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Name: Period: Practice Test, Module 1, Topic C1 1. Solve each equation or inequality. Then circle all of the problems where the solution set. DO NOT MULTIPLY Give each parentheses a zero: is included in divide by -1 to get: inequality flips Then solve each one separately: Make into 16 Multiply each side by Left side is just 100 You get: Solve as usual Answer: distribute the 2 add the 14 to the -9 divide by 2 you get: this DOES include the value of since -2 is less than 2.5 Multiply each side by 5 To get: Solve for x You get: subtract 3 from both the 0 and 9 you get: this DOES include the value of since -2 is between -3 and 6 The problems with solutions that include -2 are circled in green 2. Circle and describe the mistake(s) in the following steps used to solve: . Describe the mistake(s): This is the typo-fixed version. The only mistake is flipping the inequality when it shouldn’t be flipped. Graph the correct solution on a number line: This is the typo-fixed version. This version has an answer of so its graph would have a dot at -6 with shading to the right. 3. Fill in the chart. Write an equation or inequality Describe solution set in words example: all real numbers except 2 Write the solution in set notation Show on Number Line all real numbers less than 3 all real #’s greater or equal to -1 all real #’s less than or = to 4, or greater than or = to 6 4 6 all real numbers between 0 and 3 all real numbers equal to -2 and 2 -2 0 2 all real numbers equal to 5 The first column can repeat the solution from the set notation as its equation, or you can make it more complex by adding 1 or subtracting 2 or multiplying by 3 or whatever to both sides of the solution 4. Using the phrases from the left, fill in the blanks with the property used to convert the equation from one line to the next. Some may not be used; others may be used more than once. Associative Property Commutative Property commutative (changed order) Distributive Property distributive (got rid of parentheses) Combine like terms additive prop of equality Additive Property of Equality additive prop of equality (“additive” includes subtracting something from both sides) We often end a problem by dividing on both sides; this would be called the multiplicative prop of equality because dividing can also be done by multiplying by the reciprocal Multiplicative Prop. of Equality 5. A car salesman earns a percentage of the value of the cars he sells. He also earns $200 per month to cover expenses. Last month, the value of cars sold was $60,000 and he earned a total of $1700. a. Write an equation to find the percentage, , that he earns on the cars sold. (because the value is the only one multiplied by the percent) b. Find the percentage he earns. subtract 200 from 1700; divide that by 60,000 but the answer is interpreted as: 2.5% 6. Tax is added to some purchases in Washington. At the grocery store, I bought some items that were not taxed for $18. The rest of the items cost $25 before tax was added. I paid a total of $45. a. Write an equation to find the tax percentage, , paid on the second part of my purchase. (the 25 is the only one multiplied by the percent; the 18 stays 18) b. What is the tax percentage? subtract 18 from 45; divide by 25. You get But that is interpreted as a tax percentage of 8% for the final answer 7. Given this situation: To get a discounted movie ticket, movie goers must be either under the age of 10 (not including 10) or aged 65 and older. Complete the table: define the variable write a compound inequality show on a number line age in years Sorry, I can’t draw the graph with this program. But you’d have an open circle on 10, with shading to the left; then you’d have a dot on 65, with shading to the right. And the # line would probably count by 10’s. 8. Given this situation: To be considered a miniature poodle, a poodle’s height at its shoulders must be more than 28 cm but less than 38 cm. Complete the table: define the variable write a compound inequality show on a number line graph has open circles on 28 and 38 (# line probably counting by 2’s from mid-20’s) with shading in between. 9. A farmer needs to move some bales of hay between two barns. Right now, the larger barn has 400 bales and the smaller barn has 150 bales. How much hay does he need to move in order for the larger barn to have four times as much hay as the smaller barn? Support your answer.. group split (the split is your x’s, the group is your total) BUT you then compare: the small group started with 150 now they’re at 110, so they must have moved 40, so the answer is 40 10. A packaging company makes large and small bags of candy. Right now, the machines put 75 candies in the large bag and 45 candies in the small bag. But they want the small bag to have one third as much candy as the large bag. How many candies do they need to move from the small bag to the large bag in order to make that happen? Support your answer. one third is another way of saying “three times as many”, just from the large bag’s perspective solve and you get that x is 40, but remember, you need to compare: small bag started with 45, ended with 40 the answer is: 5 moved 11. A rectangular sandbox measuring 4 feet by 5 feet will have a shredded bark walkway around it. The walkway is of uniform width, but the width has not been chosen. There is only enough bark to cover 46 square feet of walkway. This started as 26, but that was a typo. a) Write an inequality that represents the total area of the walkway, based on the width the walkway, given that the total cannot exceed 46 square feet. of The represents the 4 corners. The 18w represents the rectangular edge pieces, which lie against the inner rectangle’s 4-by-5 perimeter. The 46 is the upper limit. b) Select all the possible values for the width of the walkway from the chart that could satisfy the square footage requirement. 1.5 foot 2 feet 2.5 feet 3 feet Test the given numbers in Start with something in the middle. Here I’d start with 2: and I would get 52. Since that’s too big, 1.5 must be the answer. 12. A small rectangular pool is going to have a concrete edge of uniform width around all of its sides. The pool is 10 feet by 20 feet, but the width of the concrete edge has not yet been chosen. To keep expenses low, the total area of the concrete cannot exceed 220 square feet. a) Write an inequality that represents the total area of the walkway, based on the width the walkway, given that the total cannot exceed 220 square feet. of again, 4 corners for the first term; perimeter times w for the second term. b) Select all the possible values for the width of the walkway that could satisfy the square footage constraint. 1.5 foot 2 feet 2.5 feet 3 feet Test the given numbers in Start with something in the middle. Here I’d start with 2: and I would get 136. So I go bigger & try either 2.5 or 3. (Actually, all of these work.) 13. Solve each equation or inequality: distribute the 3 divide by -50 combine like terms on the left it won’t go evenly, so leave it as a fraction subtract 7 from the 12 remember to flip the inequality divide by 3 (won’t go evenly --- use a fraction) distribute the -3 (watch your signs!) distribute the 2 combine like terms on the left put all the letters on one side, subtract the 6 from the 12 numbers on the other divide divide (won’t go evenly) 14. Match the solutions in the box to the equations shown below. Not all solutions will be used; some solutions may be used more than once. a) Different x’s --- not a trick problem #’s will cancel to zero --- answer will be zero a) b) Same x’s, same numbers on each side, so everything cancels That means “infinite solutions” b) c) c) the terms cancel; you have different x’s, no numbers Not a trick problem, but the answer is zero 15. Solve the following equation for x. Left side: Distribute the -1 Combine the 2 and -2 (they cancel to 0) Right side: Distribute the 2 Combine the 7 and 8 Subtract the 2x from both sides You should end with Divide to get 16. Consider the inequality: . Decide whether or not each value in the table is a solution. Solution Not a solution -15 8 10 15 Two choices --Option 1: you can solve the inequality (subtract 7 from both sides, divide by -1, flip the inequality). Consider how you would graph the result. Then look at the #’s in each row of the table. Decide if the number would be part of the shaded section of the number line graph that would result. If so, mark it as a solution. If not, mark it as not a solution. Remember, an open circle would not be part of the shaded solution. Option 2: take the number from column 1 and plug it in for x in the inequality. Multiply it by -1, add 7, and see if it does result in something larger than -3. If it does, then it’s a solution. If not, then not.