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# In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as

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Re: In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as [#permalink]
Bunuel wrote:
In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as far as Bob. Mike jumped approximately what percent farther than Bob?

A. 5%
B. 6.67%
C. 7.5%
D. 8.33%
E. 10%

We can let c = the distance of Carl’s jump, m = the distance of Mike’s jump, and b = the distance of Bob’s jump.

Since Carl jumped 75% as far as Mike:

c = 0.75m

c/0.75 = m

c/(3/4) = m

4c/3 = m

Since Carl jumped 80% as far as Bob:

c = 0.8b

c/0.8 = b

c/(4/5) = b

5c/4 = b

We need to determine approximately what percentage farther Mike jumped than Bob jumped. We use the percent difference formula:

[(Mike’s distance - Bob’s distance)/Bob’s distance] x 100:

[(4c/3 - 5c/4)/(5c/4)] x 100

[(16c/12 - 15c/12)/(15c/12)] x 100

(c/12)/(15c/12) x 100

1/15 x 100 = 0.0667 x 100 = 6.67%

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Re: In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as [#permalink]
Bunuel wrote:
In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as far as Bob. Mike jumped approximately what percent farther than Bob?

A. 5%
B. 6.67%
C. 7.5%
D. 8.33%
E. 10%

Two ways.

First, "Hidden Plug In" (there's no variable, but there is information that would be really nice to know, so we can just make it up):
Make Bob 100. Carl is 80. Mike is 320/3.
Bob is 300/3. Mike is 320/3.
(20/3)/(300/3) = 20/300 = 2/30 = 6.667%

If you get to the part in red and don't like the fraction, you can always take one step backward to give yourself easier numbers.
Make Bob 300. Carl is 240. Mike is 320.
20/300 = 2/30 = 6.667%

Second, algebra:
4C = 3M
5C = 4B
M = 4C/3 = 16C/12
B = 5C/4 = 15C/12
$$\frac{M-B}{B}$$
$$(\frac{16C}{12}-\frac{15C}{12})/\frac{15C}{12}$$
$$\frac{C}{12}/\frac{15C}{12}$$
1/15 = 6.667%
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Re: In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as [#permalink]
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Re: In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as [#permalink]
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