2018 - JAMB Mathematics Past Questions and Answers - page 6
Find the average of the first four prime numbers greater than 10
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
What is the modal age?
The modal age is the age with the highest frequency, and that is age 5 years with f of 7
Convert 0.04945 to two significant figures
Calculate 243\(_{six}\) – 243\(_{five}\) expressing your answer in base 10
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
Evaluate ∫\(^2_1\) \(\frac{5}{x}\) dx
∫\(\frac{5}{x}\) 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
Tossing a coin and rolling a die are two separate events. What is the probability of obtaining a tail on the coin and an even number on the die?
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 \(\frac{1}{2}\) = \(\frac{1}{4}\)
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
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
From a point P, R is 5km due West and 12km due South. Find the distance between P and R'.
Apply Pythagoras theorem:
PR\(^2\) = 5\(^2\) + 12\(^2\)
25 + 144 = 169
PR = √(169)= 13km
if y = 23\(_{five}\) + 101\(_{three}\) find y leaving your answer in base two
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\(_{five}\)
A box contains two red balls and four blue balls. A ball is drawn at random from the box and then replaced before a second ball is drawn. Find the probability of drawing two red balls.
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) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)