IB Questionbank Mathematical Studies 3rd edition1

1.Unit penalty applies in parts (b) (c) and (d).

(a)Angle ABC = 50(A1)

(b)

=

55

sin

25

sin50

AC

(M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their angle ABC.

UPAC = 23.4 m(A1)(ft)(G2)

(c)Area of ABC =

2

1

23.379... 25 sin 75(M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution.Follow through from their AC.

OR

Area of triangle ABC =

2

...

151

.

19

...

479

.

29

(A1)(ft)(M1)

Note: (A1)(ft) for correct values of AB (29.479) and CN (19.151).Follow through from their (a) and /or (b).Award (M1) for substitution of their values of AB and CN into the correct formula.

UPArea of ABC = 282 m2(A1)(ft)(G2)

Note: Accept 283 m2 if 23.4 is used.

(d)NM =

2

50

sin

25

(M1)(M1)

Note: Award (M1) for 25 sin 50 or equivalent for the length of CN, (M1) for dividing their CN by 2.

UPNM = 9.58 m(A1)(ft)(G2)

Note: Follow through from their angle ABC.

Notes: Premature rounding of CN leads to the answers 9.60 or 9.6.Award at most (M1)(M1)(A0) if working seen. Do not penalize with (AP).CN may be found in (c).

Note: The working for this part of the question may be in part (b).

(e)Angle NCB = 40 seen(A1)(ft)

Note: Follow through from their (a).

From triangle MCP:MP2 = (9.5756...)2 + 12.52 2 9.5756... 12.5 cos (40)(M1)(A1)(ft)

MP = 8.034... m(A1)(ft)(G3)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d).Award (G3) for correct value of MP seen without working.

OR

From right triangle MCPCP = 12.5 m seen(A1)MP2 = (12.5)2 (9.575)2(M1)(A1)(ft)MP = 8.034m(A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d).Award (G3) for correct value of MP seen without working.

OR

From right triangle MCPAngle MCP = 40 seen(A1)(ft)

12.5

MP

= sin (40) or equivalent(M1)(A1)(ft)MP = 8.034... m(A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (a).Award (G3) for correct value of MP seen without working.

The goat cannot reach point P as MP > 7 m.(A1)(ft)

Note: Award (A1)(ft) only if their value of MP is compared to 7m, and conclusion is stated.

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2.(a)(i)8.5 (cm)(A1)

(ii)120(A1)

(iii)30(A1)(C3)

(b)

=

30

sin

5

.

8

sin120

BC

(M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

BC = 14.7

2

3

17

(A1)(ft)(C3)

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3.(a)(i)Mean = (5.96 + 5.95 + 6.02 + 5.95 + 5.99) / 5 = 5.974 (5.97)(A1)

(ii)

100%

value

actual

error

error

%

=

100%

5.974

5.974

6

-

=

= 0.435%(M1)(A1)(ft)(C3)

Note:(M1) for correctly substituted formula. Allow 0.503% as follow through from 5.97 Note: An answer of 0.433% is incorrect.

(b)number is 29.45728613

(i)Nearest integer = 29(A1)

(ii)Standard form = 2.95 101(accept 2.9 101)(A1)(ft)(A1)(C3)

Notes:Award (A1) for each correct term

Award (A1)(A0) for 2.95 10

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4.(a)(i)mean = 13.7(M1)(A1)(G2)

(ii)sd = 2.52(M1)(A1)(G2)

(b)For attempting to put their numbers in order(M1)13.1(A1)(G2)

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5.(a)If the number is even and the number does not end in zero,(then) the number is not a multiple of five.(A1)(A1)(A1)

Note: Award (A1) for if(then), (A1) for the number is even and the number does not end in zero, (A1) for the number is not a multiple of 5.

(b)(i)(p

q)

r(A1)(A1)(A1)(A1)(A1) for

, (A1) for

, (A1) for p and q,(A1) for r

Note: If parentheses not present award at most (A1)(A1)(A1)(A0).

(ii)r

(p

q) OR r

(p

q)(A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for reversing the order, (A1) for negating the statements on both sides.If parentheses not present award at most (A1)(ft)(A0).Do not penalise twice for missing parentheses in (i) and (ii).

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6.(a)r =

3

1

108

36

(A1)(C1)

Note: Accept 0.333.

(b)u1

7

3

1

= 36(M1)

Note: Award (M1) for correct substitution in formula for nth term of a GP. Accept equivalent forms.

u1 = 78732(A1)(ft)(C2)

Notes: Accept 78700. Follow through from their common ratio found in part (a). If 0.333 used from part (a) award (M1)(A1)(ft) for an answer of 79285 or 79300 irrespective of whether working is shown.

(c)118096 =

-

-

3

1

1

3

1

1

78732

k

(M1)(M1)

Notes: Award (M1) for correct substitution in the sum of a GP formula, (M1) for equating their sum to 118096. Follow through from parts (a) and (b).

OR

Sketch of the function y = 78732

-

-

3

1

1

3

1

1

k

(M1)Indication of point where y = 118 096(M1)

OR

78 732 + 26 244 + 8748 + 2916 + 972 + 324 + 108 + 36 + 12 + 4= 118 096(M1)(M1)

Note: Award (M1) for a list of at least 8 correct terms, (M1) for the sum of the terms equated to 118096.

k = 10(A1)(ft)(C3)

Notes: Follow through from parts (a) and (b). If k is not an integer, do not award final (A1). Accept alternative methods.If 0.333 and 79285 used award (M1)(M1)(A1)(ft) for k = 5.If 0.333 and 79300 used award (M1)(M1)(A0).

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