Caitilin Posted January 7, 2013 Share Posted January 7, 2013 Here's the question: Sam and Terry ran on the treadmill. Sam started 10 minutes before Terry. Terry finished 5 minutes before Sam. They both ran the same distance. When Terry finished, Sam had completed 4/5 of the journey and still had 1/2 mile to go. Find Sam's speed. Find Terry's speed. I don't feel like we have enough information to do this, or if we do, it's hidden in some way I can't see. :blushing: Any help? Quote Link to comment Share on other sites More sharing options...

Kathy in Richmond Posted January 7, 2013 Share Posted January 7, 2013 Start here: "When Sam has completed 4/5 of his journey, he still has 1/2 mile to go." How much time does it take Sam to complete that last 1/2 mile? Then, knowing the time and distance for Sam's final stretch, can you figure out Sam's speed? Quote Link to comment Share on other sites More sharing options...

Kate in Arabia Posted January 7, 2013 Share Posted January 7, 2013 I needed to draw a bar diagram. Sam |----------------------------------------------------------------------| |---------------------------|-----------------------------|-------------| Terry If Sam finished 4/5 by the time Terry finished and ran for another 5 min, that means the last segment -- 1/5 -- took him 5 min. So the whole distance took 25 min. If he had 1/2 mile to go, the entire distance was 5*1/2 = 2 1/2 miles. So 2.5 miles/25 min, or 1/10 mile/min (do you need in in miles per min?) If Sam started 10 min before Terry and finished 5 min after him, and Sam's total time was 25 min, then Terry's total time was 25-15=10 min. So 2.5 miles/10 min, or 1/4 mile/min. Is that the right answer, lol? (Sorry, I can't get my bar diagram to look right..) Quote Link to comment Share on other sites More sharing options...

Dealea86 Posted January 7, 2013 Share Posted January 7, 2013 Sam ran for 15 minutes more than Terry. So, S = T + 15, where S and T refer to the time each boy spent running. Both boys ran the same distance, which I will call d. Terry ran distance d in T minutes. Sam ran distance d in S minutes. Speed = distance/time. I sill call Terry's speed t and Sam's speed s. t = d/T and s = d/S That's too many variables to solve. BUT we also know that when Sam had run for S-5 minutes, he had run 4d / 5 distance, and that distance was 0.5 miles less than d. So we have the equation 4d / 5 = d - 0.5 Let's solve for d 4d = 5d - 2.5 So d = 2.5 miles Now let's plug 2.5 in for d --> t = 2.5/T and s = 2.5/S We also remember that at S-5 minutes, Sam had run 2 miles. So we can also say that s = 2/(S-5) Therefore, 2 / (S-5) = 2.5 / S By multiplying both sides by the denominators we get that 2S = 2.5 (S-5), or 2S = 2.5S - 12.5 0.5 S = 12.5, so S = 25 minutes Plugging that into the equation, we get that s = 2.5 / 25, or that s = 0.1 miles per minute We know that S = T + 15, so therefore T = 10 minutes t = 2.5/10, or 0.25 miles per minute Hopefully I did that all correctly... I had a fussy baby and a rowdy 3yo screaming as I worked this out. ;) Quote Link to comment Share on other sites More sharing options...

AKshanmar Posted January 7, 2013 Share Posted January 7, 2013 When using Singapore, your first step is to always use a bar graph like Kate demonstrated above. Fill in the information you have, and the solution starts to fall into place. I am such a visual learner, I know I learned more, teaching my kids with this program, than I did in College Calculus! Quote Link to comment Share on other sites More sharing options...

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.