Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 Minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is -
A. 30
B. 35
C. 36
D. 37.5
E. 40
The OA is the option E.
Experts, can you help me here? I don't know what are the equations that I should set here. <i class="em em-disappointed"></i>
Anthony covers a certain distance on a bike. Had he moved
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We can let the distance = d and his rate = r. Thus, we can create the equations (notice that 40 minutes = 2/3 hour):VJesus12 wrote:Anthony covers a certain distance on a bike. Had he moved 3mph faster, he would have taken 40 Minutes less. Had he moved 2mph slower, he would have taken 40 minutes more. The distance (in miles) is -
A. 30
B. 35
C. 36
D. 37.5
E. 40
d/(r + 3) = d/r - 2/3
and
d/(r - 2) = d/r + 2/3
Subtracting the first equation from the second, we have:
d/(r - 2) - d/(r + 3) = 4/3
Multiplying by 3(r - 2)(r + 3), we have:
3d(r + 3) - 3d(r - 2) = 4(r - 2)(r + 3)
3dr + 9d - 3dr + 6d = 4(r - 2)(r + 3)
15d = 4(r - 2)(r + 3)
d = 4(r - 2)(r + 3)/15
Substituting this back into the first equation, we have:
[4(r - 2)(r + 3)/15](r + 3) = [4(r - 2)(r + 3)/15]/r - 2/3
4(r - 2)/15 = 4(r - 2)(r + 3)/(15r) - 2/3
Multiplying by 15r, we have:
4r(r - 2) = 4(r - 2)(r + 3) - 10r
4r^2 - 8r = 4(r^2 + r - 6) - 10r
4r^2 - 8r = 4r^2 + 4r - 24 - 10r
-8r = -6r - 24
-2r = -24
r = 12
So d = 4(12 - 2)(12 + 3)/15= 4(10)(15)/15 = 40 miles.
Answer: E
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