WAEC Mathematics Past Questions & Answers - Page 64

If x : y : z = 3 : 3 : 4, evaluate $\frac{9x + 3y}{6x - 2y}$

$\frac{x}{y}$ = $\frac{2}{3}$ and $\frac{y}{z}$ = $\frac{3}{4}$

Thus; x = $\frac{2}{3}T_1$ and z = $\frac{3}{5}T_1$

y = $\frac{3}{7}T_2$ and z =  $\frac{4}{7}T_2$

Using y = y

$\frac{3}{5}T_1$ = $\frac{3}{7}T_2$; $\frac{T_1}{T_2}$ = $\frac{3}{7}$ x $\frac{5}{3}$

$\frac{T_1}{T_2}$  = $\frac{15}{21}$

$T_1$ = 15 and $T_2$ = 21

Therefore;

x = $\frac{2}{5}$ x 15 = 6

y = $\frac{3}{5}$ x 15 = 9

y = $\frac{3}{7}$  x 21 = 9 (again)

z = $\frac{4}{7}$ x 21 = 12

Hence;

$\frac{9x + 3y}{6z - 2y}$ = $\frac{9(6) + 3(9)}{6(12) - 2(9)}$

$\frac{54 + 27}{72 - 18}$ = $\frac{81}{54}$ = $\frac{3}{2}$

= 1$\frac{1}{2}$

y + 2x = 4 .....(1)

y - 3x = -1 ......(2)

Subtract (2) from (1)

2x - (-3x) = 4 - (-1)

2x + 3x = 4 + 1

5x = 5

X = $\frac{5}{5}$

= 1

Substitute 1 for x in (1);

y + 2(1) = 4

y + 2 = 4

y = 4 - 2 = 2

Hence, (x + y) = (1 + 2)

= 3

F = $\frac{9}{5}$C + 32

When F = 98.6

98.6 = $\frac{9}{5}$C + 32

98.6 - 32 = $\frac{9}{5}$C

66.6= $\frac{9}{5}$C

66.6 x 5 = 9C

C = $\frac{66.6 \times 5}{9}$

= 37

$\log_{10}$(6x - 4) - $\log_{10}$2 = 1

$\log_{10}$(6x - 4) - $\log_{10}$2 = $\log_{10}$10

$\log_{10}$$\frac{6x - 4}{2}$ - $\log_{10}$10

$\frac{6x - 4}{2}$ = 10

6x - 4 = 2 x 10

= 20

6x = 20 + 4

6x = 20

x = $\frac{24}{6}$

x = 4

Population of school = 250 + 150 = 400

60% of 250 = $\frac{\text{60%}}{\text{100%}}$ x 250 = 150

40% of 150 = $\frac{\text{40%}}{\text{100%}}$ x 150 = 60

Total number of students who plays football;

150 + 60 = 210

Percentage of school that play football;

$\frac{210}{400}$ x 100% = 52.5%