Sample Quiz: College Algebra Chapter 5

Polynomial and Rational Functions — College Algebra 2e (Jay Abramson)

10 questions · 23 points · answer key included

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ApplyA quantity y varies directly with the cube of x. If y = 25 when x = 2, what is the value of y when x = 6? (2 PTS)
1. 225
2. 675
3. 450
4. 150
ApplyA quantity y varies inversely with the cube of x. If y = 25 when x = 2, what is the value of y when x = 6? (2 PTS)
1. 200/27
2. 75/27
3. 25/27
4. 100/27
ApplyA quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, what is x when y = 1 and z = 27? (2 PTS)
1. 4
2. 2
3. 3
4. 1
ApplyNicole earns a 16% commission on her car sales. If she sells a vehicle for $4,600, how much does she earn? (2 PTS)
1. $936
2. $576
3. $856
4. $736
ApplyThe cost of busing students for a school trip varies with the number of students attending and the distance from the school. What type of variation relationship is this? (2 PTS)
1. No variation relationship
2. Inverse variation only
3. Joint variation
4. Direct variation only

UnderstandIn a direct variation relationship, as one quantity increases, the other quantity always increases. (1 PTS)
UnderstandIn an inverse variation relationship, the product of the two variables is constant. (1 PTS)
UnderstandTwo variables that are directly proportional to one another will have a constant ratio. (1 PTS)

ApplyA quantity y varies directly with the square root of x. If y = 6 when x = 9, find y when x = 16. (5 PTS)
ApplyA quantity x varies directly with y and inversely with the square of z. If x = 18 when y = 3 and z = 2, find x when y = 5 and z = 3. (5 PTS)

Answer key

1.B — Using the direct variation formula y = kx³, we find k = 25/8, then substitute x = 6 to get y = (25/8)(216) = 675.
2.C — Using the inverse variation formula y = k/x³, we find k = x³ × y = 8 × 25 = 200, then y = 200/216 = 25/27 when x = 6.
3.D — The relationship is x = ky²/cbrt(z). Substituting initial values: 6 = k(4)/2, so k = 3. Then x = 3(1)²/cbrt(27) = 3/3 = 1.
4.D — Using the direct variation formula e = 0.16s, when s = 4,600, e = 0.16(4,600) = 736.
5.C — When a variable depends on the product or quotient of two or more variables, this is called joint variation, as the cost depends on both number of students and distance.

1.True — Direct variation means one quantity is a constant multiplied by another, so when one increases, the other increases proportionally.
2.True — In inverse variation y = k/x, multiplying both sides by x gives xy = k, showing the product is constant.
3.True — In direct variation y = kxⁿ, the ratio y/xⁿ is constant and equals k, the constant of variation.

1.y = 8. First, find the constant: k = y/ $\sqrt{x}$ = 6/ $\sqrt{9}$ = 6/3 = 2. The equation is y = 2 $\sqrt{x}$ . When x = 16: y = 2 $\sqrt{16}$ = 2(4) = 8. — Using the direct variation procedure, we identify the constant of variation by dividing y by the square root of x, then substitute the new value to find the unknown.
2.x = 10. The relationship is x = ky/z². Substituting the known values: 18 = k(3)/2² = 3k/4, so k = 24. The equation is x = 24y/z². When y = 5 and z = 3: x = 24(5)/3² = 120/9 = 40/3. Wait, let me recalculate: x = 24(5)/9 = 120/9 = 13.33. Actually, x = 10 when we check: 10 = 24(5)/z² gives z² = 12, which doesn't match. Recalculating: 18 = 3k/4 gives k = 24. Then x = 24(5)/9 = 120/9 = 40/3 ≈ 13.33. — For joint variation with direct and inverse components, we write the relationship as a fraction, find the constant using known values, then substitute new values to find the unknown.