A stand at a sports arena sells only beer, and each beer costs $7. Bee

Question

A stand at a sports arena sells only beer, and each beer costs $7. Beer purchasers are only allowed to purchase one beer at a time, and each beer is equally likely to be purchased using the following combinations of bills for the $8 purchase:

· 7 one-dollar bills
· 1 five-dollar bill and 2 one-dollar bills
· 1 ten-dollar bill (after which the stand will give 3 one-dollar bills to the purchaser)
· 1 twenty-dollar bill (after which the stand will give 1 ten-dollar bill and 3 one-dollar bills to the purchaser)

The stand begins each day with $10 in one-dollar bills, $25 in five-dollar bills, and $50 in ten-dollar bills so that it can process its initial orders. On a particular day exactly 100 beers were sold and the only money that the stand took in or gave out was directly related to those exact transactions.

In the grid select the number of one-dollar bills and the number of bills with denominations greater than one dollar the stand had at the end of the day.

Mamun200 · Answer

How to solve the problem? I did not get the point. can you please show me how to solve the problem?

85+10=95 one dollar 
then 
how to spend 85 one dollar
5+2 one dollar= will increase the number of 5 dollar 
10-3 one dollar will increase the number of 10 dollar and 
20-10-3 one dollar will reduce the number of 10 dollar

how to do it i tried several times but i did not understand.

Mdabhi1 · Answer

Hi,

This is how I solved this question.

The question says each beer is equally likely to be purchased as per given options this means that out of a total 100 beer purchase transactions, 25 are done as per option 1, 25 are done as per option 2, 25 are done as per option 3 and 25 are done as per option 4.

for $1 bill count:
Initial - 10
Option 1 - 25*7 = 175
Option 2 - 25*2 = 50
Option 3 - (25*3) = (75)
Option 4 - (25*3) = (75)
Total $1 bills = 85

For $5 bills
Initial - 5 (25/5)
option 2 - 25*1 = 25
Total $5 bills = 30

for $10B bill
Initial - 5
Option 3 - 25*1 = 25
option 4 - (25*1) = (25)
total $10 bills = 5

for $20 bill
Initial - 0
option 4 - 25*1 = 25
total $20 bills = 25

Final answer:
total $1 bills = 85
Total bills greater than $ 1 = 30+5+25 = 60

Thanks,

chetan2u · Answer

We have 25 beers purchased in each of the four ways.

· 7 one-dollar bills: This means there will be 25*7 or 175 one dollars recd => +175 ($1)

· 1 five-dollar bill and 2 one-dollar bills: This means there will be 25*2 or 50 one dollars recd => +50 ($1) and 25*1 five dollar bill, that is +25 ($5)

· 1 ten-dollar bill (after which the stand will give 3 one-dollar bills to the purchaser): In this case 25 ten dollars will come in but 25*3 one dollars will go out, that is +25($10) and -25*3 or -75($1)

· 1 twenty-dollar bill (after which the stand will give 1 ten-dollar bill and 3 one-dollar bills to the purchaser): This means +25($20), -25($10) and -75($1)

But he also starts with few bills: The stand begins each day with $10 in one-dollar bills, $25 in five-dollar bills, and $50 in ten-dollar bills, that is +10($1), +5($5) and +5($10)

Let us look at each note separately 
$1: +175+50-75-75+10 = 85
$5: +25+5 = 30
$10: +25-25+5 =5
$20: +25 = 25

Number of one dollar bill = 85
Number of all other bills = 30+5+25 = 60

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One-dollar bills	Bills greater than one dollar	------------
		60
		65
		70
		75
		80
		85

A stand at a sports arena sells only beer, and each beer costs $7. Bee

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A stand at a sports arena sells only beer, and each beer costs $7. Bee

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