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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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  55% (hard)

Question Stats:

67% (01:47) correct 33% (02:24) wrong based on 52 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
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
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"?

Cheers.
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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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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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 &nbs [#permalink] 22 Jun 2018, 11:52
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