Question

At noon, Garrett left Magnolia and headed North at 10 kph. At 2 p.m., Ben left Magnolia and headed North. If Ben was 15 km ahead of Garrett at 7 p.m., how fast was Ben
traveling?
Choose one answer.
O
a. 17 kph
o
b. 12 kph
c. 21 kph
Ο Ο
d. 16 kph

Answer 1

Answer:

Step-by-step explanation:

This is simple, but the extra numbers given in the form of the specific times might throw you off.

If Garrett leaves at noon and at 7 Ben is some distance ahead of him, that means that Garrett has been driving for 7 hours. Ben left at 2, so at 7 pm he has been driving for 5 hours. That's part of what's confusing. We'll put that in a table to hopefully make things easier:

              d      =       r      *        t

G                                             7

B                                             5

We also know that Garrett is driving at 10 km/h, so:

              d        =        r        *        t

G                               10       *        7

B                                 r                 5

The r is because Ben's rate is our unknown. Look at the top of the table. That is the formula we are going to use to solve this problem: d = rt.

If Garrett drives for 7 hours at 10 km/hr, then the distance he has traveled is 70 km (that's found by multiplying the rate of 10 km/h by the time of 7 hours). Ben's rate, along those same lines of reasoning, is 5r. Fill that in:

            d        =        r        *        t

G         70       =       10       *       7

B         5r        =        r        *       5

Ok now the table is filled out. Let's look at the rest of the problem. It says that at 7 pm Ben's distance is 15 km more than Garrett's distance. The words "more than" indicate addition. In words that is

"Ben's distance is Garrett's distance plus 15 km" which translates to, mathematically speaking:

5r = 70 + 15 and

5r = 85 so

r = 17

Ben's rate is 17 km/h, choice a.