Please help me solve this word problem...

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MS Math

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An electronics store has a current inventory of 50 stereo systems. The lowest priced stereo system in the store sells for $800, The lowest priced stereo system in the store sells for 800 dollars, and the highest priced stereo system sells for $3000. and the highest priced stereo system sells for 3000 dollars. Which of the following is the maximum amount of $3000 systems on hand Which of the following is the maximum amount of 3000 dollar systems on hand if the current inventory totals $111,000?

the choices were : 30, 32 , 35, 37

i picked 37 but the answer was wrong it was 32

please help me figure out why the answer was 32
 
Why did you pick 37?
Do you see why 37 is not the correct answer?
 
It is sort of a trick question although it is easily solved by algebra. If you have 37 of the 3,000 dollar systems, you have 13 more systems, EACH of which is worth AT LEAST 800.

37 * 3,000 + 13 * 800 > 110,000.

Keep on trying numbers and eventually you get to

32 * 3000 + 18 * 800 = 96,000 + 14,400 = 110,400.

So the maximum number is actually 31 unless the problem said the inventory had a sales value of 110,400.
 
MS Math said:
An electronics store has a current inventory of 50 stereo systems. The lowest priced stereo system in the store sells for $800, The lowest priced stereo system in the store sells for 800 dollars, and the highest priced stereo system sells for $3000. and the highest priced stereo system sells for 3000 dollars. Which of the following is the maximum amount of $3000 systems on hand Which of the following is the maximum amount of 3000 dollar systems on hand if the current inventory totals $111,000?

the choices were : 30, 32 , 35, 37

i picked 37 but the answer was wrong it was 32

please help me figure out why the answer was 32
Click to expand...
Why did you choose 37 to be the answer?

If you had 37 equipment priced at 3000 - then that equals to inventory of $111000! That means that store can have only 37 stereo system - all high priced and no low priced stereo. Thus violating the condition of having 50 stereo system.

If you do not know algebra - it can be solved by trial and error.

You found that if you have 37 high-priced stereo and (50-37 =) 13 low priced stereo you exceed the money allocated for total price of inventory.

So we must decrease the amount of High priced stereo(from 37) and increase the amount of low priced stereos.

Lets guess the amount is 25 (high priced) and 25 (Low priced) stereo.

The inventory price in that case is = 25*3000 + 25*800 = 95000 (short)

Must increase higher priced item.

Lets guess the amount is 30 (high priced) and 20 (Low priced) stereo.

The inventory price in that case is = 30*3000 + 20*800 = 106000 (short)

Must increase higher priced item.

Lets guess the amount is 35 (high priced) and 15 (Low priced) stereo.

and continue.....

Or you solve this by algebra.

Let us assume:

Number of high priced stereo = H

and

Number of low priced stereo = L

Assuming that the inventory consists of only two types stereos, we have:

H + L = 50 ................................................. (1) and

3000*H + 800*L = 111000 .................. (2)

Now you have two equations and two unknowns. Solve for those using your favorite method.
 
As an arithmetic problem (which you said this is), this may be intended to teach you to check your answer. It is natural at first to just divide 111,000 by 3000 and get 37, then think because it came out so nicely, it must be right! But if you then go back through the problem to check that answer, you discover what JeffM pointed out: the total would actually be at least 37 * 3000 + 13 * 800 = 121,400, which is too much by $10,400. Now you've come to a better understanding of what the problem is about!

An alternative method for solving, rather than pure "guess and check", is to observe that for each $3000 system you remove, you add an $800 system (or more), for a net reduction of up to $2200. How many times do you need to do this to reduce the total by $10,400? Divide again, and you get 10,400/2200 = 4.7. So if you take away 4 big systems, the total will not be reduced enough; you have to take away at least 5. So your answer is 37 - 5 = 32.

Now, of course, we do another check. JeffM again showed the result; the amount is now less than $111,000, so we've solved it. (To get exactly $111,000, there must be some intermediate-priced systems; if there are enough of those, the number of $3000 systems may be less than 32.)

If you are actually taking algebra, let us know what ideas you have for solving the problem that way. You'll be solving an inequality, and it will be easier to be sure of your answer.
 
Oh, thanks to Dr. P, I see i misread the problem. I needed to be comparing to 111,000. So 32 is the maximum, and there must be at least one system that sells for more than 800 and less than 3000.
 
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