2013 GCE O Level Examination - Revision Paper 27 - Answers

8/22/2019 2013 GCE O Level Examination - Revision Paper 27 - Answers

1/12

2013 GCE O Level Examination Revision Paper 27

Answers

1. (a) Express22

9as a percentage.

11

1040 % [B1]

(b) Express %4

128 as a fraction in its simplest form.

400

113[B1]

2. Calculate368.023.11

0845.02

, leaving your answer in standard form.

000671995.0

368.023.11

0845.02

=

[M1]

41072.6 = [A1]

3. Factorise fully 23 4520 xyx . )94(54520 2223 yxxxyx = [M1: Bringing out the common factor.]

)32)(32(5 yxyxx += [A1]

4. (a) Evaluate

4

3

x

44

3

3

=

x

x

81

4x= [B1]

(b) Given that 497730 = k , find k.

230 77 =k

By comparing the power of indices,230 = k

28=k [B1: Accept trial & error method.]

5. Alicia, Benny and Carina were given a sum of money in the ratio of 1 : 8 : 13. If Aliciareceived $180 less than Carina, what was the total sum of money given to them?

Method 1Letx be the total sum of money.

180

22

12 = x

330=x [A1]

Method 212 units rep $180 [M1]22 units rep $330 [A1]

Note: Accept any other reasonable methods.

[M1: Attempt to form analgebraic equation.]


2/12

6. (a) Solve the inequality 26538


3/12

162 =v

4=v [B1]

(b) Simplify )3(5)73(2 xyyx .

xyyxxyyx 155146)3(5)73(2 += [M1] yx 1921 = [A1]

(c) Simplify 3

24

18

25

6

5

y

x

yz

x .

2

34

3

24

25

18

6

5

18

25

6

5

x

y

yz

x

y

x

yz

x = [M1]

z

yx

5

322

= [A1]

10.Solve the following simultaneous equations.

01223

32

=+=ba

ba

6=a [B1] 15=b [B1]

11. The number 36, written as a product of its prime factors, is 22 3236 = .

(a) Express 204 as a product of its prime factors.

17322042 = [B1]

(b) 36 teachers, 204 female students and 384 male students attended a camp. What isthe greatest number of groups that can be formed if all the teachers, femalestudents and male students have to be distributed equally among all the groups?

323847 =

Greatest number of groups = HCF

322 =

= 12 [B1]

(c) How many teachers will there be in each group?

No. of teachers =12

36

= 3 [B1]

12. The speed of the bullet train in Japan is 300 km/h. Express this speed in metres per second.

smhkm /3600

300000/300 =

sm /3183= [A1]

13. In the diagram,A, B, CandD lie on a circle, centre O.

[M1: Give 1 mark as long as student converts 300km to30000m AND/OR 1 hour to 3600 seconds]


4/12

The tangents atA andB meet atEand 096=AOB

Find , stating your reasons clearly,

*Deduct 1 mark off whole question for no reasons and/or one or more incorrect reason.

(a)2

96180=OBA ( OAB is an isosceles .)

= 420 [B1]

(b)2

96=ACB (at centre = 2 at circumference)

= 480 [B1]

(c) 42180=BCD (opp s in a cyclic quad)

= 1380 [B1]

(d) 090== OBEOAE (tangent perpendicular to radius) [M1]

(*Students have to explain why 090== OBEOAE )

969090360 =AEB (angles in a quadrilateral)

= 840 [A1]

14. A box contains 10 blue marbles and 6 white marbles. Dylan draws a marble at random andkept it in his pocket. He then draws a second marble.

(a) Construct a probability tree diagram to show all the possible outcomes of drawing the twomarbles.

5

3

E

B

.

A

C

D

960

O

1st marble 2nd marble

Blue

White

Blue

White

Blue

White


5/12

8

5

5

2[B1: Correct diagram]

[B1: Correct probability shown]

8

3

3

2

3

1

(b) Hence or otherwise, find the probability of drawing(i) two blue marbles.

P(drawing two blue marbles) =8

3

5

3

8

5 = [B1]

(ii) two marbles of different colours,

P(drawing two marbles of different colours) =3

2

8

3

5

2

8

5 +

2

1= [B1]

15. Express 2)2(

1

2

2

+

xx

xas a single fraction in its simplest form.

22)2(

)1()2(2

)2(

1

2

2

+

=+

x

xx

x

x

x[M1: Same denominator]

2)2(

142

=

x

xx

2

)2(

5

=x

x[A1]

16. y is inversely proportional tox2. It is known thaty = 500 for a particularx. Find the value ofy

when this value ofx is doubled.

2x

ky = [B1]

Wheny = 500,2

500x

k=

500

2 k

x =

When x is doubled, 2)2( x

ky = [M1]


6/12

=500

4k

k

=k

k4

500

= 125 [A1]

17. The diagram shows the speed-time graph for the first 80 seconds of a cars journey.

(a) Calculate the acceleration during the first 20 seconds.

Acceleration30

15=

= 0.5 m/s2 [B1]

(b) Calculate the distance travelled during the first 80 seconds.

Distance travelled = area under graph

)15)(8050(2

1 += [M1]

= 975m [A1]

(c) (i) After 80 seconds, the car decelerated at a constant speed until it came to acomplete rest. If the total distance travelled by the car is 1275 m, calculate thetotal time taken for this journey.

Let tbe the total time taken.

1275)15)(50(2

1

=+ t [M1]17050 =+ t

120=t [A1]

(d) On the grid below, draw the distance time graph of the car for the whole journey.

Time (seconds)

-------------------

Speed(m/s)

15

30 80

Time (seconds)

010050

500

1000

Distance(metres)

225

975

1275

30 80 120


7/12

[1 mark for drawing the correct curve between 0 30 seconds]

[1 marks for drawing a straight line between 30 to 80 seconds]

[1 mark for drawing the correct curve between 80 and 120 seconds]

18. Seventeen boys and fifteen girls took a test.The marks are shown in the stem-and-leaf diagram.

Boys Girls

9 9 5 2 0 5 2 2 3 6 9 98 7 4 4 1 1 6 0 2 4 4

9 6 6 3 7 1 3 3 30 0 8 0

Key (Boys)

0 | 5 means 50

Key (Girls)

5 | 2 means 52

(a) Write down the mode of the girls marks.

73 [B1]

(b) Write down the median of the boys marks. 64 [B1]

Answer (c) The boys performed better because the median mark of the boys is higher than thatof the girls. [B1]

Note: The boys performed better because the mean mark of the boys is higher than that of the

girls. (This answer is accepted only if students show working to find the mean marks of the

boys and girls.)

(d) The box-and-whisker diagram below shows the marks ofall the students.

50 59 64 y 80


8/12

(d)(i) Find the value ofy.

y = upper quartile

= 73 [B1]

(d)(ii) Find the interquartile range.

Interquartile range = 73 59= 14 [B1: allow ECF if students value ofy is wrong.]

19. (a) Express 1762 + xx in the form khx + 2)( . 8)3(176 22 +=+ xxx [B1]

(b) Hence, sketch the graph of 1762 += xxy

Answer (b)

[1 mark for correct shape][1 mark for correct min. point and y-intercept]

(c) Write down the equation of the line of symmetry of 1762 += xxy

x = 3 [B1]

20. A, B and Care the points (-10, 15), (-10, -10) and (10, -20).

O

y

x

y

3

17

8

y

x

-10 -5 0 5 10

20

10

-10

-20

A

B

C


9/12

(a) Find the area of triangleABC.

Area of triangleABC= 20252

1

= 250 units2 [B1]

(b)ABCD is a parallelogram. Find the coordinates of the point D.

D = (10, 5) [B1]

(c)ABCEis a trapezium withAB parallel to CE.

The area of the trapezium is 650 units2.Find the coordinates of the pointE.

Since the area of the trapezium is 650 units2,

650)20)(25(2

1 =+CE

25 + CE= 65CE= 40 [M1]

)20,10(=E [A1]

21. Square and round stickers are arranged in the series of pattern as shown below.

Pattern 1 Pattern 2 Pattern 3

(a) Find the number of round stickers needed to form pattern 5.

20 [B1]

(b) (i) To form patternN, we need to use Ssquare stickers. Write down a formulaconnectingNand S.

S=N2 [B1]

(b) (ii) To form patternN, we need to useR round stickers. Write down a formula


10/12

connectingNandR.

R = 4N [B1]

(c) Explain why the value ofR can never be 482.

482 is not a multiple of 4. [B1]

(d)Trepresents the total number of stickers needed to form PatternN. Express Tinterms ofN.

T = N2+4N [B1]

(e) Hence, find the pattern number that uses a total of 572 stickers.N2+4N = 572 [M1]N2+4N 572 = 0

(N-22)(N+26) = 0N=22 or N=26 (rejected) [A1]

(Trial and error method is accepted)

22. The table below shows the attendance of two classes in January and February for 3events A, B and C.

January February

Class A B C A B C4E6 20 15 18 10 10 16

5N4 15 35 10 17 32 12

The information for Januarys attendance can be represented by the matrix

=103515

181520P .

The information for Februarys attendance can be represented by a matrix Q.

(a) Write down matrix Q.

=

123217

161010Q


11/12

(b) Calculate ( )QP+2

1.

( )

+

=+

123217

161010

103515

181520

2

1

2

1QP

=

226732

342530

2

1

=

115.3316

175.1215

(c) Describe what is represented by the elements of ( )QP+2

1.

It represents the average number of students who turned up for event A, B and C for bothclasses.

23. Two pointsA andB are shown below.Answer (a), (b), (c), (e)

A B. .


12/12

(a) Construct triangleABCsuch thatAC= 6cm andBC= 10cm.

(b) Construct the angle bisector of .ABC

(c) Construct the perpendicular bisector ofAB.

(d) The perpendicular bisector ofAB cuts the lineAB atD and the angle bisector of ABC atE. MeasureAE.

Answer (d) AE= ______________________cm

(e) Name the triangle whose area is twice the area of .ACD

ACB [B1]

End of Paper 1

Documents

2013 GCE O Level Examination - Revision Paper 27 - Answers