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10 Oct 2018, 13:35
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:52) correct
32%
(02:12)
wrong
based on 402
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Aaron travels from town X to town Y and then back from town Y to town X, taking different routes in each direction. If his speed when travelling from town X to town Y is 36 miles per hour, and his speed when travelling in the opposite direction is 48 miles per hour, what is his average speed for the entire journey?
1) The length of the return trip is 24% of the entire distance traveled.
2) The length of the return trip is 100 miles.
*I noted difficulty as 600-700. Maybe it is 700. Please let me know what you think the difficulty should be.
1) The length of the return trip is 24% of the entire distance traveled.
2) The length of the return trip is 100 miles.
*I noted difficulty as 600-700. Maybe it is 700. Please let me know what you think the difficulty should be.
11 Oct 2018, 16:40
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Nice solution Salsanousi.
I've elaborated below so that others can see "1 equation 1 variable we can find the answer" (as I certainly didn't when I started).
Average speed = \(\frac{Total Distance}{Total Time}\)
D1T1V1 and D2T2V2 6 variables
D1T136mph and D2T248mph 4 unknown variables (speed taken from the original condition)
1)0.76D, T1, 36mph and 0.24D, T2, 48mph
T=\(\frac{D}{V}\)
T1=\(\frac{0.76D}{36}\)
T2=\(\frac{0.24D}{48}\)
Now we have the respective Ts. Want to find the respective Ds
D1=36·\(\frac{0.76D}{36}\)=36·0.76D
D2=48·\(\frac{0.24D}{48}\)=48·0.24D
Average speed = \(\frac{Total Distance}{Total Time}\)
Average speed = \(\frac{36·0.76D+48·0.24D}{\frac{0.76D}{36}+\frac{0.24D}{48}}\)
Multiply through by 100 to get rid of the decimals
=\(\frac{3600·76D+4800·24D}{\frac{76D}{3600}+\frac{24D}{4800}}\)
Simplify the denominator
=\(\frac{\frac{D(3600·76+4800·24)}{1}}{\frac{47D}{1800}}\)
=\(\frac{1800D(3600·76+4800·24)}{47D}\)
Cancel D from the numerator and denominator
=\(\frac{1800(3600·76+4800·24)}{47}\)
Shows that average speed is a number, we just don't need to work out what that number is.
Sufficient
AD
2) D1T136mph and D2100miles48mph 3 unknown variables
We can't solve for 3 unknown variables with one equation
Anyone that reads this answer, kudos please! It took me over half an hour with all of the formatting!
I've elaborated below so that others can see "1 equation 1 variable we can find the answer" (as I certainly didn't when I started).
Average speed = \(\frac{Total Distance}{Total Time}\)
D1T1V1 and D2T2V2 6 variables
D1T136mph and D2T248mph 4 unknown variables (speed taken from the original condition)
1)0.76D, T1, 36mph and 0.24D, T2, 48mph
T=\(\frac{D}{V}\)
T1=\(\frac{0.76D}{36}\)
T2=\(\frac{0.24D}{48}\)
Now we have the respective Ts. Want to find the respective Ds
D1=36·\(\frac{0.76D}{36}\)=36·0.76D
D2=48·\(\frac{0.24D}{48}\)=48·0.24D
Average speed = \(\frac{Total Distance}{Total Time}\)
Average speed = \(\frac{36·0.76D+48·0.24D}{\frac{0.76D}{36}+\frac{0.24D}{48}}\)
Multiply through by 100 to get rid of the decimals
=\(\frac{3600·76D+4800·24D}{\frac{76D}{3600}+\frac{24D}{4800}}\)
Simplify the denominator
=\(\frac{\frac{D(3600·76+4800·24)}{1}}{\frac{47D}{1800}}\)
=\(\frac{1800D(3600·76+4800·24)}{47D}\)
Cancel D from the numerator and denominator
=\(\frac{1800(3600·76+4800·24)}{47}\)
Shows that average speed is a number, we just don't need to work out what that number is.
Sufficient
AD
2) D1T136mph and D2100miles48mph 3 unknown variables
We can't solve for 3 unknown variables with one equation
Anyone that reads this answer, kudos please! It took me over half an hour with all of the formatting!
General Discussion
10 Oct 2018, 15:11
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philipssonicare
D1 = T1 * V1
T1 = D1/V1
Avg speed = (Total distance)/(Total time)
Statement one tells us that D1 = 0.24Dtotal
D1 + D2 = Dtotal
0.24Dtotal + D2 = Dtotal
D2 = 0.76Dtotal.
T1 = 0.24Dtotal/36
T2 = 0.76Dtotal/48
Avg speed = Dtotal/ (T1+T2)
1 equation 1 variable we can find the answer
A is sufficient
Statement 2
gives us D2 but that is not enough to find anything.
Insufficient
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