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Bunuel
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A certain company only accepts payments in its own special currency. 29 Dec 2021, 06:45
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41% (02:03) correct 59% (02:27) wrong based on 81 sessions

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A certain company only accepts payments in its own special currency. The currency has the following denominations: one, ten and fifty. What is the number of different ways that a customer can pay the company if the customer's bill amounts to 107 in this currency?

(A) 15

(B) 16

(C) 17

(D) 18

(E) 19


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A certain company only accepts payments in its own special currency. 29 Dec 2021, 13:46
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Bunuel
A certain company only accepts payments in its own special currency. The currency has the following denominations: one, ten and fifty. What is the number of different ways that a customer can pay the company if the customer's bill amounts to 107 in this currency?

(A) 15

(B) 16

(C) 17

(D) 18

(E) 19

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Breaking Down the Info:

The 7 dollars must be split with seven $1, so we focus on how to break down $100.

My method is to count how many ways to fill in $10 and $50's given the number of $1's.

Start with $0 in 1's: 50 *2, 10*10, 50 + 10*5, that is 3 ways.

$10 in 1's: 50 + 40 or 10*9. 2 ways

$20 in 1's: 50 + 30, or 10*8. 2 ways (see a pattern now?)

$30 in 1's, $40 in 1's, and $50 in 1's: 2 ways.

$60~$90 in 1's: 1 way. (4 options total).

Finally, add everything up to find we have 17 ways to break down $100.

Answer: C
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A certain company only accepts payments in its own special currency. 24 Oct 2024, 08:44
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Hi, Let me know that we should also consider 100 - 1$ right. Then count to 18 ways
JerryAtDreamScore
Bunuel
A certain company only accepts payments in its own special currency. The currency has the following denominations: one, ten and fifty. What is the number of different ways that a customer can pay the company if the customer's bill amounts to 107 in this currency?

(A) 15

(B) 16

(C) 17

(D) 18

(E) 19


Breaking Down the Info:

The 7 dollars must be split with seven $1, so we focus on how to break down $100.

My method is to count how many ways to fill in $10 and $50's given the number of $1's.

Start with $0 in 1's: 50 *2, 10*10, 50 + 10*5, that is 3 ways.

$10 in 1's: 50 + 40 or 10*9. 2 ways

$20 in 1's: 50 + 30, or 10*8. 2 ways (see a pattern now?)

$30 in 1's, $40 in 1's, and $50 in 1's: 2 ways.

$60~$90 in 1's: 1 way. (4 options total).

Finally, add everything up to find we have 17 ways to break down $100.

Answer: C
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Su_bha
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A certain company only accepts payments in its own special currency. 25 Oct 2024, 04:57
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Bunuel
A certain company only accepts payments in its own special currency. The currency has the following denominations: one, ten and fifty. What is the number of different ways that a customer can pay the company if the customer's bill amounts to 107 in this currency?

(A) 15

(B) 16

(C) 17

(D) 18

(E) 19


Breaking Down the Info:

The 7 dollars must be split with seven $1, so we focus on how to break down $100.

My method is to count how many ways to fill in $10 and $50's given the number of $1's.

Start with $0 in 1's: 50 *2, 10*10, 50 + 10*5, that is 3 ways.

$10 in 1's: 50 + 40 or 10*9. 2 ways

$20 in 1's: 50 + 30, or 10*8. 2 ways (see a pattern now?)

$30 in 1's, $40 in 1's, and $50 in 1's: 2 ways.

$60~$90 in 1's: 1 way. (4 options total).

Finally, add everything up to find we have 17 ways to break down $100.

Answer: C
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Consider $50 either 0,1or max 2 for 0 you got 11 cases, for 1 you got 6 cases for 2 you will get 1 case.
for $50 =0counts x+10Y=107, we got (7,10), (17,9)....................(107,0) =11 cases similarly for $50 =1counts We get X+10Y= 57, (7,5), (17,4)............(57,0) =6 cases if we have $50=2 counts then X+10Y=7 single case X=7,y=0 total 18 cases
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A certain company only accepts payments in its own special currency. 25 Oct 2024, 05:22
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Let there be "a" ones, "b" tens, and "c" fifties.

Given information:

a + 10b + 50c = 107.

"c" can be either 0 or 1 or 2.

(1) When c = 2

a + 10b = 7.

Only 1 possibility: a = 7; b =0.

(2) When c = 1

a + 10b = 57

"a" can take any value from 7, 17, 27...till 57. The corresponding "b" values will be 50, 40, ....till 0.

Number of possibilities = ((57 - 7) / 10) + 1 = 6.

(3) When c = 0

a + 10b = 107

"a" can take any value from 7, 17, 27...till 107. The corresponding "b" values will be 100, 90, ....till 0.

Number of possibilities = ((107 - 7) / 10) + 1 = 11.


Final answer = 1 + 6 + 11 = 18.

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A certain company only accepts payments in its own special currency. 25 Oct 2024, 07:03
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A certain company only accepts payments in its own special currency. The currency has the following denominations: one, ten and fifty.

What is the number of different ways that a customer can pay the company if the customer's bill amounts to 107 in this currency?

Number of cases with 0 50's = 11
Number of cases with 1 50 = 6
Number of cases with 2 50's = 1

Total cases = 11 + 6 + 1 = 18

IMO D
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