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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 13 Jun 2017, 05:54
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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%
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B
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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 13 Jun 2017, 06:02
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0.8x = 0.75y
y= (0.8/0.75)x= 1.0667x
% ahead = ((1.0667x-x)/x) *100
=6.67

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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 13 Jun 2017, 08:01
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Bunuel
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%
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Let the distance Mike jumped = 100

Carl jumped = 75% as far as Mike = 75% of 100 = \(\frac{75}{100} * 100\) = 75

Carl jumped = 80% as far as Bob ==> 75 = 80% of B ==> 75 = \(\frac{80}{100} * B\)

Bob jumped = \(\frac{75 * 100}{80}\) = 93.75

Difference between distance jumped by Mike and Bob = 100 - 93.75 = 6.25

Required percentage = \(\frac{6.25}{93.75} * 100\) = 6.67%. Answer B...
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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 15 Jun 2017, 15:59
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Bunuel
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%
Show more

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%

Answer: B
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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 18 Jul 2022, 07:17
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Bunuel
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%
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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%
Answer choice B.

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%
Answer choice B.

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%
Answer choice B.
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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 20 Apr 2025, 07:02
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why are we equating 0.8x with 0.75y? can it be 1.8 and 1.75 instead?

apoorv2708
0.8x = 0.75y
y= (0.8/0.75)x= 1.0667x
% ahead = ((1.0667x-x)/x) *100
=6.67

Sent from my A1601 using GMAT Club Forum mobile app
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In a long-jump competition, Carl jumped 75% as far as Mike, and 80% as 20 Apr 2025, 08:12
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shisingh
why are we equating 0.8x with 0.75y? can it be 1.8 and 1.75 instead?

apoorv2708
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%

0.8x = 0.75y
y= (0.8/0.75)x= 1.0667x
% ahead = ((1.0667x-x)/x) *100
=6.67

Sent from my A1601 using GMAT Club Forum mobile app
Show more

We're not equating 0.8x with 0.75y arbitrarily. Here's why:

Let Bob's jump be x and Mike's jump be y.
We're told Carl jumped 80% of Bob's and 75% of Mike's distances.

So, Carl's jump = 0.8x = 0.75y
This equation helps us relate Mike's and Bob's jumps through Carl.

Using 1.8 and 1.75 would be incorrect, since those would mean Carl jumped 180% and 175%, which contradicts the actual percentages given.
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