Further Mathematics questions and answers

Further Mathematics Questions and Answers

Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.

321.

In the diagram, a ladder PS leaning against a vertical wall PR makes angle x° with the horizontal floor. The ladder slides down to a point QT such that angle QTR = 30° and SNT = y°. Find an expression for tan y.

A.

\(\frac{\sqrt{3} \tan x - 1}{\sqrt{3} + \tan x}\)

B.

\(\frac{\sqrt{3} \tan x}{\sqrt{3} + \tan x}\)

C.

\(\frac{\sqrt{3} \tan x + 1}{\sqrt{3} - \tan x}\)

D.

\(\frac{3 \tan x - 1}{\sqrt{3} - \tan x}\)

Correct answer is A

No explanation has been provided for this answer.

322.

In the diagram, a ladder PS leaning against a vertical wall PR makes angle x° with the horizontal floor. The ladder slides down to a point QT such that angle QTR = 30° and SNT = y°. Find the relation between x and y.

A.

\(y = \frac{1}{2}x + 30\)

B.

\(y = \frac{1}{2}x + 30\)

C.

\(y = x + 30\)

D.

\(y = x - 30\)

Correct answer is D

No explanation has been provided for this answer.

323.

A body is acted upon by forces \(F_{1} = (10 N, 090°)\) and \(F_{2} = (6 N, 180°)\). Find the magnitude of the resultant force.

A.

11.6 N

B.

11.7 N

C.

11.8 N

D.

11.9 N

Correct answer is B

\(F = F \cos \theta i + F \sin \theta j\) (Resolving F into its components)

\(F_{1} = (10 N, 090°) = 10 \cos 90 i + 10 \sin 90 j\)

= \(10 j\)

\(F_{2} = (6 N, 180°) = 6 \cos 180 i + 6 \sin 180 j\)

= \(-6 i\)

\(R = - 6 i + 10 j\)

\(|R| = \sqrt{(-6)^{2} + (10)^{2}}\)

= \(\sqrt{136}\)

= 11.7 N

324.

The ages, in years, of 5 boys are 5, 6, 6, 8 and 10. Calculate, correct to one decimal place, the standard deviation of their ages.

A.

3.2 years

B.

2.6 years

C.

1.9 years

D.

1.8 years

Correct answer is D

\(x\) 5 6 6 8 10 Total
\(x - \bar{x}\) -2 -1 -1 1 3  
\((x - \bar{x})^{2}\) 4 1 1 1 9 16

Mean (\(\bar{x}\)) = \(\frac{5 + 6 + 6 + 8 + 10}{5} \)

= \(\frac{35}{5} = 7\)

\(SD = \sqrt{\frac{\sum (x - \bar{x})^{2}}{n}}\)

= \(\sqrt{\frac{16}{5}} \)

= \(\sqrt{3.2}\)

\(\approxeq 1.8 years\)

325.

Three defective bulbs got mixed up with seven good ones. If two bulbs are selected at random, what is the probability that both are good?

A.

\(\frac{3}{7}\)

B.

\(\frac{21}{50}\)

C.

\(\frac{7}{15}\)

D.

\(\frac{49}{100}\)

Correct answer is C

\(P(\text{both bulbs are good}) = \frac{7}{10} \times \frac{6}{9}\)

= \(\frac{7}{15}\)