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# M07-24

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Math Expert
Joined: 02 Sep 2009
Posts: 43901

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15 Sep 2014, 23:35
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Difficulty:

45% (medium)

Question Stats:

68% (01:48) correct 32% (02:44) wrong based on 53 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
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43901

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15 Sep 2014, 23:35
Expert's post
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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 $$23\frac{1}{3}$$).

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Joined: 20 Jun 2014
Posts: 9
Schools: CBS '18

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10 Feb 2015, 18:47
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
Math Expert
Joined: 02 Sep 2009
Posts: 43901

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11 Feb 2015, 02:05
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

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Joined: 20 Jun 2014
Posts: 9
Schools: CBS '18

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11 Feb 2015, 10:19
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.
Intern
Joined: 20 Oct 2015
Posts: 1

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01 Mar 2016, 13:42
Bunuel - Why is it "r+10" and not "r-10"?

Cheers.
Manager
Joined: 19 Jan 2016
Posts: 78
Location: India
Concentration: Strategy

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25 Apr 2016, 10:43
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?
Math Expert
Joined: 02 Sep 2009
Posts: 43901

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26 Apr 2016, 02:59
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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Joined: 14 Oct 2012
Posts: 179

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30 Mar 2017, 19:37
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
Re: M07-24   [#permalink] 30 Mar 2017, 19:37
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# M07-24

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