On her way to work, Angela traveled p percent of the total trip at an

Question

On her way to work, Angela traveled  p  percent of the total trip at an average speed of  v  miles per hour and the rest of the distance at an average speed that was one-fifth less than  v . In terms of  p  and  v , what was Angela’s average speed for the entire commute?

(A)  \frac{400v }{ 500 - p}

(B)  \frac{500v }{ 400 - p}

(C)  \frac{500v }{ 500 - p}

(D)  \frac{500v }{ 600 - p}

(E)  \frac{600v }{ 500 - p}

KarishmaB · Accepted Answer

When options have variables, plugging in values is a quick way to do it. I love plugging in 0s and 1s. 
If p% distance is at v and rest at 4v/5, I assume that p = 0. So the entire distance was covered at speed 4v/5 so the average speed will be 4v/5. 
I am looking for an option that gives me 4v/5 when p = 0.

Only option (A) does that.

nick1816 · Answer

We can infer that average speed of the trip must be less than v. Eliminate B, C and E

Assume p=50 and v=10

Average speed = \frac{2*10*8}{(10+8)} = \frac{80}{9}

Option A-  \frac{400v}{500-p} = \frac{400*10}{450} = \frac{400}{45} = \frac{80}{9}  (answer)

nick1816 · Answer

Another way is

Assume total distance=D

Time taken to travel p% of the trip =  \frac{D*p}{100v}

Time taken to travel rest of the trip =  (1-\frac{p}{100})*(\frac{D}{0.8v})

Total time taken 
 = \frac{D*p}{100v} + (\frac{100-p}{100})(\frac{5D}{4v})

=\frac{4D*p}{400v }+ (\frac{500D-5D*p}{400v})

= \frac{500D-D*p}{400v}

Average speed = D/ (500D-D*p/400v) =   \frac{400v }{ 500 - p}

chetan2u · Answer

Substitution is a pretty safe way here. 
Let p=20, and v =50mph. 
So 20 at 50mph and 80 at 40mph. 
Total time taken =  \frac{20}{50}+\frac{80}{40}=2.4

Total distance =100 and time taken =2.4 
Average speed = 100/2.4=1000/24=125/3

Check option for this value. 
A. 400*50/(480)=5*50/6=125/3

A

Kapil1650 · Answer

Took me 5 min to solve ! Alas !

I did it by assuming the value of total distance as 100 then solving !

5 min : bad huh

Posted from my mobile device

Kinshook · Answer

Given: On her way to work, Angela traveled  p  percent of the total trip at an average speed of  v  miles per hour and the rest of the distance at an average speed that was one-fifth less than  v .

Asked: In terms of  p  and  v , what was Angela’s average speed for the entire commute?

Let the total distance be D

Time taken for Dp% distance = Dp%/v

Time taken to cover rest = D(100%-p%)/(4v/5) = D(125%-1.25p%)/v

Average speed = D/{Dp%/v + D(400%-5p%/4)/v} = v/{p% + 125% -1.25p%} = 100v / (125-.25p) = 400v/(500-p)

IMO A

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On her way to work, Angela traveled p percent of the total trip at an

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On her way to work, Angela traveled p percent of the total trip at an

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