Stuck on Math 11 Problem Solving questions.

In summary: How much water do you want in the system compared to how much is there now? That ratio, applied to the liquid generally, gives the answer.In summary, the sales rep averaged 58 miles per hour on the first half of the trip, but only 52 miles per hour on the last half because of the increased traffic. It would take 5 hours and 45 minutes to travel the 317 miles using each speed, so the total driving time would be 12 hours and 45 minutes.
  • #1
wassworth
2
0

Homework Statement



I've worked through and been able to understand each problem I've been given except for the following two, does anyone have a suggestion or an explanation on how to solve either of these problems? Thanks in advance.

1. "On the first part of a 317 mile trip, a sales representative averaged 58 miles per hour. The sales representative averaged only 52 miles per hour on the last part of the trip because of an increased volume of traffic. Find the amount of driving time at each speed if the total time was 5 hours and 45 minutes."

2. "The cooling system on a truck contains 5 gallons of coolant that is 40% antifreeze. How much must be withdrawn and replaced with 100% antifreeze to bring the coolant in the system to 50% antifreeze?"

Homework Equations



s=d/t

The Attempt at a Solution



I don't even know where to begin with these two.

Regarding #1 something to the effect of '58m/h +52m/h = 317m/5.75h. But I'm not sure to go from there.
 
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  • #2
forthe first one, try using a system of equations.
let [itex] d_1+d_2=317 [/itex] where that is the sum of the first distance plus the second distance. Then, since distance=rate X time, d/r=t. So,
[tex] \frac{d_1}{58} + \frac{d_2}{52} = 5.75 [/tex]
 
  • #3
ArcanaNoir said:
forthe first one, try using a system of equations.
let [itex] d_1+d_2=317 [/itex] where that is the sum of the first distance plus the second distance. Then, since distance=rate X time, d/r=t. So,
[tex] \frac{d_1}{58} + \frac{d_2}{52} = 5.75 [/tex]

Oh, I see! Thank you. That helps a lot. Can you offer any insight into the second problem?
 
  • #4
wassworth said:
Oh, I see! Thank you. That helps a lot. Can you offer any insight into the second problem?

How much antifreeze is in the original 5 gal (i.e., gal. of antifreeze)? How many gal. of antifreeze must the new mix contain? If you pour out x gal of the original mix, how much antifreeze are you pouring out?

RGV
 
  • #5
... or you could think about the water (that dilutes the antifreeze). How much water do you want in the system compared to how much is there now? That ratio, applied to the liquid generally, gives the answer.

But that might be sneakier than the examiner wants. :-)
 

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