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aliya0001 [1]
3 years ago
12

Please Please help me with this  problem

Mathematics
2 answers:
Mkey [24]3 years ago
5 0
The work become much simpler if you do it in a table. Hope this helps!!

Sergeu [11.5K]3 years ago
4 0
Assuming that both cyclists are traveling AWAY from each other with the same starting point, consider the following:

Let us denote the speed of the first bicyclist by s (measured in miles per hour). Then, the speed of the second bicyclist must be s-5 (miles per hour).

To verify this, assign an arbitrary number, say 17 mph, to the first bicyclist. Since the first bicyclist is always 5 mph faster than the second, we must have 17 - 5 = 12 mph as the second speed.

At the 0th hour,
- first bicyclist has traveled 0 miles
- second bicyclist has traveled 0 miles
- distance between both = 0 miles

At the 1st hour,
- first bicyclist has traveled 1\times s=s miles
- second bicyclist has traveled -1\times(s-5)=-s+5 miles (note the negative sign here. Since they are traveling in opposite directions, they must have opposite signs of each other in distances traveled)
- distance between both = s-(-s+5) = s+s-5 = 2s-5 miles

At the second hour,
- first bicyclist has traveled 2s miles
- second bicyclist has traveled -2(s-5)=-2s+10 miles
- distance between both = 2s-(-2s+10)=2s+2s-10=4s-10 miles

According to the question statement, both of them are 70 miles apart at the second hour, i.e.
                                                4s-10=70
such that solving this gives:
     4s=70+10
     4s=80
     s= \frac{80}{4}
     s=20 mph

Therefore, their rates are as follows:
- first bicyclist = 20 mph
- second bicyclist = 20 - 5 = 15 mph
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