Mtahs expo waec 2017

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1 CBBACCCBBA

11 ADBBAABCDB

21 BCADBCCAAB

31 CDCACDBACB

41 BDABCDDCAD

Theor- Answers

+ ++ + ++ + ++ + + ++ + ++ + ++ + + ++ + +

SECTION (A ) ANSWER ALL

1 a)

(y - 1 )log 4 ^ 10 = ylog 16 ^ 10

log 4^ 10 (y - 1 )= log 16 ^ y 10

4 ^ ( y - 1)= 16 y

4 ^ y - 1 = 4^ 2 y

y - 1 =2 y

- 1 = 2y = y

- 1 = y

y = - y

1 b)

let the actual time for 5km / hr be t

for 4km / hr=30 mint + t

4 km/ hr= 0. 5 + t

distance = 4( 0 . 5 +t )

= 2* 4 t

for 5km / hr, time = t

distance =5 t

1 +4 t= 5 t

t= 2 hrs

actual distance = 5 * 2= 10 km

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2 a)

2 /3 (3 x - 5 )- 3 / 5( 2x - 3)= 3 / 1

L C M = 15

10 ( 3x - 5 ) - 9 (2 x - 3 )= 45

30 x - 50 - 18 x + 27 = 45

30 x - 18 x = 45 +50 - 27

12 x - 23 =45

12 x = 45 + 23

12 x = 68

x =68 /12

x =34 /6

x =17 /3

2 b)

U ' aS= 180- (n +88 )

= 180- n - 88 = 92 - n

also , u' TQ = 18 m

80 degree + 92 - n +180- m = 180degree

80 + 92 + 180- n - m =180degree

352- n - m = 180degree

- n - m = 180- 352

- n - m = - 172

+ (n +m )= + 172

m + n= 172dgree

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3 a)

Tan 23 . 6 ° = h/ 50

Cross multiply

Tan 23 . 6 ° x h/ 50

h = 50 tan 23 . 6 °

= 21 . 844m

3 b)

Area of

A = 1 /2 bh

45 = 1/ 2 x 10 x h

45 = 5h

h = 9 cm

Area of < QTUS = 1 / 2 ( QT + US )h

= 1 /2 ( 6 + 16 ) 9

= 99 cm ^ 2

================================

4 a)

T 6= 37

T 6= a +( 6 - 1 )d

T 6= a +5 d

a +5 d =37 - - - - - (eq 1)

s 6= 147

sn= n /2 (2 a +( n- 1) d )

147= 3( 2a + 5d )

49 = 2 a+ 5 d

2 a+ 5 d= 49 - - - - ( eq2 )

a +5 d =37 - - - ( eq1 )

2 a+ 5 d= 49 - - - (eq 2)

a =12

4 b)

S 15 = 15 / 2( 2( 12 ) +14 d)

S 15 = 15 / 2( 24 + 14 d )

from( 1 )

a +5 d =37

12 + 5 d= 37

5 d= 37 - 12

5 d= 25

d =5

S 15 = 15 /2 (24 +14 (15 )

S 15 = 15 / 2 (24 + 70 )

S 15 = 15 / 2* 94

S 15 = 15 * 42

S 15 = 630

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5 a)

draw

U =20

B = y - 45

S = y - 34

B =bag

S =shoe

let n ( B)= y

n (S )= y +11

for bag only y - 45

for shoe only y - 11 - 45 =y - 34

5 b)

y - 45 + 45 + y - 34 =120

2 y - 34 = 120

2 y = 154

y =154/ 2

y =77

number of customers who bought shoe = y + 11

77 + 11 = 88

5 c )

n (bag)= 77 customers

probability = 77 /120

= 0. 642

================================

SECTION (B ) ANSWER 5 QUESTIONS ONLY

10 a )

Sin x = 5 /13

Using pythagoras rule

M ^ 2 = 13 ^ 2 - 5 ^ 2 (^ means Raise to power)

M ^ 2 = 169 - 25

M ^ 2 = 144

M = √ 144

M = 12

Hence:

Cos x - 2 sin x / 2 tan x

12 / 13 - 2 (5 /13 ) / 2 (5 / 12 )

= 12 / 13 - 10 /23 / 5/ 6

FIND LCM

= 12 - 10 /13 / 5/ 6

= 12 / 65

10 bi )

Considering < LMB

/MB / ^ 2 . = 12 ^ 2 - 9 . 6^ 2

/MB / ^ 2 = 51 . 84

/MB / = √ 51 . 84

/MB / = 7 . 2 m

From < AML

/LA /^ 2 = 2 . 8 ^ 2 + 9 . 6 ^ 2

/LA / ^ 2 = 100

/LA / = √ 100

/LA / = 10 m

10 bii )

Let the angle be. θ

From < AML

Tanθ = 9 . 6 /2 . 8

Tan θ = 3 . 4288

θ = Tan ^ - 1 ( 3. 4288)

= 73 . 74 °

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ANSWER LOADING . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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