If Car A took n hours to travel 2 miles and Car B took m hours to trav

Alternative approach:

Looking at the choices only choice A has the unit of time all other have different units. You dont need to solve anything in this question
A- (time^2/time + time) = time
B- (time + time)/(time + time) = 1 (i.e. No unit)
C- (time + time)/time^2 = 1/time
D- (time + time)/(time + time) = 1(No unit)
E - (time + time )/ (time + time) = 1(No unit)

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If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?

Speed = distance/time
Speed of car A = 2/n
Speed of car B = 3/m

Speed of car C = (2/n + 3/m)/2
= (2m/mn + 3n/mn)/2
= [(2m + 3n)/mn]/(2/1)
= [(2m + 3n)/mn](1/2)
= (2m + 3n)/2mn

Which of the following expresses the time it would take Car C to travel 5 miles?
time = distance/speed
Car C's time = 5/[((2m + 3n)/2mn)]
= 5[2mn/(2m + 3n)]
= 10mn/(2m + 3n)

Answer: A

Cheers,
Brent

We normally only discuss "average speed", which you correctly describe, when there is a single thing (person or car or plane or whatever) completing a trip in two (or more) steps, at different individual speeds. Here we don't have a single thing -- we have two cars, A and B. The question doesn't ask for their "average speed" (which wouldn't make much sense anyway, with two cars), so that concept isn't relevant in this particular question. Instead the question asks for "the average (arithmetic mean) of those rates". So if we find the two rates, we want to compute their "average", which just means we want to add them, and divide by 2.

The situation in this question is rare. Most of the time, if you see a question that talks about speeds, and an "average" of some kind, it will be a traditional "average speed" question, and you will want to use the definition, avg speed = total distance/total time. I might not be remembering correctly, but I think this is the only official question I've seen where you actually want to compute the average of the two speeds of two different cars (or people or trains, etc).

speed = distance/time, so car A's speed is 2/n, and car B's speed is 3/m. To average these two values, we add them and divide by 2, so the speed of car C is

[ 2/n + 3/m ] / 2 = (2m + 3n)/2mn

Since speed = distance/time, then at this speed, we can find the time T it would take to travel 5 miles using:

(2m + 3n)/2mn = 5/T
T = 5(2mn) / (2m + 3n)
T = 10mn/(2m + 3n)

An alternative approach which will work, though I think it's substantially more time-consuming (though a good approach if you'd find the algebra above difficult) - we can invent extremely simple numbers for m and n, solve the problem using those numbers, and then see which of the five answer choices produces the correct answer for our specific choice of numbers. One danger using this approach is that sometimes two or more answer choices both appear to give the correct answer, and then to decide between them you have to come up with another set of numbers.

But here, if n = 2, then car A's speed is 1 mile per hour, and if m = 3, car B's speed is 1 mile per hour. Then car C's speed is clearly 1 mile per hour, so it will take it 5 hours to travel 5 miles. So when n=2 and m=3, the answer to the question needs to work out to be '5'. If we now plug n=2 and m=3 into each answer choice, we find only answer A gives us a value of '5', so it must be right.

Find avg speed by adding both speeds and dividing by 2. Multiply top and bottom by nm. Distance =5miles. So our D=RT equation is going to be 5miles = Avg speed * Time. Get time on it's own side. multiply top and bottom by 2mn.

Speed A = \(\frac{2}{n}\)

Speed B = \(\frac{3}{m}\)

Find average speed: (\(\frac{2}{n}\) + \(\frac{3}{m}\)) / 2.

Multiply top and bottom by nm.

(2/n + 3/m)/2 *\(\frac{nm}{nm}\) = \(\frac{2m+3n}{2nm}\)

How much time does it take, traveling at this average speed, to go 5 miles in distance?

5 = \(\frac{2m+3n}{2nm}\) * Time

Get Time on it's own side by dividing the Speed.

Time = 5 / \(\frac{2m+3n}{2nm}\)

Multiply top and bottom by 2nm

Time = \(\frac{5*2mn}{2m+3n}\)

Time = \(\frac{10mn}{2m+3n}\)

The average of the rates of Cars A and B is:

(2/n + 3/m)/2

[2m/(mn) + 3n/(mn)]/2

[(2m + 3n)/(mn)]/2

(2m + 3n)/(2mn)

Since time = distance/rate, it would take Car C the following amount of time to travel 5 miles:

5/[(2m + 3n)/(2mn)] = 10mn/(2m + 3n)

Answer: A
_________________

why isnt the average speed = Total distance / Total Time? 5/(n+m)

It is said that C is traveling with the speed that is average of A and B. SO C's speed will be (2/n + 3/m)/2 = (2m+3n)/ 2mn.

Time taken by C = Dist/ speed = 5 / {(2m+3n)/ 2mn} = 10mn/(2m+3n)

Let's say n=2 and m=3. So the average time taken by A and B is 5/5= 1 miles/hour.

So Car see takes 5 hours to cover 5 miles.

Option A. \(\frac{10*2*3}{12}\) That's equal to 5.


PS78502.01
Quantitative Review 2020 NEW QUESTION[/quote]

Hi Brent! Totally understand your approach but I've always been thought that for avg speed (unlike avg rate) , we suppose to use Total distant/ Total time?
So, in this case, Avg Speed A and B = Avg Speed C = 2+3/n+m = 5/n+m

Please clarify if I miss any point
Thank you!!

If Car A took n hours to travel 2 miles and Car B took m hours to travel 3 miles, which of the following expresses the time it would take Car C, traveling at the average (arithmetic mean) of those rates, to travel 5 miles?

Hi..
The word is the average (arithmetic mean) of those rates, so we have to find the rates first..
Car A.. 2/n and
Car B..3/m
Average of these speed = \(\frac{1}{2}(\frac{2}{n}+\frac{3}{m})=\frac{2m+3n}{2mn}\)

Had it been the average of total ditance covered by A and B, then you would be correct.
But then the question would be something like 'If Car A took n hours to travel 2 miles and then took m hours to travel 3 miles, which of the following expresses the time it would take Car A to travel 5 miles?
_________________

Hello,
Can any math expert please tell me why the average rate here is not calculated as total distance/ total time taken= (3+2)/(m+n)= 5/(m+n) ?

Because in this question (A plane traveled k miles in its first 96 minutes of flight time. If it completed the remaining 300 miles of the trip in t minutes, what was its average speed, in miles per hour, for the entire trip?) avg speed is calculated using the method I just mentioned.
So how are these two questions different in terms of only calculating avg speed (if at all they're)?
PS- I couldn't post the link since I am a new member here but you can search this question on the forum.

Hoping to get some clarity on this concept. VeritasKarishma Bunuel

Speed = \(\frac{Distance }{ Time}\)

Car C: Speed: \(\frac{2}{n}\)

Car B: Speed: \(\frac{3}{m}\)

Average speed of both: \(\frac{\frac{2}{n}+\frac{3}{m}}{2}\)

=> \(\frac{2m+3n }{ 2mn}\)

Time: \(\frac{Distance }{ Speed}\)

Time it would take Car C : \(\frac{5}{\frac{2m+3n}{2mn}}\)

=> \(\frac{10mn }{ 2m+3n}\)

Answer A

10 second approach

Plug in:
n=2 and m=3
speed of A=1
speed of B=1
average time taken by A and B is 1 mile/hour.

So speed of Car C = 5/1=5

Option A. 10∗2∗31210∗2∗312 That's equal to 5.

The question is asking for the time it would take, so the correct answer should have units of hours.
Only answer A is in hours.
B, D, and E don't have units, and the units of answer C are (1/hours).
Unit analysis typically eliminates at least 3 answer choices in this type of problem.
Read all about it in my ebook (see link in signature below).

Make up numbers:

n = 2

That means Car A is traveling at 1MPH

m =3

That means Car B is traveling at 1MPH

Take the Average (1+1)/2=1

Car C travels 1 MPH so it will take 5 Hours.

Start Plugging in:

A. 10(NM)/(3N+2N) --> Remember n = 2 m = 3 --> we end up at 60/12 = 5 (Which is the number were looking for and where done).

It's an interesting point and hence, I will add my two cents too (though Ian has already explained it very well).

Normally, average speed is the weighted average of two or more speeds when time is the weight. It is also Total Distance/Total Time.

So average speed of a car that travels 2 miles in m hrs and 3 miles in n hrs will be 5/(m + n).
It will also be [(2/m)*m + (3/n)*n] / (m + n) = 5/(m+n) (Using weighted average concept)

They both match and of course they should. Note that this concept of Total distance/Total time makes sense for one car.

The question here is different. Here, you are given the rates of two cars and are asked to find their arithmetic mean (We use arithmetic mean only when time for which they travel is the same but the cars here travel for diff times, m and n hrs).
Hence, we find their rates and find the AM
[(2/m) + (3/n)] / 2

We do not take the weighted average as shown above because we have been asked to take the arithmetic mean of the rates.

Rate of A = 2/n

Rate of B = 3/m

Rate of C = (2/n+3/m)/2 = (2m+3n)/2mn

Time taken by C = 5/(2m+3n)/2mn = 2mn * 5/(2m+3n) => 10mn/(2m+3n) [Option A]

Don't Solve - Look at the units

a. hr.*hr./hr. = Hr., rest options don't have any unit.