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A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk? A. 20 B. 30 C. 48 D. 60 E. 72 OA after three days
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Originally posted by chetan2u on 26 Mar 2016, 04:31.
Last edited by Vyshak on 19 May 2016, 09:50, edited 1 time in total.
Added OA
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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26 Mar 2016, 04:40
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Total distance = 80
Distance = Speed * Time
Walking speed = s1 = 8
Walking time = t1
Bike speed = s2 = 16
Time traveled in bike = t2
d1 + d2 = 80
s1t1 + s2t2 = 80
8*t1 + 16*t2 = 80
t1 + 2*t2 = 10 ----- (1)
Given: t1 + t2 = 8 ----- (2)
(1) - (2) --> t2 = 2 and t1 = 8 - 2 = 6
Walking distance = s1*t1 = 8*6 = 48
Answer: C
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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26 Mar 2016, 05:36
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Vyshak wrote: Total distance = 80
Distance = Speed * Time
Walking speed = s1 = 8
Walking time = t1
Bike speed = s2 = 16
Time traveled in bike = t2
d1 + d2 = 80
s1t1 + s2t2 = 80
8*t1 + 16*t2 = 80
t1 + 2*t2 = 10 ----- (1)
Given: t1 + t2 = 8 ----- (2)
(1) - (2) --> t2 = 2 and t1 = 8 - 2 = 6
Walking distance = s1*t1 = 8*6 = 48
Answer: C Hi Vyshak, you are correct, BUT Alligation or Weighted average method..is the fastest method in such Qs It enables us to find the ratio in which two or more ingredients at the given price/%/weight must be mixed to produce a mixture of a desired price/%/weight.Lets manipulate this Q to fit into weighted average method-- here the walking speed= 8mph, so time he would take cover 80miles= 80/8=10 hr similarly the biking speed= 16mph, so time he would take cover 80miles= 80/16=5 hr But the average has to be 8hrs, so the ratio of dist by walking = \(\frac{(Avg-biking time)}{(walking time-biking time)}= \frac{(8-5)}{(10-5)}= \frac{3}{5}\) so the distance travelled = 3/5 th of total = 3/5 *80 = 48
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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19 May 2016, 09:42
Hi Chetan2u, Please add OA to the question. Thanks! 
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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19 May 2016, 10:13
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I backsolved this question Total distance= 80 miles Total time= 8 hrs First check for option 'C' 48 miles of walking means he walked for 48/8= 6 hrs Remaining 32 miles of biking means he biked for 32/16= 2 hrs Total hours traveled= 6+2 = 8hrs (as mentioned in the question stem)
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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20 May 2016, 12:28
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chetan2u wrote: A person can walk at a constant rate of 8 mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days AN ALTERNATE APPROACH
Check the answer options , the correct answer for walking must be a multiple of 8 and for bike must be multiple of 16 , so possible answers are C. 48 & E. 72 Back solve C. 48 Walk = 48 miles ; Bike = 32 Walking speed = 8 miles/hr ; Biking speed = 16 miles/hr Walking time = 6 hr ; Biking speed = 2 hr ( total time is 8 hours )E. 72 Walk = 72 miles ; Bike = 8 Walking speed = 8 miles/hr ; Biking speed = 16 miles/hr Walking time = 9 hr ; Biking speed = 1/2 hr ( total time is 19/2 hours )Hence correct answer must be (C)  
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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20 Sep 2016, 17:34
chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days time of bike: b time of walk: w b+w=8 -> b=8-w D=Rt: 8w+16b =80 substitute b: 8w+16(8-w)=80. Solving for w and get w=6. To get the total distance of walk, multiply 6 by 8 = 48. C
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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06 Oct 2016, 11:59
chetan2u wrote: Vyshak wrote: Total distance = 80
Distance = Speed * Time
Walking speed = s1 = 8
Walking time = t1
Bike speed = s2 = 16
Time traveled in bike = t2
d1 + d2 = 80
s1t1 + s2t2 = 80
8*t1 + 16*t2 = 80
t1 + 2*t2 = 10 ----- (1)
Given: t1 + t2 = 8 ----- (2)
(1) - (2) --> t2 = 2 and t1 = 8 - 2 = 6
Walking distance = s1*t1 = 8*6 = 48
Answer: C Hi Vyshak, you are correct, BUT Alligation or Weighted average method..is the fastest method in such Qs It enables us to find the ratio in which two or more ingredients at the given price/%/weight must be mixed to produce a mixture of a desired price/%/weight.Lets manipulate this Q to fit into weighted average method-- here the walking speed= 8mph, so time he would take cover 80miles= 80/8=10 hr similarly the biking speed= 16mph, so time he would take cover 80miles= 80/16=5 hr But the average has to be 8hrs, so the ratio of dist by walking = \(\frac{(Avg-biking time)}{(walking time-biking time)}= \frac{(8-5)}{(10-5)}= \frac{3}{5}\) so the distance travelled = 3/5 th of total = 3/5 *80 = 48 Thanks for the solution, I was able to solve it using traditional method and was able to follow your weighted average method. Can you pls explain that instead of using Average time taken for weighted average method, why can't we use directly the respective speeds given ( 8 mph, 16 mph etc). TIA
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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06 Oct 2016, 21:56
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chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days gauravkSpeedW - 8 mph SpeedB - 16 mph Avg speed = 80/8 = 10 mph TW/TB = (SpeedB - SpeedAvg)/(SpeedAvg - SpeedW) = (16 - 10)/(10 - 8) = 3/1 So he walked for 3/4th of the time i.e. for (3/4)*8 hrs = 6 hrs. (Note that when averaging speed, the weights will always be time taken, never distance.) In 6 hrs, he would have walked a distance of 8*6 = 48 miles Answer (C)
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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06 Oct 2016, 23:50
VeritasPrepKarishma wrote: chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days gauravkSpeedW - 8 mph SpeedB - 16 mph Avg speed = 80/8 = 10 mph TW/TB = (SpeedB - SpeedAvg)/(SpeedAvg - SpeedW) = (16 - 10)/(10 - 8) = 3/1 So he walked for 3/4th of the time i.e. for (3/4)*8 hrs = 6 hrs. (Note that when averaging speed, the weights will always be time taken, never distance.)In 6 hrs, he would have walked a distance of 8*6 = 48 miles Answer (C) Thanks a lot for the explanation, I was going wrong with the highlighted part. I was using the ratio directly on the total distance (which obviously wasn't working). Is there a mathematical derivation of the highlighted part which can help me understand this, rather than cramming it. TIA.
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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07 Oct 2016, 02:35
gauravk wrote: VeritasPrepKarishma wrote: chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days gauravkSpeedW - 8 mph SpeedB - 16 mph Avg speed = 80/8 = 10 mph TW/TB = (SpeedB - SpeedAvg)/(SpeedAvg - SpeedW) = (16 - 10)/(10 - 8) = 3/1 So he walked for 3/4th of the time i.e. for (3/4)*8 hrs = 6 hrs. (Note that when averaging speed, the weights will always be time taken, never distance.)In 6 hrs, he would have walked a distance of 8*6 = 48 miles Answer (C) Thanks a lot for the explanation, I was going wrong with the highlighted part. I was using the ratio directly on the total distance (which obviously wasn't working). Is there a mathematical derivation of the highlighted part which can help me understand this, rather than cramming it. TIA. Check here: https://www.veritasprep.com/blog/2014/1 ... -averages/
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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26 Dec 2016, 12:30
chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days similar approach by Vyshak and chetan2u but in different pattern let distance covered by walking =W & by bike=B As avg speed on travelling both by walking and bike=80km in 8 hrs==10km/hr we can arrange in avg. speed formula=total distance /total time (W+B)/{W/8+B/16}=10 (16W+16B)/(2W+B)=10 16W+16B=20W+10B B=2/3W--------------(I) as total distance W+B=80 putting value of B from (I) W+2/3W=80 W=48 Ans C
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
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27 Dec 2016, 10:15
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chetan2u wrote: A person can walk at a constant rate of 8mph and can bike at a rate of 16mph. If he wants to travel 80 miles in 8 hours using bike and walking at their constant rates, how much distance would he require to walk?
A. 20
B. 30
C. 48
D. 60
E. 72
OA after three days W=Time spent walking B=Time spent biking 80=16B+8W =>W=10-2B But, W+B=8 (given) =>W=10-2(8-W) =>W=6 => Distance walked=6*8=48miles C
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