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Suppose you have a currency, named Miso, in three denominations: 1 Mis

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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 14:41
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Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

A) 16

B) 17

C) 18

D) 19

E) 20
[Reveal] Spoiler: OA

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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 18:03
GTExl wrote:
Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

A) 16

B) 17

C) 18

D) 19

E) 20

I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).

For ones (1 Miso denomination)

107 can be divided by 1 in one way (107 ones "fit" into 107), that is:

107 ones = ONE way

For tens, first assume there are no fifties, and the remainders, composed of denomination one, will correspond with only one of the ways to use tens.

107 can be divided 10 ways (only 10 tens "fit" into 107), with whatever remainders:

10 + 97 | 60 + 47
20 + 87 | 70 + 37
30 + 77 | 80 + 27
40 + 67 | 90 + 17
50 + 57 | 100 + 7

= TEN ways

For fifties and tens, assume there is one fifty. So 107 - 50 = 57. How many tens? 57/10 allows for five tens, plus whatever remainder:

50 + 10 + 47
50 + 20 + 37
50 + 30 + 27
50 + 40 + 17
50 + 50 + 7

= FIVE WAYS

Then fifties and ones: 107/50 = allows for two fifties, with whatever remainder:

50 + 57
100 + 7

= TWO ways

Ways total: 1 + 10 + 5 + 2 = 18

This is probably some horrible combinatorics problem, or a very easy arithmetic problem. Either way, or some other way, I can't see a method other than this one. I have no idea whether or not my method is harebrained. It yields the right answer. Please correct me if I've made a mistake.

The answer is 18.

Answer C

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Re: Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 20:15
genxer123 wrote:
I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).


Too time-consuming, I am looking forward to seeing a math based solution

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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 20:39
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i think the below one is a better way-

Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively-
x+10y+50z=107

Now the possible values of z could be 0, 1 and 2.

For z=0: x+10y=107
for above equation y can range from 0 - 10, taking total 11 values. that means above condition can be satisfied by 11 ways.

For z=1: x+10y=57
for above equation y can range from 0 - 5, taking total 6 values. that means above condition can be satisfied by 6 ways.

For z=2: x+10y=7
for above equation y can take only 1 value i.e 0. that means there is only 1 way to satisfy above condition.

Total no of ways = 11+6+1=18
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Last edited by GTExl on 04 Aug 2017, 14:39, edited 2 times in total.

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Re: Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 21:03
GTExl wrote:
i think the below one is a better way-

Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively.
x+10y+50z=107

Now the possible values of z could be 0, 1 and 2.

For z=0: x+10y=107
for above equation y can range from 0 - 10, taking total 11 values. that means above condition can satisfied by 11 ways.

For z=1: x+10y=57
for above equation y can range from 0 - 5, taking total 6 values. that means above condition can satisfied by 6 ways.

For z=2: x+10y=7
for above equation y can take only 1 value i.e 0. that means there is only 1 way to satisfy above condition.

Total no of ways = 11+6+1=18

Good one.

Sent from my ONE A2003 using GMAT Club Forum mobile app

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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 03 Aug 2017, 21:39
cbh wrote:
genxer123 wrote:
I get 18.

There has to be a better way to do this problem; I just started with a denomination and used division to calculate possible ways, then considered two denominations (and checked with list of possibilities below).


Too time-consuming, I am looking forward to seeing a math based solution


Wow. Really?

Then post one.

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Suppose you have a currency, named Miso, in three denominations: 1 Mis [#permalink]

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New post 18 Aug 2017, 21:20
Scenario counting

Of the 107, the last 7 miso's can be paid only with 1's. So's let's look at how the remaining 100 can be split.

50-------------------10------------------1
0----------------10,9,8,..,0-------remaining (11 cases)
1---------------5,4,3,2,1,0---------remaining (6 cases)
2---------------0--------------------remaining (1 case)

So, 18 cases.

P.S.: While answering this question, I computed 11+6+1 as 17 and selected Option B :crazy:

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Suppose you have a currency, named Miso, in three denominations: 1 Mis   [#permalink] 18 Aug 2017, 21:20
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Suppose you have a currency, named Miso, in three denominations: 1 Mis

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