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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
Dimes, nickel.. I was not aware of the values of these.(these are the part of some currency i guess)?
In actual exam will I be provided with the values for those terms??
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
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er.arun88 wrote:
Dimes, nickel.. I was not aware of the values of these.(these are the part of some currency i guess)?
In actual exam will I be provided with the values for those terms??


Yes, you are not required to know this. On the actual exam the values will be provided.
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
Statement 1 is quite clear: 1-0.83=0.17
Statement 2: Since we know all the values of coins and number of coins of each value, it is sufficient!
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
using statement 1, we can simply calculate the required probability. SUFFICIENT
Statement 2: since we know total no. of coins, and value of different types of coins,and also no. of coins of each category, it can be calculated. SUFFICIENT.

Hence, D is anwer!
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.
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Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
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TheMastermind wrote:
Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.


Hi TheMastermind ,

I believe you understood from statement 2 that we have 5 coins of nickel.

Now, we have the following denomination.

3 coins of value $ 0.25 each.
3 coins of value $ 0.10 each.
5 coins of value $ 0.01 each.

Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25.

I cannot get $0.50 with any other combination such that only 6 coins are selected.

So, Probability of selecting atleast one quarter = 1 - P(selecting no quarter).

= 1 - 8C6/11C6.

Now, I know that this will give me value more than or equal to $0.50. So, 1- this probability will give me value less than $0.50.

I hope that makes sense. :)
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
abhimahna wrote:
TheMastermind wrote:
Can someone explain the calculation if we were supposed to find out the probability using information given in statement (2)?

Thank you.


I believe you understood from statement 2 that we have 5 coins of nickel.

Now, we have the following denomination.

3 coins of value $ 0.25 each.
3 coins of value $ 0.10 each.
5 coins of value $ 0.01 each.

Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25.

I cannot get $0.50 with any other combination such that only 6 coins are selected.

So, Probability of selecting atleast one quarter = 1 - P(selecting no quarter).

= 1 - 8C6/11C6.

Now, I know that this will give me value more than or equal to $0.50. So, 1- this probability will give me value less than $0.50.

I hope that makes sense. :)


Hi abhimahna,

regarding what you said,

"Now, note that to determine the probability of getting value more than or equal to $0.50 , I need to select atleast one coin of value $0.25."

if one coin selected from the value $0.25 and rest 5 from $0.01, then the value would not be >= $0.50

Could you please clarify this?

Thanks,
Shaheen
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
Hi Shaheen,

The value of a nickel is .05$ NOT .01$.
Now it makes sense to say that we've to select at least one coin of .25$
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
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Re: Three quarters, three dimes, and x nickels are the only coins in a jar [#permalink]
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