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Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin

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Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 25 Apr 2019, 02:12
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 12 Jan 2020, 21:59
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ManjariMishra wrote:
chetan2u Bunuel VeritasKarishma Can you please explain this .What 17 different combinations infer here ?


She has 12 coins.
We can obtain 17 distinct values with the 12 coins i.e. 17 distinct sums of money with the 12 coins.

If all 12 coins were 5 cent coins, we could have obtained a total of 60 cents in multiples of 5 i.e. 5 cents, 10 cents, 15 cents, 20 cents, ... 55 cents, 60 cents. These are just 12 distinct sums of money.
If all 12 coins were 10 cent coins, we could have obtained a total of 120 cents in multiples of 10 i.e. 10 cents, 20 cents, 30 cents, 40 cents, ... 110 cents, 120 cents. These are just 12 distinct sums of money.

To get 17 combinations, we must have a mix of 5 cent and 10 cent coins.

Now note that you can make each multiple of 5 till the max multiple of 5 as per the number of coins.
Say of the 12 coins, 10 are 5 cent coins and 2 are 10 cent coins.
Max amount = 10*5 + 2*10 = 70 cents
5c, 10c, 15c (a 10c and a 5c), 20c (2 10c coins), ... for all other sums, just keep adding 5.

Similarly, if we have 17 distinct sums, it means the maximum amount is 85 cents (so 17 distinct sums will be 5c, 10c, 15c ... 85c)
So, if x is the number of 5 cent coins and y is the number of 10 cent coins,
x + y = 12
5x + 10y = 85

Solve to get x = 7 and y = 5

Answer (C)
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 26 Apr 2019, 22:17
17 different combinations means: 5, 10, 15... up to 85.


Now, with 12 coins, we can create 85c as : 10c*5 + 5c*7

So, Total 10c coins: 5

Ans C

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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 10 Jan 2020, 10:06
Shobhit7 wrote:
17 different combinations means: 5, 10, 15... up to 85.


Now, with 12 coins, we can create 85c as : 10c*5 + 5c*7

So, Total 10c coins: 5

Ans C

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Hi,

Can you please explain in detail how you solved this?
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 12 Jan 2020, 11:36
chetan2u Bunuel VeritasKarishma Can you please explain this .What 17 different combinations infer here ?
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 12 Jan 2020, 12:47
We can make 17(>12) different values. She must have atleast 1 5-cent coin; hence she can obtain all the multiples of 5 less than or equal to 5x+10y, where x is number of 5-cent coins and y is number of 10-cent coins.

\(x= \frac{120-17*5}{5}= 7\)

\(y=12-7=5\)


Bunuel wrote:
Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin  [#permalink]

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New post 12 Jan 2020, 18:41
Hi nick1816, How do we get 120? What is that?
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Re: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin   [#permalink] 12 Jan 2020, 18:41
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