Here we are going to review and work on some other harder problems.

**Question #1: Tickets for the homecoming dance cost $20 for a single ticket or $35 for a couple. Ticket sales totaled $2280, and 128 people attended. How many tickets of each type were sold?**

**Solution:**

Let there be x tickets sold for singles and y tickets sold for couples. According to the given:

20 x + 35 y = 2280 --- equation 1

x + 2 y = 128 --- equation 2

equation 2 times 20 - equation 1 and you have 20 x + 40 y - (20 x + 35 y) = 20 x 128 - 2280 = 280

5y = 280 so y = 56 and plug in to equation 2 and you get x = 16

There will be

**16 tickets sold for singles and 56 tickets sold for couples.**

Always check your answer to see if that is right.

**Question #2: There are 20 questions on an algebra test, you got 5 points for each question you answer correctly and - 2 points for each questions you answer wrong.**

**If a student gets 79 for his score and he answers all the questions, how many questions does he answer right?**

**Solution I:**

If the student gets all the questions right, he'll get 5 x 20 = 100 points; however, he only gets 79 points.

For each questions he gets wrong, he not only doesn't have the 5 points but on top of that, he gets 2 points deducted, thus it's minus 7 total.

(100 - 79) / 7 = 3 so the student gets 3 wrong and

**17 questions right.**

Check if that's correct: 17 x 5 - 2 x 3 = 79

**Solution II:**

Let the student get x questions right, and that means he gets (20 - x) questions wrong.

5x - 2 (20 -x) = 79; 5x - 40 + 2x = 79; 7 x = 119 so

**x = 17**

**Question #3:**

**My piggy bank contains only nickels, dimes, and quarters, and contains 20 coins worth a total of $3.30. If the total value of the quarters is five times the total value of all the other coins, how many dimes are in the piggy bank?**

**Solution:**According to the given:

N + D + Q = 20 --- equation 1

The total value of the quarters is 5 times the total value of all the other coins, which mean that if the total value of N + D = x, the total value of Q is 5 x

x + 5x = 330 (Make sure to use the same unit) so x = 55 (cents)

5x = 5 x 55 = 275 and 275/ 25 = 11 so there are 11 quarters

Plug in equation 1 and you have N + D = 20 - 11 = 9

Since 5N + 10D = x = 55, D and N could be

D N

5 1

4 3

3 5

2 7... etc 2 + 7 = 9 so there are

**2 dimes**(7 nickels and 11 quarters)

Other applicable problems: (answer key below)

#1: Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than te number of nickels, then how many of each does he have?

#2: Kim has $1.65 in nickels, dimes and quarters. She has 10 coins all together and the number of quarters is equal to the number of nickels and dimes combined. Haw many of each coin does she have?

#3: A collection of

*2*nickels, dimes, and quarters is worth $3.20. There are seven more nickels than dimes. How many of each are there in the collection?

**4**#4: Continue with #3, if there are seven more dimes than nickels, how many of each are there in the collection?

Answer key:

#1:

**6 quarters, 20 nickels and 15 dimes**

#2:

**2 nickels, 3 dimes and 5 quarters**

#3:

**9 quarters, 4 dimes and 11 nickels**

#4:

**7 quarters, 12 dimes and 5 nickels**

## No comments:

## Post a Comment