Year : 
2006
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 50 of 50 Questions

# Question Ans
41.

The cost of renovating a 6 m square room is N540. What is the cost of renovating a 9 m square room?

A. N1215

B. N720

C. N1620

D. N810

Detailed Solution

Cost of removing a 6m\(^2\) room = N540
Cost of renovating a 1m\(^2\) room = 540 / 6
Cost of removing a 9m\(^2\) room
= (540 / 6) * 9
= N810
42.

The sum of the first n positive integers is

A. 1/2 n(n-1)

B. n(n+1)

C. n(n-1)

D. 1/2 n(n+1)

Detailed Solution

Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)
43.

If \(T = 2\pi \sqrt{\frac{l}{g}}\), make g the subject of the formula

A. (4π l2) / T

B. (4π2l) / T2

C. (4π2l2) / T2

D. (2π√l) / T

Detailed Solution

\(T = 2\pi \sqrt{\frac{l}{g}}\)
\(\frac{T}{2\pi} = \sqrt{\frac{l}{g}}\)
\((\frac{T}{2\pi})^{2} = \frac{l}{g}\)
\(\therefore g = \frac{4 \pi^{2} l}{T^{2}}\)
44.

If y = x\(^2\) - x - 12, find the range of values of x for which y \( \geq \) 0

A. x < -3 0r x > 4

B. x \( \leq \) -3 or x \( \geq \) 4

C. -3 < x \( \geq \) 4

D. -3 \( \leq \) x \( \leq \) 4

Detailed Solution

y = x\(^2\) - x - 12
= (x - 4)(x + 3)
∴ x = 4 or x = -3
Checking the cases for y \( \geq \) 0
We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0.
We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.
45.

How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?

A. 12

B. 10

C. 9

D. 8

Detailed Solution

3, -6, +12, -24
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9
46.

Find p, q for which \(\begin{pmatrix} 2p & 8 \\ 3 & -5q \end{pmatrix}\)\(\begin{pmatrix} 1 \\ 2\end{pmatrix}\) = \(\begin{pmatrix}24 \\ -17\end{pmatrix}\)

A. -4

B. 45

C. 4,2

D. 2

Detailed Solution

2p x 1 + 8 x 2 = 24

\(\to\) 4p = 24 - 8 = 16,
p = 4

3 x 1 + -5q x 2 = -17

\(\to\) -10q = -17 - 3

-10q = -20

q = 2
47.

If x = \(\begin{pmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1\end{pmatrix}\) and y = \(\begin{pmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & -1 & 1\end{pmatrix}\)
find 2x - y

A. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & -1 \\ -4 & 1 & 1\end{pmatrix}\)

B. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

C. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

D. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & 1 & 1 \\ -4 & -1 & 1\end{pmatrix}\)

Detailed Solution

2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4

2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1

2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
48.

Find the roots of x\(^3\) - 2x\(^2\) - 5x + 6 = 0

A. 1, -2, 3

B. 1, 2, -3,

C. -1, -2, 3

D. -1, 2, -3

Detailed Solution

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.
First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.
Then, a\(_0\) which is the coefficient void of x = 6.
The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.
The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).
= \(\pm (1, 2, 3, 6)\).
Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.
Therefore x = 1 is a root of the equation.
Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.
x\(^2\) - x - 6 = (x - 3)(x + 2).
x = -2, 3.
\(\therefore\) The roots of the equation = 1, -2 and 3.
49.

The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English and Health education as the their favourite subjects?

A. 55%

B. 52%

C. 36%

D. 22%

Detailed Solution

English = 52 students
Health education = 36 students
Total = 88 students
= \(\frac{88}{160} \times 100%\)
= 55%
50.

If \(E \subseteq G \subseteq U\), where U is the universal set, then the shaded venn diagram representing \(U - E\) or \(E^{c}\) is

A. I

B. II

C. III

D. IV

D
41.

The cost of renovating a 6 m square room is N540. What is the cost of renovating a 9 m square room?

A. N1215

B. N720

C. N1620

D. N810

Detailed Solution

Cost of removing a 6m\(^2\) room = N540
Cost of renovating a 1m\(^2\) room = 540 / 6
Cost of removing a 9m\(^2\) room
= (540 / 6) * 9
= N810
42.

The sum of the first n positive integers is

A. 1/2 n(n-1)

B. n(n+1)

C. n(n-1)

D. 1/2 n(n+1)

Detailed Solution

Let the positive integers be 1, 2, ,3, 4, .....n
∴ a = 1, d = 1 and n = n
Sn = n/2(2a + (n-1)d)
= n/2 (2 + n – 1)
= 1/2n(n + 1)
43.

If \(T = 2\pi \sqrt{\frac{l}{g}}\), make g the subject of the formula

A. (4π l2) / T

B. (4π2l) / T2

C. (4π2l2) / T2

D. (2π√l) / T

Detailed Solution

\(T = 2\pi \sqrt{\frac{l}{g}}\)
\(\frac{T}{2\pi} = \sqrt{\frac{l}{g}}\)
\((\frac{T}{2\pi})^{2} = \frac{l}{g}\)
\(\therefore g = \frac{4 \pi^{2} l}{T^{2}}\)
44.

If y = x\(^2\) - x - 12, find the range of values of x for which y \( \geq \) 0

A. x < -3 0r x > 4

B. x \( \leq \) -3 or x \( \geq \) 4

C. -3 < x \( \geq \) 4

D. -3 \( \leq \) x \( \leq \) 4

Detailed Solution

y = x\(^2\) - x - 12
= (x - 4)(x + 3)
∴ x = 4 or x = -3
Checking the cases for y \( \geq \) 0
We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0.
We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.
45.

How many terms of the series 3, -6, +12, - 24, + ..... are needed to make a total of 1-28?

A. 12

B. 10

C. 9

D. 8

Detailed Solution

3, -6, +12, -24
a = 3, r = -2
\(8n = \frac{a(1-r^n)}{1-r}\\
∴1-2^8 = \frac{3(1-(-2^{n-1}))}{1-(-2)}\\
1-2^8 = \frac{3(1-(-2^{n-1}))}{3}\)
1-28 = 1-(-2)n-1
-28 = -2n-1
8 = n-1
n = 9
46.

Find p, q for which \(\begin{pmatrix} 2p & 8 \\ 3 & -5q \end{pmatrix}\)\(\begin{pmatrix} 1 \\ 2\end{pmatrix}\) = \(\begin{pmatrix}24 \\ -17\end{pmatrix}\)

A. -4

B. 45

C. 4,2

D. 2

Detailed Solution

2p x 1 + 8 x 2 = 24

\(\to\) 4p = 24 - 8 = 16,
p = 4

3 x 1 + -5q x 2 = -17

\(\to\) -10q = -17 - 3

-10q = -20

q = 2
47.

If x = \(\begin{pmatrix} 1 & 0 & 1 \\ 2 & -1 & 0 \\ -1 & 0 & 1\end{pmatrix}\) and y = \(\begin{pmatrix} -1 & 1 & 2 \\ 0 & -1 & -1 \\ 2 & -1 & 1\end{pmatrix}\)
find 2x - y

A. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & -1 \\ -4 & 1 & 1\end{pmatrix}\)

B. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

C. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -3 & 1 \\ -4 & 1 & 1\end{pmatrix}\)

D. \(\begin{pmatrix} 3 & -1 & 0 \\ 4 & 1 & 1 \\ -4 & -1 & 1\end{pmatrix}\)

Detailed Solution

2(1) - (-1) = 3 2(2) - (0) = 4 2(-1) - (2) = -4

2(0) - (1) = -1 2(-1) - (-1) = -1 2(0) - (-1) = 1

2(1) - 2 = 0 2(0) - (-1) = 1 2(1) - (1) = 1

\(\begin{pmatrix} 3 & -1 & 0 \\ 4 & -1 & 1 \\ -4 & 1 & 1\end{pmatrix}\)
48.

Find the roots of x\(^3\) - 2x\(^2\) - 5x + 6 = 0

A. 1, -2, 3

B. 1, 2, -3,

C. -1, -2, 3

D. -1, 2, -3

Detailed Solution

Equation: x\(^3\) - 2x\(^2\) - 5x + 6 = 0.
First, bring out a\(_n\) which is the coefficient of x\(^3\) = 1.
Then, a\(_0\) which is the coefficient void of x = 6.
The factors of a\(_n\) = 1; The factors of a\(_0\) = 1, 2, 3 and 6.
The numbers to test for the roots are \(\pm (\frac{a_0}{a_n})\).
= \(\pm (1, 2, 3, 6)\).
Test for +1: 1\(^3\) - 2(1\(^2\)) - 5(1) + 6 = 1 - 2 - 5 + 6 = 0.
Therefore x = 1 is a root of the equation.
Using long division method, \(\frac{x^3 - 2x^2 - 5x + 6}{x - 1}\) = x\(^2\) - x - 6.
x\(^2\) - x - 6 = (x - 3)(x + 2).
x = -2, 3.
\(\therefore\) The roots of the equation = 1, -2 and 3.
49.

The response of 160 pupils in a school asked to indicate their favourite subjects is given in the bar chart above. What percentage of the pupils have English and Health education as the their favourite subjects?

A. 55%

B. 52%

C. 36%

D. 22%

Detailed Solution

English = 52 students
Health education = 36 students
Total = 88 students
= \(\frac{88}{160} \times 100%\)
= 55%
50.

If \(E \subseteq G \subseteq U\), where U is the universal set, then the shaded venn diagram representing \(U - E\) or \(E^{c}\) is

A. I

B. II

C. III

D. IV

D