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A cyclist traveled for two days. On the second day the

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A cyclist traveled for two days. On the second day the  [#permalink]

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New post 14 May 2012, 00:00
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Question Stats:

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A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day?

A. 5 mph
B. 10 mph
C. 20 mph
D. 30 mph
E. 40 mph
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 14 May 2012, 00:34
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A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day?

A. 5 mph
B. 10 mph
C. 20 mph
D. 30 mph
E. 40 mph

Approach #1 - Algebra:

Since on the second day the cyclist traveled 4 hours longer than she traveled on the first day and spent a total of 12 hours traveling then \(t+(t+4)=12\) --> \(t=4\). So, she traveled 4 hours on the first day and 8 hours on the second day;

Let the rate on the second day be \(r\) mile per hour, then: 4(r+10)+8r=280 --> \(r=20\).

Answer: C.

Approach #2 - Logic:

Average rate of the cyclist is \(\frac{total \ distance}{total \ time}=\frac{280}{12}=23\frac{1}{3}\). Now, since the weighted average of 2 individual averages (\(r\) and \(r+10\)), must lie between these individual averages then \(r<23\frac{1}{3}<r+10\), only answer choice B fits (rate from correct answer choice must be less than \(23\frac{1}{3}\) and that rate plus 10 must be more than \(23\frac{1}{3}\)).

Answer: C.
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 14 May 2012, 01:01
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the first day speed=V time =t
the 2nd day speed =V-10 time =t+4
total distance =280
total time=12

t+t+4=12
t=4

8(v-10)+4v=280
v=30
30-10=20
answ is C.20
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 29 Jan 2015, 18:07
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A cyclist traveled for two days. On the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day. If during the two days she traveled a total of 280 miles and spent a total of 12 hours traveling, what was her average speed on the second day?

A. 5 mph
B. 10 mph
C. 20 mph
D. 30 mph
E. 40 mph


SOLUTION:

D = 280 Mi
T = 12 hrs

Day 1 time = T1
Day 2 time = T2
T2 - T1 = 4 hrs ----- (I)
T1 + T2 = 12 hrs ----- (II)
Adding I and II, T2 = 8 hrs and T1 = 4 hrs

Day 1 Rate = R1
Day 2 Rate = R2
R1 - R2 = 10 mph
i.e. R1 = 10 + R2

280 = 8R2 + 4R1
i.e. 280 = 8R2 + 4 (10 + R2)
i.e. R2 = 20 mph

ANSWER: c
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 29 Jan 2015, 19:07
Hi All,

These types of Average Speed questions require a certain amount of organization. While you have the freedom to approach this Test in any way you choose, having a "default" way to create tables will help you to set up questions quicker over the course of the entire Exam.

Here's one way to organize the work in this question:

Attachment:
GCDrtBox.png
GCDrtBox.png [ 462.05 KiB | Viewed 3155 times ]


Note that the question asks for the Average Speed on DAY 2, so we have one more step:

R = 30
R - 10 = 20

Final Answer:

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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 04 Feb 2015, 01:36
............................. Distance .................... Time ............................ Speed

Day I .................... d ................................. 4 ..................................... \(\frac{d}{4}\) (Let "d" is the distance covered on Day I)

Day II .................. 280-d ........................... 8 ..................................... \(\frac{280-d}{8}\)

Given that \(\frac{d}{4} - \frac{280-d}{8} = 10\)

d = 120

Speed on day II \(= \frac{280-120}{8} = \frac{160}{8} = 20\)

Answer = C
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 26 Oct 2015, 20:19
Bunuel, i understood the solution, but please tell me , what is the point i m in missing in my calculation

defiantly she trvelled 4 hour on first day and 8 hours on second day
total of 280 km
d1+ d2= 280

on second day speed is 10km/h is lesser than first and trevlled for 8 hours thus covers 80 km lesser in distance on second day
d1+ d1-80 = 280
d1 = 180
and d2 = 100
so speed is 100/8 = which is greater than 10 but reaching 20
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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 09 Jan 2017, 15:07
Drawing a diagram is helpful for this problem!!!

Day 1 --> d = r x 4
Day 2 --> 280-d = (r-10)x(8)

Note: you can figure out that hours allocated to day 1 and day 2 by solving the following formula from the prompt: t+t+4=12 --> t=4 --> all you have to do then is plug the correct values for t into day 1 and day 2 based on the information provided in the prompt

Plug d from Day 1 into Day 2 and solve --> you'll find r = 30 --> making Day 2 r = 20

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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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New post 28 Sep 2018, 01:20
Let

T = t1 + t2
t1 is hours for day 1
t2 is hours for day 2

We know t2 = 4+t1 so this means t2 = 8 and t1 = 4

we are also given that v2 = x - 10
v1 = x

D1 + D2 = Dtotal
4x + 8 (x - 10) = 280
4x + 8x - 80 = 280
12x = 360
x = 30

We want speed for second day so x - 10 = 20 answer choice C

We could also solve this by using the answer choices and backsolving.

If we let the second day avg speed be 30 then the first day speed is 40

D1= 40 * 4 = 160
D2 = 30 * 8 = 240

Adding these two together makes it more than the distance. So we know the speed should be less than the picked.

We try for 20

20 * 8 = 160
30 * 4 = 120

Total 280, answer choice C

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Re: A cyclist traveled for two days. On the second day the  [#permalink]

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Re: A cyclist traveled for two days. On the second day the   [#permalink] 29 Sep 2019, 15:38
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