# Question: Nani’s Average Speed

## Comment on Nani’s Average Speed

### Hi Brent. Its also possible

Hi Brent. Its also possible to use SMART numbers right. If D= 80.
Therefore TD = (80+80)= 160
T1 = 80/40 = 2 Hrs
T2= 80/R
Therefore Average speed: TD/TT = 160/(2+(80/R)) = 30mph. Simplify and get the value of R. Perfect!!

### i dont understand how to ### I'm assuming that you're

I'm assuming that you're referring to the equation 160/(2+(80/R)) = 30

Let's first simply the denominator, 2 + (80/R)

2 + 80/R = 2/1 + 80/R (our common denominator will be R)
= 2R/R + 80/R
= (2R + 80)/R

So, 160/(2+(80/R)) = 160/[(2R + 80)/R]
= 160 x R/(2R + 80)
= 160R/(2R + 80)

This, we are told, is equal to 30, so we get: 160R/(2R + 80) = 30

The easiest thing to do here is cross multiply (see 6:15 at https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...)

When we take 160R/(2R + 80) = 30/1 and cross multiply we get: (1)(160R) = (30)(2R + 80)

Simplify: 160R = 60R + 2400

Subtract 60R from both sides: 100R = 2400

Solve: R = 24

### how did you rewrite the

how did you rewrite the fractions with common denominators. i dont get this step. i watched every single video up to this point, i dont remember you explaining rewriting with common denominators when there i a variable in one denominator. ### When it comes to equations

When it comes to equations with fractions and variables, you can either combine fractions via common denominator (see https://www.gmatprepnow.com/module/gmat-arithmetic/video/1068) or try to eliminate the fractions (more here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...).

The technique for explaining rewriting with common denominators is the same for fractions with or without variables. ### We have d/40 + d/h

We have d/40 + d/h

So, the common denominator must be a multiple of both 40 and h.

ASIDE: This is no different from adding 2/3 + 1/7. In this case, the common denominator must be a multiple of both 3 and 7. An easy option is a common denominator of 21 (the product of 3 and 7)

Likewise, a nice common denominator of 40 and h can be 40h (the product of 40 and h)

At this point, we need to create fractions that are equivalent to d/40 + d/h so that they have a common denominator of 40h

Take d/40 and multiply top and bottom by h to get: d/40 = dh/40h
Also, take d/h and multiply top and bottom by 40 to get: d/h = 40d/40h

So, d/40 + d/h = dh/40h + 40d/40h = (dh + 40d)/40h

thanks a lot

### I posted a similar question

I posted a similar question like this. After seeing this post, it helps. Thanks

### Brent, I found this to be a

Brent, I found this to be a great example of when to test answers to save time and eliminate. Starting with C obviously yields the correct answer in this case, but even starting with A tells you that h must be higher, but no a lot higher. Long-form took me way longer than testing the choices as h in the formula 30 = (d + d) / [(d/40) + (d/h)] Nice!

### I am scratching my head as to

I am scratching my head as to why using D=40 gives rate as 2.4 vs D=80 gives 24.
My dain is bramaged. :-| ### Hi Brent ,

Hi Brent ,

We can also use the formula of Avg Speed = 2AB / A+B , where A is average speed for going , and B is average speed of returning . The only caution that we have to take using this formula is that the distance should be same .

Regards ### I'm not a big fan of students

I'm not a big fan of students remembering formulas that work for a very specific case. That said, the formula is valid.

### Hi Brent. Really enjoying

Hi Brent. Really enjoying your video lessons!
Wanted to clarify the steps involved in rewriting the equations for the denominator and numerator. Am I correct in thinking the 2 steps below are used to re-write the equations from (d/40+d/h) to (dh/40h+40d/40h)?
1) Create equivalent denominators of 40h by applying LCM to (d/40+d/h).
2) Apply ad=bc rule to calculate numerator by converting (d/40+d/h) to (dh+40d). ### Hi Leon,

Hi Leon,

I'm delighted to hear you're enjoying the course!

I'm not entirely sure what you mean by step 2, so I'll go over the steps below:

We want to simplify d/40 + d/h (i.e., rewrite this as one fraction)
As you noted, the least common multiple of 40 and h is 40h.
So, we must convert d/40 and d/h to EQUIVALENT FRACTIONS with the shared denominator of 40h

Take d/40 and multiply top and bottom by h to get: dh/40h
Take d/h and multiply top and bottom by 40 to get: 40d/40h

So, d/40 + d/h = dh/40h + 40d/40h
Now that we have the same denominators, we can add the numerators to get: (dh + 40d)/40h

Does that help?

Cheers,
Brent

Very helpful indeed Brent. Thank you for clarifying! :)

### Hi Brent,

Hi Brent,
For both this video and the previous one I have simply made up a distance (a nice multiple of the two known speeds, so for this video 120m and for the previous 160m) and then proceeded to calculate.
In both cases I have achieved the correct result but at a fraction of the time.
Can I extrapolate that this is a reliable time saver?
Thank you ### In many cases, assigning a

In many cases, assigning a nice value to the distance is a great time-saver. This strategy will typically work for average speed questions in which the distance is not stated (either directly or indirectly).

However, if the question contains information that "locks in" the travel distance, then you can't assign values.
For example, if the question says "Sue traveled for 5 hours at a constant speed of 60 miles per hour," then we can't assign a distance, since the question indirectly tells us that the travel distance is 300 miles.

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

For this rate question, can you please explain why here we have to take the outcome/work as 1200 or unknown for answer A? In majority of these problems, we usually assume that output is 1 (filling a tank, completing a job etc).

https://gmatclub.com/forum/working-together-at-their-constant-rates-a-and-b-can-fill-97316.html#p750195 For questions in which the "job" isn't defined, we have the option of saying the output = 1.
For example, for the information "Joe can paint a house in 3 days," we can save the output equals 1 painted house, which means we can conclude that Joe's RATE = 1/3 of a house per day.

That said, we're not obligated to say the output is 1.
We also have the option of assigning a "nice" value (other than 1) to the job.
For example, for the "Joe can paint..." example above, we COULD say that the job of painting the house = painting all 12 rooms in the house.
If we do this then, Joe's RATE = 4 rooms per day.

In the case of the linked question, we're given the rate and output, so we really shouldn't change that.

### Hi Brent,

Hi Brent,

Can you please solve the following data sufficiency question?

Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?
1) Reiko's average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.
2) It took Reiko 20 more minutes to drive from A to B than to make the return trip ### I dont wanna deal with TWO

I dont wanna deal with TWO variables so how about let's just assume d = 40 and we can still get 24 at the end :) ### Great idea!

Great idea!
Average speed = (total distance)/(total travel time) = 30 mph

So, if we let 40 = the distance from home to work, we have:
80/(total travel time) = 30

Total travel time = (time going to work) + (time going home)

Let x = the speed travelling from work to home

Since time = distance/speed, we can write the following:
Total travel time = 40/40 + 40/x
= 1 + 40/x
= x/x + 40/x
= (40 + x)/x

Substitute into our equation to get: 80/((40 + x)/x) = 30
Rewrite as: (80)(x)/(40 + x) = 30
Multiply both sides by (40 + x) to get: 80x = 1200 + 30x
Solve to get: x = 24