DHS Equations & Inequalities (9758) (Revision Solutions)

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 1 2. Equations and Inequalities (solutions) (I) Equations 1 140dfg++= …(1) 20 20gdf dfg=++ ⇒+−= − …(2) 21 42 10 2900d fg++= …(3) Solving the 3 equations, 20, 40, 80dfg= = = 2 Let x, y, and z be the number of trays of blueberry, strawberry and chocolate cupcakes respectively. Time: 8 7 6 17 60 1020xyz++=×= Amt: 0.6 0.6 0.8 96xyz++= Price: 12 (1) 12 (0.9) 12 (0.8) 1572xy z++= Using GC, 50, 50, 45xyz= = = 3 4 Let x be no. of chickens. Let y be no. of horses. Let z be no. of sheep. 2zx= 20 0x yz⇒ + −= --------------(1) 2x + 4y + 4z = 1250 2 4 4 1250xyz⇒++= ---------(2) Case 1: If 250xyz++= -------------(3) By GC, x = 125− , y = 625, z = 250− (rejected) (Alternative): Reject 250xyz++= , because any combination of 250 animals will never have 1250 legs. (Maximum no of legs = 250 x 4 = 1000)

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 2 Case 2: If 350xyz++= -------------(3) By GC, x = 75, y = 125, z = 150 ∴ Correct number of chickens =75, horses =125, sheep = 150 5 Finding table values in SGD, Cheese/kg Chocolate/kg Candy/kg Price/SGD 4 6 6 Price/SGD 8 10 4 Price/SGD 8 5 7 Let x, y, z be the number of three kg packs bought from Denmark, England and Russia respectively. 4 8 8 84 6 10 5 85 6 4 7 77 xyz x yz xyz ++= + += ++= 5,x= 3,y= 5z= She should buy 13 packs in total. 6 Sub and into . 1 ----- (1) 8 4 2 2 -----(2) abcd abc d +++= + + += Since (2,2) is also the stationary point, h'(2) 0= . i.e. 12 4 0 ----- (3)a bc+ += Using the GC, 11 24 35 24 2 ad bd cd = −− = + =− 0 11 35 24 24 02 ab c dd d ≤   −− +     ≤− 6{ : 2 or 0} 5dd d∈ ≤− − ≤ < (1, 1) (2, 2) h( )yx= 0

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 3 7 22 22 22 : 0 02 2 44 M x y Ax By C A B ABxy C + + + +=   ⇒ + + + −−+ =     Centre of M : ,22 AB−− 2( 1)yx= −+ passes through the centre: 2122 2 4 ----- (1 ) BA AB −= −−+  += At intersection between | |yx= and M, we have 22 || || 0x x Ax B x C+ + + += At x = –2, 22 ( 2) (| 2 |) ( 2) (| 2 | ) 0 2 2 8 ----- (2) AB C A BC − +− + −+ − += − + += − At x = –8, 22 ( 8) (| 8 |) ( 8) (| 8 | ) 0 8 8 128 ----- (3) AB C A BC − +− + −+ − += − + += − Solving (2), (3) and (4) using GC: A = 8, B = –12, C = 32 22: 8 12 32 0Mx y x y++− += 8 Let x : y : z be the ratio for the servings of fish fillet, salad and fries. 150x + 15y + 5z = 4k 60x + 30y + 250z = 8k 25x + 5y + 110z = 3k where k is a constant. 150x + 15y + 5z = 4 60x + 30y + 250z = 8 25x + 5y + 110z = 3 Solving matrix or simultaneous equatio