2018 - WAEC Mathematics Past Questions and Answers - page 1

$\sqrt{108}+\sqrt{125}-\sqrt{75}$

= $\sqrt{3\times 36}+\sqrt{5\times 25}-\sqrt{3\times 25}$

= $6\sqrt{3}+5\sqrt{5}-5\sqrt{3}$

= $\sqrt{3}+5\sqrt{5}$

$[{64}^{\frac{1}{2}}+{125}^{\frac{1}{3}}{]}^2$ = $[\sqrt{64}+\sqrt[3]{125}{]}^2$ 

$[8+5{]}^2$ = $[13{]}^2$

= 169

Y $\alpha \frac{1}{x^2}\to y=\frac{k}{x^2}$

If x = 3 and y = 100,

then, $\frac{100}{1}=\frac{k}{3^2}$

$\frac{100}{1}=\frac{k}{9}$

k = 100 x 9 = 900

Substitute 900 for k in

y = $\frac{k}{x^2}$; y = $\frac{900}{x^2}$

= $y{x}^2=900$

${32}_4={22}_x$

$3\times 4^1+2\times 4^o$ = $2\times x^1+2\times x^o$

12 + 2 x 1 = 2x + 2 x 1

14 = 2x + 2

14 - 2 = 2x

12 = 2x

x = $\frac{12}{2}$

x = 6

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%

${\log }_{10}$(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

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

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

9 - 3x = -1 ......(2)

Substract (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

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$\frac{1}{2}$

$\frac{2-18m^2}{1+3m}$ = $\frac{2(1-9)m^2}{1+3m}$

= $\frac{2(1+3m)(1-3m)}{1+3m}$

= $2(1-3m)$