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

   

 

1. **Understanding Inverse Variation:** Since *y* varies inversely as the square of *x*, we can write the relationship as \\(y = \frac{k}{x^2}\\), where *k* is the constant of variation. 2. **Using the Given Values:** We are given that *x* = 3 when *y* = 100. Substitute these values into the equation: \\(100 = \frac{k}{3^2}\\) 3. **Solving for k:** Simplify and solve for *k*: \\(100 = \frac{k}{9}\\) \\(k = 100 \times 9\\) \\(k = 900\\

1. Convert the base-four number to base-ten: \\(32\_{four} = (3 \times 4^1) + (2 \times 4^0) = 12 + 2 = 14\\) 2. Convert the base-x number to base-ten: \\(22\_x = (2 \times x^1) + (2 \times x^0) = 2x + 2\\) 3. Set the base-ten equivalents equal to each other and solve for x: \\(14 = 2x + 2\\) \\(12 = 2x\\) \\(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\(\frac{1}{2}\)

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

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

= \(2(1 - 3m)\)