Year : 
1994
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

31 - 40 of 48 Questions

# Question Ans
31.

What is the value of sin(-690)?

A. \(\frac{\sqrt{3}}{2}\)

B. -\(\frac{\sqrt{3}}{2}\)

C. \(\frac{-1}{2}\)

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

Detailed Solution

Sin(-690o) = Sin(-360 -3300

sin - 360 = sin 0

∴ sin(-690o) = sin(330o)

Negative angles are measured in clockwise direction

The acute angle equivalent of sin(-330o) = sin(30o)

sin(-330o) = sin(30o)

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

= 0.5
32.

If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1

A. -1

B. 1

C. -2

D. 2

Detailed Solution

y = 3t3 + 2t2 - 7t + 3

\(\frac{dy}{dt}\) = 9t2 + 4t - 7

When t = -1

\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7

= 9 - 4 -7

= 9 - 11

= -2
33.

Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient

A. (1, 3)

B. (2, 7)

C. (0, 3)

D. (3, 15)

Detailed Solution

Equation of curve;

y = 2x2 - 2x + 3

gradient of curve;

\(\frac{dy}{dx}\) = differential coefficient

\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2

∴ \(\frac{dy}{dx}\) = 2

4x - 2 = 2

4x = 4

∴ x = 1

When x = 1, y = 2(1)2 - 2(1) + 3

= 2 - 2 + 3

= 5 - 2

= 3

coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)
34.

Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

A. 3\(\frac{2}{3}\)

B. 4

C. 4\(\frac{1}{3}\)

D. 4\(\frac{2}{3}\)

Detailed Solution

\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)

= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + c]\)

= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]

= \(\frac{8}{3}\) + 3 + -1 - C

= \(\frac{8}{3}\) + 2

= \(\frac{14}{3}\)

= \(4 \frac{2}{3}\)
35.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y - 2 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y

A. 1

B. 2

C. 3

D. 4

Detailed Solution

Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)
36.

Find the mean deviation of the set of numbers 4, 5, 9

A. zero

B. 2

C. 5

D. 6

Detailed Solution

x = \(\frac{\sum x}{N}\)

= \(\frac{18}{3}\)

= 6

\(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\)

M.D = \(\frac{|x - x|}{N}\)

= \(\frac{6}{3}\)

= 2
37.

\(\begin{array}{c|c} \text{Class Interval} & 1 - 5 & 6 - 10 & 11 - 15 & 16 - 20 & 21 - 25 \\ \hline Frequency & 6 & 15 & 20 & 7 & 2\end{array}\)
Estimate the median of the frequency distribution above

A. 10\(\frac{1}{2}\)

B. 11\(\frac{1}{2}\)

C. 12

D. 13

Detailed Solution

Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h

N = Sum of frequencies

L = lower class boundary of median class

f = sum of all frequencies below L

fm = frequency of modal class and

h = class width of median class

Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5

= 11 + (\(\frac{25 - 21}{20}\))5

= 11 + (\(\frac{(4)}{20}\))

11 + 1 = 12
38.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above

A. \(\frac{3}{2}\)

B. \(\frac{9}{4}\)

C. \(\frac{5}{2}\)

D. 3

Detailed Solution

\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)

x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{24}{8}\)

= 3

Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)

= \(\frac{18}{8}\)

= \(\frac{9}{4}\)
39.

In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?

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

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

C. \(\frac{3}{8}\)

D. \(\frac{3}{11}\)

Detailed Solution

40 = 20 - x + x + 35 - x

40 = 55 - x

x = 55 - 40

= 15

∴ P(both) \(\frac{15}{40}\)

= \(\frac{3}{8}\)
40.

The equation of the graph is

A. y = (x - 3)3

B. y = (x + 3)3

C. y = x3 - 27

D. y = -x3 + 27

Detailed Solution

y = x3 - 27, y = -27 \(\to\) (0, -27)

when y = 0, x = 3 (3, 0)
31.

What is the value of sin(-690)?

A. \(\frac{\sqrt{3}}{2}\)

B. -\(\frac{\sqrt{3}}{2}\)

C. \(\frac{-1}{2}\)

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

Detailed Solution

Sin(-690o) = Sin(-360 -3300

sin - 360 = sin 0

∴ sin(-690o) = sin(330o)

Negative angles are measured in clockwise direction

The acute angle equivalent of sin(-330o) = sin(30o)

sin(-330o) = sin(30o)

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

= 0.5
32.

If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1

A. -1

B. 1

C. -2

D. 2

Detailed Solution

y = 3t3 + 2t2 - 7t + 3

\(\frac{dy}{dt}\) = 9t2 + 4t - 7

When t = -1

\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7

= 9 - 4 -7

= 9 - 11

= -2
33.

Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient

A. (1, 3)

B. (2, 7)

C. (0, 3)

D. (3, 15)

Detailed Solution

Equation of curve;

y = 2x2 - 2x + 3

gradient of curve;

\(\frac{dy}{dx}\) = differential coefficient

\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2

∴ \(\frac{dy}{dx}\) = 2

4x - 2 = 2

4x = 4

∴ x = 1

When x = 1, y = 2(1)2 - 2(1) + 3

= 2 - 2 + 3

= 5 - 2

= 3

coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)
34.

Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

A. 3\(\frac{2}{3}\)

B. 4

C. 4\(\frac{1}{3}\)

D. 4\(\frac{2}{3}\)

Detailed Solution

\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)

= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)

= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + c]\)

= [\(\frac{4}{3}\) + 3 + c] - [4 + \(\frac{1}{3}\) + c]

= \(\frac{8}{3}\) + 3 + -1 - C

= \(\frac{8}{3}\) + 2

= \(\frac{14}{3}\)

= \(4 \frac{2}{3}\)
35.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & y + 2 & y - 2 & 2y - 3 & y + 4 & 3y - 4\end{array}\)
This table shows the frequency distribution of a data if the mean is \(\frac{43}{14}\) find y

A. 1

B. 2

C. 3

D. 4

Detailed Solution

Mean = \(\frac{\sum fx}{\sum f}\)
\(\therefore \frac{28y - 13}{8y - 2} = \frac{43}{14}\)
\(\implies 14(28y - 13) = 43(8y - 2)\)
\(392y - 182 = 344y - 86\)
\(392y - 344y = -86 + 182 \implies 48y = 96\)
\(y = 2\)
36.

Find the mean deviation of the set of numbers 4, 5, 9

A. zero

B. 2

C. 5

D. 6

Detailed Solution

x = \(\frac{\sum x}{N}\)

= \(\frac{18}{3}\)

= 6

\(\begin{array}{c|c} x & x - x & x - x \\ \hline 4 & -2 & 2\\ 5 & 1 & 1\\ 9 & 3 & 3\\ \hline & & 6\end{array}\)

M.D = \(\frac{|x - x|}{N}\)

= \(\frac{6}{3}\)

= 2
37.

\(\begin{array}{c|c} \text{Class Interval} & 1 - 5 & 6 - 10 & 11 - 15 & 16 - 20 & 21 - 25 \\ \hline Frequency & 6 & 15 & 20 & 7 & 2\end{array}\)
Estimate the median of the frequency distribution above

A. 10\(\frac{1}{2}\)

B. 11\(\frac{1}{2}\)

C. 12

D. 13

Detailed Solution

Median = L + [\(\frac{\frac{N}{2} - f}{fm}\)]h

N = Sum of frequencies

L = lower class boundary of median class

f = sum of all frequencies below L

fm = frequency of modal class and

h = class width of median class

Median = 11 + [\(\frac{\frac{50}{2} - 21}{20}\)]5

= 11 + (\(\frac{25 - 21}{20}\))5

= 11 + (\(\frac{(4)}{20}\))

11 + 1 = 12
38.

\(\begin{array}{c|c} x & 1 & 2 & 3 & 4 & 5 \\ \hline f & 2 & 1 & 2 & 1 & 2\end{array}\)
Find the variance of the frequency distribution above

A. \(\frac{3}{2}\)

B. \(\frac{9}{4}\)

C. \(\frac{5}{2}\)

D. 3

Detailed Solution

\(\begin{array}{c|c} x & f & fx & \bar{x} - x & (\bar{x} - x)^2 & f(\bar{x} - x)^2 \\ \hline 1 & 2 & 2 & -2 & 4 & 8\\ 2 & 1 & 2 & -1 & 1 & 1\\ 3 & 2 & 6 & 0 & 0 & 0\\ 4 & 1 & 4 & 1 & 1 & 1\\ 2 & 2 & 10 & 2 & 4 & 8\\ \hline & 8 & 24 & & & 18 \end{array}\)

x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{24}{8}\)

= 3

Variance (62) = \(\frac{\sum f(\bar{x} - x)^2}{\sum f}\)

= \(\frac{18}{8}\)

= \(\frac{9}{4}\)
39.

In a survey, it was observed that 20 students read newspapers and 35 read novels. If 40 of the students read either newspapers or novels, what is the probability of the students who read both newspapers and novels?

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

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

C. \(\frac{3}{8}\)

D. \(\frac{3}{11}\)

Detailed Solution

40 = 20 - x + x + 35 - x

40 = 55 - x

x = 55 - 40

= 15

∴ P(both) \(\frac{15}{40}\)

= \(\frac{3}{8}\)
40.

The equation of the graph is

A. y = (x - 3)3

B. y = (x + 3)3

C. y = x3 - 27

D. y = -x3 + 27

Detailed Solution

y = x3 - 27, y = -27 \(\to\) (0, -27)

when y = 0, x = 3 (3, 0)