Checking units, constraints, and reasonableness

Recognize. Convert both the distance and time units. Arrange the factors so the original units cancel.

Work. 72$\frac{\text{km}}{\text{h}}$ · $\frac{\text{1000 m}}{\text{1 km}}$ · $\frac{\text{1 h}}{\text{3600 s}}$ = 20$\frac{\text{m}}{\text{s}}$.

Answer: 20 meters per second.

Check. In 3,600 seconds, a speed of 20 meters per second covers 72,000 meters, or 72 kilometers.

Avoid the trap. Multiplying by 3,600 instead of dividing reverses the time conversion. Write the units in the factors to make their cancellation visible.

Recognize. Both dimensions of an area are affected by the length conversion.

Work. Since 1 m² = (100 cm)² = 10,000 cm², the panel’s area is 0.75(10,000) = 7,500 square centimeters.

Answer: 7,500 square centimeters.

Check. A 75-centimeter by 100-centimeter rectangle has area 7,500 square centimeters and also 0.75 m × 1 m = 0.75 m².

Avoid the trap. Multiplying by 100 gives 75, which is a linear conversion applied to a squared quantity. The exponent on the unit determines the exponent on the conversion factor.

Recognize. The requested decrease is measured relative to the new price, $100.

Work. The required decrease is $20, so the percent decrease is 100(20/100) = 20%.

Answer: 20%.

Check. Decreasing $100 by 20% gives $80. The original increase was 20/80 = 25%, which has a different base.

Avoid the trap. Equal dollar changes in opposite directions do not generally have equal percentage changes. State the denominator in words before dividing.

Recognize. Selection is restricted to the 30 students who walk.

Work. Of these 30 students, 12 are in grade 11. The probability is 12/30 = 2/5.

Answer: 2/5.

Check. The numerator is a subset of the denominator, and the result lies between 0 and 1.

Avoid the trap. 12/40 reverses the condition: it is the probability of walking among grade 11 students. 12/100 ignores the restriction to walkers.

Recognize. Let w be the width. The model is w(w + 5) = 84, and both side lengths must be positive.

Work. Factoring gives w² + 5w − 84 = (w + 12)(w − 7) = 0. Reject w = −12. The width is 7, so the requested length is 7 + 5 = 12.

Answer: 12 centimeters.

Check. The dimensions are 7 and 12, their difference is 5, and their product is 84.

Avoid the trap. Reporting 7 stops at the variable you chose rather than the quantity requested. Reporting a negative root fails the geometric domain even though it satisfies the polynomial equation.

Recognize. The right side must be nonnegative, so x ≥ 0. Squaring can introduce candidates that violate this sign restriction.

Work. Squaring gives x + 6 = x², so (x − 3)(x + 2) = 0. The candidates are 3 and −2. In the original equation, x = 3 gives 3 = 3, while x = −2 gives 2 = −2.

Answer: x = 3.

Check. The accepted solution satisfies both the original radicand condition and the nonnegative right-side condition.

Avoid the trap. The radicand at x = −2 is positive, but that alone does not make it a solution. The two original sides must also have equal signs and values.

Recognize. The original equation excludes x = 2. Multiplying by the denominator is valid only for allowed inputs.

Work. For x ≠ 2, multiplication by x − 2 gives x + 1 = 3, so the only candidate is x = 2. It is excluded. Therefore there are no real solutions.

Answer: 0.

Check. At every allowed input, subtracting the right side from the left gives (x − 2)/(x − 2) = 1, never 0.

Avoid the trap. A candidate produced after clearing denominators is not automatically valid. The original restriction survives every equivalent transformation on its domain.

Recognize. The multiplier after three years is 1.2³. The percent increase is based on the amount above the original, not the final multiple itself.

Work. The final multiplier is 1.2³ = 1.728. Subtract the original multiplier 1 and convert to a percent:

100(1.728 − 1) = 72.8%.

Answer: 72.8%.

Check. Starting from 100 gives 120, 144, and 172.8 after the successive years, an increase of 72.8.

Avoid the trap. The final amount is 172.8% of the original, not a 172.8% increase. Adding 20 + 20 + 20 also ignores the changing base.

Recognize. Tangent relates opposite to adjacent. The stated angle is in degrees, so numerical evaluation must use degrees.

Work. If the opposite leg is h, then tan 35° = h/12, giving h = 12 tan 35° ≈ 8.4025.

Answer: 8.4 centimeters.

Check. Because 35° < 45°, the opposite leg is shorter than the adjacent leg. Thus 0 < h < 12, consistent with the answer.

Avoid the trap. A calculator set to radians evaluates a different angle. A correct formula entered with the wrong angle unit is still the wrong computation. The bound helps flag a suspicious result but does not replace checking the mode.

Recognize. Capacity must cover every student. The count of buses is a positive integer satisfying 48b ≥ 275.

Work. The quotient 275/48 ≈ 5.729 shows that five buses are insufficient. Round upward to six buses.

Answer: 6 buses.

Check. Five buses carry at most 240 students, leaving 35 without seats. Six buses carry up to 288, which is enough.

Avoid the trap. Ordinary rounding is not the underlying rule. The required direction comes from the capacity inequality; the same kind of decimal in a maximum-budget problem may need to be rounded downward.

Recognize. The task asks for an estimate using a sample proportion, not a census count.

Work. The sample proportion is 150/240 = 5/8 = 0.625. Applying it to the school population gives 0.625(1,000) = 625.

Answer: An estimated 625 students.

Check. The sample represents 62.5% support, so the corresponding estimate should exceed half of 1,000 and be less than 1,000.

Avoid the trap. The calculation does not prove that exactly 625 students prefer the schedule. It also does not justify applying the estimate to a different school’s students. The numerical estimate and the strength of the conclusion must be kept distinct.

Recognize. The target is the completion rate among individuals in both groups. The group sizes are unequal, so the two percentages need different weights.

Work. The total number of completions is 10 + 45 = 55, out of 20 + 180 = 200 students. Thus the combined rate is 55/200 = 0.275 = 27.5%.

Answer: 27.5%.

Check. The group rates are 50% and 25%. The combined rate is much closer to 25% because the second group contains nine times as many students.

Avoid the trap. The unweighted average (50% + 25%)/2 = 37.5% gives equal weight to groups rather than to students, so it answers a different question.

Recognize. The unknown leg is opposite A, and AC is adjacent to A. Tangent uses the two legs, not the hypotenuse.

Work. The Pythagorean theorem gives BC = $\sqrt{{\text{13}}^{\text{2}}\text{−}{\text{5}}^{\text{2}}}$ = $\sqrt{\text{144}}$ = 12. Therefore tan A = BC/AC = 12/5.

Answer: 12/5.

Check. The opposite leg is longer than the adjacent leg, so A > 45° and its tangent exceeds 1. The result 12/5 is consistent.

Avoid the trap. 12/13 is sin A, not tan A. Correctly finding the missing side does not guarantee that the final ratio answers the question.

Recognize. A fixed perimeter does not fix the area. Express the area with one variable and find a bound.

Work. If one side is w, the other is 20 − w, where 0 < w < 20. Then A = w(20 − w) = 100 − (w − 10)² ≤ 100.

Answer: C, A ≤ 100.

Check. A 10 by 10 square gives area 100, while an 8 by 12 rectangle gives 96. Both satisfy the perimeter, disproving a single fixed area.

Avoid the trap. Choosing one convenient rectangle can show what is possible, not what must happen for every rectangle. The squared expression establishes the universal upper bound.

Recognize. The width must satisfy both a budget inequality and an area inequality. The target is the area, not the width. Start with the simpler constraint.

Work. The perimeter is 2w + 2(w + 3) = 4w + 6, so the budget gives

7(4w + 6) ≤ 350 w ≤ 11.

The area condition is w(w + 3) ≤ 150. Test the largest width allowed by the budget: at w = 11, the area is 11(14) = 154, which is too large. At w = 10, the area is 10(13) = 130, which is allowed. For positive w, both w and w + 3 increase as w increases, so the area increases. Thus the greatest feasible area occurs at w = 10.

Answer: 130 square meters.

Check. The dimensions 10 and 13 have perimeter 46, so fencing costs 46(7) = 322 dollars. The area 130 satisfies its limit. The next width, 11, fits the fencing budget exactly but violates the area limit.

Alternative. Solving w² + 3w − 150 ≤ 0 gives the positive boundary (−3 + $\sqrt{\text{609}}$)/2 ≈ 10.839. The greatest feasible integer is 10, not 11. Testing adjacent integers is enough once the simpler budget restricts the candidates.

Avoid the trap. Checking only one constraint would accept an invalid garden. Rounding a real boundary to the nearest integer would also fail. Finally, entering 10 would report the width instead of the requested area.