WAEC Mathematics Past Questions & Answers - Page 91

let the number of N50 notes = x let the number of N20 notes = y from the first statement x + y = 13.....(1) from the second statement 50x + 20y = 530 ......(2) dividing through by 10, we have 5x + 2y = 53 ....(20 solving simultaneously by eliminating x + y = 12....(1) 5x + 2y = 53.....(2) multiply eqn (1) by 2 multiply eqn (2) by 1 2x + 2y = 26 -5x + 2y = 53 -3x = -27 x = 9 substitute x = 9 into equation (1) x + y = 13 y = 13 - x = 13 - 9 = 4 y = 4 the number of N50 notes = 9

No explanation has been provided for this answer.

No explanation has been provided for this answer.

let the chord be AB = 7cm

radius OA = 3.7cm distance of the = OM using Pythagoras theorem

OA² = AM² + OM²

3.7² = 3.5² + OM²

13.69 = 12.25 + OM²

13.69 - 12.25 = OM²

1.44 = OM²

OM = \(\sqrt{1.44}\)

OM = 1.2cm

distance of chord = 1.2cm

diameter = 12cm

radius = \(\frac{12}{2}\)cm = 6cm

area of sector = 66cm²

area of sector = area of cone

66 = \(\pi rl\)

66 = \(\frac{22}{7} \times r \times 6\)

divide both sides by 6

11 = \(\frac{22}{7}r\)

22r = 11 x 7

r = \(\frac{77}{22}\)

= 3.5cm