JAMB Mathematics Past Questions & Answers - Page 79

391.

Find the sum to infinity of the series
\(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\),..........

A.

\(\frac{1}{2}\)

B.

\(\frac{3}{5}\)

C.

\(\frac{-1}{5}\)

D.

\(\frac{73}{12}\)

Correct answer is A

Sum to infinity

∑ = arn − 1

= \(\frac{a}{1}\) − r

a = \(\frac{1}{4}\)

r = \(\frac{1}{8}\) ÷ \(\frac{1}{4}\)

r = \(\frac{1}{s}\) × \(\frac{4}{1}\)

= \(\frac{1}{2}\)

S = \(\frac{1 \div 4}{1}\) − \(\frac{1}{2}\)

= \(\frac{1}{4}\) ÷ \(\frac{1}{2}\)

= \(\frac{1}{4}\) × \(\frac{2}{1}\)

= \(\frac{1}{2}\)

392.

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

393.

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

394.

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

395.

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