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Mathematics answer 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 - 1log 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 ================================ 2 a) 2 /3 (3 x - 5 )- 3 / 5( 2x - 3)= 3 / 1 L C M = 15 10 ( 3x - 5 ) - 9 (2 x - 345 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 ================================ 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 ================================ 5 a) draw U =20 B = y - 45 S = y - 34 B =bag S =shoe let n ( B)= y n (Sy 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 ° 13ai) given x(*)y=x y/2 i)3(*)2/5=3 2/5/2 15 2/5)*1/2 =17/5*1/2 =17/10= 1,7/10 13aii) 8(*)y=8^1/4 =8 y/2 =33/4 32 4y=66 4y=66-32 4y=34 y=34/4 y=17/2 y=8^1/2 13b) given DABC AB=(^-4/6) and AC 3/^-8) so AP =1/2(^-4/6) AP=(^-2/3) hence CP = CA AP CP= -(3/^8) (^-2/3) CP = (^-5/11)