Year : 
1983
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

31 - 40 of 48 Questions

# Question Ans
31.

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per house hold
\(\begin{array}{c|c} \text{No. of members per household} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \text{Total} \\ \hline \text{No. of households} & 3 & 12 & 15 & 28 & 21 & 10 & 7 & 4 & 100\end{array}\)

What is the median?

A. 4

B. 3

C. 5

D. 4.5

E. None

Detailed Solution

\(\begin{array}{c|c} x & f \\\hline 1 & 3\\ 2 & 13\\3 & 15\\ 4 & 28\\ 5 & 21\\ 6 & 10\\ 7 & 7\\ 8 & 4\end{array}\)

Median is half of total frequency = 50th term

4 falls in the range = 4
32.

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

A. 6.00cm2

B. 6.10cm2

C. 6cmv

D. 6.09cm2

E. 4.00cm2

Detailed Solution

Area of the paper = area of square = L x B or S2

where s = S x k

Area of the paper = (2.524)2

area of the diagram = (0.524)2

area not covered = (2.524)2 - (0.524)2

= 6.370576 - 0.274576

= 6.096

= 6.10cm2 (2 s.f)
33.

Simplify \(\sqrt[3]{\frac{27a^{-9}}{8}}\).

A. \(\frac{9a^2}{2a^3}\)

B. \(\frac{3}{2a^3}\)

C. \(\frac{2}{3a^2}\)

D. \(\frac{3a^2}{2}\)

Detailed Solution

\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\)

= \(\frac{3}{2}\) x \(\frac{1}{a^3}\)

= \(\frac{3}{2a^3}\)
34.

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60o. Find the perimeter of the figure PTRS. \(\pi\) = \(\frac{22}{7}\)

A. 25cm

B. 18cm

C. 36cm

D. 29cm

E. 25.5cm

Detailed Solution

Circumference of PRS = \(\frac{\pi}{2}\) = \(\frac{22}{7}\) x \(\frac{7}{1}\) x \(\frac{1}{2}\)

= 11cm

Side PT = 7cm, Side TR = 7cm

Perimeter(PTRS) = 11cm + 7cm + 7cm

= 25cm
35.

In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30o. Find angles P and R

A. P = 60o and R = 90o

B. P = 30o and R = 120o

C. P = 90o and R = 60o

D. P = 60o and R 60o

E. P = 45o and R = 105o

Detailed Solution

By using cosine formula, p2 = Q2 + R2 - 2QR cos p

Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\)

= \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\)

= \(\frac{3 + 12 - 9}{12}\)

= \(\frac{6}{12}\)

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

= 0.5

Cos P = 0.5

p = cos-1 0.5 = 60o

= < P = 60o

If < P = 60o and < Q = 30

< R = 180o - 90o

angle P = 60o and angle R is 90o
36.

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

A. 25.12

B. 25.30

C. 25.26

D. 25.50

E. 25.75

Detailed Solution

\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\)

mean = \(\frac{126.3}{5}\)

= 25.26
37.

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

A. 120km/hr

B. 60km/hr

C. 670km/hr

D. 40km/hr

Detailed Solution

Speed = \(\frac{distance}{time}\)

let x represent the speed, d represent distance

x = \(\frac{d}{4}\)

d = 4x

2x = \(\frac{600 - d}{3}\)

6x = 600 - d

6x = 600 - 4x

10x = 600

x = \(\frac{600}{10}\)

= 60km/hr
38.

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

A. 15

B. 10

C. 18

D. 12

E. 20

Detailed Solution

(2x - 24)° + (3x - 18)° + (2x + 12)° + (x + 12)° + x° = 360°

9x = 360° + 18°

x = \(\frac{378}{9}\)

= 42°, if x = 42°, then add maths = \(\frac{2x - 42}{360}\) x 60

= \(\frac{2 \times 42 - 24}{360}\) x 60

= \(\frac{84 - 24}{6}\)

= 10
39.

In the figure, find PRQ

A. 66\(\frac{1}{2}\)o

B. 62\(\frac{1}{2}\)o

C. 125o

D. 105o

E. 65o

Detailed Solution

Angle subtended at any part of the circumference of the circle \(\frac{125^o}{2}\) at centre = 360o - 235o = 125o

\(\bar{PQR}\) = \(\frac{125}{2}\)

= 62\(\frac{1}{2}\)o
40.

Which of the following equations represents the graph?

A. y = 1 + 2x + 3x2

B. y = 1 - 2x + 3x2

C. y = 1 + 2x - 3x2

D. y = 1 - 2x - 3x2

E. y = 3x2 + 2x - 1

Detailed Solution

The roots of the function are 1 and \(\frac{1}{3}\)

sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\)

product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\)

x2 - (sum of roots)x + (product of roots) = 0

x2 + (-\(\frac{2}{3}\))x - (-\(\frac{1}{3}\)) = 0

x2 + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0

3x2 + 2x - 1 = 0
31.

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per house hold
\(\begin{array}{c|c} \text{No. of members per household} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \text{Total} \\ \hline \text{No. of households} & 3 & 12 & 15 & 28 & 21 & 10 & 7 & 4 & 100\end{array}\)

What is the median?

A. 4

B. 3

C. 5

D. 4.5

E. None

Detailed Solution

\(\begin{array}{c|c} x & f \\\hline 1 & 3\\ 2 & 13\\3 & 15\\ 4 & 28\\ 5 & 21\\ 6 & 10\\ 7 & 7\\ 8 & 4\end{array}\)

Median is half of total frequency = 50th term

4 falls in the range = 4
32.

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

A. 6.00cm2

B. 6.10cm2

C. 6cmv

D. 6.09cm2

E. 4.00cm2

Detailed Solution

Area of the paper = area of square = L x B or S2

where s = S x k

Area of the paper = (2.524)2

area of the diagram = (0.524)2

area not covered = (2.524)2 - (0.524)2

= 6.370576 - 0.274576

= 6.096

= 6.10cm2 (2 s.f)
33.

Simplify \(\sqrt[3]{\frac{27a^{-9}}{8}}\).

A. \(\frac{9a^2}{2a^3}\)

B. \(\frac{3}{2a^3}\)

C. \(\frac{2}{3a^2}\)

D. \(\frac{3a^2}{2}\)

Detailed Solution

\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\)

= \(\frac{3}{2}\) x \(\frac{1}{a^3}\)

= \(\frac{3}{2a^3}\)
34.

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60o. Find the perimeter of the figure PTRS. \(\pi\) = \(\frac{22}{7}\)

A. 25cm

B. 18cm

C. 36cm

D. 29cm

E. 25.5cm

Detailed Solution

Circumference of PRS = \(\frac{\pi}{2}\) = \(\frac{22}{7}\) x \(\frac{7}{1}\) x \(\frac{1}{2}\)

= 11cm

Side PT = 7cm, Side TR = 7cm

Perimeter(PTRS) = 11cm + 7cm + 7cm

= 25cm
35.

In a triangle PQT, QR = \(\sqrt{3}cm\), PR = 3cm, PQ = \(2\sqrt{3}\)cm and PQR = 30o. Find angles P and R

A. P = 60o and R = 90o

B. P = 30o and R = 120o

C. P = 90o and R = 60o

D. P = 60o and R 60o

E. P = 45o and R = 105o

Detailed Solution

By using cosine formula, p2 = Q2 + R2 - 2QR cos p

Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\)

= \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\)

= \(\frac{3 + 12 - 9}{12}\)

= \(\frac{6}{12}\)

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

= 0.5

Cos P = 0.5

p = cos-1 0.5 = 60o

= < P = 60o

If < P = 60o and < Q = 30

< R = 180o - 90o

angle P = 60o and angle R is 90o
36.

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

A. 25.12

B. 25.30

C. 25.26

D. 25.50

E. 25.75

Detailed Solution

\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\)

mean = \(\frac{126.3}{5}\)

= 25.26
37.

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

A. 120km/hr

B. 60km/hr

C. 670km/hr

D. 40km/hr

Detailed Solution

Speed = \(\frac{distance}{time}\)

let x represent the speed, d represent distance

x = \(\frac{d}{4}\)

d = 4x

2x = \(\frac{600 - d}{3}\)

6x = 600 - d

6x = 600 - 4x

10x = 600

x = \(\frac{600}{10}\)

= 60km/hr
38.

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

A. 15

B. 10

C. 18

D. 12

E. 20

Detailed Solution

(2x - 24)° + (3x - 18)° + (2x + 12)° + (x + 12)° + x° = 360°

9x = 360° + 18°

x = \(\frac{378}{9}\)

= 42°, if x = 42°, then add maths = \(\frac{2x - 42}{360}\) x 60

= \(\frac{2 \times 42 - 24}{360}\) x 60

= \(\frac{84 - 24}{6}\)

= 10
39.

In the figure, find PRQ

A. 66\(\frac{1}{2}\)o

B. 62\(\frac{1}{2}\)o

C. 125o

D. 105o

E. 65o

Detailed Solution

Angle subtended at any part of the circumference of the circle \(\frac{125^o}{2}\) at centre = 360o - 235o = 125o

\(\bar{PQR}\) = \(\frac{125}{2}\)

= 62\(\frac{1}{2}\)o
40.

Which of the following equations represents the graph?

A. y = 1 + 2x + 3x2

B. y = 1 - 2x + 3x2

C. y = 1 + 2x - 3x2

D. y = 1 - 2x - 3x2

E. y = 3x2 + 2x - 1

Detailed Solution

The roots of the function are 1 and \(\frac{1}{3}\)

sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\)

product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\)

x2 - (sum of roots)x + (product of roots) = 0

x2 + (-\(\frac{2}{3}\))x - (-\(\frac{1}{3}\)) = 0

x2 + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0

3x2 + 2x - 1 = 0