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# There are 100 apples in a bag of which 98% are green and the rest red.

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Manager
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Joined: 02 Aug 2014
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There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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07 Nov 2016, 04:39
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There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need
to remove so that only 96% of the apples are green?

A. 40
B. 8
C. 15
D. 25
E. 50
[Reveal] Spoiler: OA

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Joined: 11 Sep 2015
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Re: There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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07 Nov 2016, 05:06
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ReussirleGMAT wrote:
There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need
to remove so that only 96% of the apples are green?

A. 40
B. 8
C. 15
D. 25
E. 50

There are 100 apples in a bag of which 98% are green and the rest red.
So, we can conclude that (initially) there are 98 green apples and 2 red apples in the bag.

If (after we remove some green apples) 96% of the apples are green. This means that 4% of the apples are red.
So, let's reword the question as "How many green apples do you need to remove so that 4% of the apples are red?"

We already know there are 2 red apples, and we know that 2/50 = 4%
That is, if there were 50 apples altogether, and 2 were red, then 4% of the apples would be red.
In order to get 50 apples altogether, we must remove 50 apples from the bag.

[Reveal] Spoiler:
E

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Re: There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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07 Nov 2016, 07:12
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ReussirleGMAT wrote:
There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need
to remove so that only 96% of the apples are green?

A. 40
B. 8
C. 15
D. 25
E. 50

$$\frac{98 - x}{100 - x} = \frac{96}{100}$$

Or, $$9800 - 100x = 9600 - 96x$$

Or, $$200 = 4x$$

Or, $$x = 50$$

Hence, only 50 green apples need to be removed from the bag , answer will be (E)

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Joined: 07 Dec 2014
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There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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07 Feb 2018, 18:30
AnisMURR wrote:
There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need
to remove so that only 96% of the apples are green?

A. 40
B. 8
C. 15
D. 25
E. 50

let a=total apples remaining after green apples removed
.04a=2 red apples
a=50 total apples
50-2=48 green apples remaining
98-48=50 green apples removed
Manager
Status: Turning my handicaps into assets
Joined: 09 Apr 2017
Posts: 67
Re: There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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14 Feb 2018, 02:38
98-x=(100-x)*96/100

Solving this equation,we get x-24x/25=2,hence x=50
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Re: There are 100 apples in a bag of which 98% are green and the rest red. [#permalink]

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15 Feb 2018, 09:54
AnisMURR wrote:
There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you need
to remove so that only 96% of the apples are green?

A. 40
B. 8
C. 15
D. 25
E. 50

We see that there are 98 green apples and 2 red apples. We can let n = the number of green apples to remove and create the equation:

(98 - n)/(100 - n) = 96/100

100(98 - n) = 96(100 - n)

9800 - 100n = 9600 - 96n

200 = 4n

50 = n

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Re: There are 100 apples in a bag of which 98% are green and the rest red.   [#permalink] 15 Feb 2018, 09:54
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# There are 100 apples in a bag of which 98% are green and the rest red.

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