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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
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 106
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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45% (medium)

Question Stats:

67% (02:00) correct 33% (02:12) wrong based on 571 sessions

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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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Joined: 11 Sep 2015
Posts: 3332
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
6
Top Contributor
4
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.

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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
17
6
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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##### General Discussion
VP
Joined: 07 Dec 2014
Posts: 1152
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
1
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: 127
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
1
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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Joined: 25 May 2015
Posts: 18
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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01 Apr 2018, 11:54
$$\frac{x}{(x+2)}$$= $$\frac{96}{100}$$

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
Intern
Joined: 29 Nov 2016
Posts: 35
GMAT 1: 650 Q47 V33
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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15 Apr 2018, 00:53
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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Intern
Joined: 03 Aug 2017
Posts: 27
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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27 Aug 2018, 01:13
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
Intern
Joined: 28 Jul 2018
Posts: 17
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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28 Aug 2018, 02:07
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

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?
Intern
Joined: 28 Jul 2018
Posts: 17
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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28 Aug 2018, 02:10
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.

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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
VP
Joined: 09 Mar 2016
Posts: 1287
Re: There are 100 apples in a bag of which 98% are green and the rest red.  [#permalink]

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28 Aug 2018, 04:17
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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28 Aug 2018, 05:51
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?? (Just kidding!!)
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There are 100 apples in a bag of which 98% are green and the rest red. &nbs [#permalink] 28 Aug 2018, 05:51
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# There are 100 apples in a bag of which 98% are green and the rest red.

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