On a particular trip, Jane drove x miles at 30 miles per hour and y miles at 40 miles per hour. Was her average speed for the trip greater than 35 miles per hour?

\(average \ speed = \frac{total \ distance}{total \ time}=\frac{x+y}{(\frac{x}{30}+\frac{y}{40})}=\frac{120(x+y)}{4x+3y}\)

Thus the question asks whether \(\frac{120(x+y)}{4x+3y}>35\).
After simplifying the question becomes: is \(\frac{y}{x}>\frac{4}{3}\).

(1) y > x + 60.

If x = 1 and y = 100, then y/x > 4/3.
If x = 1000 and y = 1200, then y/x < 4/3.

Not sufficient.

(2) y/x > 4/3. Directly gives an answer to our question. Sufficient.

Answer: B.

Hope it's clear.

Alternatively, notice that 35 miles per hour is halfway between 30 and 40 miles per hour. Now, for the average speed to be exactly 35 miles per hour, the amount of time spent to cover x miles (x/30 hours) must equal to the amount of time spent to cover y miles (y/40 hours). The average speed to be greater than 35 miles per hour, the amount of time spent to cover x miles at the slower speed (x/30 hours) must be less than the amount of time spent to cover y miles at the faster speed(y/40 hours):

x/30 < y/40.

Then continue as above.
_________________

Hi Bunuel

i found another explanation for this question

it says that avg speed ie 4 is one away from 3 and 4 away from 8. Therefore time spent travelling at 3 m/hr is 4/5 of total.

I know this explanation is in compliance with the weighted average formula.

Can you plz elaborate on this method how it is 4/5 and not 1/5
_________________

Hello GMATinsight

I think there is typo in 3 step: "Step 3: Distance of 3 from Average 4 is 1 AND Distance of 8 from Average 4 is 4"
_________________

120(x+y)/(4x+3y)>35

Since x and y are Positive Numbers hence you can cross multiply

120x + 120 y > 35*(4x+3y)

i.e. 120x + 120 y > 140x + 105y

i.e. 15y > 20x

i.e. 3y > 4x

i.e. (y/x) > (4/3)

I hope it clears your doubt!
_________________

Bunuel

you have mentioned "If x = 100 and y = 200, then y/x < 4/3."
In this case y/x = 2 and it is not less than 4/3.

we might need something like x = 240 and y = 320 to prove option 1 is not sufficient.
in this case y/x = 4/3 and thus not less than 4/3

Please share your thoughts.

Could anyone please explain the logic behind the highlighted part. I m not able to get how more than average speed correlates to time taken to cover specific distances, x and y.

Hello earnit

This is weighted average question
We can imagine scales:
on one side --------------------- on another side
speed 40 mph ----------------- 30mph

if both side have equal weight (in our case weight is time) when speed will be in middle:
speed 40 mph--------------------- 30mph
10 hours --------------------------10 hours

Average = 35 mph
===============================

if we shift our weight to left: more time with speed 40 mph then average will grow

speed 40 mph---------------------30mph
20 hours---------------------------0 hours

Average = 40 mph
===============================

if we shift our weight to right: more time with speed 30 mph then average will decline

speed 40 mph--------------------- 30mph
0 hours --------------------------- 20 hours

Average = 30 mph

So if we want average to be more than 35 mph we should put more weight on 40 mph scale (distance y in this question)

Try to read Karishma's articles on this topic. She like this topic and write really good posts about it.
At first weighted average was very difficult for me but when I read her articles and solve like 50 tasks this topic became very intuitive:

http://www.veritasprep.com/blog/2014/12 ... -averages/
http://www.veritasprep.com/blog/2011/04 ... ge-brutes/
http://www.veritasprep.com/blog/2011/03 ... -averages/
http://www.veritasprep.com/blog/2014/09 ... difficult/
http://www.veritasprep.com/blog/2013/12 ... -question/

There is much more articles I just post first that I find. You can use search by words "weighted average"
_________________

Bunuel,

If y=41miles and x=40 miles
the average speed will be 34,7 (less than 35!)

41 miles = ~1 hour
40 miles = ~1 hour and 20 minutes

total distance: 81 miles/ 2hours and 20 minutes = 37,7 miles.

Am I doing something wrong?

Explanation: For average speed problems, you always need to remember that the average speed is total distance divided by total time, not a simple average of the speeds (unless the times are the same). Or in weighted average terms: it is the time that affects the weighting of the two speeds when you average them together.

To answer this question, you need to determine at which speed more time was spent. If you can prove more time was spent at one speed, then you can answer the question or in other words, this question is really: is time for y greater than the time for x.

Algebraically this can be written as: is y/40>x/30 or is y/x>4/3

From this you realize clearly that statement (2) is sufficient.
For statement (1) you cannot prove this, because you do not know how big x and y are.
If x and y were really large distances, then the average might not be greater than 35, and if x and y were really small distances then it would be.

Answer is (B).

Stem:
rate*time=distance
30 * t1 = x
40 * t2 = y

Question:
(x+y)/(t1+t2) > 35, expressing time as Work/rate, (x+y)/(x/40+y/30) > 35
The question is asking whether the distance traveled at 40 mph is more than that at traveled at 30 mph.

1) y>x+60
if y = 80 and x=0, then obviously Yes since y>x
if y = 400 and x = 330 then 400/40 = 10 and 330/30 = 11, then No since x>y
Depends on values of y and x, NS

2) 3y > 4x, so y is more than x in this ratio, which is another way of saying that whatever the distance is, the part traveled at 40 mph is more than the part traveled at 30 mph, which means the Avg Speed will be >35. Sufficient.