3.4.2 Sets, SPM Paper 1 (Short Questions 5 – 7)


3.4.2 Sets, SPM Practice Paper 1 (Short Questions)

Question 5:
Diagram below shows a Venn diagram with the number of elements of set P, set Q and set R.

It is given that the universal set, ξ=PQR and n( Q' )=n( QR ).
Find the value of x.

Solution:
n(Q)’ = n(QR)
3 + 8 + 5 = x– 3 + 9
16 = x + 6
x = 10


Question 6:
Diagram below is a Venn diagram showing the number of quiz participants in set P, set Q and set R.
It is given that the universal set, ξ = P Q R , set P = {Science quiz participants}, set Q = {Mathematics quiz participants} and set R = {History quiz participants}.

If the number of participants who participate in only one quiz is 76, find the total number of the participants.

Solution:
Number of participants who participate in only one quiz = 76
(5x – 2) + (x + 6) + (2x + 8) = 76
8x + 12 = 76
8x = 64
x = 8
Total number of the participants
= 76 + 7 + 4 + 5 + 3(8)
= 116


Question 7:
Diagram below is a Venn diagram showing the number of students in set K, set L and set M.
It is given that the universal set, ξ = K L M , set K = {Karate Club}, set L = {Life Guards Club} and set M = {Martial Arts Club}.

If the number of students who join both the Life Guards Club and the Martial Arts Club is 8, find the number of students who join only two clubs.

Solution:
Number of students who join both the Life Guards Club and the Martial Arts Club = n(M) = 2 + 2x
2 + 2x = 8
2x = 6
x = 3
Number of students who join only two clubs
= x + 4 + 2x
= 3 + 4 + 2(3)
= 13

3.4.1 Sets, SPM Paper 1 (Short Questions 1 – 4)


Question 1:
List all the subsets of set P = {r, s}.
 
Solution:
There are 2 elements, so the number of subsets of set P is 2n = 22= 4.
Set P = {r, s}
Therefore subsets of set P = {r}, {s}, {r, s}, { }.


Question 2:
 
Diagram above shows a Venn diagram with the universal set, ξ = Q υ P.
List all the subset of set P.
 
Solution:
Set P has 3 elements, so the number of subsets of set P is 2n = 23 = 8.
Set P = {2, 3, 5}
Therefore subsets of set P = { }, {2}, {3}, {5}, {2, 3}, {2, 5}, {3, 5}, {2, 3, 5}.


Question 3:
It is given that the universal set, ξ = {x: 30 ≤ x < 42, x is an integer} and set P = {x: x is a number such that the sum of it its two digits is an even number}.
Find set P’.
 
Solution:
ξ = {30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41}
P = {31, 33, 35, 37, 39, 40}
Therefore P’ = {30, 32, 34, 36, 38, 41}.


Question 4:
Given that universal set ξ = {x : 3 < x ≤ 16, is an integer},
Set A = {4, 11, 13, 16},
Set B = {x : x is an odd number} and
Set C = {x : x is a multiple of 3}.

The elements of the set (A υ C)’ ∩ B are

Solution:

ξ = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}
A = {4, 11, 13, 16}
B = {5, 7, 9, 11, 13, 15}
C = {6, 9, 12, 15}
(A υ C)’ = {5, 7, 8, 10, 14}
Therefore (A υ C)’ ∩ B = {5, 7}.

3.3 Operations on Sets

3.3b Union of Sets (Part 2)
Example 1:
The Venn diagram below shows the number of elements in the universal set, ξ, set P, set Q and R.


Given n(Q) = n(P υ R)’, find n(ξ).
 
Solution:
n(Q) = n (P υ R)’
2x + 6 + 1 + 5 = 2x + 2x
2x + 12 = 4x
2x = 12
= 6
 
n(ξ) = 2x + 2x + x + 7 + 6 + 1 + 5
= 5x+ 19
= 5(6) + 19
= 30 + 19
= 49
 


Example 2:
Diagram below is a Venn diagram showing the universal set, ξ = {Form 3 students}, set = {Students who play piano} and set = {Students who play violin}.
 

 
Given n(ξ) = 60, n(A) = 25, n(B) = 12 and n(AB) = 8, find the number of students who do not play the two instruments.

Solution:
The students who do not play the two instruments are represented by the shaded region, (A υ B)’.

 
Number of students who do not play the two instruments
= n (A υ B)’
= 60 – 17 – 8 – 4
= 31
 

3.3a Intersection of Sets (Part 2)

3.3a Intersection of Sets
Example 1:
Given that A= {3, 4, 5, 6, 7}, B = {4, 5, 7, 8, 9, 12} and C = {3, 5, 7, 8, 9, 10}.
(a) Find ABC.
(b) Draw a Venn diagram to represent ABC.
 
Solution:
(a)  ABC= {5, 7}

(b)


 

4. The complement of the intersection of two sets, P and Q, represented by (PQ)’, is a set that consists of all the elements of the universal set, ξ, but not the elements of PQ.

5. The complement of set (PQ)’ is represented by the shaded region as shown in the Venn diagram.


 
 
 
 

Quadratic Equations Long Questions (Question 11 – 13)

Question 11:
Solve the following quadratic equation:
5x+3 x 2 12x =6

Solution:
5x+3 x 2 12x =6 5x+3 x 2 =612x 3 x 2 +5x+12x6=0 3 x 2 +17x6=0 ( 3x1 )( x+6 )=0 3x1=0     or     x+6=0      3x=1                   x=6        x= 1 3                   x=6


Question 12:
Diagram above shows a rectangle ABCD.
(a) Express the area of ABCD in terms of n.
(b) Given the area of ABCD is 60 cm2, find the length of AB.
 
Solution:
(a)
Area of ABCD
= (n + 7) × n
= (n2+ 7n) cm2

(b)
Given the area of ABCD = 60
n2+ 7n = 60
n2+ 7n – 60 = 0
(n – 5) (n + 12) = 0
= 5 or    n = – 12 (not accepted)
 
When n = 5,
Length of AB = 5 + 7 = 12 cm



Question 13 (4 marks):
Solve the following quadratic equation:
2 3x5 = x 3x1

Solution:
2 3x5 = x 3x1 2( 3x1 )=x( 3x5 ) 6x+2=3 x 2 5x 3 x 2 5x+6x2=0 3 x 2 +x2=0 ( 3x2 )( x+1 )=0 3x2=0     or     x+1=0 x= 2 3      or     x=1

SPM Practice 2 (Question 8 – 10)

Question 8:
The variables x and y are related by the equation y= p 3 x , where k is a constant. 
Diagram below shows the straight line graph obtained by plotting log 10 y against x.

 

  1. Express the equation y= p 3 x in its linear form used to obtain the straight line graph shown in Diagram above.
  2. Find the value of p.


Solution:


 

Question 9:
Variable x and y are related by the equation y 2 =p x q . When the graph lg y against lg x is drawn, the resulting straight line has a gradient of -2 and an vertical intercept of 0.5 . Calculate the value of p and of q.

Solution:





 

Question 10:
Variable x and y are related by the equation y= c dx . When the graph y against xy is drawn the resulting line has gradient 0.25 and an intercept on the y-axis of 1.25. Calculate the value of c and of d.

Solution:

 





SPM Practice 2 (Question 6 & 7)

Question 6:
The variables x and y are related by the equation y=p x 3 , where p is a constant. Find the value of p and n.

 

Solution:



 

Question 7:
Diagram A shows part of the curve y=a x 2 +bx .  Diagram B shows part of the straight line obtained when the equation is reduced to the linear form. Find
(a) the values of a and b,
(b) the values of p and q.


Diagram A

Diagram B

Solution:





 

SPM Practice 2 (Question 4 & 5)

Question 4:
The diagram shows part of the straight line graph obtained by plotting y x against x .


Given its original non-linear equation is  y=px+q x 3 2 .  Calculate the values of p and q.

Solution:




 

Question 5:
The diagram below shows the graph of the straight line that is related by the equation x y = 2 x +3x .


Find the values of p and k.

Solution:



 

(G) Sum to Infinity of Geometric Progressions (Part 2)

(H) Recurring Decimal
Example of recurring decimal:
2 9 = 0.2222222222222..... 8 33 = 0.242424242424..... 41 333 = 0.123123123123.....

Recurring decimal can be changed to fraction using the sum to infinity formula:
S = a 1 r


Example (Change recurring decimal to fraction)
Express each of the following recurring decimals as a fraction in its lowest terms.
(a) 0.8888 ...
(b) 0.171717...
(c) 0.513513513 ….

Solution:
(a)
0.8888 = 0.8 + 0.08 + 0.008 +0.0008 + ….. (recurring decimal)
G P , a = 0.8 , r = 0.08 0.8 = 0.1 S = a 1 r S = 0.8 1 0.1 S = 0.8 0.9 S = 8 9 check using calculator 8 9 = 0.888888....

(b)

0.17171717 …..
= 0.17 + 0.0017 + 0.000017 + 0.00000017 + …..
G P , a = 0.17 , r = 0.0017 0.17 = 0.01 S = a 1 r S = 0.17 1 0.01 = 0.17 0.99 = 17 99 remember to check the answer using calculator

(c)
0.513513513…..
= 0.513 + 0.000513 + 0.000000513 + …..
G P , a = 0.513 , r = 0.00513 0.513 = 0.001 S = a 1 r S = 0.513 1 0.001 = 0.513 0.999 = 513 999 = 19 37

3. The nth Term of Geometric Progressions (Part 2)

(D) The Number of Term of a Geometric Progression
Smart TIPS: You can find the number of term in an arithmetic progression if you know the last term

Example 2:
Find the number of terms for each of the following geometric progressions.
(a) 2, 4, 8, ….., 8192
(b) 1 4 , 1 6 , 1 9 , ..... , 16 729   
(c) 1 2 , 1 , 2 , ..... , 64  

Solution:
(a)
2 , 4 , 8 , ..... , 8192 ( Last term is given) a = 2 r = T 2 T 1 = 4 2 = 2 T n = 8192 a r n 1 = 8192 ( T h e n th term of GP, T n = a r n 1 ) ( 2 ) ( 2 ) n 1 = 8192 2 n 1 = 4096 2 n 1 = 2 12 n 1 = 12 n = 13

(b)
1 4 , 1 6 , 1 9 , ..... , 16 729 a = 1 4 , r = 1 6 1 4 = 2 3 T n = 16 729 a r n 1 = 16 729 ( 1 4 ) ( 2 3 ) n 1 = 16 729 ( 2 3 ) n 1 = 16 729 × 4 ( 2 3 ) n 1 = 64 729 ( 2 3 ) n 1 = ( 2 3 ) 6 n 1 = 6 n = 7

(c)
1 2 , 1 , 2 , ..... , 64 a = 1 2 , r = 2 1 = 2 T n = 64 a r n 1 = 64 ( 1 2 ) ( 2 ) n 1 = 64 ( 2 ) n 1 = 64 × 2 ( 2 ) n 1 = 128 ( 2 ) n 1 = ( 2 ) 7 n 1 = 7 n = 8


(E) Three consecutive terms of a geometric progression
If e, f and g are 3 consecutive terms of GP, then
g f = f e
Example 3:
If p + 20,   p − 4, p −20 are three consecutive terms of a geometric progression, find the value of p.

Solution:
p 20 p 4 = p 4 p + 20 ( p + 20 ) ( p 20 ) = ( p 4 ) ( p 4 ) p 2 400 = p 2 8 p + 16 8 p = 416 p = 52