Past papers › WAEC · 2022 · May/June · General Maths · Paper 2 › Question 1 Question WAEC General Maths 2022 Theory Sequences & series (AP, GP) Quadratics & their graphs Linear & simultaneous equations Sequences & series (AP, GP), Quadratics & their graphs, Linear & simultaneous equations
(a) Given that ( 7 − 2 x ) (7 - 2x) ( 7 − 2 x ) 9 9 9 ( 5 x + 17 ) (5x + 17) ( 5 x + 17 ) r r r x x x
(b) Two positive numbers are in the ratio 3 : 4 3 : 4 3 : 4
Worked solution (try it first) (a) In a G.P., each term divided by the one before gives the same common ratio:
9 7 − 2 x = 5 x + 17 9 \frac{9}{7 - 2x} = \frac{5x + 17}{9} 7 − 2 x 9 = 9 5 x + 17 .
Cross-multiply:
81 = ( 7 − 2 x ) ( 5 x + 17 ) 81 = (7 - 2x)(5x + 17) 81 = ( 7 − 2 x ) ( 5 x + 17 ) .
Expand:
81 = 35 x + 119 − 10 x 2 − 34 x 81 = 35x + 119 - 10x^2 - 34x 81 = 35 x + 119 − 10 x 2 − 34 x = − 10 x 2 + x + 119 = -10x^2 + x + 119 = − 10 x 2 + x + 119 .
Rearrange:
10 x 2 − x − 38 = 0 10x^2 - x - 38 = 0 10 x 2 − x − 38 = 0 .
Factorise:
( x − 2 ) ( 10 x + 19 ) = 0 (x - 2)(10x + 19) = 0 ( x − 2 ) ( 10 x + 19 ) = 0 .
So
x = 2 x = 2 x = 2 or
x = − 19 10 x = -\frac{19}{10} x = − 10 19 .
(b) The numbers are in the ratio
3 : 4 3 : 4 3 : 4 , so let them be
3 k 3k 3 k and
4 k 4k 4 k .
Thrice the first plus twice the second is 68:
9 k + 8 k = 68 9k + 8k = 68 9 k + 8 k = 68 .
So
17 k = 68 17k = 68 17 k = 68 and
k = 4 k = 4 k = 4 .
The numbers are 12 and 16.
The smaller is 12.
Watch out
To get the quadratic in (a), set the two ratios equal, T 2 T 1 = T 3 T 2 \frac{T_2}{T_1} = \frac{T_3}{T_2} T 1 T 2 = T 2 T 3 Report a problem with this question