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

my answer is 1/3.


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).

Please correct me if wrong.

50-30=20
60-50=10

Required answer is 10/(10+20) = 1/3

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

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.

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.

Answer (B)

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

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

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.

Answer: B

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.

We can use the formula of average speed = total distance/total time. We can let x and y be the first and second distances, respectively.

50 = (x + y)/(x/30 + y/60)

Multiplying by 60/60, we have:

50 = (60x + 60y)/(2x + y)

50(2x + y) = 60x + 60y

100x + 50y = 60x + 60y

40x = 10y

4x = y

Thus, the fraction of the time driving 30 mph was:

(x/30)/(x/30 + y/60)

Since y = 4x, we have:

(x/30)/(x/30 + 4x/60)

Multiplying by 60/60, we have:

2x/(2x + 4x)

2x/6x

1/3

Alternate Solution:

We can use a weighted average approach to answer this question. Had the average speed been 45 miles per hour, we would know that she traveled an equal amount of time at 30 mph and at 60 mph, since 45 is exactly halfway between 30 and 60. However, since the average speed was 50 miles per hour (which is closer to 60 mph than to 30 mph), we see that she traveled at 60 mph for a longer period of time. We see that 60 - 50 = 10 and 50 - 30 = 20, so if we break the trip into 3 segments, she traveled 2 of the 3 segments at 50 and 1 of the 3 segments at 30 mph. Thus, the fraction spent driving 30 mph is 1/3.

Answer: B

A student asked me whether it's possible to assign a "nice" value to the distance, and go from there.
So, here it goes......

Let's say the total distance traveled = 300 miles

time = distance/speed
Since the average speed for the entire trip is 50 miles per hour, the TOTAL travel time = 300/50 = 6 hours

Let t = the time (in hours) Myra spent driving 30 miles per hour
So, 6 - t = the time Mrya spent driving 60 miles per hour

We know that: (distance traveled at 30 mph) + (distance traveled at 60 mph) = 300 miles
distance = (rate)(time)
So, we get: 30t + 60(6 - t) = 300
Expand: 30t + 360 - 60t = 300
Simplify: 360 - 30t = 300
Solve: t = 2

We now know that the entire trip took 6 hours, and that 2 of those hours were spent driving 30 miles per hour.
So, the fraction of the total time spent driving 30 mph = 2/6 = 1/3

Answer: B

Cheers,
Brent

30________________________60

_____________50_____________

10________________________20



\(= \frac{1}{3}\) , Answer must be (B)

Hi ScottTargetTestPrep



Don't you mean 2 of the 3 segments at 60?

Yes, that is what I mean.

Can someone kindly guide where I missed.


k is the distance traveled in 30 miles

d
-------------- =50
(k/30)+ (d-k)/60


30*60 d
----------- = 50
60 k +30(d-k)


60d
--- = 50
k+d

60 d= 50k +50 d

d=5k

k/d= 1/5


Thanks a million in advance


yashikaaggarwal
KarishmaB
ChandlerBong
ThatDudeKnows
Bunuel
BrentGMATPrepNow
Narenn
walker
IanStewart

I think you (correctly) found what fraction of the total distance Myra drove at 30 mph, not what fraction of the total time she drove at 30 mph, which is what the question asks for.

Ian has already pointed out the error in your solution so that's that.
Also, average speed is the weighted average of speeds when weights are 'time taken' and hence this question can be directly solved using that concept.