Dunearn Prelim 2009 Em P1 Answers

DUNEARN SECONDARY SCHOOL 2009 Secondary 4 End-of-Year Exam Mathematics Paper 1 Marking Scheme 1 2

36%- 900 2500 ar as  12

M1 A1 B1B1

3(a)

12

B1

3(b)

9:16

B1

4(a) 4(b) 5(a) 5(b) 6

7(a) 7(b) 8(a) 8(b) 9(a) 9(b) 10(a) 10(b)(i) 10(b)(ii) 11(a)(i) 11(a)(ii) 11(b) 12(a) 12(b) 12(c)

13(a)(i) 13(a)(ii) 13(b)

5.12  10 8 2048

B1 B1

1

B1

b 2 4x 2 25

B1

6

1. 5 5 ) 100 =5655.74 to 5660 3.02 0.0258 27 3 ºC/h 14 37.5 1cm 2 : 2.25m 2 12500 2, 11, 26, 47 76 3n2 + 1 25  32 42 225 12 5  13 1 (6)(6) 2  36 2 1 32  9 Ans=4 5250(1 

M1 A1 B1 B1 B1 M1 A1 B1 M1 A1 B1 B1 B2 M1A1 B1 B1 B1 B1 M1 A1 B1 B1 M1 A1

2 15(a) 15(b) 15(c) 16(a) 16(b)

16(c)

17(a) 17(b) 17(c) 18(a) 18(b)

19(a) 19(b)

19(c) 20(b)(i) 21(a)

21(b)

22(a)

3 8 3 52 23 17 23 22 161  2   40 39 40 39 195   13    22   k 3  14 4 k= - 10.5   8。5     9  12 ABC  AED (given) BAC  EAD (common) 5  AreaABC  4(9)  36 27 5 1 1 (2  6)(20)  (10)(v )  150 2 2 V=14

B1

6 m= -2 c=4 k=1 x= 0.5 ( x  3) 2  7 1001050 Mean = =1001.05 1000

B1 B1 B1 B1 B1 B1

125(985) 2  240(995) 2  540(1005) 2  95(1015) 2  1002169000 SD = 8.24

M1 A1

Machine X. The SD is smaller, therefore it is more consistent

B1B1

2x =

1 y2 (2) 3 (2) x 3: 6x = y +6

B1 M1 A1 B1 M1 A1 M1 A1 B1 B1 B1 M1 A1 B1 M1

A1

M1A1

M1

3 A1 A1

1 2 y = -3

x=

22(b)

4x  4  x  1 2 4x  4  2x  2 2x  6

M1

x  3 x + 1 < 3x + 3

-2 < 2x -1 < x

M1 M1

-3

-2

-1

0

1

2

3

4

5