Seniour Four Mathematics Paper Two Solutions

SENIOUR FOUR MATHEMATICS PAPER TWO SOLUTIONS

SECTION A:

1.4½ + ⅔ = 9/2 + ⅔

4½ + ⅔ 9/2 x ⅔

=27 + 4

=31/6

18/6

2.a *b=means a third of the last digit in ab.

(a)6 * 4 = 6 x 4

=24

= x 4

(b)3 * (4*8)

4 * 8 = 4 x 8

= 32

= x 2

3 * = 3 x

= 2

= x 2 =

3.Let A receive x votes

B receives x = x

C receives x x = x

Therefore Total number of votes

=x + x + x

=8x + 6x + 3x

(ii)x = 75

x = 75 x 8

= 25 x 8

= 200

x = x 200

= 17 x 25

= 425 votes

4.Solve h + t = 1………………………….(i)

2 44

t - 3h + 1 = 0 ………………………….(ii)

8 4 2

From (i) (i) x 4

2h + t = 1

(ii) x 8 = t – 6h +4 = 0

Rearranging2h + t = 1

-6h + t = -4

8h = 5

h =

since 2h + t = 1

t = 1-2h

t = 1 - 2()

= 1-

t =

Using Similarity

h = 3.5

20+h7

20cm

h=1

20+h2

hcm

2h =20+h

h=20cm

Volume of pail=Volume of -Volume of

whole cone cut away

=⅓ π (72) (40) - ⅓ π (3.5)2(20)

=1960π-245π

3 3

=1715π cm3

=17795.03cm3

6.A(-3,1)B(2,-11)

lABl = √(-3-2)2+ (1-- 11)2

= √ 25 +144

= √ 169

= 13 units

Area= πr2

= π x 132

= 169 π cm2

= 530.66 cm2

7. /////////////////////////////////////////////

xm xm

60- 2x

Let the width be x

Since the length of the material is 60m, then length of rectangle is 60 – 2x

Therefore x (60 – 2x) = 400

60x - 2x2 = 400

2x2 - 60x + 400 = 0

x2 – 30x + 200 = 0

Using x = b± √(b2 – 4ac)

= 30 ± √ -302 - 4(200)

x = 30 ± √900 - 800

= 30 ± √100

x = 30 + 10Or x =30 - 10

2 2

x = 20Or x = 10

Therefore, x = 10m (Since x = 20m gives a square)

x y2 and y

x = K1y2 - - (i)y = ---(ii)

If y = 3 when Z = 2, x = 12

Using (i)

12 = K1(32)

K1 = =

Therefore, x = y2 - (iii)

Also using (ii)

3 = K2

K2 = 6

Therefore, y = --- (iv)

If Z = 4

From (iv)

y =

From (iii)

X = (3/2)2

= x

= 3

9.n(A) = 15n(B) = 13n(AUB) = 24n(ε) = 28

Draw a venn diagram

n(AUB)' = n(ε) - n(AUB)

= 28 - 24

= 4

Therefore, 15 – x + x + 13 – x = 24

28 – x = 24

x = 28-24

x = 4

Therefore, n (AnB') = 15 – x

= 15 – 4

= 11

10.Departure from Fortportal -0720 hrs

Arrival at Mubende - 1140 hrs

Time taken-4 hrs 20 min

Distance -180 Km

Departure from Mubende -1240 hrs

Arrival at Mityana -1340 hrs

Time taken -1 hr

Distance -140 Km

Average Speed =Total distance=180 + 140

Total time 5 hrs 20 mins

320=320 ÷ 5 ⅓

5 20/60

=320 x 3/16 = 60 Km/h

SECTION B

11.

(a)Angle of elevation of S from O is

Then tan = 70/280

tan =

= tan -1 (1/4)

=14.040

(b)The height of TB

Using Δ OAB

OB2= 2802 + 862

= 78400 + 7396

= 85796

OB=√ 85796

=292.9095m

Therefore, tan 140 =

h = 292.9095 x tan 140

h = 292.9095 x 0.2493 = 73.0305m

(c)Area of OAB= x 86 x 280

= 12040m2

Since 1000m = 1Km

(1000m)2 = 1Km2

106m2 = 1Km2

12040m2 = 12040 Km2

106

=1.2040 x 104 x 10-6

=1.2040 x 10-2 Km2

Therefore, Using a Scale 8cm : 1Km

64 cm2 : 1Km2

Y: 1.2040 x 10-2Km2

Therefore, y = 1.2040 x 10-2 x 64

=0.7056 cm2

=0.77 cm2

12.

(i)Mean =

= 117.3571

(ii)Median position = x 49

=24.5th

From graph, Median =109.5 + x 10

=116.5

13(a)Eight bulbs with 5 defective

(i)P(both defective)= x

(ii)P(One defective)=( x ) or ( x )

= +

= =

(b)P(3 defectives)

= x x

(ii)P (2 defective, 1 is not)

P(dd or dd or dd)

= (x x ) + (x x ) + ( x x)

= 3 x

(iii)P(at least one bulb is not defective)

= 1 - P(all defective)

= 1 - (from part (i))

14.

(a)BF;

BF2= 102 + 102

= 100 + 100

= 200

BF = √200

= 14.14 (4 sig.fig)

(b)BE;

BE2=BD2 + DE2

BD2=102 + 102

=100 + 100

=200

BE2=200 + 102

= 200 + 100

=300

BE=√300

=17.32 (4 sig.fig)

=0.707

10cm=tan-1 (0.707)

=35.260

D cm B

14. (d)

15.2x + 3y<6, y-2x≤2,y≥0

For 2x +3y = 6, y-2x=2

X / 0 / 3
Y / 2 / 0
x / 0 / -1
y / 2 / 0

(a)Possible values of (x,y)

x / -1 / 0 / 0 / 1 / 1 / 2
y / 0 / 0 / 1 / 0 / 1 / 0
x+y / -1 / 0 / 1 / 1 / 2 / 2

Therefore, the greatest value of x+y is 2

(b)Area = x (4 x 2)

=4 square units

16.

=, =, =, =

:=4:1

(a)(i)=

(ii) = +

= -+

= -

(iii)= +

, = + ( – )

= + –

= +

= ( +)

(iv)= +

but = , =

= ( ( + ) +

= ( + ) +

= - +

(b)= +

= +

= ( ( + ) +

=– +

=+ + 10

, Since =t+ k

Then t = and k =

17.Graphs of

y=(x+2) (3-x) and y=2x

for -3 ≤ x ≤3

For y = (x + 2) (3 – x)

x / -3 / -2 / -1 / 0 / 1 / 2 / 3
x + 2 / -1 / 0 / 1 / 2 / 3 / 4 / 5
3 - x / 6 / 5 / 4 / 3 / 2 / 1 / 0
y / -6 / 0 / 4 / 6 / 6 / 4 / 0

For y = 2x

x / -3 / -2 / -1 / 0 / 1 / 2 / 3
y / -6 / -4 / -2 / 0 / 2 / 4 / 6

(i)Points of intersection are

(-3,-6) and (2,4)

(ii)To find the roots of

6 + x – x2 = 0

Since

(x + 2) (3 – x) = 0

3x - x2 + 6 - 2x = 0

6 + x - x2 = 0

Then if y = (x + 2) (3-x) from graph

y = 0

Therefore, the solutions are x = -2 or 3

Gayaza High School Mathematics Department-ExercisesPage 1