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

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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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New post 07 Nov 2016, 05:39
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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

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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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New post 07 Nov 2016, 06:06
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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.

Answer:

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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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New post 07 Nov 2016, 08:12
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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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There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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New post 07 Feb 2018, 19:30
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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
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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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New post 14 Feb 2018, 03:38
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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New post 15 Feb 2018, 10:54
2
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

Answer: 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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New post 01 Apr 2018, 12:54
\(\frac{x}{(x+2)}\)= \(\frac{96}{100}\)

100x=96x+192

4x=192

x=\(\frac{192}{4}\)

x=48 not the answer

the number of green apple that should be removed is 98-48 = 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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New post 15 Apr 2018, 01:53
96(100-x)/100 = 100 - x - 2 .....{x is the total no. of green apples required to remove and 2 is the count of red apples}
Hence x = 50.
E is the answer
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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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New post 27 Aug 2018, 02:13
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


This is how i solved.
Total green aplles = 98 and Red aplles = 2 Total apples = 100
TArget question = how many apples to remove to get 96% apples Green.


We know that there are 2 apples which are red. They need to be 4 % of the total no of apples
4/100* X = 2
Sloving for X we get 50

Which means if there are 50 apples in total 2 red apples mean 4% and remainig 48 apples = green = 96 %

We know total green apples = 98 . How many should i remove to reach 48 apples = 98-50 = 48 apples green apples
= 50 = 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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New post 28 Aug 2018, 03:07
ScottTargetTestPrep wrote:
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

Answer: E


ScottTargetTestPrep, your equation means that if we remove 50 apples from 98apples, the result{48apples} is 96% of 100. This question doesnt make any sense to me. Pls is something wrong with this question or wrong with my concept of math? Does GMAC actualy set this sort of question that doesnt make real proportionate sense?
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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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New post 28 Aug 2018, 03:10
GMATPrepNow wrote:
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.

Answer:

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Thanks Brent but is there a typo in this question or is it total of 100 apples really? Cos it doesnt make sense to me at all
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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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New post 28 Aug 2018, 05:17
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



hey pushpitkc, niks18 :-) is my approach correct ?

We have 98 green and 2 red apples.

\(98-x = 0.96(100-x)\)

\(98-x = 96-0.96x\)

\(98-96 = x-0.96x\)

\(2 = 0.04x\)

\(x = 50\)
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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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New post 28 Aug 2018, 06:51
1
dave13 wrote:
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



hey pushpitkc, niks18 :-) is my approach correct ?

We have 98 green and 2 red apples.

\(98-x = 0.96(100-x)\)

\(98-x = 96-0.96x\)

\(98-96 = x-0.96x\)

\(2 = 0.04x\)

\(x = 50\)


Your approach is bang on dave13 - btw, what is x?? :P (Just kidding!!)
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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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Re: There are 100 apples in a bag of which 98% are green and the rest red.   [#permalink] 29 Aug 2019, 17:36
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