Em Mock Solutions

1

1.

The hard disk space in a computer is 80 gigabytes. If the average size of a document is 2.85 megabytes, find the number of documents that can be stored. Give your answer, in standard form. Solution: 80 × 109 ........... [M1] 2.85 × 106 ≈ 2.81 × 104.......... [A1]

Answer: ..............2.81 × 104..............[2] 2.

The numbers 98 and 168, written as the products of their prime factors, are 98 = 2 × 72,

168 = 23 × 3 × 7

Find (i) the largest integer which is a factor of both 98 and 168, (ii) the smallest positive integer value of m for which 98m is a multiple of 168. Solution: (i)

98 = 2 × 72 × 1 168 = 23 × 7 × 3 HCF = 2 × 7 × 1 = 14

(ii) 98m = LCM of 98 and 168 2 × 7 2 × m = 23 × 72 × 3 ⇒m =

23 × 7 2 × 3 2 × 72

= 22 × 3 = 12

CAO [1] Answer: (i) …..14…….……..... CAO [1] (ii) m = …12…….…..

2

3.

Solve the simultaneous equations. x + 2y = 4 …………… eqn 1 4x = y + 1 ……….. eqn 2 Solution: From eqn 2, y = 4x – 1 ………….. eqn 3 Sub. eqn 3 into eqn 1, x + 2(4x – 1) = 4

[M1] for any correct method to eliminate one variable

9x = 6 x= Sub. x =

2 3

[A1]

2 2 into eqn 3, y = 4( ) – 1 3 3

=1

2 3

[A1] 2 Answer: x = …… …...............….. 3

y = ….…1 4.

2 ..…..……...... [3] 3

The figures show two geometrically similar pyramids, A and B. The areas of the bases of pyramid A and B are 28 cm2 and 63 cm2 respectively. (i) Calculate the ratio of the height of pyramid A to the height of pyramid B. Each of these pyramids is made of 5 thin sheets of metal joined together using adhesive tapes. The mass of pyramid B is 729 g. Given that the mass of the adhesive tapes is negligible, calculate (ii) the mass of pyramid A.

28 cm2 Pyramid A

Solution: (i)

Pyramid B

28 63

height of pyramid A = height of pyramid B =

(ii)

63 cm2

2 3

[M1] [A1]

mass of pyramid A base area of A = mass of pyramid B base area of B

3

Mass of pyramid A =

28 × 729 63

= 324 g 2 Answer: (i) ……… ……………[2] 3

CAO (ii) …324…….…...…..g [1] 5.

Each interior angle of a regular polygon is 5 times that of each of its exterior angles. (i) Calculate the size of each exterior angle of the polygon. (ii) Hence, calculate the number of sides of the polygon. Solution: (i)

Let size of each exterior angle of the polygon be xº ⇒ each interior angle = 180º – xº (adj ∠ s on st. line) 180º – xº = 5xº

[M1]

6x = 180 x = 30

(ii)

[A1]

Number of sides =

360 30

= 12 Answer: (i) ………30…….……….° [2] CAO (ii) ……12…………...….... [1] 6.

(a) C is inversely proportional to b2 and b > 0. It is known that C = 40 for a particular value of b. Find the value of C when this value of b is doubled. (b) It is given that

x and y are in direct proportion. If the difference in the values of y

when x = 25 and when x = 4 is 42, express y in terms of x. Solution: (a) C =

k where k is a constant b2

Let b = m when C = 40 40 =

k m2

4

When b = 2m, C=

k

(2m )2

=

1⎛ k ⎞ ⎜ ⎟ 4 ⎝ m2 ⎠

=

1 × 40 4

= 10 (b) y = k x where k is a constant When x = 25, y = k 25 = 5k When x = 4, y = k 4 = 2k 5k – 2k = 42 ………………….. [M1] 3k = 42 k = 14 ∴ y = 14 x …………….……….. [A1] for correct final expression

Also accept if pupils starts with

x = ky

Answer: (a) …..10….........…..….. [2] (b) …. y = 14 x .…….... [2]

7.

(a) Express in set notation, as simply as possible, the subset shaded in the Venn diagram. ξ P Q

Answer: (a) …… P ∪ Q ' ……...…..[1] (b) There are 156 students in a tuition centre. All of them either study History, Geography or both subjects. 75 of them study both subjects and there are 9 more who study Geography than History. By drawing a Venn diagram, find the number of students who study History. Solution:

5

(b) ξ

H

G y

75

y+9

Let y be the number of students who took History only. y + y + 9 + 75 = 156 ............ [M1] y = 36 …………………[A1] Number of students who took History = 36 + 75 = 111 ................ [A1]

Answer: (b) ……111……………..[3] 8.

(a) Factorise completely 5pq – eq – 2de + 10dp. (b) Expand and simplify x2 – (x – 4)2 + 16 Solution: (a) 5pq – eq – 2de + 10dp = q (5p – e) + 2d (5p – e) ................ [M1] = (q + 2d) (5p – e) ............................ [A1] (b) x2 – (x – 4)2 + 16 = x2 – (x2 – 8x + 16) + 16 ............................. [M1] = 8x............................................................. [A1] Answer: (a) …(q + 2d) (5p – e) ....[2] (b) ……8x …….…...….. [2]

9

A company makes toy buses and toy cars. The following table is used in calculating the cost of manufacturing each toy. Labour (Hours)

Wood (Blocks)

Paint (Tins)

Bus

2

5

3

Car

4

3

2

Labour costs $5 per hour, wood costs $1 per block and paint costs $2 per tin.

⎛5⎞ ⎜ ⎟ ⎛ 2 5 3⎞ ⎟⎟ , B = ⎜ 1 ⎟ and C = AB. It is given that A = ⎜⎜ ⎝ 4 3 2⎠ ⎜ 2⎟ ⎝ ⎠ (a) (i)

Evaluate C.

6

(ii)

Explain what the numbers in your answer represent.

(b) In addition, D = (500 1000) . (i)

Evaluate DC.

(ii)

Explain what your answer represents.

Solution:

⎛5⎞ ⎛ 2 5 3 ⎞⎜ ⎟ ⎟⎟⎜ 1 ⎟ (a)(i) C = ⎜⎜ ⎝ 4 3 2 ⎠⎜ 2 ⎟ ⎝ ⎠ ⎛ 21 ⎞ = ⎜⎜ ⎟⎟ ⎝ 27 ⎠ ⎛ 21 ⎞ (b)(i) DC = (500 1000) ⎜⎜ ⎟⎟ ⎝ 27 ⎠ = (37500) ⎛ 21 ⎞ CAO Answer: (a)(i) … ⎜⎜ ⎟⎟ ………..……[1] ⎝ 27 ⎠ (a)(ii) 21 represents the cost of making one bus while 27 represents the cost of making one car.………...…….................................................................................................…...[1] CAO (b)(i) …(37500)………...…[1] (b)(ii) …Total cost of making 500 toy buses and 1000 cars……...……...……………...[1] 10.

(a) Given that 2 x ÷ 2 =

1 , find the value of x. 64

Solution: 2x ÷ 2 =

1 64

2 x – 1 = 2 – 6 ……….. [M1] x–1=–6 x = – 5 ………… [A1] Answer: (a) x = …– 5.….………….. [2] (b) Solve the inequality –20 < –4x ≤ x + 5. Show your solution on the number line in the answer space. Solution: – 20 < – 4x ≤ x + 5 – 20 < – 4x and – 4x ≤ x + 5 ……… [M1] 4x < 20 x<5

and – 5x ≤ 5 and

x≥–1

7

–1 ≤ x < 5............................. [A1] Answer: (b) inequality:… –1 ≤ x < 5... ….. [2] CAO Number line: 11.

-10

-15

-5

5

0

10

15

[1]

In the diagram, PQ is an arc of a circle with centre O and radius 3x cm while RS is an arc 1 of a circle with the same centre and radius 5x cm. Given that POˆ Q = 1 radians and the 4 R area of the shaded region PQSR = 250 cm2, calculate P (i) the value of x, 3x (ii) the perimeter of the shaded region PQSR. O Solution: (i) area of PQSR = 250 cm2

5x

Q

S

Area of sector ORS – Area of sector OPQ = 250 1 (5 x )2 ⎛⎜ 5 ⎞⎟ − 1 (3x )2 ⎛⎜ 5 ⎞⎟ = 250 ..................................[M1] 2 ⎝4⎠ ⎝4⎠ 2

[

]

1⎛5⎞ 2 2 ⎜ ⎟ (5 x ) − (3x ) = 250 2⎝ 4⎠

[

]

5 25 x 2 − 9 x 2 = 250 8

10x2 = 250 x2 = 25 x = 5................................................................... [A1] (ii) Perimeter of PQSR = 2(2x) + PQ + RS ⎛5⎞ = 4(5) + 3(5) ⎜ ⎟ +5(5) ⎝4⎠

= 20 +

⎛5⎞ ⎜ ⎟ ...... [M1] for either arc length correct ⎝4⎠

75 125 = 70 cm ............ [A1] + 4 4

Answer: (i) x = ……5..……............. [2] (ii) ……..70….…..........cm [2]

8

12(a)

(b)

A=

1 × 3 x × h + (h + 2)3 x 2

=

3 xh + (3 xh + 6 x) 2

=

3 xh + 6 xh + 12 x 2

=

9 xh + 6 x (shown) 2

14(a)

(b)

A1

2A = 3x ( 3h + 4) x=

13

M1

=

2A 3(3h + 4) c

a− 3 b2

×

A1 a2 b− 4 c3

M1

=

a5 b6 c2

A1

=

2 − h + 3k + 4 (3k + 4)(3k + 4)

M1

=

6 + 3k − h (3k + 4)(3k + 4)

A1

2y + 4 + 5 9 = y+2 y

1m – for both answer correct

2 y 2 + 9 y = 9 y + 18 2 y 2 − 18 = 0

M1

2( y 2 − 9) = 0 2( y − 3)( y + 3) = 0 y = 3 or − 3

15(a) (b)

=

8×9 + (8 − 1) 2 = 85 2

=

20 × 21 + ( 20 − 1) 2 = 571 2

p = 20, q = 571 16(a)(i)

A1 B1

B2

13m = 1300cm length =

1300 = 65cm 20

A1

9

(a)(ii)

Accept :

125cm 2 = 125 ÷ 400

M1

= 0.3125c m 2 actual area = 0.3125 ÷ 100

2

= 3.125 × 10 − 5 m 2 (b)

17(a) (b)

A1

10 × 275 110

M1

Sales Tax = £25

A1

4.5cm

A1 Accept:

cos ∠B = cos(180 D − x) = − cos x D = - 0.8

(c)

Area = =

A1

1 × 10 × 6 × sin ∠B 2 1 × 10 × 6 × 0.6 2

M1

= 18cm 2 18(a)

A1 y

B1

× x

(b)

0.00003125/ 1 3200

y

B1

× x

⎛4⎞ −⎜ ⎟ ⎝5⎠

10

(c)

y B1 × x

19(a)(i)

(a)(ii)

(b) 20(a) (b)

Mean =

236 8 ≈ 10.7 marks / 10 marks 22 11

A1

2

2818 ⎛ 236 ⎞ Std Dev = −⎜ ⎟ 22 ⎝ 22 ⎠ ≈ 3.61 School A students are smarter since they have a higher mean. B (0, − 8) BQ =

A1 B1 B1

− 8 − (−12) = −2 0−2

Accept: Using Substitution method

0 − (−8) = −2 AB = −4−0

They are collinear, hence Q lies on the line. (c)

21(a)(i)

(a)(ii)

(b)

(0 − 2) 2 + ( −8 − (−12)) 2

A1 M1

≈ 4.47 units

A1

2 9

A1

=

1 1 2 2 × + × 3 3 3 3

M1

=

5 9

A1

=

2 ⎛5 2⎞ + ⎜ × ⎟ 9 ⎝9 9⎠

M1

=

28 81

A1

Accept: 20 / 2 5

11

22.

(i)

AB = b − 2a

B1

4 2 BC = − ( 2 a ) = − a 5~ 5 ~ OC = OB + BC 4 = b− a ~ 5~ 3 3 OP = OB = b 5 5~ 4 3 2 4 PC = b − a − b = b− a ~ ~ ~ ~ 5 5 5 5~

B1

~

(ii) (iii)

(iv)

(v)

~

2 (b − 2 a ) ~ 5 ~ 2 Since PC = AB , PC is parallel to AB. 5

B1

M1 A1

PC =

B1 (working with explanation)