The football club has pratic one day.

1/4 did powerlifting

2/5 did technical training.

and the remaning 21 did cardio.

How many showed up for pratic this day?

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- Apr 20th 2018, 03:42 AM #1

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- Apr 20th 2018, 05:21 AM #2

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## Re: Football club, how many showed up for pratic.

Let X be the total number that showed up.

"1/4 did powerlifting" That's (1/4)X= X/4 people

"2/5 did technical training." That's (2/5)X= 2X/5 people

"and the remaining 21 did cardio" That's 21 people

The total is X/4+ 2X/5+ 21= X.

Solve that for X. I would start by subtracting X/4 and 2X/5 form both sides. However I think you will find that this is impossible. X does not turn out to be a whole number.

- Apr 26th 2018, 10:18 AM #3

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## Re: Football club, how many showed up for pratic.

- Apr 26th 2018, 10:36 AM #4

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- Apr 26th 2018, 10:51 AM #5

- Apr 26th 2018, 11:01 AM #6

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- Apr 26th 2018, 11:04 AM #7

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- Apr 26th 2018, 11:09 AM #8

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## Re: Football club, how many showed up for pratic.

$n = 1\cdot n = \dfrac{20}{20}\cdot n$

$\dfrac{13}{20}n + 21 = n$

$\dfrac{13}{20}n + 21 = \dfrac{20}{20}n$

$\dfrac{13}{20}n+21 - \dfrac{13}{20}n = \dfrac{20}{20}n - \dfrac{13}{20}n$ (what you do to one side, you must do to both)

$\cancel{\dfrac{13}{20}n} + 21 - \cancel{\dfrac{13}{20}n} = \dfrac{20-13}{20} n = \dfrac{7}{20}n$

$21 = \dfrac{7}{20}n$

- Apr 26th 2018, 11:12 AM #9

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- Apr 26th 2018, 11:18 AM #10

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