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.
Convert 0.04945 to two significant figures
0.040
0.049
0.050
0.49
Correct answer is B
No explanation has been provided for this answer.
If \(\frac{x}{a + 1}\) + \(\frac{y}{b}\) = 1. Make y the subject of the relation.
\(\frac{b(a + 1 - x)}{a + 1}\)
\(\frac{a + 1}{b(a - x + 1)}\)
\(\frac{a(b - x + 1)}{b + 1}\)
\(\frac{b}{a(b - x + 1)}\)
Correct answer is A
No explanation has been provided for this answer.
\(\frac{5}{6}\)
\(\frac{7}{12}\)
\(\frac{5}{12}\)
\(\frac{1}{6}\)
Correct answer is C
let the girls' initial pocket money be whole(1)
\(\frac{1}{4}\) of 1 = books, \(\frac{1}{3}\) of 1 = dress
fraction of her money left = 1 - \(\frac{1}{4}\) - \(\frac{1}{3}\)
= \(\frac{5}{12}\)
Solve for t in the equation \(\frac{3}{4}\)t + \(\frac{1}{3}\)(21 - t) = 11
\(\frac{9}{13}\)
\(\frac{7}{13}\)
5
9\(\frac{3}{5}\)
Correct answer is D
\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11
Multiply through by the LCM of 4 and 3 which is 12
12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12)
9t + 4(21 - t) = 132
9t + 84 - 4t = 132
5t + 84 = 132
5t = 132 - 84 = 48
t = \(\frac{48}{5}\)
t = 9 \(\frac{3}{5}\)
Answer is D
If y = 23\(_{five}\) + 101\(_{three}\) , find y, leaving your answer in base two
1110
10111
11101
111100
Correct answer is B
y = 23\(_{five}\) + 101\(_{three}\)
23\(_{five}\) = \(2 \times 5^1 + 3 \times 5^0\)
= 13\(_{ten}\)
101\(_{three}\) = \(1 \times 3^2 + 0 \times 3^1 + 1 \times 3^0\)
= 10\(_{ten}\)
y\(_{ten}\) = 13\(_{ten}\) + 10\(_{ten}\)
= 23\(_{ten}\)
= 10111\(_{two}\)