GED Math Practice Test 5

1. "How many \tfrac{5}{8}-yard pieces fit in 7\tfrac{1}{2} yards?" is a division question: 7\tfrac{1}{2} \div \tfrac{5}{8}.
2. Change the mixed number to an improper fraction: 7\tfrac{1}{2} = \dfrac{7 \times 2 + 1}{2} = \dfrac{15}{2}.
3. Dividing by a fraction means multiplying by its reciprocal: \dfrac{15}{2} \div \dfrac{5}{8} = \dfrac{15}{2} \times \dfrac{8}{5}.
4. Cancel before multiplying: \dfrac{15}{5} = 3 and \dfrac{8}{2} = 4, so the product is 3 \times 4 = 12.

She can make 12 rosettes.

Check: 12 \times \tfrac{5}{8} = \tfrac{60}{8} = 7\tfrac{4}{8} = 7\tfrac{1}{2} yards ✓ — the ribbon is used up exactly.

Tip: "How many groups of this size fit?" always means divide, and the total goes first. A quick estimate helps: each piece is a bit more than half a yard, so you get a bit fewer than 2 \times 7.5 = 15 pieces.

1. Power of a product: raise each factor to the 3rd power.
   • 3^{3} = 27
   • (x^{-2})^{3} = x^{-2 \cdot 3} = x^{-6}
   • (y^{4})^{3} = y^{12} So the numerator is 27x^{-6}y^{12}.
2. Divide the numbers: \dfrac{27}{9} = 3.
3. Divide the x's (subtract exponents): x^{-6 - (-5)} = x^{-6 + 5} = x^{-1}.
4. Divide the y's: y^{12 - 1} = y^{11}.
5. Put it together: 3x^{-1}y^{11} = \dfrac{3y^{11}}{x} (a negative exponent moves the factor to the denominator).

The answer is \dfrac{3y^{11}}{x}.

Check: Try x = 1, y = 1: original = \dfrac{3^{3}}{9} = 3 and \dfrac{3 \cdot 1}{1} = 3 ✓. Try x = 2, y = 1: original = \dfrac{27 \cdot 2^{-6}}{9 \cdot 2^{-5}} = 3 \cdot 2^{-1} = 1.5, and \dfrac{3}{2} = 1.5 ✓.

Tip: Power of a power → multiply exponents. Dividing like bases → subtract exponents, and write "minus a negative" as plus. Do not forget that the coefficient 3 is cubed too.

1. The routes leave together again after a number of minutes that is a common multiple of 12, 18 and 30. The first such time is the least common multiple (LCM).
2. Prime-factor each number:
   • 12 = 2^{2} \cdot 3
   • 18 = 2 \cdot 3^{2}
   • 30 = 2 \cdot 3 \cdot 5
3. Take the highest power of each prime: 2^{2} \cdot 3^{2} \cdot 5 = 4 \cdot 9 \cdot 5 = 180 minutes.
4. Convert: 180 minutes = 3 hours. 6{:}00 + 3 hours = 9{:}00.

All three leave together again at 9:00 a.m.

Check: 180 \div 12 = 15, 180 \div 18 = 10, 180 \div 30 = 6 — all whole numbers ✓. And no smaller number works: the multiples of 30 below 180 are 30, 60, 90, 120, 150, and none of them is divisible by both 12 and 18.

Tip: "Happen together again" problems use the LCM; "split into equal groups as large as possible" problems use the GCF. Listing multiples of the largest number (30, 60, 90, …) and testing each is a fast no-calculator method.

1. Money coming in (revenue): each taco sells for $3.50, so revenue is 3.50t.
2. Money going out (costs): each taco costs $1.20 to make, so ingredient cost is 1.20t. The permit is a fixed $85 per day, no matter how many tacos she sells.
3. Profit = revenue − costs: 3.50t - (1.20t + 85) = 3.50t - 1.20t - 85.
4. Combine like terms: 3.50t - 1.20t = 2.30t.

Profit is 2.30t - 85 dollars.

Check: Test t = 100 tacos. Revenue = \$350, ingredients = \$120, permit = \$85, so profit = 350 - 120 - 85 = \$145. The expression gives 2.30(100) - 85 = 230 - 85 = 145 ✓.

Tip: In a profit expression the per-item amount multiplies the variable, and the one-time (fixed) cost is a constant that is subtracted. Plugging in a friendly number like 100 is a fast way to test each answer choice.

1. Find the total cost of each way to cover 14 rides:
   • 14 single rides: 14 \times 2.75 = \$38.50
   • two 10-ride cards: 2 \times 22.50 = \$45.00 (6 rides wasted)
   • one 10-ride card + 4 single rides: 22.50 + 4 \times 2.75 = 22.50 + 11.00 = \$33.50
   • one 7-day pass: \$32.00
2. The lowest total is the 7-day pass, $32.00.
3. Cost per ride: \dfrac{\$32.00}{14 \text{ rides}} \approx \$2.2857, which rounds to $2.29.

Jamal pays about $2.29 per ride.

Check: 2.29 \times 14 = 32.06 \approx 32.00 ✓ (the small difference comes from rounding). The card alone is $2.25 per ride, but it only covers 10 rides, so it cannot be used for all 14.

Tip: A unit price is only a fair comparison when every option buys exactly what you need. When packages come in fixed sizes, compare total cost for the needed amount first, then divide.

1. Only months where ridership went up count: Feb→Mar, Mar→Apr and May→Jun. (Jan→Feb and Apr→May are decreases.)
2. Compute each percent increase with the earlier month as the base:
   • Feb→Mar: \dfrac{44 - 36}{36} = \dfrac{8}{36} \approx 0.222 = 22.2\%
   • Mar→Apr: \dfrac{50 - 44}{44} = \dfrac{6}{44} \approx 0.136 = 13.6\%
   • May→Jun: \dfrac{56 - 47}{47} = \dfrac{9}{47} \approx 0.191 = 19.1\%
3. The greatest is 22.2%, from February to March.

Check: May→Jun has the largest increase in number of riders (9 thousand), but it starts from a bigger base (47), so its percent is smaller. February's base of 36 is the smallest, which makes its 8-thousand gain the largest percent ✓.

Tip: On the GED, "greatest increase" and "greatest percent increase" are different questions. For percent change, always divide by the original (earlier) value.

1. Write the trips as points: (4,\ 13.20) and (10,\ 27.60).
2. Slope (cost per mile): \dfrac{27.60 - 13.20}{10 - 4} = \dfrac{14.40}{6} = 2.40.
3. Fixed fee (y-intercept): substitute (4,\ 13.20) into C = 2.40m + b: 13.20 = 2.40(4) + b \Rightarrow 13.20 = 9.60 + b \Rightarrow b = 3.60.
4. Equation: C = 2.40m + 3.60.

The answer is C = 2.40m + 3.60.

Check: For 10 miles: 2.40(10) + 3.60 = 24.00 + 3.60 = 27.60 ✓. For 4 miles: 9.60 + 3.60 = 13.20 ✓.

Tip: "Fixed fee plus a rate" means y = mx + b, not a proportion. Dividing one cost by its miles (13.20 \div 4) mixes the fee into the rate and only works when there is no fixed fee.

1. Rectangle: 8 \times 5 = 40 sq ft.
2. Semicircle: its diameter is the 8-ft top edge, so the radius is r = 4 ft. Area = \dfrac{1}{2}\pi r^{2} = \dfrac{1}{2}\pi(4)^{2} = 8\pi \approx 25.13 sq ft.
3. Total area: 40 + 25.13 \approx 65.13 sq ft.
4. Quarts needed: 65.13 \div 25 \approx 2.61 quarts.
5. Paint comes in whole quarts, and 2 quarts would cover only 50 sq ft, so round up: 3 quarts.

The owner must buy 3 quarts.

Check: 3 quarts cover 3 \times 25 = 75 sq ft, which is at least 65.13 ✓; 2 quarts cover only 50 sq ft ✗.

Tip: For a semicircle, halve the circle's area and use the radius (half the diameter). In "how many must you buy" problems, always round up, even if the decimal is small.

1. Let w = width. Then length = 2w + 4.
2. Area: w(2w + 4) = 96, so 2w^{2} + 4w - 96 = 0.
3. Divide every term by 2: w^{2} + 2w - 48 = 0.
4. Factor: two numbers that multiply to -48 and add to 2 are 8 and -6. So (w + 8)(w - 6) = 0.
5. w = -8 or w = 6. A width cannot be negative, so w = 6 ft.
6. Length = 2(6) + 4 = 16 ft.

The banner is 16 feet long.

Check: 16 \times 6 = 96 sq ft ✓, and 16 = 2(6) + 4 ✓.

Tip: After solving a quadratic word problem, (1) throw out answers that make no sense (negative lengths), and (2) re-read the question — it often asks for a different quantity than the variable you solved for.

1. Substitute with parentheses: f(-3) = -2(-3)^{2} + 3(-3) - 5.
2. Exponent first: (-3)^{2} = 9.
3. Multiply: -2 \cdot 9 = -18 and 3 \cdot (-3) = -9.
4. Add/subtract left to right: -18 - 9 - 5 = -32.

So f(-3) = -32.

Check: Use the calculator with the (-) key: -2 \times (-3)^{2} + 3 \times (-3) - 5 = -32 ✓. Note that the -2 is not squared — only the x is.

Tip: In -2x^{2}, the exponent belongs only to x. Always put a negative input in parentheses, both on paper and on the TI-30XS, so the calculator squares -3 instead of computing -(3^{2}).

1. Year 2: a 15% decrease leaves 100\% - 15\% = 85\%: 1{,}200 \times 0.85 = 1{,}020 students.
2. Year 3: a 15% increase gives 115\% of year 2: 1{,}020 \times 1.15 = 1{,}173 students.
3. Overall change: 1{,}173 - 1{,}200 = -27 students.
4. Percent change (divide by the original): \dfrac{-27}{1{,}200} = -0.0225 = -2.25\%.

Enrollment had a decrease of 2.25%.

Check: Use multipliers: 0.85 \times 1.15 = 0.9775, and 1 - 0.9775 = 0.0225, a 2.25% decrease ✓ — no matter what the starting enrollment was.

Tip: Equal percent decreases and increases do not cancel, because the second percent is taken of a different (smaller) amount. Multiply the growth factors (0.85 \times 1.15) to see the combined effect in one step.

1. Slope between two rows: \dfrac{8.0 - 3.2}{10 - 4} = \dfrac{4.8}{6} = 0.8 miles per minute.
2. Confirm with another pair: \dfrac{20.0 - 14.4}{25 - 18} = \dfrac{5.6}{7} = 0.8 ✓. (Also 3.2 \div 4 = 0.8, so the line passes through the origin.)
3. Convert to miles per hour: there are 60 minutes in an hour, so 0.8 \ \dfrac{\text{mi}}{\text{min}} \times 60 \ \dfrac{\text{min}}{\text{h}} = 48 \ \dfrac{\text{mi}}{\text{h}}

The train's speed is 48 miles per hour.

Check: At 48 mph the train goes 48 miles in 60 minutes, so in 25 minutes it goes 48 \times \tfrac{25}{60} = 20 miles, matching the table ✓.

Tip: The slope of a distance–time table is a unit rate in the table's units. Always compare those units with the units the question asks for, and convert at the end.

1. Order does not matter, so use combinations: _{n}C_{r} = \dfrac{n!}{r!(n - r)!}.
2. Seniors: choose 2 of 7: _{7}C_{2} = \dfrac{7 \times 6}{2 \times 1} = 21 ways.
3. Juniors: choose 2 of 5: _{5}C_{2} = \dfrac{5 \times 4}{2 \times 1} = 10 ways.
4. Each pair of seniors can go with each pair of juniors (fundamental counting principle), so multiply: 21 \times 10 = 210.

There are 210 possible groups.

Check: Without the 2-and-2 rule there would be _{12}C_{4} = 495 groups. Our answer must be smaller, because groups like "4 seniors" or "3 seniors + 1 junior" are not allowed ✓.

Tip: "This AND that" choices are multiplied. Use permutations only when positions or order matter (president/vice president, 1st/2nd place). On the TI-30XS, use the prb key to find _{n}C_{r}.

1. Multiply 2x by each term: 2x \cdot x^{2} = 2x^{3}, 2x \cdot 4x = 8x^{2}, 2x \cdot (-5) = -10x.
2. Multiply -3 by each term: -3 \cdot x^{2} = -3x^{2}, -3 \cdot 4x = -12x, -3 \cdot (-5) = +15.
3. Write all six products: 2x^{3} + 8x^{2} - 10x - 3x^{2} - 12x + 15.
4. Combine like terms:
   • x^{2} terms: 8x^{2} - 3x^{2} = 5x^{2}
   • x terms: -10x - 12x = -22x

The product is 2x^{3} + 5x^{2} - 22x + 15.

Check: Let x = 1. Original: (2 - 3)(1 + 4 - 5) = (-1)(0) = 0. Answer: 2 + 5 - 22 + 15 = 0 ✓. Let x = 2: (1)(4 + 8 - 5) = 7 and 16 + 20 - 44 + 15 = 7 ✓.

Tip: Count your products: a 2-term factor times a 3-term factor gives 2 \times 3 = 6 products before combining. Carry the sign with each term — the "-3" is negative in all three of its products.

1. Cone (formula sheet: V = \tfrac{1}{3}\pi r^{2}h) with r = 1.5 in, h = 4 in: V_{\text{cone}} = \tfrac{1}{3}\pi(1.5)^{2}(4) = \tfrac{1}{3}\pi(2.25)(4) = 3\pi \approx 9.42 in³.
2. Hemisphere (half of V = \tfrac{4}{3}\pi r^{3}): V_{\text{hemi}} = \tfrac{1}{2} \cdot \tfrac{4}{3}\pi(1.5)^{3} = \tfrac{2}{3}\pi(3.375) = 2.25\pi \approx 7.07 in³.
3. Total: 3\pi + 2.25\pi = 5.25\pi \approx 16.49 in³.

The total volume is about 16.5 cubic inches.

Check: A full sphere of radius 1.5 would be 4.5\pi \approx 14.1 in³; half of it (7.07) plus a cone smaller than a cylinder of the same size (\pi(2.25)(4) \approx 28.3, and one third of that is 9.42) gives 16.5 ✓.

Tip: For composite solids, find each piece with its own formula and add. Read the figure carefully: 1.5 in is a radius, and the cone's 1/3 and the hemisphere's 1/2 are easy to forget.

1. The trend line passes through the marked points (2, 8) and (10, 28).
2. Slope = \dfrac{\text{change in } y}{\text{change in } x} = \dfrac{28 - 8}{10 - 2} = \dfrac{20}{8} = 2.5.
3. Units: y is points per game, x is hours of practice per week, so the slope is 2.5 points per game for each hour of weekly practice.
4. In words: each additional hour of practice per week is associated with about 2.5 more points per game.

Answer: Each additional hour of weekly practice is associated with about 2.5 more points per game.

Check: The y-intercept is 8 - 2.5(2) = 3, so the line is y = 2.5x + 3. At x = 10: 2.5(10) + 3 = 28 ✓. The 3 is the intercept (predicted points with 0 hours), not the slope.

Tip: Interpret a slope as "y-units per one x-unit." Keep the order: points per hour, not hours per point. A trend line shows an association, not proof that practice alone causes the points.

1. Let p = number of pages.
   • Shop A: 0.12p + 18
   • Shop B: 0.18p + 6
2. Set them equal: 0.12p + 18 = 0.18p + 6.
3. Subtract 0.12p from both sides: 18 = 0.06p + 6.
4. Subtract 6: 12 = 0.06p.
5. Divide by 0.06: p = 12 \div 0.06 = 200.

The shops charge the same for 200 pages.

Check: Shop A: 0.12(200) + 18 = 24 + 18 = \$42. Shop B: 0.18(200) + 6 = 36 + 6 = \$42 ✓.

Tip: "When are they equal?" problems: set the two expressions equal and solve. A quick sense check: Shop B starts $12 cheaper but costs 6¢ more per page, so it takes 12 \div 0.06 = 200 pages to use up that $12 head start.

1. Since \tfrac{1}{4} inch represents 1 foot, 1 inch represents 4 feet.
2. Actual length: 7.5 \times 4 = 30 ft.
3. Actual width: 6 \times 4 = 24 ft.
4. Actual area: 30 \times 24 = 720 sq ft.

The classroom's area is 720 square feet.

Check: The blueprint area is 7.5 \times 6 = 45 sq in. Because each length is multiplied by 4, the area is multiplied by 4^{2} = 16: 45 \times 16 = 720 ✓.

Tip: A scale factor for lengths is squared for areas (and cubed for volumes). The safest method is to convert each length to real units before computing area.

1. A relation is a function when no input (x) is paired with two different outputs (y). Repeated outputs are allowed.
2. Check each set:
   • \{(1, 4), (2, 4), (3, 5), (4, 5)\}: the inputs 1, 2, 3, 4 are all different. The outputs repeat (4, 4 and 5, 5), but that is fine. Function.
   • \{(2, 3), (2, 5), \ldots\}: input 2 gives both 3 and 5. Not a function.
   • \{(3, 1), \ldots, (3, -1), \ldots\}: input 3 gives both 1 and -1. Not a function.
   • \{(4, 2), (4, -2), \ldots\}: input 4 gives both 2 and -2. Not a function.

The function is \{(1, 4), (2, 4), (3, 5), (4, 5)\}.

Check: Think of a vending machine: each button (input) must give one specific snack (output). Two buttons giving the same snack is fine; one button giving two different snacks is not ✓.

Tip: Scan only the first numbers. If any first number repeats with a different second number, it is not a function. On a graph, this is the vertical line test.

1. Walk along two sides: 100 + 60 = 160 yd.
2. Diagonal: the length and width are the legs of a right triangle, so by the Pythagorean theorem (formula sheet): c^{2} = 100^{2} + 60^{2} = 10{,}000 + 3{,}600 = 13{,}600 c = \sqrt{13{,}600} \approx 116.62 yd.
3. Savings: 160 - 116.62 = 43.38 \approx 43.4 yd.

Malik's walk is about 43.4 yards shorter.

Check: The diagonal must be longer than the longest side (100) but shorter than the two sides together (160), and 100 < 116.6 < 160 ✓.

Tip: Read what is asked: the diagonal (116.6) is a step, not the answer. On the TI-30XS, press 2nd x² (√), type 100 x² + 60 x² under the root, and press enter. If the screen shows an exact root (20\sqrt{34}), press toggle to see 116.6…

1. Set the equation equal to 0: 3x^{2} + 2x - 8 = 0.
2. Factor (look for (3x + ?)(x + ?)): try (3x - 4)(x + 2). Expand to check: 3x^{2} + 6x - 4x - 8 = 3x^{2} + 2x - 8 ✓.
3. Zero-product rule: 3x - 4 = 0 or x + 2 = 0.
4. Solve: x = \dfrac{4}{3} or x = -2.

The solutions are x = -2 and x = \tfrac{4}{3}.

Check: x = -2: 3(4) + 2(-2) = 12 - 4 = 8 ✓. x = \tfrac{4}{3}: 3 \cdot \tfrac{16}{9} + \tfrac{8}{3} = \tfrac{16}{3} + \tfrac{8}{3} = \tfrac{24}{3} = 8 ✓. Quadratic formula check: x = \dfrac{-2 \pm \sqrt{4 + 96}}{6} = \dfrac{-2 \pm 10}{6}, giving \tfrac{8}{6} = \tfrac{4}{3} and -2 ✓.

Tip: The zero-product rule only works when one side is 0. When the leading coefficient isn't 1, remember to divide by it: 3x - 4 = 0 gives x = \tfrac{4}{3}, not 4.

1. Before: each of n students pays \dfrac{480}{n} dollars.
2. After: there are n + 4 students, so each pays \dfrac{480}{n + 4} dollars.
3. Savings per student = before − after: \dfrac{480}{n} - \dfrac{480}{n + 4}.
4. Common denominator n(n + 4): \dfrac{480(n + 4)}{n(n + 4)} - \dfrac{480n}{n(n + 4)} = \dfrac{480n + 1920 - 480n}{n(n + 4)} = \dfrac{1920}{n(n + 4)}

Each student pays \dfrac{1920}{n(n + 4)} dollars less.

Check: Let n = 20. Before: 480 \div 20 = \$24. After: 480 \div 24 = \$20. Savings: $4. Formula: \dfrac{1920}{20 \cdot 24} = \dfrac{1920}{480} = 4 ✓.

Tip: To subtract fractions with different denominators, multiply each by the missing factor — never subtract (or add) the denominators. Testing a friendly number such as n = 20 quickly eliminates wrong choices.

1. Early-bird players: e. Since there are 84 players in all, regular players: 84 - e.
2. Money from early-bird players: 35e.
3. Money from regular players: 40(84 - e).
4. Total money: 35e + 40(84 - e) = 3{,}150.

The equation is 35e + 40(84 - e) = 3{,}150.

Check: Solve it: 35e + 3{,}360 - 40e = 3{,}150 \Rightarrow -5e = -210 \Rightarrow e = 42. Then 42 early-bird and 42 regular: 35(42) + 40(42) = 1{,}470 + 1{,}680 = 3{,}150 ✓.

Tip: When two groups make up a known total, write the second group as "total minus the first" (84 - e). Each price must multiply the number of people who paid that price, and the equation must be about one kind of quantity (here, dollars).

1. The light-rail slope is \tfrac{2}{3}.
2. A perpendicular slope is the negative reciprocal: flip \tfrac{2}{3} to \tfrac{3}{2} and change the sign: m = -\tfrac{3}{2}.
3. Substitute the point (6, -1) into y = mx + b: -1 = -\tfrac{3}{2}(6) + b \Rightarrow -1 = -9 + b \Rightarrow b = 8.
4. Equation: y = -\tfrac{3}{2}x + 8.

The bike path is y = -\tfrac{3}{2}x + 8.

Check: The point: -\tfrac{3}{2}(6) + 8 = -9 + 8 = -1 ✓. Perpendicular: \tfrac{2}{3} \times \left(-\tfrac{3}{2}\right) = -1 ✓.

Tip: Parallel lines have equal slopes; perpendicular lines have slopes whose product is -1 (flip and change sign). Always find b by substituting the given point — do not reuse the original line's intercept.

1. Train X's rate: the graph is a straight line through the origin and (15, 12), so its rate is \dfrac{12}{15} = 0.8 miles per minute.
2. Train Y's rate: in d = 0.75t the coefficient is the rate, 0.75 miles per minute.
3. Distances in 40 minutes:
   • Train X: 0.8 \times 40 = 32 miles (the graph also shows (40, 32)).
   • Train Y: 0.75 \times 40 = 30 miles.
4. Difference: 32 - 30 = 2 miles.

Train X travels 2 miles farther.

Check: Difference in rates: 0.8 - 0.75 = 0.05 mi/min; over 40 minutes, 0.05 \times 40 = 2 miles ✓.

Tip: To compare proportional relationships given in different forms, turn each one into a unit rate (y per 1 unit of x). For a graph, divide y by x at any point on the line.

1. The opposite (complement) of "late on at least one day" is "on time every day."
2. Days are independent, so multiply: P(\text{on time all 5}) = 0.85^{5}.
3. Compute: 0.85^{2} = 0.7225, 0.85^{3} \approx 0.6141, 0.85^{4} \approx 0.5220, 0.85^{5} \approx 0.4437.
4. Complement: P(\text{late at least once}) = 1 - 0.4437 = 0.5563 \approx 55.6\%.

The probability is about 55.6%.

Check: The answer must be more than the 15% chance for a single day (more days, more chances to be late) but less than 100% ✓. Adding 5 \times 15\% = 75\% overcounts days with more than one late bus.

Tip: For "at least one" questions, use 1 - P(\text{none}). On the TI-30XS, type 0.85 ^ 5 then subtract the result from 1.

1. Distribute -2: 5 - 6x - 2 \le 4x - 17.
2. Combine like terms on the left: 3 - 6x \le 4x - 17.
3. Collect x terms (subtract 4x): 3 - 10x \le -17.
4. Subtract 3: -10x \le -20.
5. Divide by -10 and flip the sign: x \ge 2.

The solution is x \ge 2.

Check: Try x = 3 (in the solution): 5 - 2(10) = -15 and 4(3) - 17 = -5; -15 \le -5 ✓. Try x = 0 (not in it): 5 - 2 = 3 and -17; 3 \le -17 is false ✓. Boundary x = 2: 5 - 14 = -9 and 8 - 17 = -9, equal ✓.

Tip: Flip the inequality only when you multiply or divide both sides by a negative. Moving terms by adding or subtracting never flips it. Test one number from your answer in the original inequality.

1. Muffin sales: 340 \times \$2.75 = \$935.00.
2. Bread sales: 125 \times \$5.40 = \$675.00.
3. Total sales: 935 + 675 = \$1{,}610.00.
4. Labor: 3 \times 22 \times 14.50 = 66 \times 14.50 = \$957.00.
5. Total costs: 412.50 + 957.00 = \$1{,}369.50.
6. Profit: 1{,}610.00 - 1{,}369.50 = \$240.50.

The bakery's profit was $240.50.

Check: Work in another order: 1{,}610 - 412.50 = 1{,}197.50 left after ingredients; 1{,}197.50 - 957 = 240.50 ✓.

Tip: In multi-step money problems, list every source of income and every cost before computing. Watch for amounts that apply "each" — here the hourly pay applies to each of 3 workers.

1. x-intercept (set y = 0): 6x - 5(0) = 30 \Rightarrow 6x = 30 \Rightarrow x = 5. Point (5, 0).
2. y-intercept (set x = 0): 6(0) - 5y = 30 \Rightarrow -5y = 30 \Rightarrow y = -6. Point (0, -6).
3. Plot (5, 0) and (0, -6) and draw the line through them.

The students should plot (5, 0) and (0, -6).

Check: 6(5) - 5(0) = 30 ✓ and 6(0) - 5(-6) = 0 + 30 = 30 ✓. The slope from (0, -6) to (5, 0) is \tfrac{6}{5}, matching y = \tfrac{6}{5}x - 6 (the equation solved for y) ✓.

Tip: The intercept method is the fastest way to graph an equation in standard form Ax + By = C. An x-intercept always has y = 0, so it is written (\text{number}, 0) — do not swap the coordinates.

1. Let a = adult tickets and s = student tickets.
   • Tickets: a + s = 412
   • Money: 8a + 5s = 2{,}555
2. From the first equation, a = 412 - s.
3. Substitute: 8(412 - s) + 5s = 2{,}555 \Rightarrow 3{,}296 - 8s + 5s = 2{,}555.
4. Simplify: 3{,}296 - 3s = 2{,}555 \Rightarrow -3s = -741 \Rightarrow s = 247.
5. Then a = 412 - 247 = 165.

The school sold 247 student tickets (and 165 adult tickets).

Check: 165 + 247 = 412 tickets ✓. 8(165) + 5(247) = 1{,}320 + 1{,}235 = 2{,}555 dollars ✓.

Tip: Solve for the variable the question asks about, or solve for both and double-check which is which. Plugging the answer choices into both equations is a fast backup on the GED.

1. Company A: rate $0.45 per mile, fixed fee $39.
2. Company B's rate (slope from the table): \dfrac{84 - 64}{100 - 50} = \dfrac{20}{50} = \$0.40 per mile.
3. Company B's fixed fee: 64 = 0.40(50) + b \Rightarrow 64 = 20 + b \Rightarrow b = \$44.
4. When are the costs equal? 0.45m + 39 = 0.40m + 44 \Rightarrow 0.05m = 5 \Rightarrow m = 100 miles. Both cost $84 there.
5. So Company B charges less per mile ($0.40 vs $0.45) but has a higher fixed fee ($44 vs $39), and the two cost the same at 100 miles.

Answer: Company B charges less per mile, and the two companies charge the same for a 100-mile trip.

Check: Company A at 100 miles: 0.45(100) + 39 = 45 + 39 = 84, which matches Company B's table value of $84 ✓. At 50 miles, A costs 22.50 + 39 = \$61.50 while B costs $64, so they are not equal at 50 miles.

Tip: To compare functions in different forms, pull the same two features out of each: rate of change (slope) and starting value (intercept). A table's intercept is not usually listed — work it out from the slope and one row.