Cumulative Assessment

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Exercises 1 First, we will use the percent proportion, WholePart​=100Percent​​ to find what part equals 37.5% of 48. By substituting 48 (the total number of trails) for the whole and 37.5 (the percentage of beginner trails) for the percent, we can solve for the part. WholePart​=100Percent​Whole=48, Percent=37.548Part​=10037.5​LHS⋅48=RHS⋅48Part=10037.5​⋅48ca​⋅b=ca⋅b​Part=1001800​Calculate quotientPart=18 There are 18 beginner trails in this park, so we can immediately exclude options A and D. Since the rest of the trails are divided evenly between intermediate and expert trails, option C doesn't fit as we would have more expert than intermediate trails. Therefore, the correct answer has to be B.
Exercises 2 To determine which equations are equivalent to cx−a=b, we will consider each of them separately.cx−a+b=2b In this equation, we can see that on the left-hand side we have an extra b, so let's add b to both sides of the original equation and check if we will get this equation. cx−a=bLHS+b=RHS+bcx−a+b=2b We got the same equation so cx−a+b=2b⇔cx−a=b.0=cx−a+b This equation stands out with 0 on the left-hand side, so let's rearrange the original equation to have 0 on one of the sides. We can do it by subtracting b from both sides. cx−a=bLHS−b=RHS−bcx−a−b=0Rearrange equation0=cx−a−b We didn't get the same equation so the equation is not equivalent to the original.2cx−2a=2b​ Here, we can see that 2cx−2a is the original left-hand side, multiplied by 2, so let's multiply the original equation by 2. cx−a=bLHS⋅2=RHS⋅22cx−2a=2b These equations are not equivalent, because the right-hand sides are different.x−a=cb​ In this equation, we see that instead of cx we have x, so let's divide the original equation by c. cx−a=bLHS/c=RHS/cccx−a​=cb​Write as a difference of fractionsccx​−ca​=cb​Simplify quotientx−ca​=cb​ The equations are not equivalent.x=ca+b​ In this equation, x is isolated so let's isolate x in the original one and see if the right-hand sides are the same. cx−a=bLHS+a=RHS+acx=b+aLHS/c=RHS/cx=cb+a​ We got the same equation, so they are equivalent. x=cb+a​⇔cx−a=b.b+a=cx Here, we have b+a on the left-hand side, so let's add a to the original equation, so that we have b+a on the right-hand side and see if we get the same equations. cx−a=bLHS+a=RHS+acx−a+a=b+aAdd termscx=b+aRearrange equationb+a=cx We got the same equation, so they are equivalent. b+a=cx⇔cx−a=b.
Exercises 3 aTo find the number of solutions for the equation when a=3, we will substitute this value into the equation and try to solve for x. 3(x−a)=3x−6a=33(x−3)=3x−6 Solve for x Distribute 33x−9=3x−6LHS−3x=RHS−3x -9≠-6 We have reached a contradiction which means this equation has no solution. Therefore, the number of solutions N is less than 1: N<​1.bWe can do the same thing as in the previous exercises. Substitute a=-3 and try to solve the equation. 3(x−a)=3x−6a=-33(x−(-3))=3x−6 Solve for x Remove parentheses and change signs3(x+3)=3x−6Distribute 33x+9=3x−6LHS−3x=RHS−3x 9≠-6 We got a contradiction, so this equation has no solution. Therefore, the number of solutions N is less than 1: N<​1.cWe can do the same thing as in the previous exercises. Substitute a=2 and try to solve the equation. 3(x−a)=3x−6a=23(x−2)=3x−6 Solve for x Distribute 33x−6=3x−6LHS−3x=RHS−3x -6=-6 We have reached an identity which is always true no matter the value of x. This equation has infinitely many solutions and therefore, the number of solutions N is more than 1: N>​1.dWe can do the same thing as in the previous exercises. Substitute a=-2 and try to solve the equation. 3(x−a)=3x−6a=-23(x−(-2))=3x−6 Solve for x Remove parentheses and change signs3(x+2)=3x−6Distribute 33x+6=3x−6LHS−3x=RHS−3x 6≠-6 We got a contradiction, so this equation has no solution. Therefore, the number of solutions N is less than 1: N<​1.eWe can do the same thing as in the previous exercises. Substitute a=x and try to solve the equation. 3(x−a)=3x−6a=x3(x−x)=3x−6 Solve for x Subtract term3⋅0=3x−6Multiply0=3x−6LHS+6=RHS+66=3xLHS/3=RHS/32=xRearrange equation x=2 x=2 is the solution to this equation so the number of solutions N is 1: N=​1.fWe can do the same thing as in the previous exercises. Substitute a=-x and try to solve the equation. 3(x−a)=3x−6a=-x3(x−(-x))=3x−6 Solve for x Remove parentheses and change signs3(x+x)=3x−6Add terms3⋅2x=3x−6Multiply6x=3x−6LHS−3x=RHS−3x3x=-6LHS/3=RHS/3 x=-2 x=-2 is the solution to this equation so the number of solutions N is 1: N=​1.
Exercises 4 aIn general terms, our equation will be: (Cost of white×Number of white cans)+(Cost of blue×Number of blue cans)​Total cost​ Our variable must be x because it is the only variable in the given list. If we let x be the number of cans of white paint, then we have x fewer than 5 cans of blue paint, White cans: Blue cans: ​x5−x,​ because we are told that we bought 5 cans in total. We also know that white paint costs $24 per can, blue paint costs $28 per can, and the total cost is $132. When we add these to our original equation, we get: 24x+28(5−x)=132.bBefore we can consider the cost had we swapped room colors, we need to solve for the number of cans we purchased for the current color configuration. Let's begin by solving for x in our equation. 24x+28(5−x)=132Distribute 2824x+140−28x=132 Solve for x Subtract term-4x+140=132LHS−140=RHS−140-4x=-8Change signs4x=8LHS/4=RHS/4 x=2 This means that for the current color scheme, a white dining room and a blue living room, we bought 2 cans of white paint and 3 cans of blue paint. But what if we swapped the colors of the rooms? What would the total cost have been if we had instead painted the living room white and the dining room blue? The new equation would be: 3⋅24+2⋅28=$128. This paint job would have only cost us $128, saving us 132−128=$4.
Exercises 5 Equations are equivalent when they have the same solution, To begin, we will name each of the given equations.NameEquation A6x+6=-14 B5x+3=-7 C8x+6=-2x−14 D7x+6=2x−13 We will now solve each equation by isolating the variable. thatNameEquationSolution A6x+6=-14x=-35​ B5x+3=-7x=-2 C8x+6=-2x−14x=-2 D7x+6=2x−13x=-519​ Notice that B and C have the same solution. Thus, they are equivalent.
Exercises 6 The perimeter of a shape is calculated by adding the lengths of all the sides. It is given that the perimeter of the triangle is 13 inches. We can write the following equation. 13=(x−5)+21​x+6 Solving the equation will give us the value of x. Then we can determine the length of the shortest side. 13=(x−5)+21​x+6Add terms13=23​x+1LHS−1=RHS−112=23​xLHS⋅32​=RHS⋅32​12⋅32​=xMultiply8=xRearrange equationx=8 Solving the equation gives x=8. One side of the triangle is 6 inches. We must solve for the other two sides. To do this, we will substitute x=8 into the given equations.Equationx=8Side length x−58−53 inches 21​x21​(8)4 inches The shortest side of the triangle measures 3 inches.
Exercises 7 aTo begin, we will define a variable for this situation. Let x represent the number of months for which cable TV is purchased. Since we pay 45 dollars per month, the following expression gives my total cost: 45x. Our friend paid 99 dollars for a satellite dish and pays 36 per month. The following expression then represents our friend's total cost. 36x+99 We can set the equations equal to each other. Solving the equation will tell us when we have paid the same amount as our friend. 45x=36x+99LHS−36x=RHS−36x9x=99LHS/9=RHS/9x=11 The solution x=11 means that we've paid the same total amount as our friend at 11 months of service.bTo determine which service costs more for one year (or twelve months) we can substitute x=12 into the expressions from Part A and evaluate.ServiceExpressionx=12Cost for one year Cable45x45⋅12540 dollars Satellite36x+9936⋅12+99531 dollars From our work in the table, we can see that satellite costs less than cable. Thus, our friend is correct.
Exercises 8 We have a mix of absolute value equations and linear equations to choose from. Let's deal with them one at a time.Absolute value equation An absolute value equation can have 0, 1 or 2 solutions:0 solutions: Absolute values are always non-negative. Therefore, whenever you have an absolute value equation that equals a negative number, it won't have a solution. 1 solution: Absolute value equations shows a distance from a midpoint on a numberline. If that distance is 0, there is only 1 solution. 2 solutions: If the distance is greater than 0, there are 2 solutions, one on either side of the midpoint. Out of the nine equations, three of them are absolute value equations. Before we can classify them, we have to make sure they are all written on the form ∣ax+b∣=c: 09​=∣x+13∣+2=3∣2x−11∣​​⇔2∣x+13∣=-2⇔∣2x−11∣=3​ Now we can classify each equation.EquationSolutions ∣8x+3∣=01 ∣x+13∣=-20 ∣2x−11∣=32 Linear equation A linear equation can have 0, 1 or infinite solutions.0 solutions: When the graphs of two linear equations are parallel, they never intersect which means there is no solution. 1 solution: When the graphs of two linear equations have different slopes, there is 1 solution. Infinite solutions: When the graphs of two linear equations are identical, they overlap and therefore have infinitely many solutions.To classify the linear equations, we first have to simplify the left-hand and right-hand side in the equations where we can. 3x−12-4(x+4)7−2x​=3(x−4)+1=-4x−16=3−2(x−2)​​⇔-3x−12=3x−11⇔-4x−16=-4x−16⇔x-7−2x=7−2x​ Now we can determine how many solutions the equations have.EquationSolutions -2x+4=2x+41 12x−2=10x−81 -6=5x−91 3x−12=3x−110 -4x−16=-4x−16Infinite 7−2x=7−2xInfinite Placing our equations Now we can complete the table from the exercise.No solutionOne solutionTwo solutionsInfinitely many solutions 0=∣x+13∣+2, 3x−12=3(x−4)+1∣8x+3∣=0, 12x−2=10x−8, -2x+4=2x+4, -6=5x−99=3∣2x−11∣-4(x+4)=-4x−16, 7−2x=3−2(x−2)
Exercises 9 The distance d the car traveled was 1000 feet and the time t it took was 12.5 seconds. We can find the rate r using the distance formula: d=rt. Let's do that now. d=rtd=1000 feet, t=12.5 seconds1000 feet=r⋅12.5 secondsLHS/12.5 seconds=RHS/12.5 seconds12.5 seconds1000 feet​=rRearrange equationr=12.5 seconds1000 feet​ba​=b/12.5a/12.5​r= second80 feet​ Note that, as long as the ratio stays the same, 80 feet:1 second, we can express this rate in a couple different ways. Let's look at the choices we have been given.FractionCorrect?Why? 80feetsecond​Noa⋅cb​=ca⋅b​ The units do not match the correct numbers.  second80 feet​YesThis is exactly what we found when solving for rate. 80secondfeet​Yesa⋅cb​=ca⋅b​ 80 feetsecond​YesEven though it is flipped, the ratio is still correct with 1 second and 80 feet.