JAMB Mathematics Past Questions & Answers - Page 7

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

Volume of the cylinder = \(\frac{θ}{360} \times \pi r^2h\)

θ = 360\(^o\) - 90\(^o\) = 270\(^o\)

∴ Volume of the cylinder = \(\frac{270}{360} \times \frac{22}{7} \times \frac{14^2}{1} \times \frac{32}{1} = \frac{37,255,680}{2520} = 14,784cm^3\)

Principal (P) = ₦15,700

Rate (R) = 8

Number of years (t) = 2

A = P \((1+\frac{R}{100})^t\)

⇒ A = 15700 \((1+\frac{8}{100})^2\)

⇒ A = 15700 (1 + 0.08)\(^2\)

⇒ A = 15700 (1.08)\(^2\)

⇒ A = 15700 x 1.1664

⇒ A = ₦18,312.48

Total amount, A = ₦18,312.48

A = P + CI

⇒ CI = A - P

⇒ CI = 18,312.48 - 15,700

∴ CI = ₦2,612.48