IGCSE Additional Mathematics (0606)
Straight-line graphs Banker
13 questions · about 6% of the marks we indexed · examined in 8 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Linear form (log graphs) · Core · Paper 2 style · calculator · 4 marks- (a) The variables x and y are related by y = Abˣ, where A and b are constants. When lg y is plotted against x, a straight line is obtained that passes through (1, 0.9) and (5, 2.1). Find the value of A and the value of b, giving each correct to 2 significant figures. [4]
Show answer
- (a) A = 4.0, b = 2.0
lg y = lg A + x lg b B1; gradient lg b = 1.2 ÷ 4 = 0.3, b = 10^0.3 = 2.0 M1 A1; intercept lg A = 0.9 − 0.3 = 0.6, A = 10^0.6 = 4.0 A1
Q2Linear form (log graphs) · Core · Paper 1 style · no calculator · 4 marks- (a) When ln y is plotted against x, a straight line is obtained passing through the points (2, 5) and (6, 13). Find ln y in terms of x. [2]
- (b) Hence express y in terms of x. [2]
Show answer
- (a) ln y = 2x + 1
gradient = 8 ÷ 4 = 2 M1; ln y = 2x + 1 A1
- (b) y = e^(2x + 1) = e·e^(2x)
y = e^(2x+1) M1 A1
Q3Linear form (log graphs) · Core · Paper 1 style · no calculator · 5 marks- (a) The variables x and y are such that when xy is plotted against x², a straight line is obtained passing through (1, 7) and (5, 19). Express y in terms of x. [4]
- (b) Find the value of y when x = 2. [1]
Show answer
- (a) y = 3x + 4/x
gradient = 12 ÷ 4 = 3 M1; xy = 3x² + 4 A1; divide by x M1; y = 3x + 4/x A1
- (b) y = 8
6 + 2 = 8 B1
Q4Linear form (log graphs) · Core · Paper 2 style · calculator · 5 marks- (a) The variables x and y are related by y = Axⁿ. When ln y is plotted against ln x, a straight line of gradient −0.5 is obtained, passing through the point (2, 3). Find the value of n and the exact value of A. [3]
- (b) Find the value of y when x = 4, giving your answer correct to 3 significant figures. [2]
Show answer
- (a) n = −0.5, A = e⁴
ln y = n ln x + ln A B1; n = −0.5 B1; ln A = 3 + 0.5(2) = 4, A = e⁴ B1
- (b) y = 27.3
y = e⁴ × 4^(−0.5) M1 = 27.3 A1
Q5Linear form (log graphs) · Core · Paper 1 style · no calculator · 5 marks- (a) When lg y is plotted against lg x, a straight line is obtained passing through (1, 4) and (3, 0). Express y in terms of x. [4]
- (b) Find the value of y when x = 100. [1]
Show answer
- (a) y = 10⁶x⁻²
gradient = −4 ÷ 2 = −2 M1; lg y = −2 lg x + 6 A1; y = 10⁶ × x⁻² M1 A1
- (b) y = 100
10⁶ ÷ 10⁴ = 100 B1
Q6Linear form (log graphs) · Core · Paper 1 style · no calculator · 4 marks- (a) The variables x and y are such that when √y is plotted against x, a straight line is obtained passing through (0, 2) and (3, 8). Express y in terms of x. [3]
- (b) Find the value of y when x = 1.5. [1]
Show answer
- (a) y = (2x + 2)²
gradient = 6 ÷ 3 = 2 M1; √y = 2x + 2 A1; y = (2x + 2)² A1
- (b) y = 25
(3 + 2)² = 25 B1
Q7Linear form (log graphs) · Standard · Paper 2 style · calculator · 6 marks- (a) The variables x and y are related by y = Axⁿ, where A and n are constants. When lg y is plotted against lg x, a straight line is obtained that passes through the points (0.4, 1.3) and (1.2, 2.9). Find the value of n and the value of A, giving A correct to 3 significant figures. [4]
- (b) Find the value of y when x = 5. [2]
Show answer
- (a) n = 2, A = 3.16
lg y = n lg x + lg A B1; n = (2.9 − 1.3) ÷ (1.2 − 0.4) = 2 M1 A1; lg A = 1.3 − 2(0.4) = 0.5, A = 10^0.5 = 3.16 A1
- (b) y = 79.1
y = 3.162 × 5² M1 = 79.1 A1 (3 s.f.)
Q8Linear form (log graphs) · Standard · Paper 2 style · calculator · 7 marksThe table shows experimental values of x and y, which are believed to be related by y = kxⁿ, where k and n are constants.x: 2, 4, 6, 8, 10y: 7.1, 20.0, 36.7, 56.6, 79.1
- (a) Calculate the values of lg x and lg y for each pair of readings, giving values correct to 2 decimal places. [2]
- (b) Plot lg y against lg x and draw a straight line. Use your graph to estimate the value of n and the value of k. [4]
- (c) Estimate the value of y when x = 5. [1]
Show answer
- (a) lg x: 0.30, 0.60, 0.78, 0.90, 1.00; lg y: 0.85, 1.30, 1.56, 1.75, 1.90
B2 all correct (B1 for at least 7 correct values)
- (b) n ≈ 1.5, k ≈ 2.5 (accept n 1.45–1.55, k 2.4–2.6)
lg y = n lg x + lg k B1; gradient = (1.90 − 0.85) ÷ (1.00 − 0.30) ≈ 1.5 M1 A1; intercept lg k ≈ 0.40, k ≈ 2.5 A1
- (c) y ≈ 28 (accept 27–29)
2.5 × 5^1.5 = 27.95 B1
Q9Linear form (log graphs) · Standard · Paper 2 style · calculator · 6 marksIn a laboratory in Karachi the number N of bacteria in a sample is counted every 2 hours. It is believed that N = N₀e^(kt), where t is the time in hours and N₀ and k are constants.t: 0, 2, 4, 6, 8N: 150, 302, 608, 1225, 2467
- (a) Explain why a graph of ln N against t should be a straight line. [1]
- (b) Using the first and last readings, estimate the value of k and the value of N₀. [3]
- (c) Estimate the time at which the number of bacteria first reaches 5000. Give your answer correct to 1 decimal place. [2]
Show answer
- (a) ln N = ln N₀ + kt, which has the form Y = mX + c
B1 taking logs correctly
- (b) k ≈ 0.35, N₀ ≈ 150
gradient = (ln 2467 − ln 150) ÷ 8 = 0.350 M1 A1; intercept ln N₀ = ln 150, N₀ = 150 A1
- (c) t ≈ 10.0 hours
ln 5000 = ln 150 + 0.35t M1; t = 10.0 A1
Q10Linear form (log graphs) · Standard · Paper 2 style · calculator · 6 marksThe value, V rupees, of a car in Lahore t years after it was bought is modelled by V = pqᵗ, where p and q are constants. When lg V is plotted against t, a straight line is obtained passing through (1, 6.42) and (5, 6.10).
- (a) Find the value of p and the value of q, giving p correct to 3 significant figures and q correct to 3 decimal places. [4]
- (b) Find the age of the car when its value is Rs 1 500 000. Give your answer correct to 1 decimal place. [2]
Show answer
- (a) q = 0.832, p = 3 160 000 (3.16 × 10⁶)
lg V = lg p + t lg q B1; lg q = (6.10 − 6.42) ÷ 4 = −0.08 M1, q = 0.832 A1; lg p = 6.42 + 0.08 = 6.50, p = 3.16 × 10⁶ A1
- (b) t = 4.0 years
lg 1 500 000 = 6.50 − 0.08t M1; t = 4.0 A1
Q11Linear form (log graphs) · Standard · Paper 2 style · calculator · 7 marksIt is thought that the variables x and y are related by y = a ln x + b. Values recorded are:x: 1, 2, 4, 8, 16y: 2.0, 4.08, 6.16, 8.24, 10.32
- (a) State which two variables should be plotted to obtain a straight line, and say how a and b can be found from the line. [2]
- (b) Find the values of a and b. [3]
- (c) Find the value of x for which y = 9, correct to 3 significant figures. [2]
Show answer
- (a) Plot y against ln x; gradient = a, intercept on the y-axis = b
B1 variables; B1 gradient a, intercept b
- (b) a = 3, b = 2
ln x values 0, 0.693, 1.386, 2.079, 2.773 M1; gradient = (10.32 − 2.0) ÷ 2.773 = 3.00 A1; intercept b = 2 A1
- (c) x = 10.3
ln x = 7/3 M1; x = 10.3 A1
Q12Linear form (log graphs) · Challenge · Paper 2 style · calculator · 7 marksThe table shows values of x and y which are related by y = a ÷ (x + b), where a and b are constants.x: 1, 2, 3, 4, 5y: 3.0, 2.4, 2.0, 1.714, 1.5
- (a) Show that, if y = a ÷ (x + b), then a graph of 1/y against x is a straight line, and state the gradient and intercept in terms of a and b. [2]
- (b) Use the table to find the value of a and the value of b. [4]
- (c) Find the value of x for which y = 0.8. [1]
Show answer
- (a) 1/y = (1/a)x + b/a; gradient 1/a, intercept b/a
invert: 1/y = (x + b)/a M1; gradient 1/a, intercept b/a A1
- (b) a = 12, b = 3
1/y values: 0.333, 0.417, 0.5, 0.583, 0.667 M1; gradient = 1/12 so a = 12 M1 A1; intercept 0.25 = b/12 so b = 3 A1
- (c) x = 12
12 ÷ (x + 3) = 0.8 gives x + 3 = 15, x = 12 B1
Q13Linear form (log graphs) · Challenge · Paper 2 style · calculator · 7 marks- (a) The variables x and y are related by y = Ae^(bx²). When ln y is plotted against x², a straight line passes through (1, 2.2) and (4, 0.7). Find the value of b and the value of A, giving A correct to 3 significant figures. [4]
- (b) Find the positive value of x for which y = 5, giving your answer correct to 3 significant figures. [3]
Show answer
- (a) b = −0.5, A = 14.9
ln y = ln A + bx² B1; b = (0.7 − 2.2) ÷ 3 = −0.5 M1 A1; ln A = 2.7, A = e^2.7 = 14.9 A1
- (b) x = 1.48
ln 5 = 2.7 − 0.5x² M1; x² = 2.181 A1; x = 1.48 A1
Q14Linear form (log graphs) · Challenge · Paper 2 style · calculator · 7 marksA space agency records the orbital radius r (in millions of km) and the orbital period T (in days) of five moons of a distant planet. It is believed that T = krⁿ.r: 0.5, 1.0, 1.5, 2.0, 3.0T: 1.13, 3.2, 5.88, 9.05, 16.63
- (a) By plotting lg T against lg r, find the value of n and the value of k, each correct to 2 significant figures. [5]
- (b) Another moon of the same planet has an orbital period of 12 days. Estimate its orbital radius. [2]
Show answer
- (a) n = 1.5, k = 3.2
T = kr^n so lg T = n lg r + lg k B1; lg r: −0.301, 0, 0.176, 0.301, 0.477; lg T: 0.053, 0.505, 0.769, 0.957, 1.221 M1; gradient ≈ 1.5 M1 A1; intercept 0.505 so k = 3.2 A1
- (b) r ≈ 2.4 × 10⁶ km
12 = 3.2r^1.5 M1; r = 3.75^(2/3) = 2.41 A1
Q15Lines, midpoints and distances · Core · Paper 1 style · no calculator · 5 marks- (a) The points A and B have coordinates (−3, 4) and (5, −2). Find the coordinates of the midpoint of AB. [1]
- (b) Find the length of AB. [2]
- (c) Find the equation of the line AB in the form ax + by = c, where a, b and c are integers. [2]
Show answer
- (a) (1, 1)
B1
- (b) 10
√(8² + 6²) M1 = 10 A1
- (c) 3x + 4y = 7
gradient −6/8 = −3/4 M1; y − 4 = −¾(x + 3), 3x + 4y = 7 A1
Q16Lines, midpoints and distances · Core · Paper 1 style · no calculator · 3 marks- (a) The point P has coordinates (2, k) and the point Q has coordinates (8, 1). Given that PQ = 10 units, find the possible values of k. [3]
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- (a) k = 9 or k = −7
6² + (k − 1)² = 100 M1; (k − 1)² = 64 A1; k = 9 or −7 A1
Q17Lines, midpoints and distances · Core · Paper 2 style · calculator · 5 marksOn a map of a town near Islamabad, with axes in kilometres, a school is at S(2, 3) and a hospital is at H(10, 9). A straight road joins S and H.
- (a) Find the distance between the school and the hospital. [2]
- (b) A bus stop is to be placed at the midpoint M of SH. Find the coordinates of M. [1]
- (c) A post box T is on SH, one quarter of the way from S to H. Find the coordinates of T. [2]
Show answer
- (a) 10 km
√(8² + 6²) M1 = 10 A1
- (b) M(6, 6)
B1
- (c) T(4, 4.5)
S + ¼(8, 6) M1 = (4, 4.5) A1
Q18Lines, midpoints and distances · Standard · Paper 2 style · calculator · 5 marks- (a) Find the equation of the line l which passes through (4, −1) and is parallel to the line 2x − 5y = 3. [2]
- (b) The line l meets the x-axis at P and the y-axis at Q. Find the area of triangle OPQ, where O is the origin. [3]
Show answer
- (a) 2x − 5y = 13
gradient 2/5 B1; y + 1 = (2/5)(x − 4), 2x − 5y = 13 B1
- (b) 8.45 square units
P(6.5, 0), Q(0, −2.6) M1 A1; area = ½ × 6.5 × 2.6 = 8.45 A1
Q19Lines, midpoints and distances · Standard · Paper 1 style · no calculator · 5 marks- (a) The points A and B have coordinates (1, 2) and (7, 5). The point C lies on AB such that AC : CB = 2 : 1. Find the coordinates of C. [2]
- (b) Find the exact distance OC, where O is the origin. [1]
- (c) Find the equation of the line through C parallel to OA. [2]
Show answer
- (a) C(5, 4)
C = A + ⅔(B − A) = (1 + 4, 2 + 2) M1 A1
- (b) √41
√(25 + 16) B1
- (c) y = 2x − 6
gradient of OA = 2 B1; y − 4 = 2(x − 5) B1
Q20Lines, midpoints and distances · Standard · Paper 2 style · calculator · 5 marks- (a) The lines 3x + 2y = 12 and y = x − 4 intersect at the point P. Find the coordinates of P. [2]
- (b) Find the area of the triangle enclosed by the two lines and the y-axis. [3]
Show answer
- (a) P(4, 0)
3x + 2(x − 4) = 12 M1; x = 4, y = 0 A1
- (b) 20 square units
y-intercepts 6 and −4 B1; base 10, height 4 M1; area 20 A1
Q21Lines, midpoints and distances · Challenge · Paper 1 style · no calculator · 6 marksThe points A(−2, 1), B(4, 3), C(6, 9) and D(0, 7) are the vertices of a quadrilateral ABCD.
- (a) Show that ABCD is a rhombus. [3]
- (b) Find the area of ABCD. [3]
Show answer
- (a) All four sides equal √40 (and diagonals bisect each other at (2, 5))
AB = √(6² + 2²) = √40, BC = √(2² + 6²) = √40 M1; CD = √40, DA = √40 A1; all sides equal so rhombus A1
- (b) 32 square units
AC = 8√2, BD = 4√2 M1 A1; area = ½ × 8√2 × 4√2 = 32 A1
Q22Lines, midpoints and distances · Challenge · Paper 1 style · no calculator · 6 marksThe points A(−1, 2), B(5, 4) and C(3, −2) form a triangle.
- (a) Find the area of triangle ABC. [3]
- (b) Hence find the exact perpendicular distance from C to the line AB, giving your answer in the form k√10. [3]
Show answer
- (a) 16 square units
e.g. shoelace: ½|−1(4 + 2) + 5(−2 − 2) + 3(2 − 4)| M1 = ½ × 32 A1 = 16 A1
- (b) (8/5)√10
AB = √40 = 2√10 B1; ½ × 2√10 × h = 16 M1; h = 16/√10 = (8/5)√10 A1
Q23Perpendicular lines and bisectors · Core · Paper 1 style · no calculator · 4 marks- (a) Find the equation of the perpendicular bisector of the line joining A(2, 5) and B(6, −3). Give your answer in the form ax + by = c. [4]
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- (a) x − 2y = 2
midpoint (4, 1) B1; gradient AB = −2 B1; perpendicular gradient ½ M1; y − 1 = ½(x − 4), x − 2y = 2 A1
Q24Perpendicular lines and bisectors · Core · Paper 1 style · no calculator · 3 marks- (a) Find the equation of the line through (3, −2) which is perpendicular to 4x + 3y = 7. Give your answer in the form ax + by = c. [3]
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- (a) 3x − 4y = 17
gradient of given line −4/3 B1; perpendicular gradient 3/4 M1; 3x − 4y = 17 A1
Q25Perpendicular lines and bisectors · Standard · Paper 1 style · no calculator · 6 marks- (a) Find the coordinates of the foot of the perpendicular from P(7, 4) to the line y = 2x − 5. [4]
- (b) Hence find the exact shortest distance from P to the line. [2]
Show answer
- (a) (5, 5)
perpendicular gradient −½ B1; y − 4 = −½(x − 7) M1; solve with y = 2x − 5 M1; (5, 5) A1
- (b) √5
√(2² + 1²) M1 = √5 A1
Q26Perpendicular lines and bisectors · Standard · Paper 2 style · calculator · 4 marksThe points A(0, 4), B(6, 0) and C(10, 6) form a triangle.
- (a) Show that angle ABC is a right angle. [2]
- (b) Find the area of triangle ABC. [2]
Show answer
- (a) gradient AB = −2/3, gradient BC = 3/2, product −1
both gradients M1; product = −1 so perpendicular A1
- (b) 26 square units
AB = BC = √52 M1; area = ½ × 52 = 26 A1
Q27Perpendicular lines and bisectors · Standard · Paper 1 style · no calculator · 6 marks- (a) Find the equation of the perpendicular bisector of PQ, where P is (−1, 3) and Q is (5, 7). [3]
- (b) The perpendicular bisector meets the x-axis at R and the y-axis at S. Find the exact area of triangle ORS, where O is the origin. [3]
Show answer
- (a) y = −(3/2)x + 8 (or 3x + 2y = 16)
midpoint (2, 5) B1; perpendicular gradient −3/2 M1; 3x + 2y = 16 A1
- (b) 64/3 square units
R(16/3, 0), S(0, 8) B1 B1; area = ½ × 16/3 × 8 = 64/3 B1
Q28Perpendicular lines and bisectors · Challenge · Paper 1 style · no calculator · 5 marks- (a) Find the coordinates of the image of the point P(4, 5) after reflection in the line x + 2y = 4. [5]
Show answer
- (a) (0, −3)
perpendicular gradient 2 B1; line y = 2x − 3 M1; meets x + 2y = 4 at (2, 1) M1 A1; image = 2(2, 1) − (4, 5) = (0, −3) A1
Q29Perpendicular lines and bisectors · Challenge · Paper 1 style · no calculator · 6 marks- (a) The points A(1, 2) and C(7, 6) are opposite vertices of a square ABCD. Find the coordinates of B and D. [5]
- (b) Find the area of the square. [1]
Show answer
- (a) (2, 7) and (6, 1)
midpoint M(4, 4) B1; MC = (3, 2) M1; perpendicular vector (−2, 3) M1; B, D = (4 ∓ 2, 4 ± 3) A1 A1
- (b) 26 square units
side √(1² + 5²) = √26, area 26 B1
Q30Perpendicular lines and bisectors · Challenge · Paper 2 style · calculator · 8 marksThe line l₁ has equation y = 3x − 2. The point A is (8, 1). The line l₂ passes through A and is perpendicular to l₁, and meets l₁ at B. The point C lies on l₂ such that B is the midpoint of AC.
- (a) Find the coordinates of B. [4]
- (b) Find the coordinates of C. [2]
- (c) Find the length AC, correct to 3 significant figures. [2]
Show answer
- (a) B(1.7, 3.1)
l₂ gradient −1/3 B1; y − 1 = −⅓(x − 8) M1; solve with y = 3x − 2 M1; B(1.7, 3.1) A1
- (b) C(−4.6, 5.2)
C = 2B − A M1 A1
- (c) 13.3
AC = 2AB = 2√44.1 M1 = 13.3 A1