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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
2
Cant the problem be done like this :

Let the initial total no of marbles be x 'Given 60% of them are red which means 3x/5 of the marbles are red

Later 20 marbles are added out of which 3 are red which means total marble count is x+20 and total red marble count is 3x/5+3= (3x+15)/5

given (3x+15)/5(x+20)=50/100

=>6x+30=5x+100
=>x=70 the ans sud be 70 ?
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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
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adrijatina wrote:
=>x=70 the ans sud be 70 ?

70 is correct value for x. But x represents red marbles before addition of 20, question asked for new value. So x+20 = 70+20 = 90

I made the same mistake.
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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
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Here's another method.
If initial 60 percent is converted to 50 percent upon adding 15 percent of red, then the fraction of 15 percent red collection is 2/9.
Now 15 percent of 2/9 of total is 3.
i.e
3/20 * 2/9 * x = 3
So x = 90
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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
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Initial number of marbles = Y

Initial number of red marbles =60/100Y = 3/5Y
Number of non red marbles = 40/100y = 2/5Y

If 20 marbles added and 3 goes to red = 3/5Y + 3

Non red = 2/5Y + 17
If new red marbles in jar is 50% let new number of marbles be X
So X/2 =3/5Y + 3
Non red will be
X/2 = 2/5Y + 17
Simultaneous equation becomes
5X - 4Y = 170
5X - 6Y = 30
Y = 70
substituting for Y gives new number of red and non red marbles to be 45
Answer = 90.
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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
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Re: 60% of the marbles in a jar are red. Twenty marbles.... [#permalink]
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