Correct Answer: C. \(135a^{4}\)

Explanation

\((a + 3x)^{6}\).

The coefficient of \(x^{2}\) is:

\(^{6}C_{4}(a)^{6 - 2} (3x)^{2} = \frac{6!}{(6 - 4)! 4!} (a^{4})(9x^{2})\)

\(15 \times a^{4} \times 9 = 135a^{4}\)

Correct Answer: A. 16

Explanation

(a)(i) \(p > q .... (1)\)

\(p + q = 3(p - q) ...... (2)\)

From (2), \(p + q = 3p - 3q\)

\(p - 3p = - 3q - q \implies - 2p = - 4q\)

\(p = 2q\)

(ii) \(\frac{p^{2} + q^{2}}{pq}\)

= \(\frac{(2q)^{2} + q^{2}}{2q \times q}\)

\(\frac{4q^{2} + q^{2}}{2q^{2}}\)

= \(\frac{5q^{2}}{2q^{2}}\)

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

(b) Selling price = \(\frac{100 \times 66}{25}\)

= \(N264.00\)

Using \(\frac{SP - CP}{CP} = % gain\)

\(\frac{264 - CP}{CP} = \frac{32}{100}\)

\(100(264 - CP) = 32CP\)

\(26400 = 32CP + 100CP = 132CP\)

\(CP = \frac{26400}{132}\)

= \(N200\)

When he sells 20 for N50, Selling price = \(\frac{100 \times 50}{20} = N250.00\)

Hence, he made gain.

\(% gain = \frac{250 - 200}{200} \times 100%\)

\(\frac{1}{4} \times 100%\)

= \(25% gain\)

Correct Answer: A. 5.0

Explanation

Diagonal |AC| = (2x + 1)cm

In the diagram,area of \(\Delta\)ABC

is \(\frac{110}{2}\) = \(\frac{1}{2}\) x |AC| x |HB|

55 = \(\frac{1}{2}\) x (2x + 1) x 10

55 = (2x + 1)5

55 = 10x + 5

55 - 5 = 10x

50 = 10x

x = \(\frac{50}{10}\)

= 5.0

Correct Answer: D. -4 log 2

Explanation

\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)

= \(\log 2^{-3} + \log 2^{-1}\)

= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)

Correct Answer: D. 1800°

Explanation

number of sides = \(\frac{360^o}{\theta} = \frac{360^o}{306o}\)

n = 12^o

Sum of interior angle = (n - 2) 180^o

(12 - 2) 180v = 10 x 180^o

= 1800^o

Correct Answer: C. \(\frac{5}{4}\)

Explanation

(1 + \(\frac{1}{2} x )^5\)

1(1)\(^3\)(\(\frac{1}{2} x)^0\) - 5(1)\(^4\)(\(\frac{1}{2} x)^1\)

+ 10 (10)\(^3\)(\(\frac{1}{2} x )^2\) + 10(1)\(^2\) (\(\frac{1}{2} x\))\(^3\)

Coefficient of x\(^3\)

10(\(\frac{1}{2} x)^3\) = \(\frac{10}{8} x^3\) =

= \(\frac{5}{4}\)

Correct Answer: C. 15

Explanation

Rearranged: 12,13,15, 36, 39, 41, 41, 43, 43, 47 and 49

Lower quartile or First quartile = \(\frac{total number of items}{4}\)

If odd: items numbers + 1 → \(\frac{11 items + 1}{4}\)

= \(\frac{12}{4}\)

Lower quartile = 3rd iems →15

Explanation

(a) (i) Let :

L = { all good literature students in the school}

A = {all students in the General Arts class}

Then the Venn diagram below represents the statement : 'All good Literature students in a school are in the General Arts class'.

where \(\in\) = {all students in the school}.

(ii) (1) Invalid conclusion

(2) Invalid conclusion

(3) Valid conclusion

(b) (i) c = h + y

where \(y \propto n \implies y = kn\)

\(\implies c = h + kn\)

When n = 600 bricks, c = GH¢ 950.00

\(950.00 = h + 600k ....... (1)\)

When n = 1000 bricks, c = GH¢ 1,030.00

\(1,030.00 = h + 1000k ....... (2)\)

Subtracting (1) from (2), we have

\(1,030 - 950 = 1000k - 600k\)

\(80 = 400k\)

\(k = \frac{80}{400} = \frac{1}{5}\)

\(\therefore c = h + \frac{1}{5}n\)

From (1), \(950 = h + 600k\)

\(950 = h + (600 \times \frac{1}{5})\)

\(950 = h + 120\)

\(h = 950 - 120 = 830\)

\(c = 830 + \frac{1}{5}n\)

(ii) Using

\(c = 830 + \frac{1}{5}n\), we have

\(c = 830 + \frac{1}{5} \times 500\)

= \(830 + 100\)

= \(GH¢ 930.00\)

Hence, the cost of producing 500 bricks = GH¢ 930.00.

Correct Answer: C. 24