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Sanjana jogged uphill for a while at an average speed of 3
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06 Oct 2013, 03:41
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63% (02:59) correct 37% (02:39) wrong based on 410 sessions
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Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill? A. 1 mile B. 1 1/3 miles C. 1 1/2 miles D. 1 3/5 miles E. 1 2/3 miles my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2
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Re: Sanjana jogged uphill for a while
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06 Oct 2013, 04:09
piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
a. 1 mile b. 1 1/3 miles c. 1 1/2 miles d. 1 3/5 miles e. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr D1/3+D2/8=40/60 hrs =>8D1+3D2=16Eq(1) also given she travelled uphill and downhill at an avg of 4 miles/hr =>total distance/Avg speed=40/60hrs =>(D1+D2)/4=2/3 =>3D1+3D2=8Eq(2) solving equations 1 and 2 D1=8/5 mile and D2=16/15 miles =>D1=1 3/5 miles the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case.. Hope it is helpful




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Re: Sanjana jogged uphill for a while
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06 Oct 2013, 05:49
piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
a. 1 mile b. 1 1/3 miles c. 1 1/2 miles d. 1 3/5 miles e. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Here we can't use the S1*S2/ (S1+S2) formula because the uphill and downhill distances are not same.
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Re: Sanjana jogged uphill for a while
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26 Nov 2013, 16:02
kusena wrote: piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
a. 1 mile b. 1 1/3 miles c. 1 1/2 miles d. 1 3/5 miles e. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr D1/3+D2/8=40/60 hrs =>8D1+3D2=16Eq(1) also given she travelled uphill and downhill at an avg of 4 miles/hr =>total distance/Avg speed=40/60hrs =>(D1+D2)/4=2/3 =>3D1+3D2=8Eq(2) solving equations 1 and 2 D1=8/5 mile and D2=16/15 miles =>D1=1 3/5 miles the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case.. Hope it is helpful But aren't the uphill and downhill distances actually the same? Are we assuming is the same hill no? Cheers J



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Re: Sanjana jogged uphill for a while
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26 Nov 2013, 23:45
jlgdr wrote: kusena wrote: piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
a. 1 mile b. 1 1/3 miles c. 1 1/2 miles d. 1 3/5 miles e. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Let the uphill distance be D1 with speed S1=3miles/hr and downhill distance be D2 with speed 8 miles/hr D1/3+D2/8=40/60 hrs =>8D1+3D2=16Eq(1) also given she travelled uphill and downhill at an avg of 4 miles/hr =>total distance/Avg speed=40/60hrs =>(D1+D2)/4=2/3 =>3D1+3D2=8Eq(2) solving equations 1 and 2 D1=8/5 mile and D2=16/15 miles =>D1=1 3/5 miles the formula for average speed s1*s2/s1+s2........can be used only when uphill distance and downhill distances are same which is not in this case.. Hope it is helpful But aren't the uphill and downhill distances actually the same? Are we assuming is the same hill no? Cheers J Question says average speed of both the journeys is 4 miles/hr Where as uphill speed 3miles/hr and downhill speed 8 miles/hr If both uphill and down hill distances are same then the avg speed will be (v1*v2)/(v1+v2)=8*3/(8+3)=24/11 miles/hr which is not equal to 4 miles/hr.hence the the distances traveled uphill and downhill aren't the same. Hope its clear



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Re: Sanjana jogged uphill for a while at an average speed of 3
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31 Dec 2013, 11:15
piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
A. 1 mile B. 1 1/3 miles C. 1 1/2 miles D. 1 3/5 miles E. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Easy question 3x + 8y / x+y = 4 x + y = 40 So solving y = 8 x = 32 So D is out correct answer choice Hope it helps Provide kudos! Cheers J



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Re: Sanjana jogged uphill for a while at an average speed of 3
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14 Feb 2014, 12:57
jlgdr wrote: piyush272 wrote: Sanjana jogged uphill for a while at an average speed of 3 miles per hour, then jogged downhill for a while at an average speed of 8 miles per hour. If Sanjana jogged the uphill and downhill stretches in a total of 40 minutes at an average speed of 4 miles per hour, how far did she jog uphill?
A. 1 mile B. 1 1/3 miles C. 1 1/2 miles D. 1 3/5 miles E. 1 2/3 miles
my question is in such problems why cant we use the formula for average speed s1*s2/s1+s2 Easy question 3x + 8y / x+y = 4 x + y = 40 So solving y = 8 x = 32 So D is out correct answer choice Hope it helps Provide kudos! Cheers J Just found another way of doing it with weighted average. Let's call x the uphill distance and y the downhill distance Also we know that average speed is 4mph. So x is at 3mph and y is at 8mph. By differentials x = 4y. So if total time is 40 minutes then the time spent uphill must be 4 times that spend downhill Therefore x = 32 min. Now, (3)(32/60)=1 3/5 Hope you enjoyed Cheers J



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Sanjana jogged uphill for a while at an average speed of 3
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09 Sep 2015, 02:38
I used weighted avg to get uphill to downhill ratio. U/D = 4/1. i.e. U = 4D. Uphill to total will be 4/5.
Does 4/5 ratio only apply to the total time OR does it also apply to the total distance. I mean, can the total distance also be split in this ratio?
I got total dist as 8/3 miles. (total time 40/60 * avg speed 4mph) will the uphill distance be 4/5 of 8/3...It does not seem so...
Could somebody pls clarify whether 4/5 ratio applies only to the time and why should it not be applied to the total distance?
Thanks
PS  In the above post we have "if total time is 40 minutes then the time spent uphill must be 4 times that spend downhill". I would like to know whether the same ratio applies to total distance also or just the time.



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Re: Sanjana jogged uphill for a while at an average speed of 3
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03 Oct 2015, 00:15
add times to get 2/3 hours (40min.) and consider full distance (2/3)*4=8/3:
x/3 + (8/3)x)/8=2/3
5x=8
x=8/5 => 1+3/5
D



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Re: Sanjana jogged uphill for a while at an average speed of 3
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03 Oct 2015, 14:08
let D=uphill distance D/3 hours=uphill time (4 mph avge speed)(2/3 hour total time)=8/3 miles total distance 8/3D=downhill miles (8/3D)/8=downhill time D/3+(8/3D)/8=2/3 D=8/5➡1+3/5 miles



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Re: Sanjana jogged uphill for a while at an average speed of 3
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04 Mar 2016, 18:14
i used only 1 variable to solve this question... uphill 3mph  time 2/3x  distance 23x downhill 8mph  time x  distance 8x
total distance 8/3 miles
so 23x+8x=8/3 5x=8/3  6/3 5x = 2/3 x=2/15
now, distance uphill is 23x 22/5 = 1 and 3/5
D



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Sanjana jogged uphill for a while at an average speed of 3
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05 Mar 2016, 12:10
let t=uphill time 3t+8(2/3t)=8/3 total miles t=8/15 hours (3 mph)(8/15 hours)=1 3/5 uphill miles



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Re: Sanjana jogged uphill for a while at an average speed of 3
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19 Dec 2016, 10:18
Apply mixture and alligation method. Please refer the attached diagram. time spent at the rate of 3 miles/hr: time spent at the rate of 8 miles/hr = 4:1 So reqd time = 4/5*40 = 32 minutes So distance traveled = 32/60 * 3 = 8/5 = 1 3/5 miles.
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Re: Sanjana jogged uphill for a while at an average speed of 3
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21 Jul 2017, 19:10
D=Distance R=Rate T=Time T1= Time 1 = Uphill T2 = Time 2 = Downtown
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Re: Sanjana jogged uphill for a while at an average speed of 3
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10 Sep 2018, 01:28
chetan2u, VeritasKarishma, Why can we not apply weighted average ratio of 4/5 on total distance traveled (ie8/3 miles) directly instead of applying it on total time taken of 40 minutes and then multiplying it to the uphill speed?




Re: Sanjana jogged uphill for a while at an average speed of 3 &nbs
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10 Sep 2018, 01:28






