Mixed practice: Problem-Solving Strategies and Challenging Applications

8 hours

Undo the tax on the entire subtotal: 79.92/1.08 = 74. The hourly charges total 74 − 18 = 56, so the rental lasted 56/7 = 8 hours. The complete model is 1.08(18+7h) = 79.92. A final check gives 1.08(18+56) = 79.92. Subtracting the 18-dollar pretax fee from the taxed total would produce a different, incorrect model. See Example 1.01.

43

The requested expression is 3(5a − 2b) + 4. Replace the known group by 13: 3(13) + 4 = 43. There is no unique value of a or b to discover, but the target is fixed by the given relationship. One possible check is a = 3, b = 1, giving 45 − 6 + 4 = 43. See Example 3.02.

B, 0.84p

A 30% decrease multiplies by 0.70. The later 20% increase multiplies the new value by 1.20. Thus the final price is p(0.70)(1.20) = 0.84p. Testing a positive original price of 100 gives 100 70 84, eliminating the other choices. Adding −30% and +20% incorrectly treats both changes as percentages of the original price. See Example 2.02.

B, 12

The parabola’s axis is x = 6. Distinct inputs with the same output lie equally far from that axis, so their average is 6 and their sum is 12. Algebra confirms this: (r − 6)² = (s − 6)² gives (r − s)(r + s − 12) = 0. Since r ≠ s, r + s = 12. The vertical shift −20 does not change horizontal symmetry. See Example 3.12.

4

Rearrange to (k − 4)x = −6. For k = 4, this becomes 0 = −6, which is impossible. Every other value of k gives one solution, x = −6/(k − 4). The parameter must be classified before division by k − 4; otherwise the no-solution case disappears from the calculation. See Example 4.02.

3/4

The adjacent leg has length $\sqrt{{\text{15}}^{\text{2}}\text{−}{\text{9}}^{\text{2}}}$ = $\sqrt{\text{144}}$ = 12. Tangent is opposite divided by adjacent, so tan θ = 9/12 = 3/4. Using 9/15 would find sine instead. The answer being less than 1 is reasonable because the opposite leg is shorter than the adjacent leg. See Example 5.13.

4/7

The phrase “from those who borrowed fiction” restricts the sample space to 42 students. Of these, 24 are in the first year. The required conditional probability is 24/42 = 4/7. The fraction 24/60 instead asks what fraction of first-year students borrowed fiction, reversing the conditioning. The fraction 24/100 describes a joint event in the entire group. See Example 5.04.

120 minutes

Convert the volume first: 1.8 m³ = 1,800 liters. At 15 liters per minute, filling time is 1,800/15 = 120 minutes. A dimensional check is (1,800 L)/(15 L/min) = 120 min. The conversion is supplied for volume; do not cube 1,000 again. See Example 5.01.

B, 6

Zero cannot satisfy the given equation because its constant term is 7. Therefore division by x is legal: x² − 6x + 7 = 0 becomes x − 6 + 7/x = 0. Hence x + 7/x = 6. Computing the two roots 3 ± $\sqrt{\text{2}}$ adds work without changing the target, which has the same value at both roots. See Example 3.06.

B, 48 kilometers per hour

The travel times are 30/40 = 3/4 hour and 30/60 = 1/2 hour. Total distance divided by total time gives 60/(5/4) = 48 kilometers per hour. The midpoint of the speeds, 50, is not correct because the slower segment takes longer. Equal distances do not mean equal time weights. See Example 1.06.

7

At a = 2, the quadratic term vanishes and 6x + 3 = 0 has one solution. For a ≠ 2, a repeated quadratic root requires 6² − 4(a − 2)(3) = 0. Thus 60 − 12a = 0 and a = 5. At that value the equation is 3(x + 1)² = 0. The sum of all qualifying parameter values is 2 + 5 = 7. See Example 4.11.

144

Combining eight equal spheres multiplies volume by 8, so the radius multiplies by $\sqrt[\text{3}]{\text{8}}$ = 2. The new radius is 6 centimeters. Its surface area is 4π(6²) = 144π, so A = 144. Alternatively, set $\frac{\text{4}}{\text{3}}$πR³ = 8($\frac{\text{4}}{\text{3}}$π · 3³). Volume is conserved, but surface area is not: eight original surface areas would total 288π, not the new 144π. See Example 1.12.

C, 7

The observations give ab = 12 and ab³ = 108. Their ratio eliminates a: b² = 9. Since b > 0, b = 3, and then a = 4. Because f(0) = a, the requested result is 4 + 3 = 7. A check gives 4(3) = 12 and 4(27) = 108. Reporting either constant alone misses the composite target. See Example 4.10.

42 meters

Let width be w > 0 and length w +5. The area equation is w(w +5) = 104, or (w +13)(w −8) = 0. Only w = 8 is a physical width, so the length is 13. The perimeter is 2(8 + 13) = 42 meters. Both the positive-dimension condition and the requested quantity matter; neither the algebraic root −13 nor the valid width 8 is the final answer. See Example 5.05.

8

Substitute x = 1 into the identity to obtain 3(a + b) = 5 + 13 + 6 = 24. Hence a + b = 8. Coefficient comparison provides a check: a = 5, 2b = 6, so b = 3; the middle coefficient is 2a + b = 13. Choosing x = −2 gives only 0 = 0, which cannot determine the target. See Example 2.15.

25

Let the original positive value be P. The final value is 1.20P(1 − r/100) = 0.90P. Cancel P and divide by 1.20 to get 1 − r/100 = 0.75. Therefore r = 25. With P = 100, the process is 100 120 90; the second change is 30/120 = 25%, not 30%. See Example 2.13.

30

Add the two equations exactly as written: (x − y) + (2x + 3y) = 5 + 25. This directly produces the requested expression 3x + 2y = 30. Finding x = 8 and y = 3 is valid but unnecessary. The most useful elimination operation is sometimes the one that creates the target, not the one that removes every unrequested variable. See Example 3.03.

B, an estimate of 2,832

The sample proportion is 118/200 = 0.59. Applying it to the sampled population gives an estimated count of 0.59(4,800) = 2,832. Random sampling supports an estimate for this town’s registered voters, not exact knowledge of every voter, a claim about neighboring towns, or a causal conclusion about changed opinions. The word “estimate” is essential, not a hedge to discard. See Example 5.11.

57

The root sum is r + s = 9 and the root product is rs = 12. Consequently r² + s² = (r + s)² −2rs = 81−24 = 57. The discriminant is 81 − 48 = 33 > 0, so the described real roots exist. There is no need to calculate (9 ± $\sqrt{\text{33}}$)/2 individually. See Example 3.09.

−32

The center is (4, −2) and the radius is 6. Horizontal tangents touch the top and bottom of the circle, so their heights are k = −2 + 6 = 4 and k = −2 − 6 = −8. Their product is −32. A check sets x = 4 in the circle equation and obtains (y + 2)² = 36. Do not read the center’s y coordinate as +2. See Example 4.09.

C, {5}

The square root is nonnegative, so any solution must satisfy x ≥ 0. Squaring gives 3x + 10 = x², or (x − 5)(x + 2) = 0. The candidates are 5 and −2, but only 5 can meet the sign requirement. Direct checks give $\sqrt{\text{25}}$ = 5 and $\sqrt{\text{4}}$ = 2 ≠ −2. Squaring generates candidates, not automatically valid solutions. See Example 2.14.

55 square centimeters

The perimeter constraint is 2[w + (2w + 1)] ≤ 35, or 6w + 2 ≤ 35. Thus w ≤ 5.5, and integer width forces w ≤ 5. At w = 5, the length is 11, area is 55, and perimeter is 32; both constraints hold. Area increases for positive integer widths, so no smaller width can improve it. Width 6 would have allowable area 78 but forbidden perimeter 38, illustrating why both conditions must be checked. See Example 5.15.

24

Totals, rather than unweighted means, combine: 12(74) + 86n = 82(12 + n). This simplifies to 888 + 86n = 984 + 82n, so 4n = 96 and n = 24. Check: (888 + 2,064)/36 = 82. Since 82 is closer to 86 than to 74, it is reasonable that the second group is larger. In fact, the distances 8 and 4 imply a 1-to-2 count ratio. See Example 6.07.

A, 0

The domain excludes x = 5. For allowed inputs, cancellation gives x + 5 = 2x − k, so the only candidate is x = k + 5. This candidate is excluded exactly when k = 0. For any nonzero k, the candidate is not 5 and satisfies the original equation. Canceling a factor simplifies the rule but does not fill the original hole. See Example 4.12.

B, 4 values

The leading coefficient is always 1. Two distinct real roots require Δ = 4(k − 1)(k − 3) > 0, or k < 1 or k > 3. Both roots are positive precisely when their sum 2k and product 4k − 3 are positive, provided they are real. Together these require k > 3/4. Thus the real-parameter ranges are 3/4 < k < 1 or k > 3. Among integers from −2 through 7, only 4, 5, 6, and 7 qualify. See Example 4.15.

17 liters

The salt amount is 10(0.30) + 15(0.50) = 10.5 liters in a total of 25 liters of solution. Pure water changes the denominator, not the salt amount. If w liters are added, 10.5/(25 + w) = 0.25, giving 25 + w = 42 and w = 17. The simple average of 30% and 50% is not the mixture’s concentration because the original volumes differ. See Example 1.07.

14

At intersections, x² − 4x + 7 = −x² + 8x − 5, so x² − 6x + 6 = 0. The two x coordinates sum to 6. This equation also gives x² = 6x − 6, so each corresponding y coordinate equals (6x − 6) − 4x + 7 = 2x + 1. Their sum is therefore 2(6) + 2 = 14. This avoids finding the individual coordinates. The positive discriminant 12 confirms two distinct intersections. See Example 3.03 and Example 3.06.

B, 10/3

Combine the fractions: a/b + b/a = (a² + b²)/(ab). The numerator is (a + b)² − 2ab = 64 − 24 = 40, and the denominator is 12. The ratio is 40/12 = 10/3. As a check, positive values 2 and 6 satisfy both given relationships and yield 1/3 + 3 = 10/3. The sum of two fractions is not obtained by adding their denominators. See Example 3.09.

C, 45 units

The area scale factor from PQR to XYZ is 75/12 = 25/4. The corresponding length scale factor is its positive square root, 5/2. Perimeter scales as a length, so the new perimeter is 18(5/2) = 45. Multiplying 18 directly by 25/4 would incorrectly apply an area factor to a one-dimensional measurement. See Example 2.03.

3 values

Solving |x − 3| = k gives candidates x = 3 − k and x = 3 + k. At k = 0, these coincide and give one point. For k = 1, 2, 3, both are in [0, 8], so there are two points; at k = 3, the left endpoint x = 0 is included. For k = 4, 5, only the right point remains in the domain. At k = 6, 7, neither does. Exactly one point occurs for k = 0, 4, 5, or three values. See Example 4.14.

68 inches

Let the positive frame width be w. Both dimensions gain 2w, so (10 + 2w)(16 + 2w) = 280. Expanding and simplifying gives w² + 13w − 30 = 0, or (w + 15)(w − 2) = 0. The physical width is 2 inches. Outside dimensions are 14 and 20, yielding perimeter 2(14 + 20) = 68 inches. Reporting 2 would answer a different question. See Example 1.08.

64

Divide the given population values: N(10)/N(4) = b⁶ = 640/80 = 8. The requested ratio is N(13)/N(1) = b¹² = (b⁶)² = 8² = 64. The common starting factor a cancels in both ratios. No annual growth-rate calculation or estimate of a is needed. See Example 1.09 and Example 3.11.

4/3

Let u = x². Then u² −10u +9 = (u −1)(u −9) = 0, so x = ±1 or ±3. The two positive roots are 1 and 3, whose reciprocal sum is 1 + 1/3 = 4/3. A structural alternative uses r² + s² = 10 and rs = 3; then (r + s)² = 16, so r + s = 4 and (r + s)/(rs) = 4/3. Positivity selects the positive square root. See Example 3.13.

5/9

Group A has 60(0.75) = 45 supporters, and group B has 90(0.40) = 36. The selection is made from all 45 + 36 = 81 supporters, not from all 150 people. The required probability is 45/81 = 5/9. The fraction 60/150 ignores support, while 45/150 describes joint membership and support without conditioning on support. See Example 1.10 and Example 5.04.

18

Equating the two y expressions yields x² + 8x + c − 1 = 0. No intersection means no real solution, so the discriminant must be negative: 64 − 4(c − 1) = 68 − 4c < 0. Thus c > 17, and the least possible integer is 18. At c = 17, the graphs are tangent, which is still one intersection and therefore not allowed. See Example 4.08.

B, $10

Let n be the number of 2-dollar increases. Predicted revenue is (40+2n)(300−10n) = 12,000+200n −20n² = −20(n − 5)² + 12,500. Its maximum occurs at n = 5, an allowed integer that leaves 250 predicted sales. The requested dollar increase is 2(5) = 10, not the number of increases 5, the new price 50, or the maximum revenue 12,500. See Example 1.15 and Example 6.01.