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M07-24

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M07-24  [#permalink]

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New post 16 Sep 2014, 00:35
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Question Stats:

73% (02:23) correct 27% (02:48) wrong based on 144 sessions

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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 M07-24  [#permalink]

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New post 16 Sep 2014, 00:35
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Official Solution:

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 \(t+(t+4)=12\), so \(t=4\). 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: \(4(r+10)+8r=280\), so \(r=20\).

Approach 2 - Logic:

Average rate of the cyclist is \(\frac{\text{total distance}}{\text{total time}}=\frac{280}{12}=23\frac{1}{3}\). Now, since the weighted average of 2 individual averages (\(r\) and \(r+10\)), must lie between these individual averages then \(r \lt 23\frac{1}{3} \lt r+10\), 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: M07-24  [#permalink]

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New post 10 Feb 2015, 19:47
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Hey Bunuel,

I did the problem a little differently. I made "r" equal to the rate of the first day and "r-10" equal to the rate of the second day and I did everything else the same as you, but i got the wrong answer. Can you please show me how to do the problem with the assignments of "r" as rate of day 1 and "r-10" as rate of day 2?

Thanks
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Re: M07-24  [#permalink]

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New post 11 Feb 2015, 03:05
psavalia wrote:
Hey Bunuel,

I did the problem a little differently. I made "r" equal to the rate of the first day and "r-10" equal to the rate of the second day and I did everything else the same as you, but i got the wrong answer. Can you please show me how to do the problem with the assignments of "r" as rate of day 1 and "r-10" as rate of day 2?

Thanks


Please show your work.
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Re: M07-24  [#permalink]

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New post 11 Feb 2015, 11:19
Bunuel,

Upon looking at my work the second time, I found my mistake, it looks like I copied the problem wrong which lead me to the wrong answer. Thanks for your help though.
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Re: M07-24  [#permalink]

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New post 01 Mar 2016, 14:42
Bunuel - Why is it "r+10" and not "r-10"?

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Re: M07-24  [#permalink]

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New post 25 Apr 2016, 11:43
Bunuel wrote:
Official Solution:

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 \(t+(t+4)=12\), so \(t=4\). 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: \(4(r+10)+8r=280\), so \(r=20\).



Instead, let the rate on the first day be r mile per hour, the rate on the second day be (r-10), then
\(4(r) + 8(r-10) = 280\), so \(r=30\). What's the wrong with this approach?
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Re: M07-24  [#permalink]

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New post 26 Apr 2016, 03:59
atturhari wrote:
Bunuel wrote:
Official Solution:

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 \(t+(t+4)=12\), so \(t=4\). 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: \(4(r+10)+8r=280\), so \(r=20\).



Instead, let the rate on the first day be r mile per hour, the rate on the second day be (r-10), then
\(4(r) + 8(r-10) = 280\), so \(r=30\). What's the wrong with this approach?


You found the rate for the first day, while we need the rate for the second day, which would be 30 - 10 = 20.

Hope it's clear.
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Re: M07-24  [#permalink]

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New post 30 Mar 2017, 20:37
My 2 cents,
I did this question as follows:
Day-1) time = t & speed = s
Day-2) time = (t+4) & speed = (s-10)
Total time traveled = 12 hours
t + t + 4 = 12
2t = 8
t = 4 hrs

D = 280 = d1 + d2
280 = st + (s-10)*(t+4)
280 = 4s + (s-10)*8
280 = 4s + 8s - 80
360 = 12s
s = 30

d2 = (s-10)*(t+4) = 20*8
Avg S2 = d2/t2 = 20*8/8 = 20 mph | C
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Re M07-24  [#permalink]

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New post 22 Jun 2018, 11:52
I think this is a high-quality question and I agree with explanation.
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Re M07-24  [#permalink]

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New post 25 Feb 2019, 13:43
I think this is a high-quality question and I agree with explanation.
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Re: M07-24  [#permalink]

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New post 24 Nov 2019, 15:08
Bunuel wrote:
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


Another approach: You can also solve this one with another sort of logic and number sense. Focus first on the time. Namely, 12 hours across 2 days can be broken down readily enough to 6 hours per day on average. Since we know there was a 4-hour difference between days one and two, we can take the 4 and divide it by 2 and skew our 6-hour split by going 2 hours in either direction:

6 - 2 hours = 4 hours (Day 1)
6 + 2 hours = 8 hours (Day 2)
--------------------------
12 hours altogether over two days, with the correct split

If the question asks about the second day, pick a number in the middle, say, (B), 10 mph:

10 mph * 8 hours = 80 miles (Day 2)
20 mph * 4 hours = 80 miles (Day 1)
-------------------------------
12 hours altogether over two days, but only 160 total miles traveled--TOO LOW. Out with (A) and (B), onto (D). Why (D)? Because regardless of whether it ends up being the answer, we will know definitively whether it is too high or too low.

30 mph * 8 hours = 240 miles (Day 2)
40 mph * 4 hours = 160 miles (Day 1)
--------------------------------
12 hours altogether over two days, but with 400 miles traveled--TOO HIGH. Thus, (D) and (E) are out, and (C), 20 mph, must be the answer.

This mental math is simple enough and takes about a minute to work through, maybe a minute and a half in a Day 2-type situation.

Cheers,
Andrew
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Re: M07-24   [#permalink] 24 Nov 2019, 15:08
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