Lower Elementary:
Question: Billy has a stack of 5 nickels, 1 dime, and 4 quarters. Mandy has a stack of 7 dimes, and 3 quarters. Who has more money and by how much?
Answer: Mandy, by 10¢
Solution: First, let’s find out how much money Billy has. Billy has 5 nickels. A nickel is worth 5 cents. 5, 10, 15, 20, 25. The nickels have a total value of 25¢. He has 1 dime. A dime is worth 10¢. He has 4 quarters. A quarter is worth 25 cents. 25, 50, 75, 100. The quarters have a total value of $1. Altogether, Billy has $1.35 (25¢ + 10¢ + $1.00 = $1.35). Now, let’s find out how much money Mandy has. Mandy has 7 dimes. 10, 20, 30, 40, 50, 60, 70. The dimes have a total value of 70¢. She has 3 quarters. 25, 50, 75. The quarters have a total value of 75¢. Altogether, Mandy has $1.45 (70¢ + 75¢ = $1.45). From $1.35 up to $1.45 is 10¢. So, Mandy has more money by 10¢.
Upper Elementary:
Question: Lance has a pin collection that he started with 45 pins from his father. He then went to a trading convention where he bought 7 new bags of pins. Each bag contained 6 new pins. He then traded one of his pins to get two new ones. How many pins does Lance have now?
Answer: 88 pins
Solution: Lance started off with 45 pins. He bought 7 bags of pins with 6 pins in each bag. 7 × 6 = 42. He got a total of 42 new pins from the bags. That means Lance now has a total of 87 pins. He then trades away one of his pins to get two new ones. That means he gives away one of his 87 pins (87 – 1 = 86) to gain two new ones to have a total of 88 pins. (86 + 2 = 88).
Middle School:
Question: If Tammy types “2 + 12 • 5 – 3 (10 ÷ 2)2 + 5” into the calculator, what number will display?
Answer: –8
Solution: We have to solve this problem in order of PEMDAS.
2 + 12 • 5 – 3 (10 ÷ 2)2 + 5
First, we do what is inside the parenthesis.
2 + 12 • 5 – 3 (5)2 + 5
Next, we do the exponents.
2 + 12 • 5 – 3 (25) + 5
The next step is to do multiplication and division from left to right.
2 + 60 – 75 + 5
Finally, we do addition and subtraction from left to right.
62 – 75 + 5
–13 + 5
–8
The calculator will display –8.
Algebra and Up:
Question: Five hats and two bracelets cost $41. Two hats and three bracelets cost $34. How much does each hat cost? How much does each bracelet cost?
Answer: One hat is $5 and one bracelet is $8
Solution: Let h represent the cost for one hat and let b represent the cost of one bracelet. If we write the problem as an algebraic equation, we have:
5h + 2b = 41
2h + 3b = 34
One way to solve this system of equations is to use elimination. Let’s eliminate the b in the equations by multiplying the top row by 3 and the bottom row by 2.
15h + 6b = 123
4h + 6b = 68
Let’s subtract the equations.
11h = 55
Divide both sides by 11.
h = 5.
So, one hat is $5. To find the cost of one bracelet, let’s take the value we found for h and substitute into one of the equations. Let’s substitute it in the equation 2h + 3b = 34.
2(5) + 3b = 34
10 + 3b = 34
Subtract 10 from both sides.
3b = 24
Divide both sides by 3.
b = 8
So, one bracelet costs $8. We can check out answer by using the other equation.
5h + 2b = 41
5(5) + 2(8) = 25 + 16 = 41
The equation is true, so the cost of one hat is $5 and the cost of one bracelet is $8.
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