__MATHEMATICS ANSWERS__**OBJ:**

1-10: DBABCBAEDD

11-20: BEBACABCCB

21-30: DACAEDBDEA

31-40: BEEDDDEBBB

41-50: CAADCDDDBC

**ESSAY:**

(1B)

X={2,4,6,8}

Y={2,3,7,9}

(XUY) = {2,3,4,6,7,8,9}

Y={2,3,7,9}

Z={4,5,6,7,8}(YUZ) = {2,3,4,5,6,7,8,9}

{XUY} n {YUZ}={2,3,4,6,7,8,9}ANS

…………………………….

(2A)

A=35

X=7

Y=4

A proportion X 1/Y^2

A = XK/Y^2

When K is Constant

35 = 7K/42

35= 7K/16

35*16=7K

560 = 7K

K = 560/7

K=80

.:.

A = XK/Y^2

A=5 * 80/8^2

A=400/64

A=6.25

(2B)

If S = ut + 1/At^2 Make T the subject of formula of the equation.

S = ut + 1/At^2

SAt^2 = ut + 1

SAt^2 – ut =1

SA – u (t^2-t) =1

t^2 – t = 1/SA – u

t = 1/SA-u

(3) Using Ruler And a Pair of Compasses Only,construct triangle ABC with /AB/=10cm, /BC/=12cm and =120 degree, construct the locus l of ponts equidistant from A and B.

DRAW IT LIKE THE ABOVE IMAGE ON YOUR ANSWER SHEET.

(4A) Using Completing The Square Method Solve The equation. X^2 – X – 12 = 0

X^2-X-12=0

X^2-X/2 – (1/2)^2 = 12/2 + 1/4

(X^2 – 1/2)^2 = 24+1/4 = 25/4

(X-1/2)^2 = √25/4

X-1/2 = ± √25/4

X – 1/2 = ± 5/2

X = 1/2 + 5/2OR1/2 – 5/2

X = 1+5/2OR1-5/2

X = 6/2 OR -4/2X=3OR-2ANS

**SECTION B:**Answer (5) Question in This Section.

5th term = T5 = a + 4d

14th term = T14 = a + 13d

T5 = a + 4d = 28……………….(equa 1)

T14 = a + 13d = 82 …………..(equa 2)

Solving the simultaneous equationsSubtract equation (1) from equation (2)

a + 4d =28………(1)

a + 13d = 82 …..(2)

9d = 54

:. d = 54/9

d=6

Substitute (d) in equation (1)

a + 4 (6) = 28

a + 24 = 28

a = 28 – 24

a=4

To find the 21st term

T21 = a + 20d

=4 + 20 (6)

= 4 + 120

= 124

12A)

Let the distance be a,c

In ∆ABC ^A=90-205=115°(A FROM glasses emoticon

also ABC=180-205=25°

:.B=60-25=35

:.C=180-(35+115)=180-150=30°

USING SINE RULE

(i)a/sinA =b/sineB

a=B sine a/sine B= 15 sin 115°/sin30°=15×0.9404/0.9880=14.181/0.9880

=14.20a=14.20kmii)similarlySin b=C sin B/sin C =15 sin

35°/sin30°=15×0.4281/0.9880

=6.4215/0.9880

=6.499km

:.distance of B from A =14.20km

:.bearing of B from A= 6.499 km

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