Explanation

(a) \((p + qx)^{7} = p^{7} + 7p^{6} qx + 21p^{5} (qx)^{2} + 35p^{4} (qx)^{3} + 35p^{3} (qx)^{4} + 21p^{2} (qx)^{5} + 7p(qx)^{6} + (qx)^{7}\)

The coefficient of \(x^{2}\) = \(21p^{5} q^{2}\)

The cofficient of \(x^{3}\) = \(35p^{4} q^{3}\)

\(\therefore 21p^{5} q^{2} = 35p^{4} q^{3}\)

\(\implies 21p = 35q\)

\(q = \frac{21}{35} p = \frac{3}{5} p\).

(b) N180 + N260 + N340 + ...

This sequence is a linear sequence with first term = a = 180, d = 80.

\(S_{16} = ?\)

\(S_{n} = \frac{n}{2} (2a + (n - 1) d)\)

\(S_{16} = \frac{16}{2} (2(180) + (16 - 1) 80\)

= \(8(360 + 1200)\)

= \(8(1560)\)

= \(N12,480\)

He would have contributed N12,480 by the 16th week.

Correct Answer: A. -i - 5j

Explanation

\(\overline{PQ}\) = 5i - 2j; \(\overline{QR}\) = 4i + 3j

\(\overline{RP}\) = \(\overline{PQ}\) - \(\overline{QR}\)

= 5i - 2j - [4i + 3j] = 5i - 2j - 4i - 3j

= i - 5j

Correct Answer: D. 86°

Explanation

< OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58°

< PSR = 36° + 58° = 94°

< PSR + < PQR = 180°

94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°

Correct Answer: C. 0.9

Explanation

log\(_x\)2 = 0.3

log\(_x\)8 = log\(_x\)2\(^3\) = 3 log\(_x\)2

= 3 x 0.3

= 0.9

Correct Answer: A. (30 - n)years

Explanation

Let Tom's present agr be x.

Then x = 25 - n

If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5

= (30 - n)

Explanation

(a)

Let T be the tension in the rope and W the weight of the bucket of water.

T - W = net force = ma

\(T - mg = ma\)

\(150 - 80 = 8a\)

\(a = \frac{70}{8} = 8.75 m/s^{2}\).

(b) m = 12kg ; u = 15 m/s.

v = 25 m/s ; s = 50m ; a = ?

\(v^{2} = u^{2} + 2as\)

\(625 = 225 + 100a \implies 100a = 400\)

\(a = 4m/s^{2}\)

\(t = ?\)

\(s = ut + \frac{1}{2} at^{2}\)

\(50 = 15t + \frac{1}{2}(4t^{2})\)

\(50 = 15t + 2t^{2} \implies 2t^{2} + 15t - 50 = 0\)

\(2t^{2} + 20t - 5t - 50 = 0\)

\(2t(t + 10) - 5(t + 10) = 0 \implies (t + 10)(2t - 5) = 0\)

\(t = \frac{5}{2}s \text{or t = -10s}\)

Since time cannot be negative, t = 2.5 seconds.

\(F = m\frac{(v - u)}{t}\)

= \(12(\frac{25 - 15}{2.5})\)

= \(48N\)

\(2as = v^{2} - u^{2}\)

\(2as = 35^{2} - 15^{2}\)

\(2(4)(s) = 1000\)

\(s = 125m\)

Correct Answer: B. x ≤ -1, 1 ≥ x ≥ 2

Explanation

Let \(x^{\frac{1}{3}} = b\) so that the equation is

\((x^{\frac{1}{3}})^{2} - 5x^{\frac{1}{3}} + 6 = 0\)

= \(b^{2} - 5b + 6 = 0\)

\(b^{2} - 3b - 2b + 6 = 0\)

\(b(b - 3) - 2(b - 3) = 0 \implies b = \text{2 or 3}\)

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

\(x^{\frac{1}{3}} = 3 \implies x = 3^{3} = 27\)

Correct Answer: C. 7m

Explanation

Volume of a cylinder = \( \pi r^2\)h

385 = \(\frac{22}{7}\) x \(r^2\) x 10

385 x 7 = 22 x \(r^2\) x 10

\(r^2\) = \(\frac{385 \times 7}{22 \times 10}\)

= 12.25

r = \(\sqrt{12.25}\)

= 3.5m

Hence, diameter of tank = 2r

= 2 x 3.5 = 7m

Correct Answer: C. 62