Question Library

(159.75)10 = (y)6 \( \begin{array}{c|c} 6 & 159 \\ \hline 6 & 26 \text{ rem 3} \\ 6 & 4 \text{ rem 2} \\ & 0 \text{ rem 4} \\ \end{array}\uparrow \) 15910 = 4236 Then Convert 0.75 to base 6 0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5 0.5 x 6 = 3.00 423 + 1.5 =424.5 159.75 = (424.5)6 (159)10 = (424.5)6

Percentage Loss = (actual loss ) ÷ (Cost price) × 100 Actual loss = Cost Price − Selling Price (sold price) = N800 − N520 = N280 Percentage loss = (280 ÷ 800) × 100 = 35% ∴ His loss percent = 35%

6½ − 2¾ + 1½ 73/12 − 11/4 + 3/2 [(73 − 33 + 18) ÷ (12)] 58/12 29/6 \( 4\frac{5}{6}\)

No expanded working has been attached for this question yet. The verified answer is Option D.

1st G.P. = a =2 2nd G.P. = ar\(^{2 −1}\) = 54 2(r) = 54 r = 54/2 = 27 r = 27 3rd term = ar\(^2\) = (2) (27)\(^2\) 2 \times 27 \times 27 = 1458

S = {0, 1, 2, 3, 4, 5} T = {− 1, 0, 1, 2} S \(\cup\) T = {− 1, 0, 1, 2, 3, 4, 5 } ⇒ − 2 < x ≤ 5

[Return ÷ Investment] as a ratio ; i.e The Ratio is Return : Investmen [(Return1÷ Investment1 ) = (Return2÷ Investment2)] R1 = N 25000 R2 =? I1 = N450,000, I1 = N 700000 (25000 ÷ 450000) = (R2 ÷ 700000) R2 = [(25000 × 700000 ) ÷ 450000] = N38,888.90K ∴ The return on a investment of Y = N38888.90K

log717 = [log 17 ÷ log7] = [1.2304 ÷ 0.8451] [100.0899 ÷ 101.9270] = 1.455(antilog)

Convert the binary to base 10 and they convert back to base two 100112 + xxxxx2 + 111002 + 1012 = 10011112 (1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 1 × 20) + xxxxx2 +(1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 0 × 20) + (1 × 22 + 0 × 21 + 1 × 20) = (16 + 0 + 0 + 2 + 1) + xxxxx2 + (16 + 8 + 4 + 0 + 0 ) + (4 + 0 + 1) =(64 + 0 + 0 + 8 + 4 + 2 + 1) 19 + xxxxx2 + 33 = 79 xxxxx2 + 52 = 79 xxxxx2 = 79 − 52 xxxxx2 = 2710 \( \begin{array}{c|c} 2 & 27 \\ \hline 2 & 13 \text{ rem 1}\\ 2 & 6 \text{ rem 1}\\ 2 & 3 \text{ rem 0}\\ 2 & 1 \text{ rem 1}\\ & 0 \text{ rem 1}\\ \end{array}\uparrow \) 2710 = 110112 Therefore xxxxx2 = 2710 = 110112

\(\log_{5}(y^{2} x^{5} \div 125b^{\frac{1}{2}})\) = \(\log_{5} y^{2} + \log_{5} x^{5} - [\log_{5} 125 + \log_{5} b^{\frac{1}{2}}\) = \(2\log_{5} y + 5\log_{5} x - \log_{5} 5^{3} - \frac{1}{2} \log_{5} b\) = \(2\log_{5} y + 5\log_{5} x - 3 - \frac{1}{2}\log_{5} b\)

\(\int (5x^{3} + 7x^{2} - 2x + 5) \mathrm d x\) = \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - \frac{2x^{2}}{2} + 5x + c\) = \(\frac{5x^{4}}{4} + \frac{7x^{3}}{3} - x^{2} + 5x + c\)

Find the slant height \( l^2 = h^2 + r^2(h = 4cm,r = 3cm)\) \( l^2 = 4^2 + 3^2 = 16 + 9 = 25 \) \( l^2 = √ 25 \) Squaring both sides l = 5cm The area of curved surface (s) =π(3)(5) 15π = 15 × 3.14 = 47.1cm2

[1 \div (x+1)] + [1 \div (x − 1)] = ((x − 1) + [(x + 1)) \div (x+1)(x − 1)] Using the L.C.M. = (x − 1 + x + 1) \div (x + 1)(x − 1) = (x + 2 − 1 + 1) \div (x + 1)(x − 1) = 2x \div (x + 1)(x − 1) =2x \div (x + 1)(x − 1)

Area of a circle (A) = πr2 Radius = 4cm π = 3.142 A = 3.142 × (4)2 = 3.142 × 16 = 50.272 A = 50.3cm2 (1dp)

To find the minor axis of an ellipse: Given: - Area \(A = 132 \, \text{cm}^2\) - Major axis = \(14 \, \text{cm}\) Semi-major axis: \(a = \frac{14}{2} = 7 \, \text{cm}\) Area formula: \(A = \pi \times a \times b \implies 132 = \pi \times 7 \times b\) Solving for \(b\): \(b = \frac{132}{\pi \times 7} \approx \frac{132}{21.98} \approx 6 \, \text{cm}\) Length of minor axis: \(\text{Minor axis} = 2b \approx 2 \times 6 = 12 \, \text{cm}\)