📝 Inverse proportion functions (12 MCQs)

📖 From Calculus • 1. Basics before calculus • 12 questions available

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What is Inverse proportion functions?

Definition:
Inverse proportion functions express a relationship where one variable is inversely proportional to another, typically written as $y = k/x$ or $y = kx^{-1}$ , meaning as $x$ increases, $y$ decreases, and vice versa, with constant product $xy = k$ .

Example:
If $y = 12/x$ , then when $x=3$ , $y=4$ ; when $x=6$ , $y=2$ , showing the product $3 \cdot 4 = 6 \cdot 2 = 12$ .

Reason:
This models real-world situations like speed and time for a fixed distance, or pressure and volume in Boyle's law, highlighting trade-offs.

Easy

Medium

Hard

📝 All Inverse proportion functions MCQs

Q1. Which of the following best defines an inverse proportion between variables $x$ and $y$ ?

y = kx

for some constant

k

y = \dfrac{k}{x}

for some positive constant

k

✅

y = k - x

y = k + x

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The definition of inverse proportionality states that the product $xy$ is a constant $k>0$ , which can be written as $y = k/x$ . Option B captures this relationship, while the other options describe direct or linear relations, not the characteristic reciprocal behavior of inverse proportion.

Q2. In a data set the product $x y$ is constantly $20$ . If the value of $x$ is changed from $4$ to $8$ , what is the new value of $y$ ?

A.10

B.5

C.2.5 ✅

D.20

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Since the product $xy$ remains $20$ , we solve $8y = 20$ giving $y = 20/8 = 2.5$ . This demonstrates how the inverse relationship forces the second variable to adjust inversely to maintain the constant product, confirming the answer choice C.

Q3. Suppose $y$ is inversely proportional to $x$ and $y=3$ when $x=5$ . Which statement correctly describes the effect on $y$ if $x$ is increased by 50%?

A.y becomes 2 ✅

B.y becomes 2.5

C.y becomes 2.0

D.y becomes 2.4

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: First find the constant: $k = xy = 3 \times 5 = 15$ . Increasing $x$ by 50% gives $x' = 1.5 \times 5 = 7.5$ . The new $y$ is $k/x' = 15/7.5 = 2$ . Thus $y$ decreases to 2, matching option A.

Q4. Which of the following graphs represents an inverse proportional relationship between $x$ and $y$ ?

A.A straight line through the origin

B.A hyperbola in the first quadrant ✅

C.A parabola opening upward

D.A horizontal line

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: An inverse proportional relationship $y = k/x$ produces a rectangular hyperbola that lies in the first quadrant for positive $k$ . The hyperbolic shape distinguishes it from linear, parabolic, or constant graphs, making option B the correct representation.

Q5. Given two relationships: $y = \dfrac{6}{x}$ and $y = 2x$ . Which statement correctly compares the behavior of $y$ as $x$ grows large?

A.Both

y

values approach zero

B.The first

y

approaches zero while the second

y

grows without bound ✅

C.Both

y

values grow without bound

D.The first

y

grows without bound while the second

y

approaches zero

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For $y = 6/x$ , as $x \to \infty$ the term $6/x$ tends to zero. Conversely, $y = 2x$ increases linearly without bound. Therefore the first function approaches zero and the second diverges, which is described by option B.

Q6. Consider the function $y = \frac{k}{x}$ with $k>0$ . If the area under the curve between $x = a$ and $x = b$ (where $0<a<b$ ) is equal to the area of a rectangle with sides $k$ and $\ln(b/a)$ , which of the following statements is true?

A.The statement holds for any positive

k

✅

B.It holds only when

k=1

C.It holds only when

ab = 1

D.It never holds

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The integral $\int_a^b \frac{k}{x}\,dx = k\ln(b/a)$ gives the exact area under the curve. A rectangle with sides $k$ and $\ln(b/a)$ also has area $k\ln(b/a)$ . Hence the equality is true for every positive constant $k$ , confirming option A.

Q7. If the speed $v$ of a car is inversely proportional to the time $t$ it takes to travel a fixed distance, what happens to the speed when the travel time is halved?

A.Speed doubles ✅

B.Speed halves

C.Speed remains unchanged

D.Speed triples

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The relationship $v = k/t$ implies that halving $t$ multiplies $v$ by two, because the product $vt$ stays constant. Therefore the speed doubles, which corresponds to option A.

Q8. A photographer uses the relation $\text{Exposure} = \dfrac{k}{\text{Aperture}}$ . If the aperture is increased from f/2 to f/4, how does the exposure change?

A.It doubles

B.It halves ✅

C.It remains the same

D.It quarters

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Increasing the denominator from 2 to 4 doubles the denominator, so the overall fraction is halved. Consequently the exposure is reduced to one‑half of its original value, matching option B.

Q9. In Boyle’s law, pressure $P$ and volume $V$ are inversely proportional. If a gas initially at $P = 2 \text{ atm}$ and $V = 5 \text{ L}$ is compressed to $V = 3 \text{ L}$ , what is the new pressure (assuming temperature constant)?

A.1.2 atm

B.3.33 atm ✅

C.2.5 atm

D.4 atm

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Boyle’s law gives the constant $k = PV = 2 \times 5 = 10$ . With the new volume $V' = 3$ L, the pressure becomes $P' = k/V' = 10/3 \approx 3.33$ atm, which is option B.

Q10. A certain biological reaction rate $R$ is inversely proportional to the concentration $C$ of an inhibitor, i.e., $R = \frac{k}{C}$ . Given that when $C = 0.5$ mol/L the rate is 2 units, what is the highest concentration of inhibitor allowed while still keeping the reaction rate at least 0.8 units?

A.0.5 mol/L

B.0.625 mol/L

C.0.8 mol/L

D.1.25 mol/L ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: From the known condition, $k = R C = 2 \times 0.5 = 1$ . To maintain $R \ge 0.8$ , we need $1/C \ge 0.8$ → $C \le 1/0.8 = 1.25$ mol/L. Therefore the greatest permissible concentration is 1.25 mol/L, corresponding to option D.

Q11. If a set of measurements obeys $xy = 15$ and one measurement is $(x, y) = (3, 5)$ , which of the following pairs could NOT be part of the same data set?

A.(5, 3)

B.(1.5, 10)

C.(2, 7.5)

D.(4, 4) ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: All listed pairs except (4,4) satisfy the product condition $xy = 15$ . The pair (4,4) yields a product of 16, violating the constant‑product rule, so it cannot belong to the same data set. Hence option D is the correct choice.

Q12. Two variables $x$ and $y$ satisfy $y = \dfrac{9}{x}$ . If we plot $y$ against $x$ on a log‑log graph, what is the slope of the line?

A.-1 ✅

B.0

C.1

D.9

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Taking logarithms gives $\log y = \log 9 - \log x$ . This is a straight line with slope $-1$ (the coefficient of $\log x$ ). Therefore the log‑log plot has slope $-1$ , which corresponds to option A.

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