# Odd question

• Sep 29th 2011, 04:00 PM
pvpkillerx
Odd question
10 apples = 1cents
5 oranges = 2cents
2 pear = 5cents

Using at least 1 of each, get 100 fruits and 1 dollar. WITHOUT GUESSING.
• Sep 29th 2011, 06:03 PM
pickslides
Re: Odd question
Here's a start

Making A = number of apples, O = number of oranges and P = number of pears

A+O+P=100

and

0.01A+0.02O+0.05P=$1 • Sep 29th 2011, 07:05 PM pvpkillerx Re: Odd question Quote: Originally Posted by pickslides Here's a start Making A = number of apples, O = number of oranges and P = number of pears A+O+P=100 and 0.01A+0.02O+0.05P=$1

That won't work cause its 3 variables for 2 equations right?
• Sep 29th 2011, 07:07 PM
pickslides
Re: Odd question
Yep, you'll have to find a third.
• Sep 29th 2011, 08:32 PM
Soroban
Re: Odd question
Hello, pvpkillerx!

Quote:

$\displaystyle \begin{array}{ccc}\text{10 apples} &=& \text{1 cent} \\ \text{5 oranges} &=& \text{2 cents} \\ \text{2 pears} &=& \text{5 cents} \end{array}$

Using at least one of each, get 100 fruits and 1 dollar ... without guessing.
The unit prices are: .$\displaystyle \begin{bmatrix}\text{apples} & 0.1\,\rlap{/}c \\ \text{oranges} & 0.4\,\rlap{/}c \\ \text{pears} & 2.5\,\rlap{/}c \end{bmatrix}$

Let: .$\displaystyle \begin{Bmatrix}A &=& \text{no. apples} \\ \theta &=& \text{no. oranges} \\ P &=& \text{no. pears} \end{Bmatrix}$

There are 100 fruits: .$\displaystyle A + \theta + P \;=\;100 \;\;[1]$

They cost 100 cents: .$\displaystyle 0.1A + 0.4\theta + 2.5P \;=\;100 \;\;[2]$

Multiply [2] by 10: .$\displaystyle A + 4\theta + 25P \;=\;1000$
. . . . Subtract [1]: . $\displaystyle A +\; \theta +\; P \;\;\;=\;\;\;100$

. . . . And we have: . . . $\displaystyle 3\theta + 24P \:=\:900$

. . $\displaystyle P \:=\:\frac{900 - 3\theta}{24} \quad\Rightarrow\quad P \;=\;37 + \frac{4-\theta}{8}$

Since $\displaystyle P$ is an integer, $\displaystyle 4-\theta$ must be divisible by 8.
. . Hence: .$\displaystyle \theta \,=\,4 \quad\Rightarrow\quad P \,=\,37$

Substitute into [1] and solve for A.