A hiker walked for two days. On the second thay the hiker 2 hours longer and at an average speed 1 mile per hour (mph) faster than he walked the first day. If during the two days he walked a total of 64 miles and spent a total of 18 hours walkingm what was his average speed on the first day?
A. 2 mph
B. 3 mph
C. 4 mph
D. 5 mph
E. 6 mph
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The key concern here is a) to translate correctly informations int equations and b) not to let the variables number become excessive, in order to minimize resolution time - three or more equations can become complicated.
Let us start with «A hiker walked for two days» we would normally need variables T1, T2 for the walking hours in day-1 and day-2, L1, L2 for the distances and V1, V2 for the speeds (V=L/T). We'll try to reduce that 6-variable number right away.
«On the second thay the hiker 2 hours longer»
OK, so we eliminate T2, by substituting it with T2 = T1+2
« and at an average speed 1 mile per hour (mph) faster than he walked the first day.»
Great, V2 = V1+1
«If during the two days he walked a total of 64 miles»
Now, L1+L2 = 64
« and spent a total of 18 hours walking»
T1 + T2 = 18
But we have already found that T2 = T1+2, so T1 + T2 = T1+(T1+2) = 18, yielding T1 = 16/2 = 8 hours. T2 is 10 hours, then.
Now they say what they want:
« what was his average speed on the first day? »
Lets put in a system of equations the information we gathered so far:
V1 = L1/T1 => L1 = 8 x V1
V2 = L2/T2 => L2 = 10 x V2 = 10 x (1 + V1)
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.....(c)..... L1+L2 = 8V1 + 10x(1+V1) = 18xV1 + 10
We also know the total walked distance is L1 + L2 = 64. So substituting this sum in expression (c), above, yields:
64 = 18 x V1 + 10 => V1 = 54 / 18 = 3 mph
Correct answer is then (B). No swet! ... ;-)
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