# Simple algebra problem (sanity check)

Discussion in 'Homework Help' started by strantor, Dec 11, 2012.

1. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
4,302
1,989
This question comes from my overseas niece's homework. I suspect the question is wrong.
Here it goes:
Here's what I got:

200php bills = x
500php bills = y
1000php bills = z

z = 2x
y = z - 5
substitute z-value:
y = 2x - 5

200x + 500y + 1000z = 30,000

Substitute to get all in terms of x:
200x + 500(2x-5) + 1000(2x) = 30,000

Distribute:
200x + 1000x - 2500 + 2000x = 30,000

Combine like terms:
3200x - 2500 = 30,000

3200x = 32,500

divide both sides by 3200:
x = 10.15625

# of bills needs to be a whole number...

So I put it into excel, and using the first 2 equations with only whole numbers of bills, I get can totals of 29,800 and 31,700, but not 30,000.

2. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
This is a trick question, Denise hate calamity victims... I kid, I kid....

Anyway, it would seem the question is flawed.....

here's how I did it:
X = # 200php
Y = # 500php
Z = # 1000php

30000 = 200X + 500Y + 1000Z

X = 2Z
Y = Z - 5

30000 = 200(2Z) + 500(Z-5) + 1000Z

30000 = 400Z + 500Z - 2500 + 1000Z

30000 = 1900Z - 2500

32500 = 1900Z

Z = 17.10526

Y = 12.10526

X = 34.21053

200X + 500Y + 1000Z = 29999.996

Apparently, we are dealing with fractional currency

3. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
4,302
1,989
Ok, your numbers match with what I put in excel. I got z=2x and x=2z mixed up in my algebra, but the bills are still fractional. Thanks for the confirmation tshuck.

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
The problem can be solved a couple of ways as a system of 3 linear equations:

Having solved it and noted that the total number of bills is not an integer, one wonders what changes to the problem would lead to a solution with an integer number of bills. Perhaps thereby discovering if a simple mistake had been made in the statement of the problem--one digit off; that sort of thing.

Making this change in the problem statement leads to an integer solution:

Denise organized a fund raising project to help calamity victims. She collected a total of 30,000Php consisting 200Php, 500 php and 1000Php Bills. The number of 200 is twice the number of 1000 and the number of 1000 is five less than the number of 200. How many of each bill did she have?

I suspect this is the mistake.

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5. ### strantor Thread Starter AAC Fanatic!

Oct 3, 2010
4,302
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where did your values for the matrix come from? (the 6 values above 200, 500, and 1000)

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
The question is flawed, but so is your solution attempt. With x being the number of 200Php notes, the equation should have been:

200x + 500(x/2-5) + 1000(x/2) = 30,000

Let's divide everything by 100 first

2x + 5(x/2-5) + 10(x/2) = 300

2x + (5/2)x - 25 + 5x = 300

Multiply everything by 2

4x + 5x - 50 + 10x = 600

19x = 650

x = 34.21, which can't be.

7. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Okay, this makes no sense. I looked at this thread for the first time less than ten minutes ago and there were only two posts. Then I posted mine and there three others and my first thought was, "my, what a coincidence that we are all posting at the same time." But the posts are hours older than mine. Wonder what gives.

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,300
335
Follow the rules for multiplication of a matrix times a vector. The first row of the matirix is [1 0 -2]. The vector of unknowns is:

[ Php200 ]
[ Php500 ]
[ Php1000 ]

This multiplication leads to the equation:

1*Php200 + 0*Php500 - 2*Php1000 = 0 which is the same as the problem statement "The number of 200 is twice the number of 1000".

Multiplication of the second row of the matrix times the unknown vector leads in similar fashion to the equation:

0*Php200 - 1*Php500+ 1*Php1000 = 5 which is the same as the problem statement "the number of 500 is five less than the number of 1000".