4.10 SPM Practice (Long Questions)

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Question 11 (5 marks):
Table 8 shows the information of books purchased by Maslinda.
 
Table 8

Maslinda purchased x history books and y Science books. The total number of books purchased is 5. The total price of the books purchased is RM17.

(a) Write two linear equations in terms of x and y to represent the above information.

(b) Hence, by using matrix method, calculate the value of x and of y.


Solution:
(a)
x + y = 5
4x + 3y = 17

(b)
( 1    1 4    3 )( x y )=(  5 17 )             ( x y )= 1 1( 3 )−4( 1 ) (   3    −1 −4       1 )(  5 17 )             ( x y )=−1( 3( 5 )+( −1 )( 17 ) −4( 5 )+( 1 )( 17 ) )             ( x y )=−1( 15−17 −20+17 )             ( x y )=−1( −2 −3 )             ( x y )=( 2 3 ) x=2 and y=3



Question 12 (6 marks):
 Given A=( 4    −2 3    −1 ), B=m( −1    n −3    4 ) and I=( 1    0 0    1 ).
(a) Find the value of m and of n if AB = I.
(b) Write the following simultaneous linear equation as matrix equation:
4x – 2y = 3
3x – y = 2
Hence, by using matrix method, calculate the value of x and of y.

Solution:
(a)
If AB=I, then B= A −1 A −1 = 1 4( −1 )−( −2 )( 3 ) ( −1    2 −3    4 ) A −1 = 1 2 ( −1    2 −3    4 ) By comparison: B= A −1 m( −1    n −3    4 )= 1 2 ( −1    2 −3    4 ) m= 1 2  ; n=2


(b)
4x−2y=3 3x−y=2 ( 4    −2 3    −1 )( x y )=( 3 2 )                ( x y )= 1 2 ( −1    2 −3    4 )( 3 2 )                ( x y )= 1 2 ( ( −1 )( 3 )+( 2 )( 2 ) ( −3 )( 3 )+( 4 )( 2 ) )                ( x y )= 1 2 (   1 −1 )                ( x y )=(    1 2 − 1 2 ) x= 1 2  and y=− 1 2


4.6 SPM Practice (Long Questions)

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Question 13 (5 marks):
(a) State whether the following statements are true statement or false statement.
( i ) {  }⊂{ S,E,T } ( ii ) { 1 }⊂{ 1,2,3 }={ 1,2,3 }

(b) Diagram 7 shows the first three patterns of a sequence of patterns.

Diagram 7

It is given that the diameter of each circle is 20 cm.
(i) Make a general conclusion by induction for the area of the unshaded region.

(ii) Hence, calculate the area of the unshaded region for the 5th pattern.


Solution:
(a)(i) True

(a)(ii) False

(b)(i)
Area of unshaded region (first)
= (20 × 20) – π(10)2
= 400 – 100π
= 100 (4 – π)

Area of unshaded region (second)
= 4 × 100 (4 – π)
= 400 (4 – π)

Area of unshaded region (third)
= 9 × 100 (4 – π)
= 900 (4 – π)
100 (4 – π), 400 (4 – π), 900 (4 – π), …
102 (4 – π), 202 (4 – π), 302 (4 – π), …

General conclusion = n2 (4 – π)
n = 10, 20, 30, …


(b)(ii)
Area of unshaded region (5th)
= 502 (4 – π)
= 2500 (4 – π)


4.6 SPM Practice (Long Questions)

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Question 11 (5 marks):
(a) State whether the following compound statement is true or false.
2 > 3 and (–2)3 = –8

(b) Write down two implications based on the following statement:
a > b if and only if a – b > 0

(c) Table 1 shows the number of sides and the number of axes of symmetry for some regular polygons.

Table 1

Make a conclusion by induction by completing the statement in the answer space:
The number of axes of symmetry for a regular polygon with n sides is __________.


Solution:
(a)
Palsu. Nilai bagi 2 adalah kurang daripada 3, 2 < 3.

(b)
Implication 1: Jika a > b, maka a – b > 0
Implication 2: Jika a – b > 0, maka a – b > 0

(c)
The number of axes of symmetry for a regular polygon with n sides is
Number of sides of regular polygon = Number of axes of symmetry of regular polygon



Question 12 (5 marks):
(a) State whether the following statement is true or false.


(b) Write down the converse of the following implication:
 

(c) Write down Premise 2 to complete the following argument:
Premise 1   :  If x is an odd number, then x is not divisible by 2.
Premise 2   : ………………………………………………………
Conclusion : 24 is not an odd number.

(d) Based on the information below, make one conclusion by deduction for the surface area of sphere with radius 9 cm.



Solution:
(a)
True

(b)
If an angle is an acute angle, then the angle lies between 0o and 90o.

(c)
24 is divisible by 2.

(d)
4π(9)2 = 324π
Surface area of the sphere is 324π.


5.7 SPM Practice (Long Questions 6)

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Question 11 (5 marks):
Diagram 4 shows a parallelogram drawn on a Cartesian plane which represents the locations of Rahman's house, a cinema, a school and a shop.

It is given that the scale is 1 unit : 1 km.
(a) Calculate the distance, in km, between Rahman's house and the school.
(b) Find the equation of the straight line that links the school to the cinema.


Solution:
(a)
2y = 3x + 15
When y = 0
3x + 15 = 0
3x = –15
x = –5

Rahman's house = (–5, 0)
School = (3, 0)

Distance, between Rahman's house and the school
= 3 – (– 5)
= 8 units
= 8 km

(b)
2y=3x+15 y= 3 2 x+ 15 2 Thus m= 3 2 At point ( 3, 0 ),  y 1 =m x 1 +c 0= 3 2 ( 3 )+c 9 2 +c=0 c=− 9 2 Thus, the linear equation is y= 3 2 x− 9 2 2y=3x−9



Question 12 (6 marks):
Diagram 6 shows two parallel straight lines, JK and LM, drawn on a Cartesian plane.
The straight line KM is parallel to the x-axis.

Diagram 6

Find
(a) the equation of the straight line KM,
(b) the equation of the straight line LM,
(c) the value of k.



Solution:
(a)
The equation of the straight line KM is y = 3

(b)
Given, equation of JK: 2y=4x+3 y=2x+ 3 2 Thus,  m JK =2 m LM = m JK =2 y=mx+c At M( 5, 3 ) 3=2( 5 )+c 3=10+c c=−7 ∴ Equation of the straight line LM is y=2x−7.


(c)
Substitute ( k,0 ) into y=2x−7 0=2( k )−7 7=2k k= 7 2

Quadratic Equations Long Questions (Question 14 & 15)

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Question 14 (4 marks):
An aquarium has the length of (x + 7) cm, the width of x cm and the height of 60 cm.
The total volume of the aquarium is 48000 cm3. The aquarium will be filled fully with water.
Calculate the value of x.

Solution:


Volume of aquarium=48000  cm 3 So, ( x )( x+7 )( 60 )  cm 3 =48000  cm 3 ( x )( x+7 )= 48000 60 x 2 +7x=800 x 2 +7x−800=0 ( x−25 )( x+32 )=0 x−25=0   or   x+32=0 x=25   or   x=−32( not accepted ) So the value of x=25 cm



Question 15 (4 marks):
Diagram 3 shows a garden path with a rectangular shape. There are 8 similar circular stepping stone built in the path.


Given the area of the path is 32 m2, find the diameter, in m, of one piece of stepping stone.

Solution:
Given the area of the path =32  m 2 =320000  cm 2 x( x+4 )=320000 x 2 +4x−320000=0 a=1, b=4, c=−320000 x= −b± b 2 −4ac 2a x= −4± 4 2 −4( 1 )( −320000 ) 2( 1 ) x= −4± 1280016 2 x= −4+1131.38 2    or    −4−1131.38 2 x=563.69   or   −567.69 ( ignored ) Diameter of one piece of stepping stone=x =563.69 cm =5.64 m

2.5 SPM Practice (Long Questions)

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Question 13 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y= 30 x  by writing down the values of y when x = 2 and x = 5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of  y= 30 x  for 0≤x≤7.  

(c) From the graph in 12(b), find
(i) the value of y when x = 2.6,
(ii) the value of x when y = 17.5.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation  30 x =−5x+30 for 0≤x≤7.
State the values of x.

Answer:
Table 2


Solution:
(a)



(b)



(c) From the graph
(i) When x = 2.6; y = 11.5
(ii) When y = 17.5; x = 1.7

(d)
Given, 30 x =−5x+30   y=−5x+30 From the graph; x=1.25 and 4.75



2.5 SPM Practice (Long Questions)

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Question 12 (12 marks):
(a) Complete Table 2 in the answer space, for the equation y = –x3 + 4x + 10 by writing down the values of y when x = –2 and x = 1.5.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 10 units on the y-axis, draw the graph of y = –x3 + 4x + 10 for –3 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –2.5,
(ii) the positive value of x when y = 4.

(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation x3 – 14x + 5 = 0 for –3 ≤ x ≤ 4.
State the values of x.

Answer:




Solution:
(a)
y = –x3 + 4x + 10
When x = –2
y = –(–2)3 + 4(–2) + 10
y = 8 – 8 + 10
y = 10

When x = 1.5
y = –(1.5)3 + 4(1.5) + 10
y = –3.375 + 6 + 10
y = 12.625

(b)




(c) From graph
(i) When x = –2.5; y = 16
(ii) When y = 4; x = 2.5

(d)
x3 – 14x + 5 = 0
–x3 + 14x – 5 = 0
–x3 + (4x + 10x) + (10 – 15) = 0
–x3 + 4x + 10 = –10x + 15
Thus, y = –10x + 15


From graph, the values of x are 0.35 and 3.5.


2.5 SPM Practice (Long Questions)

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Question 11 (12 marks):
(a) Complete Table 12 in the answer space, for the equation y = –x2 + 2x + 10 by writing down the values of y when x = –1 and x = 2.

(b) For this part of the question, use the graph paper. You may use a flexible curve ruler.
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graph of y = –x2 + 2x + 10 for –3.5 ≤ x ≤ 4.

(c) From the graph in 12(b), find
(i) the value of y when x = –1.5,
(ii) the positive value of x when y = 8.2.
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(d) Draw a suitable straight line on the graph in 12(b) to find the values of x which satisfy the equation 7 – x = x2 for –3.5 ≤ x ≤ 4.
State the values of x.

Answer:



Solution:
(a)
y = –x2 + 2x + 10
When x = –1
y = –(–1)2 + 2(–1) + 10
y = –1 – 2 + 10
y = 7

when x = 2
y = –(2)2 + 2(2) + 10
y = –4 + 4 + 10
y = 10

(b)



(c) From graph
(i) When x = –1.5; y = 4.6
(ii) When y = 8.2; x = 2.7

(d)
y = –x2 + 2x + 10 ……. (1)
0 = –x2 – x + 7 ………. (2)
(1) – (2): y = 3x + 3 



From graph, the values of x are –3.2 and 2.2.


4.10 SPM Practice (Long Questions)

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Question 9:
 (a) Given  1 s ( −4 2 −5 3 )( t −2 5 −4 )=( 1 0 0 1 ), find the value of s and of t.

(b) Using matrices, calculate the value of x and of y that satisfy the following matrix equation:
( −4 2 −5 3 )( x y )=( 1 2 )


Solution:
(a) 1 s ( t −2 5 −4 )= ( −4 2 −5 3 ) −1 = 1 ( −4 )( 3 )−( 2 )( −5 ) ( 3 −2 5 −4 ) = 1 −2 ( 3 −2 5 −4 ) ∴s=−2, t=3

(b) ( −4 2 −5 3 )( x y )=( 1 2 )   ( x y )= 1 −2 ( 3 −2 5 −4 )( 1 2 )   ( x y )= 1 −2 ( ( 3 )( 1 )+( −2 )( 2 ) ( 5 )( 1 )+( −4 )( 2 ) )   ( x y )= 1 −2 ( −1 −3 )   ( x y )=( 1 2 3 2 ) ∴x= 1 2 ,  y= 3 2



Question 10 (5 marks):
During the sport day, students used coupons to buy food and drink. Ali and Larry spent RM31 and RM27 respectively. Ali bought 2 food coupons and 5 drinks coupons while Larry bought 3 coupons and 1 drinks coupon.
Using the matrix method, calculate the price, in RM, of a food coupon and of a drinks coupon.

Solution:
Ali spent RM31. He bought 2 food coupons and 5 drinks coupons.
Larry spent RM27. He bought 3 food coupons and 1 drinks coupons.
x = price of one food coupon
y = price of one drinks coupon

( 2    5 3    1 )( x y )=( 31 27 )             ( x y )= 1 2( 1 )−5( 3 ) ( 1    −5 −3     2 )( 31 27 )             ( x y )= 1 2−15 ( 1( 31 )+( −5 )( 27 ) −3( 31 )+2( 27 ) )             ( x y )= 1 −13 ( −104 −39 )             ( x y )=( 8 3 ) x=8 and y=3 Thus, the price for a food coupon is RM8 and the price for drink coupon is RM3.

4.10 SPM Practice (Long Questions)

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Question 7:
(a) Find the inverse matrix of ( 3 2 5 4 ).
(b) Ethan and Rahman went to the supermarket to buy cucumbers and carrots. Ethan bought 3 cucumbers and 2 carrots for RM9. Rahman bought 5 cucumbers and 4 carrots for RM16.
By using matrix method, find the price, in RM, of a cucumber and the price of a carrot. 


Solution:
(a) Inverse matrix of ( 3 2 5 4 ) = 1 12−10 ( 4 −2 −5 3 ) = 1 2 ( 4 −2 −5 3 ) =( 2 −1 − 5 2 3 2 )

(b) 3x+2y=9.................( 1 ) 5x+4y=16...............( 2 ) ( 3 2 5 4 )( x y )=( 9 16 )               ( x y )=( 2 −1 − 5 2 3 2 )( 9 16 )               ( x y )=( ( 2 )( 9 )+( −1 )( 16 ) ( − 5 2 )( 9 )+( 3 2 )( 16 ) )               ( x y )=( 18−16 − 45 2 +24 )               ( x y )=( 2 3 2 ) x=2,  y= 3 2 ∴Price of a cucumber=RM2    Price of a carrot=RM1.50



Question 8:
The inverse matrix of   ( 4 −1 2 5 ) is t( 5 1 −2 n ).
(a) Find the value of n and of t.
(b) Write the following simultaneous linear equations as matrix equation:
4x – y = 7
2x + 5y = –2
Hence, using matrix method, calculate the value of x and of y.


Solution:
(a) t( 5 1 −2 n )= ( 4 −1 2 5 ) −1 = 1 ( 4 )( 5 )−( −1 )( 2 ) ( 5 1 −2 4 ) = 1 22 ( 5 1 −2 4 ) ∴t= 1 22 , n=4

(b) ( 4 −1 2 5 )( x y )=( 7 −2 )   ( x y )= 1 22 ( 5 1 −2 4 )( 7 −2 )   ( x y )= 1 22 ( ( 5 )( 7 )+1( −2 ) ( −2 )( 7 )+( 4 )( −2 ) )   ( x y )= 1 22 ( 35−2 −14−8 )   ( x y )= 1 22 (  33 −22 )   ( x y )=(   3 2 −1 ) ∴x= 3 2 ,  y=−1