Mathematics questions and answers

Mathematics Questions and Answers

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.

741.

Find the sum of the range and the mode of the set of numbers 10, 9, 10, 9, 8, 7, 7, 10, 8, 10, 8, 4, 6, 9, 10, 9, 7, 10, 6, 5

A.

16

B.

14

C.

12

D.

10

Correct answer is A

Range = Highest Number - Lowest Number

Mode is the number with highest occurrence
10, 9, 10, 9, 8, 7, 7, 10, 8, 4, 6,, 9, 10, 9, 7, 10, 6, 5

Range = 10 − 4 = 6

Mode = 10

Sum of range and mode = range + mode = 6 + 10

= 16

742.

Given the quadrilateral RSTO inscribed in the circle with O as centre. Find the size angle x and given RST = 60o

A.

100o

B.

140o

C.

120o

D.

10o

Correct answer is C

If RST = 60o

RXT = 2 × RST

(angle at the centre twice angle at the circumference)

RXT = 2 × 60

= 120o

743.

The locus of a point which is equidistant from the line PQ forms a

A.

circle centre P

B.

pair of parallel lines each opposite to PQ

C.

circle centre Q

D.

perpendicular line to PQ

Correct answer is D

The locus of points at a fixed distance from the point P is a circle with the given P at its centre.

The locus of points at a fixed distance from the point Q is a circle with the given point Q at its centre

The locus of points equidistant from two points P and Q i.e line PQ is the perpendicular bisector of the segment determined by the points

Hence, The locus of a point which is equidistant from the line PQ forms a perpendicular line to PQ

744.

Simplify 3 \(^{n − 1}\) ×  \(\frac{27^{n + 1}}{81^n}\)

A.

3\(^{2n}\)

B.

9

C.

3n

D.

3 \(^{n + 1}\)

Correct answer is B

3\(^{n - 1}\) × \(\frac{27^{n + 1}}{81^n}\)

= 3\(^{n - 1}\) ×  \(\frac{3^{3(n + 1)}}{3^{4n}}\)

= 3\(^{n - 1 + 3n + 3 − 4n}\)

= 3\(^{4n − 4n − 1 + 3}\)

= \(3^{2}\)

= 9

745.

Given m = N\(\sqrt{\frac{SL}{T}}\) make T the subject of the formula

A.

\(\frac{\text{NSL}}{M}\)

B.

\(\frac{N^2SL}{M^2}\)

C.

\(\frac{N^2SL}{M}\)

D.

\(\frac{NSL}{M^2}\)

Correct answer is B

M = N \(\sqrt{\frac{SL}{T}}\),

make T subject of formula square both sides

M\(^{2}\) = \(\frac{N^2SL}{T}\)

TM\(^{2}\) = N\(^{2}\)SL

T = \(\frac{N^2SL}{M^2}\)