1st Class Maths Predicted Paper June 24 2H

Question 1

Write as a single fraction in its simplest form

Answer 1

1) common denominator: 6 & 8 is 24; x & x^3 is x^3; y & y^2 is y^2; therefore 24x^3y^2
2) 6x^3 to 24x^3y^2 is times by 4y
3) 8xy^2 to 24x^3y^2 is times by 3x^2
4) 7(4y) & 1(3x^2)
5) answer is 28y - 3x^2 over 24x^3y^2

Question 2

Here is some information about similar solids X, & Y. Complete the table

Answer 2

1) common height for X & Y is 6cm & 7.2cm respectively
2) scale factor is 7.2 / 6 which is = to 1.2
3) therefore length scale factor is 1.2
4) area scale factor is 1.2^2
5) & volume scale factor is 1.2^3
6) SA for X is 240cm^2; so SA for Y is 240 x 1.2^2 which is = to 345.6cm^2
7) Volume for Y is 432cm^3; so SA for X is 432 / 1.2^3 which is = to 250cm^3

Question 3

Muammar recorded how many minutes he spent revising each day for 15 days. Complete the table

Answer 3

1) median is the (15 + 1) / 2 value; so 8th value
2) LQ is the (15 + 1) / 4 value; so 4th value
3) UQ is the ((15 + 1) / 4) x 3 value; so 12th value

Question 4

The graph of y = f(x) is shown on the grid below. Draw the graph of y = f(x - 1) + 2 onto the grid above

Answer 4

1) -1 reperesents moving one to the right
2) +2 represents moving two units up

Question 5

Point A(1, -3) is on the graph y = f(x). When the graph is transformed to the graph with equation y = f(-x) the point A is mapped to point B. Write the coordinates of point B.

Answer 5

1) f(-x) is a reflection in the y axis
2) so (-1,-3)

Question 6

Here is a solid metal cylinder. The mass of the cylinder is 8300kg (correct to 2 significant figures). The height of the cylinder = 2.3m (correct to 1 decimal place). The radius of the cylinder = 0.7m (correct to 1 decimal place). Dylan calculates the density of the cylinder to be 2.9g/cm^3. Show clearly that Dylan must be incorrect

Answer 6

1) volume of a cylinder is π x r^2 x h
2) density is mass / volume
3) we want to find to upper (maximum) bound of the density, so we divide the upper bound of the mass by the lower bound of the volume
4) UB of mass is 8300 + 50 = 8350kg = 8350000g
5) LB of height is 2.3 - 0.05 = 2.25
6) LB of radius is 0.7 - 0.05 = 0.65
7) therefore LB of volume is π x 0.65^2 x 2.25, which you times by 100^3 to convert to cm^3
8) UB of density = 8350000 / ((π x 0.65^2 x 2.25) x 100^3) = 2.79…
9) 2.79… is the maximum density
10) Dylan is wrong as 2.9 > 2.79…

Question 7

A, B & C are points on the circumference of a circle, centre O. Work out the area of triangle ABC. Give your answer to 2 decimal places

Answer 7

1) extend to make a cyclic quadrilateral
2) angle at circumference is half than at centre so 130 / 2 = 65
3) Angle ABC = 180 - 65 = 115; as opposite angles in a cyclic quadrilateral add up to 180
4) Angle BAC = 180 - 115 - 20 = 45
5) use sine rule to find missing length
6) area of non-right angled triangle to find area

Question 8

Here are the first 4 terms of a quadratic sequence -> 387 k 892 1167; work out the value of k

Answer 8

1) find 1st difference by taking away last number by penultimate number & so on
2) therefore 1st differences in ascending order are k - 387, 892 - k, & 275
3) find the second difference by using the same method as to find the first difference
4) therefore 2nd differences in ascending order are (remember to put in brackets): (892 - k) - (k - 387) & (275) - (892 - k)
5) expanding the first bracket, we get 892 - k - k + 387 = 1279 - 2k
6) expanding the second bracket, we get 275 - 892 + k = k - 617
7) second differences are equal, so solve for k
8) k = 632

Question 9

Answer 9

1) fg(x) = (x - 3)^2 - 9 = x^2 - 6x
2) h^-1(x) = 3x + 8
3) solve the inequality & use quadratic formula to solve for x
4) 3/7 < x < 8