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3 years ago

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It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular

train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?
(A) z(y – x)/x + y
(B) z(x – y)/x + y
(C) z(x + y)/y – x
(D) xy(x – y)/x + y
(E) xy(y – x)/x + y

1 answer:

STatiana [176]3 years ago

7 0

Answer:

B

Step-by-step explanation:

To solve this, we use ratio.

Firstly, we need to know the number of hours traveled. The total number of hours traveled = x+y

Ratio of this used by high speed train = x/(x +y).

Total distance traveled before they meet = [x/(x + y)] × z

For low speed train = [y/(x + y)] × z.

The difference would be distance by high speed train - distance by low speed train.

= z [ (x - y)/x + y)]

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The first step that we must take before attempting to solve the problem is to understand what the problem is asking us to do and what is given to us to help accomplish that goal. Although it does not explicitly state that we must solve for t, this is usually what the problem statement would be asking if we just receive and expression like this. What is given to us to accomplish that goal is the expression $4.9t+0.3=5.6t+0.72$ .

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