Mathematica may june 1999

Answer

(a) If ε is the set 1,2,3,…,19,20 and A, B and C are subsets of ε such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;

(b) Find : (i) A∩B ; (ii) A∩C ; (iii) B∪C.

(c) Using your results in (b), show that (A∩B)∪(A∩C)=A∩(B∪C).

Answer

(a)(i) A = {5, 10, 15, 20}

(ii) B = {4, 8, 12, 16, 20}

(iii) C = {3, 6, 9, 12, 15, 18}

(b) (i) A∩B=20
(ii) A∩C=15
(iii) B∩C=12
(c) (A∩B)∪(A∩C)=A∩(B∪C)
(A∩B)∪(A∩C)=15,20
A∩(B∪C)=5,10,15,20∩3,4,6,8,9,12,15,16,18,20
= 15,20
∴(A∩B)∪(A∩C)=A∩(B∪C)

In the diagram, ABCD is a trapezium in which AD∥BC and <ABC is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :

(a) |AB| ;

(b) the area of the triangle BCD ;

(c) |CD| ;

(d) perimeter of the trapezium.

Answer

(a) |AB| =  √ |BD|²−|AD|2
= √ 172−152  = √289−225
= √64 = 8cm

(b) Area of triangle BCD = 1/2 × 9 × 8
= 36cm2

(c) |EC| = 6cm
     |CD| = √82+62 = √100
             = 10 cm.

(d) Perimeter of the trapezium =
AD + AB + BC + CD  = 15 cm + 8 cm + 9 cm + 10 cm

= 42 cm.

(a) Solve the simultaneous equations 3y – 2x = 21 ; 4y + 5x = 5.

(b) Six identical cards numbered 1 – 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.

Answer

(a) 3y − 2x = 21.. .(1)
      4y + 5x = 5…. (2)

Multiply (1) by 4 and (2) by 3 so we have,

12y−8x=84…(3)
12y+15x=15…(4)

(3) – (4) : −8x−15x = 69 ⟹ −23x = 69
x = 69 − 23 = −3

Put x =  -3 in (1),

3y−2(−3) = 21
3y + 6 = 21 ⟹  3y = 21−6 = 15
y = 15 / 3 = 5

x, y = −3,5.

(b)(i) Probability of losing

Probability of winning = 1 /6
∴ Probability of losing = 1− 1 / 6 = 5 /6.

(ii) Losing $20.00 after two picks = picking the card numbered 1 twice.

= 1/ 6 × 1/ 6 =   1 / 36