2018 - JAMB Mathematics Past Questions and Answers - page 6

Prime numbers are numbers that has only two factors (i.e 1 and itself). They are numbers that are only divisible by 1 and their selves. First four Prime numbers greater than 10 are 11, 13, 17 and 19

  Average = sum of numbers / number

  = \(frac{(11 + 13 + 17 + 19)}{4}\)

  = \(\frac{60}{4}\)

  = 15

The modal age is the age with the highest frequency, and that is age 5 years with f of 7

0.04945 to 2s.f is 0.05

Since they are of different base, convert to base 10

  243\(_{six}\) = (2 x 62) + (4 x 61) + (3 x 60)

  = 72 + 24 + 3 = 99 base 10

  243\(_{six}\) = 2 x 52 + 4 x 51 +3 x 50

  50 + 20 + 3 = 73 base 10

  Subtracting them, 99 - 73

  = 26

∫\(\frac{5}{x}\) dx = 5 ∫\(\frac{1}{x}\) = 5Inx

  Since the integral of \(\frac{1}{x}\) is Inx

  ∫\(^2\) \(_1\)∫ \(\frac{5}{x}\) dx = 5

  dx = 5 (In<2 – InIn1)

  = 3.4657

  = 3.47

P( tail on a coin) = \(\frac{1}{2}\)

  Even numbers on a care 2, 4 and 6

  P( even number on a die) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

  P( tail on a coin and even number on a die) = \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{1}{4}\)

Words having 3 consonants and 2 vowels out of 7 consonants and 4 vowels, this implies that the number of such words is 7/3C x 4/2C = 35 x 6

  = 210

Apply Pythagoras theorem:

  PR\(^2\) = 5\(^2\) + 12\(^2\)

  25 + 144 = 169

  PR = √(169)= 13km

Convert the numbers to base ten

  23\(_{five}\)= 2 x 51 + 3 x 50

  = 10 + 3 = 13

  101\(_{five}\) = (1 x 32) + (0 x 31) + (1 x 30)

  = 9 + 0 + 1 = 10

  So, y = 13 + 10 = 23

Y = 23

  = 10111\(_{five}\)

Total number of balls = 2 + 4 = 6

  P(of picking a red ball) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

  P(of picking a blue ball) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

  With replacement,

  P( picking two red balls) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)