Explanation

(a) \(\log_{10} x + \log_{10} y = 4 ... (1)\)

\(\log_{10} x + 2 \log_{10} y = 3 ...(2)\)

\((1) \implies \log_{10} (xy) = 4 \therefore xy = 10^{4} ... (3)\)

\((2) \implies \log_{10} (xy^{2}) = 3 \therefore xy^{2} = 10^{3} ... (4)\)

Divide (4) by (3) :

\(y = 10^{-1} = 0.1\)

Put y in (3), we have

\(\frac{x}{10} = 10^{4} \implies x = 10^{5}\)

\((x, y) = (10^{5}, 10^{-1})\)

(b) (i) \(t \propto \frac{V}{P}\)

\(t = \frac{kV}{P}\)

when t = 10, V = 20 and P = 5.

\(10 = \frac{20k}{5} \implies 50 = 20k\)

\(k = \frac{5}{2}\)

\(\therefore t = \frac{5V}{2P}\)

(ii) When V = 50, P = 2, we have

\(t = \frac{5(50)}{2(2)} = \frac{250}{4}\)

= \(62.5 minutes\)

(iii) when V = 40, t = 20 minutes

\(20 = \frac{5(40)}{2P} \implies 40P = 200\)

\(P = \frac{200}{40} = 5 pumps\)

Correct Answer: B. 6

Correct Answer: B. \(\pm 2\)

Explanation

\(T_{n} = ar^{n - 1}\) (Exponential sequence)

\(T_{4} = ar^{3} = 2 .... (1)\)

\(T_{6} = ar^{5} = 8 ......(2)\)

Divide (2) by (1),

\(\frac{ar^{5}}{ar^{3}} = \frac{8}{2} \)

\(r^{2} = 4\)

\(r = \pm 2\)

Correct Answer: A. 16

Explanation

2x - 3y = -11 --- (i)

3x + 2y = 3 --- (ii)

Multiply equation (i) by 3 and equation (ii) by 2

6x - 9y = -33 --- (iii)

6x + 4y = 6 --- (iv)

Subtract equation (iii) from (iv)

13y = 39

y = \(\frac{39}{13}\) = 3

substitute (3) for y in equation (ii)

3x + 2(3) = 3

3x + 6 = 3

3x = 3 - 6

3x = -3

x = \(\frac{-3}{3}\) = - 1

Now,

\((y - x)^2 = (3 - (-1))^2\)

= \((3 + 1)^2\)

= \(4^2\)

= 16

Correct Answer: A. 96\(^o\)

Explanation

< O\(Q\)P = 48\(^o\) (base angles of isoceles \(\triangle\))

m = 48\(^o\) + 48\(^o\) = 96\(^o\) (exterior angle of a \(\triangle\))

Correct Answer: B. 9

Explanation

(a) Import duty : \(\frac{15}{100} \times 400 = GH¢ 60.00\)

New cost price : GH¢400.00 + GH¢ 60 = GH¢ 460.00

Sales tax : \(\frac{10}{100} \times 460 = GH¢ 46.00\)

Total cost price : \(GH¢(460 + 46) = GH¢ 506.00\)

Profit = GH¢(660 - 506)

= GH¢ 154.00

\(% profit = \frac{154}{506} \times 100%\)

= \(30.43%\).

(b) Let the number of girls be t.

Therefore the total number of students = (1000 + t) students.

The number that passed the exam = \(\frac{48}{100} \times (1000 + t)\)

= \(0.48(1000 + t)\)

\(\implies 0.48(1000 + t) = (\frac{50}{100} \times 1000 + \frac{40}{100} \times t)\)

\(0.48(1000 + t) = 500 + 0.4t\)

\(480 + 0.48t = 500 + 0.4t \implies 0.48t - 0.4t = 500 - 480\)

\(0.08t = 20\)

\(t = \frac{20}{0.08} = 250\)

There are 250 girls in the school.

Correct Answer: C. \(\frac{x(x+5)}{2(x+2)}\)

Correct Answer: C. y = ±17

Explanation

log (y + 8) + log (y - 8) = 2log 3 + 2log 5

⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b)

⇒ log (y\(^2\) -8y + 8y - 64) = log 3\(^2\) + log 5\(^2\)

⇒ log (y\(^2\) - 64) = log 3\(^2\) + log 5\(^2\)

⇒ log (y\(^2\) - 64) = log 9 + log 25

⇒ log (y\(^2\) - 64) = log (9 × 25)

⇒ log(y\(^2\) - 64) = log 225

⇒ y\(^2\) - 64 = 225

⇒ y\(^2\) = 225 + 64

⇒ y\(^2\) = 289

⇒ y = ±√289

∴ y = ±17

Correct Answer: D. \((\frac{\pi}{3}, \frac{5\pi}{3})\)

Explanation

\(2\cos x - 1 = 0 \implies 2\cos x = 1\)

\(\cos x = \frac{1}{2}\)

\(x = \cos^{-1} (\frac{1}{2})\)

= \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\)