WASSCE 2020 Mathematics Past Questions and Answers

Answer: B. 50.28
75.0785 - 34.624 + 9.83 = 50.2845

75.0785 - 34.624 + 9.83 ≈ 50.28 (two decimal places)

Answer: C. {1, 2, 3, 4, 6}
µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

x:x < 7 is read as x is such that x is less than 7.

X = {1, 2, 3, 4, 5, 6}

Note: since X is a subset of the universal set (µ), members of X should be found in the universal set (µ)

y:y is a factor of 24 is read as y is such that y is a factor of 24.

Y = {1, 2, 3, 4, 6, 8}

Note: since Y is a subset of the universal set (µ), members of Y should be found in the universal set (µ)

Intersection (∩) are elements that can be found in both sets

X ∩ Y = {1, 2, 3, 4, 6}

Answer: B. $\frac{3}{8}$
${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:

1. 16 = 4 x 4 = 4$^{2}$

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

2. 9 = 3 x 3 = 3$^{2}$

Expressed this way in order to cancel the denominator 2 of the $\frac{-3}{2}$ power

3. 16 = 2 x 2 x 2 x 2 = 2$^{4}$

Expressed this way in order to cancel the denominator 4 of the $\frac{-3}{4}$ power

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{(\frac{{4}^{2}}{{3}^{2}})}^{\frac{-3}{2}}x{({2}^{4})}^{\frac{-3}{4}}]}^{\frac{1}{3}}$

From the law of indices,

${(\frac{a}{b})}^{c}$ = $\frac{{a}^{c}}{{b}^{c}}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{{4}^{2 x \frac{-3}{2}}}{{3}^{2 x \frac{-3}{2}}}x{2}^{4 x \frac{-3}{4}}]}^{\frac{1}{3}}$

Notes:
1. The power 2 cancel the denominator 2 of the power $\frac{-3}{2}$

2. The power 4 cancel the denominator of 4 of the $\frac{-3}{4}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[\frac{{4}^{-3}}{{3}^{-3}}x{2}^{-3}]}^{\frac{1}{3}}$

$\frac{{4}^{-3}}{{3}^{-3}}$ = ${(\frac{4}{3})}^{-3}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${[{(\frac{4}{3})}^{-3}x{2}^{-3}]}^{\frac{1}{3}}$

The $\frac{1}{3}$ affects each of the powers in the bracket.

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${(\frac{4}{3})}^{-3x\frac{1}{3}}x{2}^{-3x\frac{1}{3}}$

The power 3 cancels the denominator 3 in $\frac{1}{3}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = ${(\frac{4}{3})}^{-1}x{2}^{-1}$

Notes:
1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).

2. ${(\frac{4}{3})}^{-1}$ = $\frac{3}{4}$

3. Every number is being divided by 1 but it is not written.

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

4. 2$^{-1}$ = $\frac{1}{2}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{3}{4}$ x $\frac{1}{2}$

${[{(\frac{16}{9})}^{\frac{-3}{2}}x{16}^{\frac{-3}{4}}]}^{\frac{1}{3}}$ = $\frac{3 x 1}{4 x 2}$ = $\frac{3}{8}$

Answer: A. 4
The least is 9 dividing the number 1 time with a remainder of 7

4*x* = 9 x 1 + 7

4*x* = 9 + 7

4*x* = 16

Divide both sides by 4

x = $\frac{16}{4}$ = 4

Answer: D. log$_{10}$90
Notes

1. log$_{b}$b = 1

log$_{10}$10 = 1

2. blog c = log c$^{b}$

2log$_{10}$3 = log$_{10}$3$^{2}$

3. log b + log c = log b x c

1 + 2 log$_{10}$3 = log$_{10}$10 + log$_{10}$3$^{2}$

1 + 2 log$_{10}$3 = log$_{10}$10 + log$_{10}$9

1 + 2 log$_{10}$3 = log$_{10}$10 x 9

1 + 2 log$_{10}$3 = log$_{10}$90