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# Myra drove at an average speed of 30 miles per hour for some time and

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Math Expert
Joined: 02 Sep 2009
Posts: 46264
Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 07:25
2
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Difficulty:

55% (hard)

Question Stats:

66% (01:48) correct 34% (02:06) wrong based on 271 sessions

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Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

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Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 07:51
2
2
Bunuel wrote:
Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

Let's name first part of distance F and second part S
We can make weighted average equation:
0.3F + 0.6S = 0.5(S + F)
0.1S=0.2F

$$\frac{S}{F}=\frac{2}{1}$$

So First part equal to $$\frac{1}{3}$$ of all time

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Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 08:00
2

Let the total distance be 1. The distance covered on 30mph= x; so the distance covered on 60 will be 1-x

<--------------------------------|---------->
X (1-x)

Equating times,

x/30 + 1-x/60 = 1/50;

Solving we get, X= 1/5.= so distance covered at 30mph=1/5 and distance covered at 60mph=4/5

Time calculation,

Time taken to cover 1/5 jounery at 30mph= 1/150.

Total time taken for the entire jouney= 1/50.

Fraction of the time.

1/150/1/50 = 1/3 (option B).

Bunuel wrote:
Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

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Joined: 07 Aug 2011
Posts: 565
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Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 08:57
1
Bunuel wrote:
Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

very simple application of weight average

30------50---60

W. Avg is 20 unit far from 30 and 10 unit far from 60 so

So at 60 mph she drove 2/3 of distance and at 30 mph she drove 1/3 of the distance.

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Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 09:04
1
50-30=20
60-50=10

Required answer is 10/(10+20) = 1/3
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Manager
Joined: 17 Mar 2015
Posts: 121
Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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06 Apr 2015, 16:10
1
I concluded that time could serve as a variable that would let us divide the journey in fractions. Lets same time = 1. X - the fraction.
We can then conclude that, since the journey is the same and the total time is the same.
$$30*x + 60*(1-x) = 50*1$$
$$30*x = 10$$
$$x = 1/3$$
Which means the answer is B
Manager
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Posts: 100
Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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07 Apr 2015, 04:30
1
Average speed of 150:

30 + 60 + 60 = 150
150 / 3 = 50

1 hour of 30 miles an hour
2 hours of 60 miles an hour

30 miles an hour 1/3 of the time.

B.
Math Expert
Joined: 02 Sep 2009
Posts: 46264
Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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13 Apr 2015, 07:46
Bunuel wrote:
Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:

Solution: We know the average speed and must find the fraction of time taken at a particular speed.

t1/t2 = (A2 – Aavg)/(Aavg – A1)

t1/t2 = (60 – 50)/(50 – 30) = 1/2

So out of a total of 3 parts of the journey time, she drove at 30 mph for 1 part and at 60 mph for 2 parts of the time. Fraction of the total time for which she drove at 30 mph is 1/3.

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Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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07 Nov 2015, 21:40
An alternate approach:

When i cannot figure out a direct way, I try the plug in approach, whcih usually means picking one of the answers and plugging in to your model. I learnt this approach from Princeton Review's Cracking the GMAT which, IMO, is the only saving grace of Princeton Review's book.

fraction of time spent driving at 30 mph= x
fraction of time spent driving at 60 mph = y
Total time = T
d1 = distance travelled at 30 mph
d2 = distance travelled at 60 mph

chose 1/3 first because it just felt about right because avg of 50 mph is 20 more than 30mph and 10 less than 60...so I figured the portion of time spent driving at 60 mph should be longer.

which means

x/T = 1/3 (fraction of time spent driving at 30mph)
y/T= 2/3 (fraction of time spent driving at 60 mph)
Assume T to be 3 hours, then x = 1 hour, y= 2 hours (this works because we just need fraction, not exact or actual time)
which would make the distances =
d1 = 30 mph x 1 hr = 30 miles
d2 = 60 mph x 2 hrs = 120 miles
Total distance = 150 miles

finally confirm the answer with, avg speed = total distance/total time = 150/3 = 50 mph

which means 1/3 is correct
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Joined: 07 Dec 2014
Posts: 1018
Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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08 Nov 2015, 13:57
1
let t=total time
x=time @ 30mph
d=total distance
30x+60(t-x)=d
30x=60t-d
we know that d/50=t ➡ d=50t
substituting for d,
30x=10t
x/t=1/3
Manager
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Posts: 110
Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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24 Nov 2015, 23:58
Bunuel wrote:
Myra drove at an average speed of 30 miles per hour for some time and then at an average speed of 60 miles/hr for the rest of the journey. If she made no stops during the trip and her average speed for the entire journey was 50 miles per hour, for what fraction of the total time did she drive at 30 miles/hour?

(A) 1/5
(B) 1/3
(C) 2/5
(D) 2/3
(E) 3/5

Kudos for a correct solution.

We don't need to get into calculations for solving this questions. We can use the concept of weighted averages.

One thing for sure we know that if the average speed for the entire journey is 50, means she drove at 60mph for a longer duration.

20 10
30----- 50----60

This shows that you can divide the entire journey in 3 parts. Thus, 2/3rd parts she drove at 50 mph and 1/3rd part she drove at 30 mph.

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Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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12 Feb 2016, 19:23
pretty straight forward question...
suppose x is the time she drove for 30 miles/hour. total distance = 30x.
suppose t is total time she drove. we have average speed = 50 miles. thus, total distance = 50t.
she drove 60 miles/hour at t-x time. thus, she drove a distance of 60t-60x miles at 60 miles/hour.
we know the fact that:
30x+60t-60x=50t
10t=30x
t=3x.
ok, so the total time is 3x. she drove x at 30mph. thus, she must have driven x/3x or 1/3 of her time at 30mph.
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Re: Myra drove at an average speed of 30 miles per hour for some time and [#permalink]

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20 May 2017, 07:55
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Re: Myra drove at an average speed of 30 miles per hour for some time and   [#permalink] 20 May 2017, 07:55
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