Mrs. Lewis needs to buy two types of grain, oats, and barley to mix as a feed supplement for her cattle. She has $4275 to spend on grain and wants the mixture to be 3 parts oats and 2 parts barley. She can buy oats for $1.10 per bushel and barley for $2.10 per bushel. Mrs. Lewis needs 7,000 bushels of grain. How many bushels of barley should she buy?

## Word Problem

# 5 Answers

Mrs. Lewis cannot buy 7,000 bushels of grain, for that would cost her at least 7,000*$1.10=$7,700, but she only has $4275.

Let X=bushels of oat and Y=bushels of barley. 3 parts oats and 2 parts barley means 2X=3Y, or X=1.5Y. Her cost is 4275=1.1X+2.1Y. Now substitute X=1.5Y and simplify: 4275=1.1(1.5Y)+2.1Y=3.75Y. Therefore, Y=4275/3.75=1140, so she should buy 1140 bushels of barley.

This problem is not a math problem, but an economic's problem. What do you do when you're over budget? There isn't a simple answer to that!

7000 bushels of grain, even if all the grain is the less expensive $1.10/bushel of oats will cost a lot more then $4275.00!

It's time to sell some livestock.

O = # of bushels of oats

B = # of bushels of barley

O+B = 7,000 ......(1)

O = (3/5)*7,000 = 4,200 bushels of oats

B = (2/5)*7,000 = 2,800 bushes of barley

But these cost 4,200*1.10 + 2,800*2.10 = $10,500. So, she needs to borrow 10,500-4,275 = $6,225 to buy 7,000 bushes of grain.

Let's assume, that Mrs. Lewis will buy "x" bushels of oats, and "y" bushes of barley, then:

x + y = 7000 .......... (1)

1.1x + 2.1y = 4275 ....... (2)

*xy < 0*

* (x, y)* e

**Ø**Let x be the oats and y be the barley. Keep the mix requirements with the grain and the $$$ with the $$$.

She needs a total of 7000 bushels of grain which is composed of 3 parts oats and 2 parts barley, thus:

3x + 2y = 7,000

In addition you are told that she has a finite dollar amount she can spend, $4275 where the oats are $1.10/bushel & $2.10/bushel, thus:

1.1x + 2.1y

Now you have 2 unknown values and equations:

3x+2y=7000 (Eq 1)

1.1x+2.1y=4275 (Eq2)

There are a couple of ways to go about this...find a common coefficient and subtract one equation or solve either equation for x or y and plug it into the other equation. Either way there will be fractions. I am going to solve for y in Eq 1 since it has whole numbers and plug that into Eq 2:

3x+2y=7000 -> 2y=7000-3x -> y= (7000-3x)2 -> y=3500-1.5x **(**)**

Now plug y into Eq 2

1.1x+2.1y=4275 -> 1.1x + 2.1(3500-1.5x) -> 1.1x + 7350 - 3.15x = 4275 -> 7350 -2.05x = 4275

-> 2.05x = 3075 -> x = 3075/2.05 -> *x=1500*

Now that you have **x=1500** plug that into Eq 1 or Eq 2 to find...you already solved Eq 1 for y earlier
**(**)** so that would be easiest to use:

**(**)** y = 3500 - 1.5x -> y = 3500 - 1.5(1500) -> y = 3500 - 2250 -> y = 1250

1500 bushels of oats and 1250 bushels of barley. Plug these numbers into either equation to verify. I don't have a calculator on me (hence my taking the whole number path!) so I will leave this to you.

Good luck and I hope it helped!

v/r

Daniel

## Comments

Joseph, you put big smile on my face!!! That is the solution for some math problems, redirect to Wall Street ....

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