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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 [#permalink] New post   12 Jan 2021, 07:23
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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) \(\frac{600v }{ 500 - p}\)
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Re: On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   13 Jan 2021, 03:41
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sjuniv32 wrote:
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) \(\frac{600v }{ 500 - p}\)


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.
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Re: On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   12 Jan 2021, 14:44
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)




sjuniv32 wrote:
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) \(\frac{600v }{ 500 - p}\)
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Re: On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   12 Jan 2021, 14:54
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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) = \( \frac{400v }{ 500 - p}\)


sjuniv32 wrote:
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) \(\frac{600v }{ 500 - p}\)
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Re: On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   12 Jan 2021, 19:09
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sjuniv32 wrote:
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) \(\frac{600v }{ 500 - p}\)



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

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Re: On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   12 Jan 2021, 23:18
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
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On her way to work, Angela traveled p percent of the total trip at an [#permalink] New post   11 Feb 2022, 18:02
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 [#permalink]
11 Feb 2022, 18:02
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