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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?

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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20 May 2016, 11:28

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This post received KUDOS

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) _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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, 16:34

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

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).

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?

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

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, 22:50

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?

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

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.

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?

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

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.

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, 11:30

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

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, 09:15

1

This post was BOOKMARKED

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?

Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]

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29 Dec 2017, 05:53

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