.4}=50$$
Thanks.
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by ErikaPrepScholar » Wed Jan 24, 2018 6:46 am
Hey AAPL, this method definitely works. You could also approach it by building an equation.

Each person walks for the same amount of time before they meet: let's call that time t. This is what we are trying to solve for.

Using Distance = Rate * Time, this means that Frederick walked a total of $$\frac{6}{10}t\ blocks$$ while Bernard walked a total of $$\frac{8}{10}t\ blocks$$

If Frederick and Bernard walked 70 blocks in total, this means that
$$\frac{6}{10}t+\frac{8}{10}t\ =\ 70$$
Then we can simplify $$\frac{14}{10}t\ =\ 70$$ $$14t\ =\ 700$$ $$t\ =\ 50$$

So the time before Frederick and Bernard meet is 50 minutes, or answer choice D.
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by Jeff@TargetTestPrep » Fri Jan 26, 2018 9:55 am
AAPL wrote:At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Bernard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes
This is a converging rate problem in which we can use the following formula:

distance of Fredrick + distance of Bernard = total distance

Since Fredrick walks 6 blocks every 10 minutes, his rate is 6/10 = 3/5 blocks per minute. Since Bernard walks 8 blocks every 10 minutes, his rate is 8/10 = 4/5 blocks per minute. Since distance = rate x time, and if we let t = the time they travel before they meet, Frederick's distance is (3/5)t and Bernard's distance is (4/5)t. Thus:

(3/5)t + (4/5)t = 70

7t/5 = 70

7t = 350

t = 50

Alternate Solution:

Initially, there is a distance of 70 blocks between Frederick and Bernard. In 10 minutes, Frederick travels 6 blocks and Bernard travels 8 blocks; therefore the number of blocks between them decreases by 6 + 8 = 14 blocks every 10 minutes. It will take 70/14 = 5 of these 10 minute walks for the distance to become zero; thus it will take 5 x 10 = 50 minutes for them to meet.

Answer: D

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