Mathematics Questions and Answers

\(\frac{5}{8} - \frac{3}{4} ÷ \frac{5}{12} x \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{3}{4} ÷ \frac{5}{12}) \times \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\)

⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\)

⇒ \(\frac{5}{8} - \frac{9}{20}\)

∴ \(\frac{7}{40}\)

Total cost price = ₦9,200.00 + ₦1,500.00 = ₦10,700.00

Selling price = ₦13,400.00

Gain = ₦13,400.00 - ₦10,700.00 = ₦2,700.00

∴ % gain = \(\frac{₦2,700.00}{₦10,700.00}\times\) 100% = 25.23%

Using Pythagoras theorem

⇒ \((20\sqrt2)2 = x^2 + x^2\)

⇒ 800 = \(2x^2\)

⇒ 400 = \(x^2\)

⇒ x = \(\sqrt400\) = 20 cm

∴ Area of a square = \(x^2 = 20^2 = 400 cm^2\)

∴ Perimeter of a square = 4x = 4 x 20 = 80 cm

Volume of the composite solid = Volume of A + Volume of B

Volume of a cuboid = length x breadth x height

Volume of A = 6 x 26 x 8 = 1248 cm\(^3\)

Volume of B = 6 x 10 x 22 = 1320 cm\(^3\)

∴ Volume of the composite solid = 1248 + 1320 = 2568 cm\(^3\)

Total possible outcome = 6 x 6 = 36

Required outcome = 15

∴ Pr(E) = \(\frac{15}{36} = \frac{5}{12}\)

An exterior angle of a n-sided regular polygon = \(\frac{360}{n}\)

For (n - 1) sided regular polygon = \(\frac{360}{n - 1}\)

For (n + 2) sided regular polygon = \(\frac{360}{n + 1}\)

⇒ \(\frac{360}{n - 1} - \frac{360}{n + 2}\) = 6 9Given)

⇒ \(\frac{360(n + 2) - 360(n - 1)}{(n - 1)(n + 2)}\)

⇒ \(\frac{360n + 720 - 360n + 360}{(n - 1)(n + 2)}\)

⇒ \(\frac{1080}{(n - 1)(n + 2)} = \frac{6}{1}\)

⇒ 1080 = 6 (n - 1)(n + 2)

⇒ 180 = (n - 1)(n + 2)

⇒ 180 = n\(^2\)+ 2n - n - 2

⇒ 180 = n\(^2\) + n - 2

⇒ n\(^2\)b+ n - 2 - 180 = 0

⇒ n\(^2\) + n - 182 = 0

⇒ n\(^2\) + 14n - 13n - 182 = 0

⇒ n (n + 14) - 13 (n + 14) = 0

⇒ (n - 13) (n + 14) = 0

⇒ n - 13 = 0 or n + 14 = 0

⇒ n = 13 or n = -14

∴ n = 13 (We can't have a negative number of side)

% error = \(\frac{Error}{Actual Value}\) x 100%

Student's Value = 18mL

Actual Value = 18.4mL

Error = 18.4 - 18 = 0.4

∴ % error = \(\frac{0.4}{18.4}\) x 100% = 2.17%

Average performance rating of Team B = \(\frac{7+9+1+9+6}{5} = \frac{32}{5}\) = 6.4

Average performance rating of Team A = \(\frac{5+3+6+10+2}{5} = \frac{26}{5}\) = 5.2

∴ The difference in the average performance ratings between Team B and Team A = 6.4 - 5.2 = 1.2

16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\)

⇒ 1.654 x 10\(^{-4}\) - 6.76 x 10\(^{-8}\) + 2.3 x 10\(^{-7}\)

⇒ 1.654 x 10\(^{-4}\) - 0.000676 x 10\(^{-4}\) + 0.0023 x 10\(^{-4}\)

⇒ (1.654 - 0.000676 + 0.0023) x 10\(^{-4}\)

∴ 1.655624 x 10\(^{-4}\) ≃ 1.66 x 10\(^{-4}\)

T\(_3\) = 6

T\(_5\) = 12

S\(_{12}\) = ?

T\(_n\) = a + (n - 1)d

⇒ T\(_3\) = a + 2d = 6 ----- (i)

⇒ T\(_5\) = a + 4d = 12 ----- (ii)

Subtract equation (ii) from (i)

⇒ -2d = -6

⇒ d\(\frac{-6}{-2}\) = 3

Substitute 3 for d in equation (i)

⇒ a + 2(3) = 6

⇒ a + 6 = 6

⇒ a = 6 - 6 = 0

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

⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\)

⇒ S\(_{12}\) = 6(0 + 11 x 3)

⇒ S\(_{12}\) = 6(33)

∴ S\(_{12}\) = 198