# ACT problem

• Sep 17th 2009, 01:20 PM
mimib1230
ACT problem
Hey everyone! You may expect to see more of these...plenty more. I can't figure these out :/

2. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

12. Grace has 16 jellybeans in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of jellybeans she must take out of her pocket to ensure that she has one of each color?

A. 4
B. 8
C. 12
D. 13
E. 16
• Sep 17th 2009, 01:42 PM
tedii
Quote:

Originally Posted by mimib1230
Hey everyone! You may expect to see more of these...plenty more. I can't figure these out :/

2. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

3. Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

A. 12 minutes
B. 15 minutes
C. 21 minutes
D. 23 minutes
E. 28 minutes

For the first one you need to first figure out how far the first guy has traveled when the second cyclist starts biking. That is 3 hours times 6 miles per hour = 18miles. Now you set up the two equations for the bicyclist.

1. y=6x+30 : y=miles, x=hours
2. y=10x : " "

There are many methods to solve from here I will perform substitution.

10x=6x+30
4x=30
x=30/4=7.5

So E is your answer 7.5 hours.

I have to clock into work now so I'll have to leave the second question to someone else. Hope this helps and don't forget to thank (Happy)!
• Sep 17th 2009, 01:55 PM
mimib1230
Thanks!

I don't get where you get the equations though.

The ACT makes me feel so ridiculously stupid. I'm in AP Calculus with an A. And I can't do this stuff. Something about it just gets me stuck :/
• Sep 17th 2009, 02:10 PM
skeeter
Quote:

Originally Posted by mimib1230

2. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? A. 2 hours
b. 4 ½ hours
c. 5 ¾ hours
d. 6 hours
e. 7 ½ hours

rate times time = distance ... They both travel the same distance.

biker 1 ...

6t = d

biker 2 ...

10(t-3) = d

10(t-3) = 6t

solve for t

3. Jim can fill a pool carrying buckets of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

A. 12 minutes
b. 15 minutes
c. 21 minutes
d. 23 minutes
e. 28 minutes

jim's job rate is (1 job)/(30 min)

sue's job rate is (1 job)/(45 min)

tony's job rate is (1 job)/(90 min)

together ... (1/30) + (1/45) + (1/90) = (1 job)/(15 min)

...
• Sep 18th 2009, 04:54 AM
tedii
Quote:

Originally Posted by tedii

1. y=6x+30 : y=miles, x=hours
2. y=10x : " "

There are many methods to solve from here I will perform substitution.

10x=6x+30
4x=30
x=30/4=7.5

y=6x+30 literally is (bike 1) rides (Y) miles when they ride 6 miles per hour times (X) plus 30 miles.

y=10x literally is (bike 2) rides (Y) miles when they ride 10 miles per hour times (X).

10x=6x+30 is after sticking Y=10x into y=6x+30

Does this help?
• Sep 18th 2009, 05:50 AM
Wilmer
Quote:

Originally Posted by mimib1230
2. Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
A. 2 hours
B. 4 ½ hours
C. 5 ¾ hours
D. 6 hours
E. 7 ½ hours

.......18 miles........C1........x miles.............>t hours (6mph)
C2....................x + 18 miles...................>t hours (10mph)

When C2 starts, C1 is 18 miles ahead : 3 hours at 6 mph

Question is: how long after C2 starts

distance = speed * time; so:
C1: x = 6t
C2: x+18 = 10t ; x = 10t - 18
10t - 18 = 6t
4t = 18
t = 18/4
t = 4 1/2 hours ; not 7 1/2