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# During a trip, Yogi Bear traveled 62.5% of the total distance at an

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During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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Updated on: 17 Aug 2015, 01:22
8
00:00

Difficulty:

55% (hard)

Question Stats:

70% (03:14) correct 30% (03:05) wrong based on 104 sessions

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During a trip, Yogi Bear traveled 62.5% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of $$x$$ miles per hour. In terms of $$x$$, what was Yogi Bear's average speed for the entire trip?

A. $$\frac{{320x}}{{3+200x}}$$

B. $$\frac{{x+80}}{{120x}}$$

C. $$\frac{{320x}}{{5x+120}}$$

D. $$\frac{{320x+5}}{{200x}}$$

E. $$\frac{{120x}}{{x+80}}$$

Again, from OG 13 with different conditions.

Originally posted by mejia401 on 09 Aug 2015, 14:41.
Last edited by Bunuel on 17 Aug 2015, 01:22, edited 4 times in total.
RENAMED THE TOPIC.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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09 Aug 2015, 19:42
2
1
I think question have some discrepancy.
the answer for the question is 320x/5x+120.

So, the given option C is 320x/3x+200. It happens only first 62.5% distance covered for xmiles/hour and remaining distance 37.5%
covered by 40 miles/hours.

Average speed of first distance m covered for phours and remaining distance n covered for q hours.
given by the formula.

= (m+n)pq/mq+np.

from given question is

= m+n= 100% distance

=40x/67.5%*x+37.5%*40

=40x/5x/8+15

=320x/5x+120.

suppose there is interchange in distance or speed then
=40x/3x/8+25=320x/3x+200
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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10 Aug 2015, 02:03
Corrected. Thank you.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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10 Aug 2015, 04:26
1
1
mejia401 wrote:
During a trip, Yogi Bear traveled 62.5% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of $$x$$ miles per hour. In terms of $$x$$, what was Yogi Bear's average speed for the entire trip?

A. $$\frac{{320x}}{{3+200x}}$$

B. $$\frac{{x+80}}{{120x}}$$

C. $$\frac{{320x}}{{5x+120}}$$

D. $$\frac{{320x+5}}{{200x}}$$

E. $$\frac{{120x}}{{x+80}}$$

Again, from OG 13 with different conditions.

The fastest way is to assume x = 1 and then see which of the options satisfy this.

Average speed = $$\frac{d}{\frac{0.625}{40}+\frac{0.375}{x}}$$= 2.56.

Only option C gives this value and is thus the correct answer.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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10 Aug 2015, 05:10
mejia401 wrote:
During a trip, Yogi Bear traveled 62.5% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of $$x$$ miles per hour. In terms of $$x$$, what was Yogi Bear's average speed for the entire trip?

A. $$\frac{{320x}}{{3+200x}}$$

B. $$\frac{{x+80}}{{120x}}$$

C. $$\frac{{320x}}{{5x+120}}$$

D. $$\frac{{320x+5}}{{200x}}$$

E. $$\frac{{120x}}{{x+80}}$$

Again, from OG 13 with different conditions.

I prefer to take Numbers in Such questions to make the expressions simpler

e.g. Let Distance = 1000
i.e. 62.5% of Distance = 625

Average Speed = Total Distance / Total Time

Total Distance Covered = 1000

Total Time = Time taken to cover 625 miles @40 miles/hr +Time taken to cover (1000-625) = 375 miles @x miles/hr

Total Time = (625/40)+(375/x)

i.e. Average Speed = 1000/ [(625/40)+(375/x)] = $$\frac{{320x}}{{5x+120}}$$

Answer: Option C
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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11 Aug 2015, 14:19
The easiest way would be to chose 40mph for the rest of the trip, but this leaves you with option C and E. On the test day I would probably go with the 50/50 chance because of the time.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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16 Aug 2015, 16:02
mejia401 wrote:
During a trip, Yogi Bear traveled 62.5% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of $$x$$ miles per hour. In terms of $$x$$, what was Yogi Bear's average speed for the entire trip?

A. $$\frac{{320x}}{{3+200x}}$$

B. $$\frac{{x+80}}{{120x}}$$

C. $$\frac{{320x}}{{5x+120}}$$

D. $$\frac{{320x+5}}{{200x}}$$

E. $$\frac{{120x}}{{x+80}}$$

Again, from OG 13 with different conditions.

Similar question to practice: during-a-trip-francine-traveled-x-percent-of-the-total-94933.html
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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16 Aug 2015, 19:31
Why are you not reducing 320x/5x+120 to 64x/x+24? My teacher would have marked this wrong.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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16 Aug 2015, 19:36
gracie wrote:
Why are you not reducing 320x/5x+120 to 64x/x+24? My teacher would have marked this wrong.

You are correct and it's a good habit to have . Whenever you want to report an answer in fractions, you should be reducing both the numerator and denominator. But if the options do not have anything reduced, then you do not have to. You can not challenge the options provided and you have to play within the confines of the question statement and the options provided.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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16 Aug 2015, 20:49
got 40x/0.625x+15 after multiplying by 8/8 get = 320x/5x+120

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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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07 Feb 2016, 09:18
mejia401 wrote:
During a trip, Yogi Bear traveled 62.5% of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of $$x$$ miles per hour. In terms of $$x$$, what was Yogi Bear's average speed for the entire trip?

A. $$\frac{{320x}}{{3+200x}}$$

B. $$\frac{{x+80}}{{120x}}$$

C. $$\frac{{320x}}{{5x+120}}$$

D. $$\frac{{320x+5}}{{200x}}$$

E. $$\frac{{120x}}{{x+80}}$$

Again, from OG 13 with different conditions.

I find it somehow strange that the answer choice is not written is the most simplified form!!!

62.5% can be rewritten as 5/8

so we have 5d/8 - distance, and 40 mph - speed.
we then have = 3d/8 - distance, and x mph - speed.

to find the average speed, we need to find the total time.

time for first leg: 5d/8 * 1/40 = d/64
time for second leg: 3d/8 * 1/x = 3d/8x

total time: d/64 + 3d/8 | multiply first by x, second by 8.
(xd+24d)/64x
factor d in the numerator:
d(x+24)/64x
is the total time.
total distance = d.

to find the average speed, divide distance by time.

d*64x / d(x+24)

simplify by D

64x/(x+24).
right away, we can eliminate B and D.
let's multiply what we have by 5.
320x/(5x+120).

still don't get why the answer wouldn't be in the most simplified way.
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an  [#permalink]

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28 Jul 2018, 03:32
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Re: During a trip, Yogi Bear traveled 62.5% of the total distance at an &nbs [#permalink] 28 Jul 2018, 03:32
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