A Level Maths Short Paper B
Covers Vectors, Sampling and Data Presentation, Probability and the Binomial Distribution and 4 more.
Questions
Question 1 [3 marks]
Vectors
Points A and B have position vectors a = 3i + 2j and b = 9i - 4j relative to a fixed origin O.
Find the position vector of the point P that divides AB such that AP:PB = 1:2.
Question 2 [3 marks]
Kinematics
A cyclist accelerates uniformly from 4 m/s over a distance of 50 m, reaching speed v m/s, with a constant acceleration of 0.6 m/s^2.
Find v, giving your answer to 3 significant figures.
Question 3 [4 marks]
Forces and Newton's Laws
A resultant force of 18 N acts on a body of mass 3 kg, starting from rest.
Find the acceleration produced, and find the speed of the body after 5 seconds.
Question 4 [4 marks]
Sampling and Data Presentation
The regression line of y on x for a set of data is y = 3.2 + 1.5x, where x is the number of hours of sunshine and y is ice cream sales in hundreds of pounds, valid for 2 <= x <= 10.
Use the regression line to estimate the sales when x = 6, and interpret the gradient of the line in context.
Question 5 [4 marks]
The Normal Distribution and Hypothesis Testing
The random variable X ~ N(64, 5^2).
Find the values between which the middle 80% of the distribution lies, symmetric about the mean.
Question 6 [5 marks]
Moments
A uniform rod AB has length 5 m and weight 40 N. It is pivoted at a point C, 2 m from A. A weight of 25 N is hung from A.
Find the weight that must be hung from B for the rod to balance horizontally.
Question 7 [5 marks]
Probability and the Binomial Distribution
A random variable X ~ B(25, 0.16).
Find the mean and variance of X, and find P(X = 3), giving your answer to 3 significant figures.
Question 8 [6 marks]
Kinematics
A ball A is projected vertically upwards from ground level with speed 24.5 m/s. At the same instant, a ball B is dropped from rest from a point 40 m directly above the launch point of A. Using g = 9.8 m/s^2,
find the time at which the two balls collide, and find the height above the ground at which they collide.
Question 9 [6 marks]
Kinematics
A car travels along a straight road. It accelerates uniformly from speed u m/s to 20 m/s in 5 seconds, covering 62.5 m in this time. It then decelerates uniformly from 20 m/s to rest in a further 8 seconds.
Find the value of u, and find the total distance travelled by the car.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| finding AB = b - a = 6i - 6j | M1 |
| finding AP = (1/3)AB = 2i - 2j | M1 |
| OP = a + AP = 5i | A1 |
| Final answer: OP = 5i (or 5i + 0j) | |
| Question 2[3 marks] | |
|---|---|
| Answer or working | Marks |
| using v^2 = u^2 + 2as | M1 |
| substituting u = 4, a = 0.6, s = 50 to get v^2 = 76 | M1 |
| v = 8.72 m/s (3 sf) | A1 |
| Final answer: 8.72 m/s (3 sf) | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| using F = ma | M1 |
| a = 6 m/s^2 | A1 |
| using v = u + at | M1 |
| v = 30 m/s | A1 |
| Final answer: a = 6 m/s^2; v = 30 m/s | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| substituting x = 6 into the regression equation | M1 |
| y = 12.2 (sales of 1220 pounds) | A1 |
| interpreting the gradient: for each extra hour of sunshine, sales increase on average by 1.5 hundred pounds (150 pounds) | B1 |
| noting the estimate is reliable, since x = 6 lies within the given data range 2 to 10 | B1 |
| Final answer: y = 12.2 (sales of 1220 pounds); each extra hour of sunshine is associated with an average increase in sales of 150 pounds | |
| Question 5[4 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the z-value such that P(Z < z) = 0.9, i.e. z = 1.2816 | M1 |
| forming 64 - 1.2816 x 5 and 64 + 1.2816 x 5 | M1 |
| the lower value 57.6 (3 sf) | A1 |
| the upper value 70.4 (3 sf) | A1 |
| Final answer: The middle 80% lies between 57.6 and 70.4 (3 sf) | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| identifying the rod's centre of mass at its midpoint, 2.5 m from A, i.e. 0.5 m from C on the B side | M1 |
| taking moments about C: 25 x 2 = 40 x 0.5 + W x 3 | M1 |
| 50 = 20 + 3W | A1 |
| rearranging to 3W = 30 | M1 |
| W = 10 N | A1 |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| the mean = np = 25 x 0.16 | M1 |
| the mean = 4 and the variance = np(1 - p) = 25 x 0.16 x 0.84 = 3.36 | A1 |
| using the binomial formula C(25,3)(0.16)^3(0.84)^22 for P(X = 3) | M1 |
| evaluating to an unrounded value of approximately 0.2033 | A1 |
| P(X = 3) = 0.203 (3 sf) | A1 |
| Final answer: mean = 4, variance = 3.36; P(X = 3) = 0.203 (3 sf) | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| the height of A above the ground: h_A = 24.5t - 4.9t^2 | M1 |
| the height of B above the ground: h_B = 40 - 4.9t^2 | M1 |
| setting h_A = h_B and simplifying to 24.5t = 40 | M1 |
| t = 1.63 s (3 sf) | A1 |
| substituting t into h_B = 40 - 4.9t^2 | M1 |
| height = 26.9 m (3 sf) | A1 |
| Final answer: t = 1.63 s (3 sf); height above the ground = 26.9 m (3 sf) | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| using s = ((u + v)/2) x t on the first stage: 62.5 = ((u + 20)/2) x 5 | M1 |
| u = 5 m/s | A1 |
| using s = ((u + v)/2) x t on the second stage: ((20 + 0)/2) x 8 | M1 |
| the second stage distance = 80 m | A1 |
| the total distance = 62.5 + 80 | M1 |
| the total distance = 142.5 m | A1 |
| Final answer: u = 5 m/s; total distance = 142.5 m | |