Let V, be the rate in still water and let C = rate river current
If the boat is going :
upstream, its rate is V-C and if going
downstream, its rate is V+C,
But V = 5C, then
Upstream Rate: 5C - C = 4 C
Downstream rate: 5C+C = 6C
Time = distance/Rate, (or time = distance/speed) , then:
Upstream time 12/4C and
Downstream time: 12/.6C
Upstream time +downstream time:= 2h30 ' then:
12/4C + 12/.6C = 2.5 hours
3/C + 2/C = 5/2 (2.5 h = 5/2)
Reduce to same denominator :
5C = 10 and Rate of the current = 2 mi/h
(for x=0) 3 & (for x=4) 19
To solve this problem you must keep on min the information given in the problem shown above:
1- You have that
<span>
- The height reached by amal's rocket was 1/4 of the height reached by min's rocket.
- Amal's rocket REAched a height of 15 meters.
</span>
2. Therefore, let's call "x" to the <span>height reached by min's rocket. Then, you have:
</span>
(1/4)x=15
x/4=15
<span> Which equation could be used to find the height reached by min's rocket?
</span>
The answer is: x/4=15
