Cumulative End-of-Course Comprehensive Practice Exam - Seventh Grade Mathematics

This cumulative examination synthesizes all major mathematical domains of seventh grade mathematics, providing comprehensive review across the number system, proportional relationships, algebraic modeling, geometry, statistics, and probability.

Exam Overview and Instructions

This cumulative examination comprises eight structured multi-step problems designed to test conceptual understanding, computational precision, and mathematical modeling across every core module of seventh grade mathematics:

               EXAMINATION DOMAIN MAP
   Problem 1: The Number System & Rational Operations
   Problem 2: Ratios & Proportional Relationships (Unit Rates & Tables)
   Problem 3: Percents & Multi-Step Financial Applications
   Problem 4: Expressions & Multi-Step Linear Equations
   Problem 5: Linear Inequalities & Boundary Modeling
   Problem 6: Two-Dimensional Geometry (Circles & Composite Shapes)
   Problem 7: Three-Dimensional Geometry (Surface Area & Volume)
   Problem 8: Statistics & Probability (MAD & Compound Events)

Read each problem thoroughly, sketch diagrams where appropriate, write out complete algebraic setups, and solve independently before reviewing the step-by-step solution explanations.

Exercises

Work through each problem carefully in your notebook before reading the provided solutions.

Exercise 1. Evaluate the following multi-step rational expression following the standard order of operations: negative five-sixths plus the quantity two-thirds minus negative one-fourth multiplied by negative eight-ninths.

Exercise 2. A commercial baker mixes dough where the amount of water in cups (y) is proportional to the amount of flour in cups (x). A batch using five cups of flour requires two and one-half cups of water. Write the equation relating water to flour in the form y equals kx, and determine how many cups of flour are needed for fifteen cups of water.

Exercise 3. A consumer electronics store purchases a tablet computer at a wholesale cost of one hundred sixty dollars. The store marks up the wholesale cost by forty-five percent to establish the retail sticker price. During a holiday clearance event, the tablet is discounted by twenty percent. If the local sales tax rate is seven percent, calculate the final checkout price paid by a customer.

Exercise 4. A mobile auto mechanic charges a flat travel dispatch fee of fifty-five dollars plus forty-two dollars per hour of on-site labor. A customer receives a total invoice of two hundred twenty-three dollars. Write an algebraic equation to model this scenario, solve for the number of hours worked, and check your solution.

Exercise 5. A student has a fifty-dollar gift card for an online media store. The student purchases an album for twelve dollars and fifty cents and plans to spend the remainder on digital mystery book downloads costing three dollars and seventy-five cents each. Write and solve an inequality to find the maximum number of digital books the student can purchase.

Exercise 6. A running track surrounds a central field shaped like a rectangle measuring eighty meters long by forty meters wide, with semicircular regions attached to both forty-meter ends. Using 3.14 for pi, calculate the total area enclosed within the track and the complete distance of one lap around the perimeter of the track.

Exercise 7. A shipping company manufactures a closed metal storage container shaped like a right triangular prism. The triangular base is a right triangle with legs measuring nine feet and twelve feet, and a hypotenuse measuring fifteen feet. The height of the prism between the triangular ends is twenty feet. Calculate both the volume and the total exterior surface area of the container.

Exercise 8. A bag contains four blue chips and six red chips (ten chips total). A player draws two chips sequentially without replacement. In addition, the player rolls a standard six-sided die. Calculate the probability of drawing two consecutive blue chips and rolling a number strictly greater than four.

Solutions

Solution to Exercise 1. Begin by simplifying the operation inside the grouping parentheses: two-thirds minus negative one-fourth. Rewrite subtraction as addition of the opposite: two-thirds plus positive one-fourth. Find the common denominator, twelve: two-thirds is eight-twelfths, and one-fourth is three-twelfths. Add: eight-twelfths plus three-twelfths equals eleven-twelfths. Next, perform multiplication before addition according to the order of operations: multiply eleven-twelfths by negative eight-ninths. Cross-cancel the common factor of four between twelve and negative eight: twelve divided by four is three, and negative eight divided by four is negative two. Multiply numerators: eleven times negative two equals negative twenty-two. Multiply denominators: three times nine equals twenty-seven. The product is negative twenty-two twenty-sevenths. Now add negative five-sixths plus negative twenty-two twenty-sevenths. The least common denominator of six and twenty-seven is fifty-four. Convert negative five-sixths: multiply numerator and denominator by nine to get negative forty-five fifty-fourths. Convert negative twenty-two twenty-sevenths: multiply numerator and denominator by two to get negative forty-four fifty-fourths. Combine numerators: negative forty-five plus negative forty-four equals negative eighty-nine fifty-fourths. As a mixed number, this simplifies to negative one and thirty-five fifty-fourths.

Solution to Exercise 2. Find the constant of proportionality k by dividing water y by flour x: two and one-half cups divided by five cups. Convert two and one-half to five-halves: five-halves divided by five equals five-halves times one-fifth, which simplifies to one-half cup of water per cup of flour (k = 0.5). The governing equation is y equals one-half x (or y = 0.5x). To find the flour needed for fifteen cups of water, substitute fifteen for y: fifteen equals one-half x. Multiply both sides by two: x equals fifteen times two, which equals thirty cups of flour.

Solution to Exercise 3. Step 1: Compute the retail sticker price after forty-five percent markup using the multiplier one point forty-five: one hundred sixty times one point forty-five equals two hundred thirty-two dollars ($232.00). Step 2: Compute the discounted sale price after twenty percent discount using the multiplier zero point eighty (1 - 0.20): two hundred thirty-two times zero point eighty equals one hundred eighty-five dollars and sixty cents ($185.60). Step 3: Compute the final checkout price including seven percent sales tax using the multiplier one point zero seven: one hundred eighty-five point sixty times one point zero seven equals one hundred ninety-eight dollars and fifty-nine cents ($198.59). The customer pays one hundred ninety-eight dollars and fifty-nine cents.

Solution to Exercise 4. Let h represent the number of hours worked. The variable labor cost is forty-two h, and the fixed travel dispatch fee is fifty-five dollars. Write the equation: forty-two h plus fifty-five equals two hundred twenty-three. Subtract fifty-five from both sides: forty-two h equals one hundred sixty-eight. Divide both sides by forty-two: h equals one hundred sixty-eight divided by forty-two, which equals four. The mechanic worked four hours. Check: forty-two times four is one hundred sixty-eight; one hundred sixty-eight plus fifty-five equals two hundred twenty-three. The solution is verified.

Solution to Exercise 5. Let b represent the number of digital mystery books purchased. The total spending is three point seventy-five b plus twelve point fifty dollars. The budget constraint is fifty dollars: three point seventy-five b plus twelve point fifty is less than or equal to fifty. Subtract twelve point fifty from both sides: three point seventy-five b is less than or equal to thirty-seven point fifty. Divide both sides by three point seventy-five: thirty-seven point fifty divided by three point seventy-five equals ten. The student can purchase at most ten digital mystery books.

Solution to Exercise 6. To find the enclosed area, divide the figure into a rectangle and two semicircles. The two semicircles of diameter forty meters combine to form one complete circle of radius twenty meters. Calculate rectangular area: eighty times forty equals three thousand two hundred square meters. Calculate circle area: 3.14 times twenty squared equals 3.14 times four hundred, which equals one thousand two hundred fifty-six square meters. Total enclosed area: three thousand two hundred plus one thousand two hundred fifty-six equals four thousand four hundred fifty-six square meters. To find the distance of one lap around the track, sum the two straight eighty-meter sides and the two semicircular curves (one full circle circumference). Straight sections: eighty plus eighty equals one hundred sixty meters. Circumference: pi times diameter equals 3.14 times forty, which equals one hundred twenty-five point six meters. Total lap distance: one hundred sixty plus one hundred twenty-five point six equals two hundred eighty-five point six meters.

Solution to Exercise 7. First calculate the volume: V equals base area B times prism height H. The base is a right triangle with legs nine feet and twelve feet: B equals one-half times nine times twelve, which equals fifty-four square feet. Multiply by prism height twenty feet: fifty-four times twenty equals one thousand eighty cubic feet of volume. Next calculate total surface area: SA equals twice base area B plus lateral area. Twice base area is two times fifty-four equals one hundred eight square feet. Lateral area is base perimeter P times prism height H: perimeter is nine plus twelve plus fifteen equals thirty-six feet; thirty-six times twenty equals seven hundred twenty square feet. Total surface area: one hundred eight plus seven hundred twenty equals eight hundred twenty-eight square feet.

Solution to Exercise 8. The problem combines a dependent marble draw and an independent die roll. For the first chip draw, four out of ten chips are blue: P(first blue) equals four-tenths, or two-fifths. For the second chip draw without replacement, three blue chips remain out of nine total chips: P(second blue) equals three-ninths, or one-third. Multiply for two consecutive blue chips: two-fifths times one-third equals two-fifteenths. For rolling a die strictly greater than four, the favorable outcomes are 5 and 6, so two outcomes out of six: P(greater than 4) equals two-sixths, or one-third. Because the die roll is independent of the chip draw, multiply the probabilities: two-fifteenths times one-third equals two forty-fifths (2/45, or approximately 4.44%). The combined probability is two forty-fifths.