Question Library

x10 = 12145 = x10 = 1 x 53 + 2 x 52 + 1 * 51 + 4 x 50 = 1 x 125 + 2 x 25 + 1 x 5 + 4 x 1 = 125 + 50 + 5 + 4 = 184

\(\frac{(0.5625)^2) - (0.4375)^2}{0.04}\) \(\frac{(0.5625 + 0.4375)(0.5625 - 0.4375)}{0.04}\) \(\frac{1.000\times 0.125}{0.04}\) = 3.125 = 3.13

2(32x-1) = 162 32x-1 = 162/2 32x-1 = 81 32x-1 = 32 2x - 1 = 4 (equating the indices) 2x = 5 5/2 x =

\(\frac{3}{5} \div \left(\frac{2}{7} \times \frac{4}{3} \div \frac{4}{9}\right)\\ =\frac{3}{5} \div \left(\frac{2}{7} \times \frac{4}{3} \times \frac{9}{4}\right)\\ =\frac{3}{5} \div \frac{6}{7}\\ =\frac{3}{5} \times \frac{7}{6}\\ =\frac{7}{10}\)

log1012.5 = log10 121/2 = log1025/2 = log1025 - log102 = log1052 - log102 = 2log105 - log102 = 2log1010/2 - log102 = 2(1-x)- x = 2 - 2x – x = 2 – 3x

5% profit = 100 + 5 = 105% 26% profit = 100 + 26 = 126% ∴ 105% → N60,000 1% → \(\frac{60000}{105}\) 126% = \(\frac{60000}{105}\) x \(\frac{126}{1}\) = N72,000.

P = {1, 3, 5, 7, 9, 11} Q = {2, 4, 6, 8, 10, 12} P∩Q = {1, 3, 5, 7, 9, 11}∩{2, 4, 6, 8, 10, 12} = {} = ∅

101122 - 10122 = (1x23 + 0x22 + 1x21 + 1x20)2 - (1x22 + 0x21 + 1x22)2 (1x8 + 0x4 + 1x2 + 1x1) 2 - (1x4 + 0x2 + 1x1) 2) = (8 + 0 + 2 + 1) 2 - (4 + 0 + 1) 2 = 112 - 52 = 16 x 6 = 96 9610 to base 2 2/96 = 48 R 0 2/48 = 24 R 0 2/24 = 12 R 0 2/12 = 6 R 0 2/6 = 3 R 0 2/3 = 1 R 1 2/1 = 0 R 1 11000002

All the perfect squares < 40 X = {1, 4, 9, 16, 25, 36} All the odd numbers from 1 to 15 Y = {1, 3, 5, 7, 9, 11, 13, 15} X ∩ Y = {1, 9}

Length of an arc = \(\frac{\theta}{360}\) x 2\pir \(\frac{90}{360}\) x 2 x \pi x 7 = \(\frac{7\pi}{2}\)cm\(^2\)

If the line 3y = 4x – 1 is parallel t[ line qy = x + 3 Implies gradient of 3y = 4x – 1 Y = 4/3x - 1/3 ∴Gradient = 4/3 Gradient of line qy = x + 3 Y = 1/qx + 3/q ∴ Gradient = 1/q 4/3 = 1/q 4q = 3 Q = 3/4

x = 30o (alternate ∠s b/c PQ//Sr) Y = 50o (vertical opposite ∠s) r = 30o (alternate ∠s b/c PQ//SR) But = y + q + t = 180o (∠s on a straight line) 50 + q + 30 = 180 q + 80 = 180 q = 180 -80 q = 100o

Volume of bowl \(\frac{2}{3}\pi r^3\\ 718\frac{2}{3}=\frac{2}{3}\pi r^3\\ \frac{2156}{3}=\frac{2}{3} \times \frac{22}{7} \times r^3 ∴r^3 = \frac{2156 \times 3 \times 7}{3 \times 2 \times 22}\\ r^3 = 343\\ r = \sqrt[3]{343}\\ r= 7.0cm\)

When a particle P moves such that the angle ∠SPT is constant while ST remains fixed, it traces a circular path. Specifically, the locus of P is a semicircle because points on the circle subtended by the line segment ST maintain that angle.

If the point of intersection is (1, -2), it implies that x = 1 and y = -2 when the two equation are solved simultaneously. ∴ substitute x = 1 and y = -2 in any of the equations 2y - k x + 2 = 0 2(-2) - k(1) + 2 = 0 -4 - k + 2 = 0 -4 + 2 = k -2 = k