2007 - JAMB Mathematics Past Questions and Answers - page 4

M + L = 28,

M : L = 600 : 800

= 3 : 4

\(\frac{M}{L}\) = \(\frac{3}{4}\) \(\to\) M = \(\frac{3}{4}\)L

\(\frac{3}{4}\)L + L = 28

\(\frac{7L}{4}\) = 28

L = \(\frac{4 \times 28}{7}\)

= 16

x₁₀ = 1214 ₅, 1 x 5³ + 2 x 5² + 1 x 5¹ + 4 x 5^o

= 125 + 50 + 5 + 4 184¹⁰

x = 184

log₁₀ 12.5 = log₁₀\(\frac{25}{2}\)

= log₁₀\(\frac{100}{8}\) = log₁₀ 100 - log₁₀8

= log₁₀ 10² - log₁₀2

= 2log₁₀ 10 - 3log₁₀²

= 2 - 3x

Total fruits 6 + 11 + 13 = 30,

prob. (Grape) = \(\frac{6}{30}\)

prob. (Banana) = \(\frac{11}{30}\)

= \(\frac{11}{30}\) + \(\frac{6}{30}\) = \(\frac{17}{30}\)

No. of students that made the trip

= 2 + 3 = 5 = 6 = 8 = 4 + 1 = 29

The roots of the graph are -2, -1 and 1

y = (x + 2)(x + 1)(x - 1) = (x + 2)(x² - 1)

= x³ + 2x² - x - 2

Arrange in ascending order
2, 5, 5, 6, 6, ,6, 6, 8, 8
Position of median = ^N/₂ = ⁸/₂ = 4th and 5th
∴Median = (6+6) / 2 = ¹²/₂ = 6
Mode the item that repeat itself most = 6
∴Product of modal and median number = 6 x 6 = 36

Number of grapes = 6
Number of banana = 11
Number of oranges = 13
Total = 30
P(Grape) = ⁶/₃₀, P(Banana) = ¹¹/₃₀
P(Orange) = ¹³/₃₀
∴P(Either grape or banana) = ⁶/₃₀ + ¹¹/₃₀
= ¹⁷/₃₀

Number of itineraries = ⁷P₄
=\(\frac{7!}{(7-4)!}\\ =\frac{7!}{3!}\\ =\frac{7 \times 6 \times 5 \times 4 \times 3!}{3!}\\ =840\)