How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME.
\(\frac{2}{3}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{9}\)
Correct answer is D
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) = \(\frac{1}{3}\) × \(\frac{1}{3}\) = \(\frac{1}{9}\)
if y = 23\(_{five}\) + 101\(_{three}\) find y leaving your answer in base two
1110
10111
11101
111100
Correct answer is B
First we 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
To convert 23 to base 2 (as in the diagram above)
Y = 23
= 10111\(_{five}\)
Answer is B
From a point P, Q is 5km due West and R is12km due South of Q. Find the distance between P and R.
5km
12km
13km
17km
Correct answer is C
Using Pythagoras theorem
PR\(^2\) = 5\(^2\) + 12\(^2\)
25 + 144 = 169
PR = √(169)= 13km
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
210
1050
21400
25200
Correct answer is A
To form words having 3 consonants and 2 vowels out of 7 consonants and 4 vowels, the number of such words is 7/3C x 4/2C = 35 x 6
= 210
\(\frac{1}{16}\)
\(\frac{1}{6}\)
\(\frac{1}{4}\)
\(\frac{3}{8}\)
Correct answer is C
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 \(\frac{1}{2}\) = \(\frac{1}{4}\)