1 a ) Tabulate x - 1 , 2 , 3 , 4 1 - 1 , 2 , 3 , 4 2 - 2 _ , 4, 0 _ , 2 _ 3 - 3 , 0 _ , 3 , 0 _ 4 - 4 _ , 2_ , 0 _ , 4 1 b ) I = PRT/ 100 , p =N 15000 R = 10 % and I = 3 years A = P+ I where I = 15000* 10 * 3 /100 = N4500 A = 4500 + 15000 = N19500 NO 1 a , b IMAGE ================ 2 a ) using sine rule b / sin20 = 6/ sin30 bsin30 = 6 sin120 b 6 sin120/ sin30 b = 6 x 0 . 2511 /0 . 4540 b = 5 . 7063 / 0. 4540 b = 12 . 57 ≠ 12 . 6 cm 2 bi) the diagram is euivalent triangles . where | AX |/ |BC| = | BY |/ |AC | = | XY |/ |YC | XY = 9 , BY = 7 YC = 18 - 7 = 11 9 / 11 = 7 / |AC | 9 | AC | = 77 | AC | = 77 /9 | AC | = 8 cm 2 bii ) XY / AB = BY /AC 9 / |AB | = 7 / 8 . 6 | AB | = 9 x 8 . 6 / 7 | AB | = 11 cm ================= 3 ) let the son age be x man = 5 x son = x 4 yrs ago; the man age = 5 x - 4 the son age = x - 4 the product of their ages ( 5 x - 4 ) ( x - 4 ) = 448 ================= 7 a ) 3 ^ 2 n + 1 - 4 ( 3 ^ n + 1 ) + 9= 0 3 ^ 2 - 3 - 4 ( 3 ^ n - 3 ) + 9 = 0 ( 3 ^n )^ 2 - 3 - 4 ( 3 ^n - 3 ) + 9 = 0 let 3 ^ n = p p ^ 2 - 3 - 4 ( p - 3 ) + 9 = 0 3 p ^ 2 /3 - 12 p / 3 + 9 /3 = 0 p ^ 2 - 4 p + 3 = 0 p ^ 2 - 3 p - p + 3 = 0 p ^ 2 p ( p - 3 ) - 1 ( p - 3 ) = 0 ( p - 1 ) ( p - 3 ) = 0 p - 1 = 0 or p - 3 = 0 p = 1 or 3 Recall 3 ^ n = p when p =1 3 ^ n = 3^ 0 n = 0 when p = 3 3 ^ n = 3^ 1 n = 1 7 b ) log( x ^ 2 + 4) = 2 + logx - log^ 20 log( x ^ 2 + 4) = log^ 100 = log^ x - log^ 20 ( x ^2 + 4 ) = log( xx ) x ^ 2 + 4 = 5 x x ^ 2 - 5 x + 4 = 0 x ^ 2 - 4 x - x + 4 = 0 x ( x - 4 ) - 1 ( x - 4 ) = 0 ( x - 1 ) ( x - 4 ) = 0 x - 1 = 0 or x - 4 = 0 x = 1 or 4 4a ) volume of fuel = cross - sectional area of X depth of fuel rectangular tank 30 , 000litres = 7 . 5* 4 . 2 * d m ^ 3 but ; 1000 litres = 1m ^ 3 therefore ;30 ( M ^ 3 ) = 7 . 5 * 4 . 2 * d( M ^ 3 ) 30 = 31 . 5d ====> d = 30 /31 . 5 = 0 . 95 (2 d. p) 4b ) to fill the tank / volume of fuel needed = 7. 5 * 4 . 2 * 1. 2 = 37 . 8m ^ 3 = 37 , 800 litres addition fuel = 37 , 800- 30 , 000 = 7, 800 litres therefore , 7 , 800 more litres would be needed ================================== 5a ) sector for building project = 48000 / 144000* 360= 120degree sector for education = 32 , 000/144000* 360=80 degree sector for saving = 19200 /144000* 360=48 degree sector for maintenance = 12000 / 144000* 360= 30 degree sector for miscellaneous = 7200 / 144000* 360= 18 degree sector for food items = 360- (120+ 80 + 48 + 30 +18 ) =360- 296 =64 degree 5b ) amount spent = 144000- [ 48 , 000+ 32000 + 19200 + 12000 + 7200 ] =144000- 118400 =N 25600 6 ) tabulate No 41 . 02 sqr 0 . 7124 42 . 81 0 . 207 0 . 0404 antilog ( 0 . 6616 ) = 4 . 587 log 1 . 6130 ===> 1 . 6130 T . 8527 / 2 = 2 ^ - + 1 . 8527 / 2 + T + 0 . 9264 ===> T . 9264 1. 5394 1 . 6321 T . 3160 + 2 ^ - . 6064 T . 5545 ===> T . 5545 1 . 9849 ===> 1 . 9849 / 3 = 0. 6616 ========= ( 9 a ) Let the lens digit x and unit digit be y , therefore x - y = 5 - - - ( 1 ) 3 xy - ( 10 x + y )=14 - - - - ( 2 ) 3 xy - 10 x - y = 14 - - - - - ( 3 ) frm eq ( 1); x = 5 + y - - - ( 4 ) therefore , 5( 5 + y ) ( y ) - 10 ( 5 + y ) - y = 14 ( 15 + 3 y ) y - 50 - 10 y - y = 14 3 y ^ 2 + 4 y - 50 - 14 = 0 3 y ^ 2 + 4 y - 64 = 0 3 y ^ 2 - 12 y + 16 y - 64 = 0 3 y ( y - 4 ) + 16 ( y - 4 )= 0 ( 3 y + 16 ) ( y - 4 )=0 y = - 16 / 3 or 4 therefore from eqn ( 1 ); x + 4 = 5 x = 5 + 4 = 9 the number is 94 ( 9 b ) ( 3 - 2 x ) / 4 + ( 2 x - 3 ) / 3 ( 3 ( 3 - 2 x ) + 4 ( 2 x - 3 )) / 12 ( 9 - 6 x + 8 x - 12 ) /12 = ( 2 x - 3 ) /12 ........... Maths obj complete and verified======== =========== 1=10 deacdacbbe 11=20 cbaeebdacc 21=30 deedacbbce 31=40 abdcebaeec 41=50 caadedbaea 51=60 edeacbacdd ================ NECO NIGERIA POWERED BY NOLLYWAP.TK...INVITE UR FRIENDS AND PLZZ DROP UR COMMENTS..GOODLUCK.