# Add Maths Perfect Score Module Form 4 Topical

April 18, 2018 | Author: Yeow Pow Choo | Category: Mode (Statistics), Maxima And Minima, Quadratic Equation, Cartesian Coordinate System, Mean

#### Description

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PERFECT SCORE 2010 F4

CHAPTER 1 : FUNCTIONS 1. Given that f : x  4 x  m and f

1

: x  nx 

3 , find the values of m and n. 4 Answer:- m = – 3 ; n =

1 4

2. Given that f : x  2 x  1 , g : x  4 x and fg : x  ax  b , find the values of a and b . Answer:- a = 8 ; b = –1 2 2 3. Given that f : x  x  3 , g : x  a  bx and gf : x  6 x  36x  56 , find the values of a and b . Answer:- a = 2 ; b = 6

1 4. Given that g : x  m  3x and g : x  2kx 

4 , find the values of m and k. 3 Answer:- k =

5. Given the inverse function f

1

( x) 

1 ;m=4 6

2x  3 , find 2

(a) the value of f(4), (b) the value of k if f –1 (2k) = – k – 3 .

6. Given the function f : x  2 x  1 and g : x  (a) f –1 (x) , (b) f – 1 g(x) , (c) h(x) such that hg(x) = 6x – 3 . Answer:-(a) 7. Diagram 1 shows the function g : x  g

x

11 1 (b)  2 2

x  2 , find 3

x 1 2

(b)

1 1 x 6 2

(c) 18x + 33

p  3x , x  2 , where p is a constant. x2

p  3x x2

7

5 Diagram 1 Find the value of p.

2

PERFECT SCORE 2010 F4

8. x 4

y 4

z 4

2

2

2

0

0

0 1

2

2 2 Diagram 2 Diagram 2 shows the mapping of y to x by the function g : y  ay + b and mapping 6 b to z by the function h : y  , y  . Find the, 2y  b 2 (a) value of a and value of b, (b) the function which maps x to y, (c) the function which maps x to z. 18 10  y Answer:- (a)a= –6, b=10 (b) (c)  y  20 6 9. In the Diagram 3, function h mapped x to y and function g mapped y to z. x

h

y

g

z 8

5

Diagram 3

2 Determine the values of, (a) h1(5), (b) gh(2) Answer:- (a)2 (b)8 10. Given function f : x  2  x and function g : x  kx + n. If composite function gf is given as gf :  3x2  12x + 8, find (a) the value of k and value of n, (b) the value of g2(0). Answer:-(a) k = 3 ,n = –4 (b)44 2

3

11.

PERFECT SCORE 2010 F4

The following information refers to the functions f and g. g (x) = 4 – 3 x fg (x) = 2 x + 5

Find f (x).

23  2 x 3

12. (a) Function f, g and h are given as f : x  2x 3 g:x ,x2 x2 h : x  6x2  2. (i) Determine the function fh(x). At the same axis, sketch the graphs of y = f(x) and y = fh(x). Hence, determine the number of solutions for f(x) = fh(x). (ii) Find the value of g1(2). (b) Function m is defined as m : x  5  3x. If p is another function and mp is defined as mp : x  1  3x2, determine function p. Answer:-(a)(i)12x2 – 4 (b) p  x   2  x 2 13. Given function f : x  4  3x. (a) Find (i) f 2(x), (ii) (f 2)1(x). (b) Hence, or otherwise, find (f 1)2(x) and show

(f 2)1(x) = (f 1)2(x).

(c) Sketch the graph of f 2(x) for the domain 0  x  2 and find it’s corresponding x 8 range. Answer:-(a)9x – 8 (b) 9

14. A function f is defined as f : x 

px , for all values of x except x = h and p 3  2x

are constants. (a) Determine the value h. (b) Given value 2 is mapped to itself by the function f. Find the (i) value p, (ii) another value of x which is mapped to itself,

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PERFECT SCORE 2010 F4

(iii) value of f 1(1). Answer:-(a) h  

3 (b)(i)p =12(ii)x = –3 (iii)–5 2

CHAPTER 2 : QUADRATIC EQUATIONS 1. One of the roots of the quadratic equation Find the possible values of p.

is twice the other root. Answer ; p  5, 7

2. If one of the roots of the quadratic equation find an expression that relates .

is two times the other root, Answer : 2b2  9ac

3. Find the possible value of m , if the quadratic equation roots.

4. Straight line y = mx + 1 is tangent to the curve x2 + y2  2x + 4y = 0. Find the possible values of m. 1 Answer :  or 2 2   and are roots of the equation k x(x  1) = 2m  x. 2 2 If  +  = 6 and   = 3, find the value of k and of m.

5. Given

1 3 ,m= 2 16

6. Find the values of  such that the equation (3  )x2  2( + 1)x +  + 1 = 0 has equal roots. Hence, find the roots of the equation base on the values of  obtained. Answer :  =  1; roots:  = 1, x = 1;  = 1, x = 0

5

PERFECT SCORE 2010 F4

CHAPTER 3 : QUADRATIC FUNCTIONS 1.

Diagram 1 shows the graph of the function y  2  x  p   5 , where p is constant. y 2

x

0 ( 0, –3 )

( 4 , –3 )

Find, Diagram 1 (a) the value of p , (b) the equation of the axis of symmetry, (c) the coordinate of the maximum point. Answer:- (a) p = 2 (b) x = 2 (c) ( 2, 5 )

2. y

0

( 0, –2 )

x

f  x    p  1 x 2  2 x  q

Diagram 2 Diagram 2 shows the graph of the function f  x    p  1 x 2  2 x  q . (a) State the value of q . (b) Find the range of values of p . Answer:-(a) q = – 2 (b) p 

1 2

6 3.

PERFECT SCORE 2010 F4

y y  x 2  bx  c

( 0, 9)

x K 0

Diagram 3

Diagram 3 shows the graph of the function y  x 2  bx  c that intersects the y- axis at point ( 0, 9 ) and touches the x- axis at point K. Find, (a) the value of b and c , (b) the coordinates of point K. Answer:-(a) b = – 6 , c = 9 (b)

4.

 3,0 

y

( 0, 23) ( 2, 3) 0

x

Diagram 4 In Diagram 4 above point ( 2, 3 ) is the turning point on the graph which has equation of the form y = p(x + h)2 + k. Find the, (a) values of p, h and k, (b) equation of the curve formed when the graph as shown is reflected at the x–axis. (c) equation of the curve formed when the graph as shown is reflected at the y–axis. Answer :- (a) p = 5 , h = 2, k = 3

(b) y = 5(x  2)2  3 (c) y = 5(x + 2)2 + 3

7

5.

PERFECT SCORE 2010 F4

Function f  x   x 2  8kx  20k 2  1 has a minimum value of r 2  4k , where r and k are constants. (a) (b)

By using the method of completing the square, show that r  2k  1 . Hence or otherwise, find the values of k and r if the graph of the function is symmetrical about x  r 2  13 . Answer:-(b)

6.

k = 3 , –1 and r = –3 , 5

The function f  x    6  x  2  x   h has a maximum value of 10 and h is a constant. (a) Find the value of h. (b) Sketch the graph of f  x    6  x  2  x   h for the value of h that is determined in (a) above. (c) Write the equation of the axis of symmetry. Answer:- ( a) h = --6 (c) x = –2

7.

Given y = x2 + 2kx + 3k has minimum value 2. (a) Without using the method of differentiation, find the two possible values of k. (b) With these values of k, sketch on the same axis, two graphs for y = x2 + 2kx + 3k. (c) State the coordinates of the minimum point for y = x2 + 2kx + 3k . Answer:- (a) k =1 , 2

(c) (1, 2), (2, 2)

****************************************************************************** CHAPTER 4 : SIMULTANEOUS EQUATIONS 1. Solve the simultaneous equations 3x + 2y = 1 and 3x2 – y2 = 5x + 3y. Answer: x = -7/3 , y = 4 ; x = 1, y = -1 2. Solve the simultaneous equations x y 

xy  3 2x  5 y and  2 3 5

Answer: x = 41/10, y = 16/3 ; x = 1, y = -5 3. Solve the simultaneous equations 2x - y = 4 and 2x2 + xy - 3x = 7. Give your answers correct to three decimal places. Answer: x = 2.461, y = 0.922; x = -0.711, y = -5.422

8

PERFECT SCORE 2010 F4

CHAPTER 5 : INDICES AND LOGARITHMS

625 x  2 

1

1.

Solve the equation

2.

Solve the equation 2 x.8 x  45 x3

3.

Show that 3 x2  3 x1  5 3 x is divisible by 13.

5 x .25 x 1

2 5

 

 

Answer : 13 3 x 1 .4.

Solve the equation 64

2 x 3

5. 6.

23 12

Solve the equation 3 x 2 2 x1  10 Answer: 0.6477

Solve the equation log 4 [log 2 2 x  3]  log 9 3

Solve the equation log( x  2)  log( 4 x  1)  log

8.

Solve the equation log2x - 4 logx16 = 0

1 x

 

1 16

Solve the equation 2 7 x1  5 x Answer: x  3.7232

10. 11.

Solve 3

log2 x

 81

Answer : x  16 2x+1

Solve the equation 3

- 2 (3

x+0.5

12.

Solve the equation 102x+1 - 7 (10x) = 26 Answer:0.3010

13.

Given that log 3 5  m and log 9 2  n , express log 3 50 in terms of m and n Answer:- 2m  2n

14.

Given that log x 2  k and log x 7  h , express log

x

3.5 x in terms of k and h

Given that 2 log 2 x  y   3  log 2 x  log 2 y , show that x 2  y 2  6 xy

If log2a + log2b = 4, show that log4ab = 2 and that log8ab = 4/3. If log2a + log2b = 4, show that log4ab = 2 and that log8ab = 4/3.

9

PERFECT SCORE 2010 F4

CHAPTER 6 :COORDINATE GEOMETRY 1.

The following information refers to the equations of two straight lines, AB and CD which are parallel to each other. AB : 2y = p x + q CD : 3y = (q + 1) x + 2 Where p and q are constants

Express p in terms of q.

p =

2 (q  1) 3

2.

The triangle with vertices A(4,3), B(-1,1) and C(t , -3) has an area 11 unit 2. Find the possible values of t. Answer: t = 0 , -22

3.

The points P(3, p), B(-1, 2) and C(9,7) lie on a straight line. If P divides BC internally in the ratio m : n , find (a) m : n , (b) the value of p. Answer:(a) 2 : 3 (b) p = 4

4.

(a) A point P moves such that its distance from point A (1,– 4) is always 5 units. Find the equation of the locus of P. Answer: x 2  y 2  2 x  8 y  8  0 (b) The point A is (1, -3) and the point B is (4, 6). The point Q moves such that QA : QB = 2 : 3. Find the equation of the locus of Q. Answer : 5x2+5y2+14x+102y-118 =0 (c) A point R moves along the arc of a circle with centre A(2, 3). The arc passes through Q(-2, 0). Find the equation of the locus of R. Answer : x2+y2-4x-6y-12 =0 (d) A point S moves such that its distance from point A(–3,4) is always twice its distance from point B(6,-2). Find the equation of the locus of S. Answer: x 2  y 2  18 x  8 y  45  0 (e) The point M is (2, –3) and N is (4, 5). The point T moves such that it is always equidistance from M and from N. Find the equation of locus of T. Answer : e) x+4y = 7 (f) Given point A (1,2) and point B (4, –5). Find the locus of point W which moves such that  AWB is always 900. Answer: x 2  y 2  5 x  3 y  6  0

10

PERFECT SCORE 2010 F4

Solutions to question no 5, 6 and 7 by scale drawing will not be accepted. 5.

In Diagram 1, the straight line PR cuts y-axis at Q such that PQ : QR = 1 : 3. The equation of PS is 2y = x + 3. y

R

S

Q( 0, 4 )

P(–3, 0 ) (a) Find

x

O Diagram 1 the coordinates of R, the equation of the straight line RS, the area  PRS.

(i) (ii) (iii)

(b) A point T moves such that its locus is a circle which passes through the points P, R and S. Find the equation of the locus of T. Answer: a)(i) R = (9 , 16) (ii) y = – 2x + 34 (iii) 80 unit 2 b) x2 + y2 – 6x – 16y – 27 = 0

6.

Diagram 2 shows the straight line graphs PQS and QRT in a Cartesian plane. Point P and point S lies on the x-axis and y-axis respectively. Q is the mid point of PS. y S y  3x = 4

Q R(0, 1) P x O T Diagram 2

11

PERFECT SCORE 2010 F4

(a) Find, (i) coordinates of the point Q, (ii) area of the quadrilateral OPQR. (iii) The equation of the straight line which is parallel to QT and passes through S. (b) Given 3QR = RT, calculate the coordinates of the point T. 1 (c) A point moves in such a way that it’s distance from S is it’s distance from the point T. 2 (i) Find the equation of locus of the point T. (ii) Hence, determine whether the locus cuts the x-axis or not. . 2 5 3 Answer: (a)(i) ( , 2) (ii) (iii) y   x  4 (b) (2, 2) (c)(i) 3x2 + 3y2 + 4x  36y +56 = 0 2 3 3 (ii) No

7.

y K P J

Q R x

O L Diagram 3 1 In Diagram 3, P(2, 9), Q(5, 7) and R(4 , 3) are the mid point of the straight line JK, KL and 2 LJ such that JPQR form a parallelogram. (a) Find, (i) the equation of the straight line JK, (ii) the equation of the perpendicular bisector of the straight line LJ. (b) Straight line KJ is extended until it intersects the perpendicular bisector of the straight line LJ at the point S. Find the coordinates of the point S. (c) Calculate the area of PQR and consequently the area of JKL. Answer: (a)(i) y = 8x  7 (ii) 4y = 6x  15

1  (b)  ,3  2 

1 (c) 6 ; 26 2

12

PERFECT SCORE 2010 F4

CHAPTER 7 :STATISTICS 1. Table 1 shows the results obtained by 100 pupils in a test. Marks

< 20

< 30

< 40

< 50

< 60

< 70

< 80

< 90

Number of pupils

3

8

20

41

65

85

96

100

Table 1 (a) Based on Table 1, complete the table below. 10 – 19

Marks Frequency

(b) Without drawing an ogive, estimate the interquartile range.

2.

Answer:-(b)Interquartile range = 22.62 The mean and standard deviation of a set of integers 2 , 4 , 8 , p and q are 5 and 2 respectively. (a) Find the values of p and of q . (b) State the mean and variance of the set integers 7, 11, 9 , 2p + 3 and 2q + 3 Answer:-(a) p  5,q  6 or p  6,q  5 (b) Mean =11 Variance = 8

3. The histogram in Diagram 1 shows the marks obtained by 40 students in Mathematics test. Number of students 10 8 6 4 2 0 15.5 20.5 25.5 30.5 35.5 40.5 Diagram 1 (a) (b)

Marks

Without drawing an ogive , calculate the median mark. Calculate the standard deviation of the marks. Answer:(a) 27.17 (b) 6.595

13

4.

PERFECT SCORE 2010 F4

Table 2 shows the frequency distribution of the Chemistry marks of a group of students. Marks 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60

Number of students 2 3 5 10 p 2 Table 2

(a) (b)

(c)

If the median mark is 34.5 , calculate the value of p . By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 2 students on the vertical axis, draw a histogram to represent the frequency distribution of the marks. Find the modal mark. What is the modal mark if the mark of each student is increased by 8 ? Answer:- (a) p = 6 (b) Mode = 36.5 (c) 44.5

5.

The scores, x , obtained by 32 students of Class 5 Alfa in a test are summarized as  x  2496 and  x 2  195488. The mean and the standard deviation of the scores, y , obtained by 40 students and Class 5 Beta in the test are 66 and 6 respectively. (i)

y

(ii)

 y2

(a)

Find

(b)

Calculate the mean and the standard deviation of the scores obtained by all the 72 students. Answer:-(a)(i) 2640 (ii)175680

6.

(b)Mean = 71.33 , S Deviation= 8.1652

A set of data consists of 10 numbers. The sum of the numbers is 120 and the sum of the squares of the numbers is 1650. (a)

Find the mean and variance of the set of data,

(b)

A number a is added to the set of data and the mean is increased by 2, find (i) the value of a, (ii) the standard deviation of the new set of data. Answer:-(a)

Mean = 12 , Variance = 21 (b)(i) a = 34 (ii) S Deviation = 7.687

14

PERFECT SCORE 2010 F4

CHAPTER 8 : CIRCULAR MEASURES 1.

A

T

B O

Diagram 1 Diagram 1 shows a circle with centre O and OA = 10 cm. Straight line AT is a tangent to the circle at point A, and AOT is a triangle. Given that the area of triangle OAT = 60 cm2, find the area of sector OAB. Answer:- 43.80 cm 2 2. A

O B

Diagram 2 Given that the area of a sector OAB in Diagram 2 with centre O and radius 20 cm is 240 cm2. Calculate (a) the length of arc AB (b) area of shaded region Answer:- : (a) 24 cm (b) 53.60 3. P

 O

R

Q

Diagram 3 In the Diagram 3, POQ is a circular sector with centre O and a radius of 17 cm. Point R is on the straight line ORQ such that RQ = 5 cm. Calculate (a) the value of  in radian (b) the area of the shaded region, in cm2 Answer:- (a) 0.7871 (b) 41.49

15

PERFECT SCORE 2010 F4

4.

Q

S1 P

O

S

R

Diagram 4 Diagram 4 shows a semicircle with centre O and radius of 10 cm. Given that QS is the length of arc with centre P and QPS 

6

(a) the length of OS. (b) the area of S1

Answer:-(a) 7.32 cm (b) 35.23 cm 2

5.

2r

C

C

Diagram 5 Two identical circles of radius 2r are drawn with their centres, C on the circumference of each circle as shown in the Diagram 5. Show that the area of shaded region A cm 2 , is given by 2 2 r 4  3 3 . 3

6. Q R

O

8cm

P 3cm A

Diagram 6 Diagram 6 shows two circles with centres O and A. The respective radii are 8 cm and 3 cm. A  tangent touches the circles at the points Q and R. Given that QOP = radians, find 3 (a) the length of QR (b) the perimeter of the shaded region (c) the area of the shaded region Answer : a) 9.80 cm b) 24.46 cm c) 10.96 cm 2

16

PERFECT SCORE 2010 F4

CHAPTER 9 : DIFFERENTIATION 1.(a) (i) Given y = 3x2 + 5, find

dy by using the first principle. dx

4  3 with the first principle. x

(ii) Differentiate y =

d  1   . dx  2 x  1 (ii) Given f(x) = 4x(2x  1)5, find f’(x). (iii) Differentiate 3x2(2x  5)4 with respect to x.

(b) (i) Find

(iv) Given f(x) = (2x  3)5, find f (x). (v) Given f(x) =

(c) (i) Given h(x) = (ii) Given f(x) =

1  2x3 , find f ‘(x). x 1

1 , find the value h’’(1). 2 (3x  5)

x

 2 , find f '(0). 1  3x 5

2

limit  n 2  4   . n2 n2  4  3r (ii) Given f (r) = . Find the limit of f (r) when r  1. 5  2r

(d) (i) Find the limit of

d2y dy  x  12 in terms of x, in the simplest form. 2 dx dx d2y dy Hence, find the value of x which satisfy the equation y 2  x  12 = 0. dx dx

(e) Given y = x(3  x), express y

Answer : (a) (i) 6x 4 (ii)  x2

(b)( i) 

2

(2 x  1) 2 (ii)[4(2x  1)4)(12x  1)]

iii )6 x(6 x  5)(2 x  5) 3 iv) 80(2x  3)3

v)

 4x3  6x 2  1

( x  1) 2 27 c) (i) 8 (ii) 96

d) i) 4 7 ii) 3

e) 12  3x; x=4

17

PERFECT SCORE 2010 F4

2. (a) Given the function of the graph f(x) = hx3 + f ‘ (x) = 3x2 

k x2

, which has a gradient function of

96

, where h and k are constant. Find, x3 (i) the value of h and the value of k, (ii) the coordinate x of the turning point of the graph. (b) The point P lies on the curve y  ( x  5) 2 . It is given that the gradient of the 1 normal at P is  . Find 4 (i) the coordinates of P. (ii) the equation of the normal to the curve at point P. (c) A curve with the gradient function 2 x 

2 x2

has a turning point at ( k , 8 ) .

(i) Find the value of k . (ii) Determine whether the turning point is a maximum or a minimum point .

Answer: 2(a) (i) h=1, k= 48 (ii) 2

(b) (i) (7, 4) (ii)4y + x = 23

(c) i) k = 1, ii) Minimum

2 . x Given that y increases at a constant rate of 4 units per second, find the rate of change of x when x = 2.

3. (a) Two variables, x and y, are related by the equation y = 3x +

(b) On a certain day, the rate of increase of temperature, , with respect to time, t s, is d 1 given by = (12  t). dt 2 (i) Find the value of t at the instant when  is maximum. (ii) Given  = 4 when t = 6, find the maximum value of . Answer: (a) 1.6 units-1

(b) i) 12

4. (a) Given y = t  2t2 and x = 4t + 1. dy in terms of x. dx (ii) If x increases from 3 to 3.01, find the corresponding small increment in t.

(i) Find

ii)

13

18

PERFECT SCORE 2010 F4

(b) Given y = 2x3  5x2 + 7, find the value of

dy at the point (2, 3). dx

Hence, find (i) the small change in x which causes y to decrease from 3 to 2.98. (ii) the rate of change of y when x = 2 if the rate of change of x is 0.6 unit per second. (c) Given y 

value of

16 dy , find the value of when x = 2. Hence, find the approximate 4 dx x

16 . (198 . )4

2 x (ii) 0.0025 4

(b) (i) 0.005 (ii) 2.4 unit s1

(c) 1.04

5. Diagram 1 shows a composite solid made up of a cone resting on a cylinder with radius x cm.

x cm Diagram 1

16   The total surface area of the solid, A cm2, is given by the equation A = 3  x 2   . x  (a) Calculate the minimum value of the surface area of the solid. (b) Given the surface area of the solid is changing at a rate of 42 cm2 s1. Find the rate of change of radius at the instant when the radius is 4 cm. (c) Given the radius of the cylinder increases from 4 cm to 4.003 cm. Find approximate increment in the surface area of the solid Answer: b) 36

(c)

2 (d) 0.063

19

PERFECT SCORE 2010 F4

CHAPTER 10 : SOLUTION OF TRIANGLES

1.

P

8 cm 6.5 cm Diagram 1

50 Q

R

Diagram 1 shows a PQR. (a) Calculate the obtuse angle PRQ. (b) Sketch and label another triangle different from PQR in the diagram above, so that the lengths of PQ and PR and the angle PQR remain unchanged. (c) If the length of PR is reduced whereas the length of PQ and angle PQR remain unchanged, calculate the length of PR so that only one PQR can be formed Answer: (a) 109 28’ or 109.47

(c) 6.128 cm

2.(a) Diagram 2 shows a pyramid VABCD with a square base ABCD. VA is vertical and the base ABCD is horizontal. Calculate, (i) VTU, (ii) the area of the plane VTU. V

8 cm

A Diagram 2

B

6 cm

U 2 cm D

T

4 cm

C

Answer: (a)(i) 84 58’ or 84.97 (ii) 22.72 cm2

20

PERFECT SCORE 2010 F4

(b) . Q

R 4 cm K L

P

Diagram 3

S 6 cm

J

8 cm

Diagram 3 shows a cuboid.

M (i) JQL, (ii) the area of JQL.

Calculate,

Answer: (i) 75 38’ or 75.640

(ii) 31.24 cm2

3.

A

.O A

5 cm D

50 4 cm

B 8 cm

C Diagram 4 Diagram 4, ABCD is a cyclic quadrilateral of a circle centered O. Calculate (a) the length of AC, correct to two decimal places, (b)  ACD, (c) the area of quadrilateral ABCD. Answer: (a) 11.85 cm (b) 115.01 (c)36.08cm2 4. Diagram 5 shows two triangles PQT and TRS. Given that PQ = 24 cm, TS = 12 cm, TPQ  320 , PT = TQ and PTS and TRQ are straight lines. P 32o

24 cm Diagram 5 R

T 12 cm S

Q

21 (a) (b) (c) (d)

PERFECT SCORE 2010 F4

Find the length, in cm, of PT, If the area of triangle PQT is three times the area of triangle TRS, find the length of TR. Find the length of RS. (i) Calculate the angle TSR. (ii) Calculate the area of triangle QRS. Answer: (a) 14.15 cm, (b) 5.563 cm, (c) 10.79 cm, (d)(i) 27.60 o (ii) 46.31 cm2

5. Diagram 6 shows a triangle PQR. R

P 10 cm

7 cm

Diagram 6

75o Q (a) Calculate the length of PR. (b) A quadrilateral PQRS is now formed so that PR is the diagonal, PRS = 40o and PS = 8 cm. Calculate the two possible values of PSR . (c) Using the obtuse PSR in (b), calculate (i) the length of RS, (ii) the area of the quadrilateral PQRS. Answer: (a) 10.62 cm (b) 58.57o ; 121.43o (c) (i) 3.964 cm ; (ii) 47.34 cm2 6. Diagram 7 shows a camp of the shape of pyramid VABC. The camp is built on a horizontal triangular base ABC. V is the vertex and the angle between the inclined plane VBC with the base is 60. V

C A

Diagram 7

B Given VB = VC = 25 cm and AB = AC = 32 cm and BAC is an acute angle. Calculate (a) BAC if the area of ABC is 400 cm2, (b) the length of BC, (c) the lengths of VT and AT, where T is the midpoint of BC, (d) the length of VA, (e) the area of VAB Answer: (a) 51.38o

(b) 27.74o

(c) 20.80 cm; 28.84 cm (d) 25.78 cm (e) 315.33 cm2

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PERFECT SCORE 2010 F4

CHAPTER 11 : INDEX NUMBER 1. Table 1 shows the price indices and percentage usage of four items, P, Q, R, and S, which are the main ingredients of a type biscuits. Item

Price index for the year 1995 based on the year 1993 135 x 105 130 Table 1

P Q R S

Percentage of usage (%) 40 30 10 20

Calculate, (a) (i) the price of S in the year 1993 if its price in the year 1995 is RM37.70 (ii) the price index of P in the year 1995 based on the year 1991 if its price index in the year 1993 based in the year 1991 is 120. (b) The composite index number of the cost of biscuits production for the year 1995 based on the year 1993 is 128. Calculate, (i) the value of x, (ii) the price of a box of biscuit in the year1993 if the corresponding price in the year 1995 is RM 32. Answer: (a)(i) RM29 (ii) 162 (b)(i)125 (ii) RM25 2.

Daily Usage ( RM) 47 34 22 12 3

Component V

W

X

Y

Z

Diagram 1 A technology product consists of five components, V, W, X, Y and Z. Diagram 1 shows a bar chart showing the daily usage of the components used to produce the technology product. The following table shows the prices and the price indices of the components.

23 Component V W X Y Z (a)

PERFECT SCORE 2010 F4

Price in the year 2001 Price in the year 2003 Price index in 2003 (RM) (RM) based on 2001 13.00 16.25 y 12.50 17.25 138 2.50 x 106 14.90 22.35 150 z 24.50 140

Find the values of x, y and z.

(b) Calculate the composite index representing the cost of the technology product in the year 2003 using the year 2001 as the base year. (c) If the total monthly cost of the components in the year 2001 is RM1.5 million, find the total monthly cost of the components in the year 2003. (d) If the cost of each component rises by 23% from the year 2003 to 2004, find the composite index representing the cost of the technology product in the year 2004 based on the year 2001. Answer:

3. (a) In the year 1995, price and price index for one kilogram of certain grade of rice is RM2.40 and 160 respectively. Based on the year 1990, calculate the price per kilogram of rice in the year 1990.

Item Timber Cement Iron Steel

Price index in the year 1994 180 116 140 124

Change of price index from the year 1994 to the year 1996 Increased 10 % Decreased 5 % No change No change Table 2

Weightage 5 4 2 1

(b) Table 2 shows the price index in the year 1994 based on the year 1992, the change in price index from the year 1994 to the year 1996 and the weightage respectively. Calculate the composite price index in the year 1996. . Answer : (a) 1.50 (b) ITimber = 198, ICement = 110.2; 152.9

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PERFECT SCORE 2010 F4

4. Table 3 shows the price indices and the weightages of Azizan’s monthly expenses in the year 2005 based in the year 2004.

(a) (b) (c) (d)

Expenses

Price index in 2005 based on 2004

Weightage

Rental

108

3

Food

120

4

Car installment

102

2

Miscellaneous

112

1

Table 3 If the expenses for miscellaneous in the year 2005 was RM 1 456 , find the miscellaneous expenses in the year 2004. If the rental increases by 10% from the year 2005 to the year 2006,find the price index for the rental in the year 2006 based on the year 2004. Calculate the composite index for the expenses in the year 2005 based on the year 2004. The price index for food in the year 2006 based on the year 2205 is 105. If the expenses on food in the year 2006 were RM3150, find the expenses on food in the year 2004. Answers : a) 1300 b) 118.8 c) 112 d) 2500