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A person can walk at a constant rate of 8mph and can bike at a rate of

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chetan2u
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A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink] New post  Updated on: 19 May 2016, 09:50
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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 Kurtosis on 19 May 2016, 09:50, edited 1 time in total.
Added OA
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chetan2u
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink] New post  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] New post  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] New post  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] New post  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] New post  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) :-D :lol:
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink] New post  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] New post  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


chetan2u

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] New post  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


gauravk

SpeedW - 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] New post  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


gauravk

SpeedW - 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)


VeritasPrepKarishma

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] New post  07 Oct 2016, 02:35
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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


gauravk

SpeedW - 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)


VeritasPrepKarishma

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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html/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] New post  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] New post  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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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink] New post  08 Apr 2019, 07:05
Distance = Speed * Time
Let speed and time for walking be s1 and t1 respectively
Let speed and time for biking be s2 and t2 respectively
t1+t2=8
t2=8-t1
(s1*t1)+(s2*t2)=80
Substituting values
(8*t1)+(16{8-t1})=80
t1=6
Distance travelled by walking = 8*6= 48 i.e C
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink] New post  12 Dec 2022, 17:20
\(8\frac{m}{h}*(8h-x) + 16\frac{m}{h}*x = 80m\)
\(8\frac{m}{h}*x = 16m\)
\(x = 2h\)

Distance walked on foot: \(8\frac{m}{h}*(8h-2h) = 48m\)
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Re: A person can walk at a constant rate of 8mph and can bike at a rate of [#permalink]
12 Dec 2022, 17:20
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