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.
N676.50
N820
N1,640
N4,920
Correct answer is C
No explanation has been provided for this answer.
Make T the subject of the relation.
T = \(\frac{R + P3}{15Q}\)
T = \(\frac{R - 15P^3}{Q}\)
T =R - \(\frac{15P^3}{Q}\)
T = \(\frac{15R + Q}{P^3}\)
Correct answer is C
P = (\(\frac{Q( R - T )}{ 15})^\frac{1}{3}\)
take cube of both sides
\(P^3 =\frac{Q( R - T)}{ 15}\)
cross multiply
\(15P^3 = Q( R - T)\)
\(\frac{15P^3}{Q}\) = R - T
T = R - \(\frac{15P^3}{Q}\)
25
1
\(\frac{1}{5}\)
\(\frac{1}{25}\)
Correct answer is A
\(\frac {25^{\frac{2}{3}} \div 25^{\frac{1}{6}}} {(\frac{1}{5})^{\frac{7}{6}} \div (\frac{1}{5})^{\frac{1}{6}}}\)
= \(\frac{25^{\frac{2}{3} - \frac{1}{6}}}{(\frac{1}{5})^{\frac{7}{6} - \frac{1}{6}}}\)
= \(\frac{25^{\frac{1}{2}}}{(\frac{1}{5})}\)
= \(5 \div \frac{1}{5}\)
= 25
14.67cm
73.33 cm
101.33cm
513.33cm
Correct answer is C
Length of the arc = \(\frac{\theta}{360} \times 2\pi r\)
= \(\frac{300}{360} \times 2 \times \frac{22}{7} \times 14\)
= \(\frac{220}{3}\)
= 73.33 cm
Perimeter of the sector = Length of the arc + 2r
= 73.33 + 2(14)
= 101.33 cm
\(\frac {1}{10}\)
\(\frac {2}{10}\)
\(\frac {9}{20}\)
\(\frac {11}{20}\)
Correct answer is D
Pr(both John and James passed)
= \(\frac {3}{4}\) x \(\frac {3}{5}\)
= \(\frac {9}{20}\)
Pr(john and james failed)= 1- Pr(john and james passed)
= 1 – \(\frac {9}{20}\)
= \(\frac {11}{20}\)
Answer is D