Simple Algebra help

G

Guest

Guest
Ok, so I have a question that goes like this:

Three different articles A, B and C can be purchased from a shop. Article A is sold at the rate of 8 for $1. Article B is sold for $1 each and article C is sold for $10 each. In total, 100 items are purchased for $100. The purchase includes all three types of articles. How many of each article are purchased?
Short of guessing the answers I have no idea how to do this question. I've worked out that A/8 + B + 10C = 100, but have no idea how to find the answers from this.

My teacher has told me that there's two answers, one is trivial and one is slightly harder to find.

Any help would be great. :)

Cheers.
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,467
Beasticly said:
I've worked out that A/8 + B + 10C = 100, but have no idea how to find the answers from this.
Well, you also know that A + B + C = 100; C = 100 - A - B; substitute:

A/8 + B + 10(100 - A - B) = 100; multiply by 8:
A + 8B + 80(100 - A - B) = 800 ; simplify:

B = (7200 - 79A) / 72

Since A is divisible by 8, you should get B quickly...
 

TchrWill

Full Member
Joined
Jul 7, 2005
Messages
856
Beasticly said:
Ok, so I have a question that goes like this
Three different articles A, B and C can be purchased from a shop. Article A is sold at the rate of 8 for $1. Article B is sold for $1 each and article C is sold for $10 each. In total, 100 items are purchased for $100. The purchase includes all three types of articles. How many of each article are purchased?
Short of guessing the answers I have no idea how to do this question. I've worked out that A/8 + B + 10C = 100, but have no idea how to find the answers from this.

My teacher has told me that there's two answers, one is trivial and one is slightly harder to find.
1--A + B + C = 100
2--A/8 + B =+ 10C = 100
3--Multiplying (2) by 8 yields A + 8B + 80C = 800
4--Ssubtracting (1) from (3) yields 7B + 79C = 700
5--Dividing through by 7 yields B + 11C + 2C/7 = 100
6--2C/7 must be an integer as does 8C/7
7--Dividing by 7 yields C + C/7
8--C/7 must be an integer k making C = 7k
9--Substituting back into (4) yields 7B + =553 = 700 making B = 100 -- 79k
10--k can only ne0 or 1
11--k.....0......1
.....C.....0......7
.....B...100...21
.....A....0.....72
Sum...100..100
Since k = 0 does not result in at least one of each item being bought, there appears to be only one viable answer.

Checking: 72(1)/8 + 21(1) + 7(10) = 9 + 21 + 70 = $100
 
G

Guest

Guest
Denis said:
Beasticly said:
I've worked out that A/8 + B + 10C = 100, but have no idea how to find the answers from this.
Well, you also know that A + B + C = 100; C = 100 - A - B; substitute:

A/8 + B + 10(100 - A - B) = 100; multiply by 8:
A + 8B + 80(100 - A - B) = 800 ; simplify:

B = (7200 - 79A) / 72

Since A is divisible by 8, you should get B quickly...
Sorry, I'm a tad confused. I understand this bit:

A + B + C = 100; C = 100 - A - B
But when you get to here:

A/8 + B + 10(100 - A - B) = 100
I'm wondering why the (100 - A - B) is in there and what happened to the C?

Then, I think I understand this:

A + 8B + 80(100 - A - B) = 800
You wanted to get rid of the /8 on the A so you multiplied everything by 8, which makes 8B, 80 and 800?

But when you get to here:

B = (7200 - 79A) / 72
You've completely lost me. :?

Cheers.
 
G

Guest

Guest
1--A + B + C = 100
With you. :p

2--A/8 + B =+ 10C = 100
Still with you.

3--Multiplying (2) by 8 yields A + 8B + 80C = 800
With you, getting rid of the /8 on the A by multiplying everything by 8?

4--Ssubtracting (1) from (3) yields 7B + 79C = 700
Lost you, how can you just take one of everything on the left side but take 100 off the 800 on the right side?

5--Dividing through by 7 yields B + 11C + 2C/7 = 100
With you.. I think. Getting rid of the 7 on the B by deviding everything by 7, but 7 doesn't go into 79 exactly so you need the 2C/7 for the remainder.

6--2C/7 must be an integer as does 8C/7
Don't really know what you mean here. Where'd you get 8C/7 from?

7--Dividing by 7 yields C + C/7
Lost you. How does 11C + 2C/7 become C + C/7?

8--C/7 must be an integer k making C = 7k
Don't know what you mean there.

9--Substituting back into (4) yields 7B + =553 = 700 making B = 100 -- 79k
Really, really lost here.

10--k can only ne0 or 1
Even more lost, don't know what ne0 means.

11--k.....0......1
.....C.....0......7
.....B...100...21
.....A....0.....72
Sum...100..100
Where did K come from?

Since k = 0 does not result in at least one of each item being bought, there appears to be only one viable answer.
Again, where did K come from?

Checking: 72(1)/8 + 21(1) + 7(10) = 9 + 21 + 70 = $100
Well that answer seems to work, but I'm totally stumped how you got it. :p

Cheers.[/quote]
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,467
OK, let's go back; here's your equation:
A/8 + B + 10C = 100 ; multiply by 8:
A + 8B + 80C = 800 [1]

We also have:
A + B + C = 100;
C = 100 - A - B [2]

Now substitute [2] into [1]:
A + 8B + 80(100 - A - B) = 800 ; OK?
A + 8B + 8000 - 80A - 80B = 800
8000 - 79A - 72B = 800
79A + 72B = 7200 ; still with me?
72B = 7200 - 79A
B = (7200 - 79A) / 72 : see it now?

OK; now you need an "A" that makes 7200-79A divisible by 72:
can you finish it?

So you know: this cannot be directly solved because we have 3 unknowns
but only 2 equations; so to solve, we need to go trial and error...

...g'nite; going to bed in bad humor: my team the Phoenix Suns just lost :cry:
 
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