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A salesperson travels weekly to a client's office and returns...

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A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 21 Jun 2017, 19:54
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A salesperson travels weekly to a client's office and returns by the same route. If the salesperson travels to the client's office and then completes 25% of the return trip, what percent of the round-trip has been completed?

A 12.5%
B 25%
C 62.5%
D 75%
E 81%
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Re: A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 21 Jun 2017, 20:10
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Tking1167 wrote:
A salesperson travels weekly to a client's office and returns by the same route. If the salesperson travels to the client's office and then completes 25% of the return trip, what percent of the round-trip has been completed?

A 12.5%
B 25%
C 62.5%
D 75%
E 81%



Hi..
He has travelled 100% of initial journey and then 25% of the 100% of return journey..

Since both ways distance is SAME, he travels (100+25)% of (100+100)%..
So the overall % covered is \(\frac{100+25}{100+100}*100=\frac{125}{200}*100=62.5%\)
C
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A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 21 Jun 2017, 20:45
1
Tking1167 wrote:
A salesperson travels weekly to a client's office and returns by the same route. If the salesperson travels to the client's office and then completes 25% of the return trip, what percent of the round-trip has been completed?

A 12.5%
B 25%
C 62.5%
D 75%
E 81%

Assume distance to and from client is 20 miles each way. (You can choose any number you want because distances are equal. I chose 20 because it's divisible by four.)

The salesperson travels 20 miles to get there. She travels only 25% of the way back; \(\frac{1}{4}\) of 20 miles is 5 miles.

So she's traveled 25 miles of a round trip of 40 miles.

\(\frac{25}{40}\) = .625 * 100 = 62.5%

Answer C
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Re: A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 21 Jun 2017, 20:51
let onwards journey be '4' miles, so return journey is also '4' miles. Total journey = 8 miles

This particular journey, total trip = 4 + 25% of 4 = 4 + 1 = 5 miles

So, required percentage = 5/8 * 100 = 62.5 %

Hence C answer
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Re: A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 21 Jun 2017, 22:10
Tking1167 wrote:
A salesperson travels weekly to a client's office and returns by the same route. If the salesperson travels to the client's office and then completes 25% of the return trip, what percent of the round-trip has been completed?

A 12.5%
B 25%
C 62.5%
D 75%
E 81%


Let's assume distance to the client's office is = 4x
so, Total distance traveled in a round trip = 8x
Sales covered 25% distance of return trip = 25% of 4x = 1x
Distance covered by Salesman = 4x + 1x = 5x
% distance travelled = 5x/8x = 62.5%
Hence Answer C
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A salesperson travels weekly to a client's office and returns... [#permalink]

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New post 03 Aug 2017, 21:56
1
Tking1167 wrote:
A salesperson travels weekly to a client's office and returns by the same route. If the salesperson travels to the client's office and then completes 25% of the return trip, what percent of the round-trip has been completed?

A 12.5%
B 25%
C 62.5%
D 75%
E 81%


Let the distance of one way trip to client's office be \(= 100\) km

Therefore total distance of round trip \(= 100 + 100 = 200\) km

Salesperson completes the one way trip, ie; \(= 100\) km

Then Salesperson completes \(25\)% of return trip, ie; \(= 25\)% of \(100 = \frac{25}{100} * 100 = 25\) km

Required percentage \(=\) ( Total distance completed \(/\) Total distance )\(* 100\)

Required percentage \(= \frac{100 + 25}{200} * 100 = \frac{125}{200} * 100 = 62.5\)%

Answer (C)...

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A salesperson travels weekly to a client's office and returns...   [#permalink] 03 Aug 2017, 21:56
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