Just a few math questions!! Please help!!!!?

Question

1.) I have 30 coins consisting of nickels and quaters. The total balue of coins is $4.10. How many of each kind do I have.

2.) Rectangular cards, 1 inches by 3 inches, are cut from a rectangular sheet 2 feet by 3 feet. What is the greatest number of cards that can be cut from the sheet?

3.) In three bowling games, alice scores 139, 143 and 144. What score will alice need in a fourth game in order to have an average score of 145 for four games?

Thanks to all who helps me!!

Answer 1

Hi there! I can help with the first question.

Let x = the no. of nickels and y = the no. of quarters.

You will have 2 equations:

1) x + y = 30 (since you have 30 coins total)

Next, we need to represent the value of each coin:

2) .05x + .25y = 4.10 (since a nickel is 5 cents, a quarter is 25 cents and the total amount of money is $4.10)

Now we need to solve one of the equations:

x + y = 30 is easiest
We get x = 30 - y (or y = 30 - x)

So plug that value in for x in the second equation:

.05(30-y) + .25y = 4.10
1.50 - .05y + .25y = 4.10
1.50 + .20y = 4.10
.20y = 2.60
y=13

So we have 13 quarters (remember, y is the no. of quarters).
Now use that value of y to find the value of x:

Go back to the equation x + y = 30.
We know y:

x + 13 = 30

Then x = 17.
So there are 17 nickels.

Let's check it:

.05x + .25y = 4.10
.05(17) + .25(13) = 4.10
.85 + 3.25 = 4.10
4.10 = 4.10

It's correct! I hope I didn't totally confuse you! Best wishes to you!

Answer 2

1)All questions like these are solved the same way. Choose a variable to represent the values you are trying to find. Then write algebra equations that say the same thing that the problem sentences say.

How many of each coin (nickels, quarters) do I have
n= number of nickels I have
q=number of quarters I have

I have 30 coins consisting of nickels and quarters.
n+q=30

The total value of the coins is $4.10.
5*n + 25 * q = 410

solve the two equations

n=30-q

5*(30-q)+25*q=410
150-5q+25q=410
20q=260
q=13
n=17

got it?

2) This is not a strictly algebraic problem. It requires a little spatial geometry-like analysis.

Each card is 1*3= 3in^2 and the whole sheet is 12*36=432 in^2
If we could have cards any shape so we had no waste we could make 432/3=144 cards 3in^2 cards, but we have to make only rectangular 1x3 cards so their may be waste. We have to see if there is a way to make it work out.

Notice that the card both card dimensions divide into 1 and 3. Therefore we will be able to cut cards that use all the card.

Picture twelve rows of cards with their 1" dimension on the bottom. Twelve in each row makes it 12 inches wide, twelve rowes each three inches tall makes it 36" tall. Twelve tall by twelve wide gives us 144 cards, just like the original calculation.

3. a=alices last score

(a+139+143+144)/4 = 145

multiply both sides by "4"
a+426=580
a=154

voila
keep at it
math is power

Answer 3

1) let q= the number of quarters and n= the number of nickels

30 = q + n
4.10 = .25q + .05n

so q = 30 - n
sub it in to the second equation:

4.10 = .25(30 - n) + .05n
4.10 = 7.5 - .25n + .05n = 7.5 - .2n
.2n = 7.5 - 4.1 = 3.4
n = 3.4/.2 = 17

17 nickels and 30-17=13 quarters

2) 2ft x 3ft = 24in x 36 in

so if the 36in side is vertical, you can cut 36/3 = 12 vertical strips from it. this is for every column 1 in wide so since their are 24 columns, you can cut 12 * 24 = 288 strips from it

3) average score = (game 1 + game 2 + game 3 + game 4) / 4
so if ave score = 145 and x is the score in game 4
we can say 145 = (139 + 143 + 144 + x) / 4
145 * 4 = 139 + 143 + 144 + x
580 = 139 + 143 + 144 + x
580 - 139 - 143 - 144 = x
154 = x

Answer 4

1. 13 Quarters, 27 Nickels
2. 288
3. 154

Answer 5

1) x+y=30,5x+25y=410. Solve for x and y. x=17, y=13.

Yea, so, I only gave you one answer, but the other two are pretty much even easier than that one, and I'm sure you've got a trillion answers by now, so good luck, hope you ace your homework.