Direct and Inverse Proportion With Non-Linear Powers - Worksheets, Questions and Revision

11 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 4)Direct and Inverse Proportion With Non-Linear Powers revision guide

This topic is chapter 6 of IGCSE Maths Practice Book 1.

Next in this strand

2.5 Direct and Inverse Proportion With Non-Linear Powers

EDEXCEL 4MA1 · Calculator allowed · about 55 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer all questions. Full sentences are required in 2 questions (where an explanation is asked). Allow 60 minutes.

Given that T is proportional to v³ for a model where thrust T increases with the cube of speed v, and T = 54 N when v = 3 m/s, find the constant of proportionality k in T = kv³.

(Total for Question 1 is 1 mark)

Given that y is proportional to 1/√x (y = k / x^1/2) for cooling time y depending inversely on the square root of mass x, and y = 10 when x = 4, find k and state the proportionality equation.

(Total for Question 2 is 2 marks)

A circular pond has water depth y that varies directly as the square of the radius r, y = kr², in an experimental model. Given that y = 18 when r = 3 m, form the equation and hence find y when r = 5 m.

(Total for Question 3 is 2 marks)

A thin metal plate has heat loss rate H that varies as the square root of its area A, H = k A^1/2. An experiment gives H = 12 W when A = 16 cm². Find k and then find H when A = 100 cm².

(Total for Question 4 is 4 marks)

Given that y is proportional to x³, y = kx³, and y = 16 when x = 2, find k and then hence find x when y = 432.

(Total for Question 5 is 3 marks)

Show that if the period T of a simple pendulum is proportional to the square root of its length L, T = k L^1/2, and T = 1.8 s when L = 0.81 m, then k = 2. Verify the period when L = 2.25 m.

(Total for Question 6 is 3 marks)

A recipe says that the volume V of a loaf varies directly as the 3/2 power of the amount of yeast x used, V = k x^3/2. If using 8 g of yeast gives volume 1440 cm³, find k and then the volume when 2 g of yeast is used. Give the final volume to the nearest whole number.

(Total for Question 7 is 4 marks)

Show that if the period P of oscillation of a particle varies inversely as the square of amplitude a, P = k / a², and P = 0.5 s when a = 0.4 m, then k = 0.5*(0.4)² = 0.08. Hence find P when a = 0.2 m and explain briefly whether P increases or decreases when amplitude halves.

(Total for Question 8 is 4 marks)

A charity fundraiser states that the amount raised A varies directly as the square of the number of attendees n, A = k n². At a small event with n = 30 attendees, A = pound 2700. Use this to estimate how many attendees would be needed to raise pound 3840, assuming the same k. Give your answer to the nearest whole person and show your working in words.

(Total for Question 9 is 5 marks)

A physics experiment models intensity I of radiation as I = k / r³, inversely proportional to the cube of distance r. If I = 0.125 units at r = 2 m, find k and hence find r when I = 0.001 units. Give r to 2 decimal places.

(Total for Question 10 is 4 marks)

A scale model ship's resistance R varies as the square of its speed v and inversely as the cube root of its length L, R = k v² / L^1/3. In a test tank, R = 0.96 N when v = 3 m/s and L = 0.125 m. Find k and then find R when v = 5 m/s and L = 1.0 m. Give R to 3 significant figures.

(Total for Question 11 is 5 marks)

Mark scheme · 2.5 Direct and Inverse Proportion With Non-Linear Powers

Question 1

B1 k = 54/27 = 2 stated
Answer: k = 2

Question 2

M1 substitutes values into y = k/x^1/2: 10 = k/2 or equivalent
A1 k = 20 and equation y = 20 / x^1/2 or y = 20 x^-1/2 cao
Answer: k = 20, so y = 20 / x^1/2

Question 3

M1 finds k from 18 = k*3², k = 2
A1 y = 2*5² = 50 stated
Answer: y = 50 (depth units)

Question 4

M1 finds k from 12 = k*(16)^1/2 so k = 12/4 = 3
M1 uses k to get H = 3*(100)^1/2 = 3*10
A1 H = 30 W cao
B1 units W stated with final numeric answer
Answer: k = 3, H = 30 W

Question 5

M1 finds k from 16 = k*(2³) so k = 16/8 = 2
M1 substitutes k into 432 = 2*x³ and rearranges to x³ = 216
A1 x = 6 cao
Answer: k = 2 and x = 6

Question 6

M1 substitutes 1.8 = k*(0.81)^1/2 and finds √0.81 = 0.9 so k = 1.8/0.9 = 2
M1 uses k = 2 to find T = 2*(2.25)^1/2 and evaluates √2.25 = 1.5
A1 T = 2*1.5 = 3.0 s cao
Answer: k = 2 and T = 3.0 s when L = 2.25 m

Question 7

M1 substitutes 1440 = k*(8)^3/2. Recognises 8^1/2 = 2*√2 or better to compute 8^3/2 = (√8)³ = (2.828427...)³ = 22.627... or 8^3/2= (8¹)*(8¹/2)=8*2.828...
M1 calculates k = 1440 / 8^3/2 = 1440 / 22.627416997 = approx 63.64, or uses exact form k = 1440 / (8*√8) = 45*√2
M1 uses k to find V at x=2: V = k*(2)^3/2 and evaluates
A1 V = 180 cm³ (nearest whole number) cao
Answer: k approx 63.64 (exact 45*√2), V at 2 g = 180 cm³

Question 8

M1 substitutes to find k: 0.5 = k / 0.4² so k = 0.5 * 0.16 = 0.08
M1 uses k to find P at a = 0.2: P = 0.08 / 0.2² = 0.08 / 0.04 = 2
A1 P = 2.0 s cao
B1 brief explanation: when amplitude halves, denominator a² becomes one quarter, so P becomes four times larger, so P increases, oe
Answer: k = 0.08, P = 2.0 s; P increases (becomes 4 times larger) when amplitude halves

Question 9

M1 finds k from 2700 = k*30² so k = 2700/900 = 3
M1 sets 3840 = 3*n² and rearranges to n² = 1280
M1 takes square root to get n = √1280 and evaluates
A1 n approx 35.777... and rounded to nearest whole person gives 36 cao
B1 brief wording: states using the same k, n = √3840/3 approx 35.78, so 36 attendees needed, oe
Answer: n approx 35.78, so 36 attendees (nearest whole person)

Question 10

M1 finds k from 0.125 = k / 2³ so k = 0.125 * 8 = 1
M1 uses k = 1 and sets 0.001 = 1 / r³ to get r³ = 1000
M1 takes cube root to find r = 10, or evaluates r = 1000^1/3
A1 r = 10.00 m cao
Answer: k = 1 and r = 10.00 m

Question 11

M1 substitutes 0.96 = k*(3²) / (0.125)^1/3 and recognises 3² = 9 and cube root 0.125 = 0.5
M1 computes k = 0.96 * 0.5 / 9 = (0.48)/9 = 0.053333... or 0.0533333
M1 substitutes k into R = k*v² / L^1/3 with v=5 and L=1.0 giving R = k*25 / 1 = 25k
M1 calculates R = 25 * 0.053333... = 1.333333...
A1 R = 1.33 N to 3 s.f. cao
Answer: k = 0.053333... (16/300), R = 1.33 N (3 s.f.)

Direct and Inverse Proportion With Non-Linear Powers - Worksheets, Questions and Revision

Next in this strand

Related topics

2.5 Direct and Inverse Proportion With Non-Linear Powers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Mark your answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11