Year : 
2001
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

31 - 40 of 50 Questions

# Question Ans
31.

If y = x sinx, find dy/dx when x = π/2.

A. -π/2

B. -1

C. 1

D. π/2

Detailed Solution

y = x sin x
dy/dx = 1 x sinx + x cosx
= sinx + x cosx
At x = π/2, = sin (π/2) + (π/2) cos (π/2)
= 1 + (π/2) x (0) = 1
32.

Find the dimensions of a rectangle of greatest area which has a fixed perimeter p.

A. square of sides p

B. square of sides 2p

C. square of sides (p/2)

D. square of sides (p/4)

Detailed Solution

Let the rectangle be a square of sides p/4.

So that perimeter of square = 4p
4 x (p/4) = p.
33.

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode.

A. 49

B. 121

C. 25

D. 64

Detailed Solution

Mode = score with highest frequency = 11.
Square of 11 = 121
34.

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is

A. 7.0

B. 8.7

C. 9.5

D. 11.0

Detailed Solution

Mean = ∑fx/∑f
Mean = (4x3) + (7x5) + ... + (8x1) all divided by 20
Mean = 174/20 = 8.7
35.

Teams P and Q are involved in a game of football. What is the probability that the game ends in a draw?

A. 2/3

B. 1/2

C. 1/3

D. 1/4

Detailed Solution

P (games end in draw)
=> Team P wins and Q wins
P(P wins) = 1/2
P(Q wins) = 1/2
Therefore P (games ends in draw) = 1/2 x 1/2 = 1/4
36.

If 6Pr = 6, find the value of 6Pr+1

A. 30

B. 33

C. 35

D. 15

Detailed Solution

6Pr = 6 Thus r = 1

N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\)

= \(\frac{\text{6!}}{\text{5!}}\)

= \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6

6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\)

= \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\)

= 30
37.

Find the variance of 2, 6, 8, 6, 2 and 6

A. 6

B. 5

C. √6

D. √5

Detailed Solution

mean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5
X → 2,6,8,6,2,6
X - x = -3, 1, 3, 1, 3, 1
(X - x)2 = 9, 1, 9, 1, 9, 1
∑(X-x) = 9+1+9+1+9+1 = 30
Variance = ∑(X-x)2/n = 30/6 = 5
38.

Find the number of ways of selecting 8 subjects from 12 subjects for an examination

A. 490

B. 495

C. 496

D. 498

Detailed Solution

Combination is the number (n) of ways of selecting a number of (m) of objects from n
\(^{12}C_{8}\frac{12!}{8!(12-8)!}\\=\frac{12!}{8!4!}\\\frac{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\After\hspace{1mm} cancelling \hspace{1mm}out \hspace{1mm}we\hspace{1mm} have\\11\times 5\times 9 = 495 \)
39.

Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3

A. 3/4

B. 5/6

C. 7/6

D. 4/3

Detailed Solution

1/6, 1/3, 3/2, 2/3, 8/9 and 4/3
L.C.M of the denominators = 36
The fraction now becomes
6/36, 12/36, 54/36, 24/36, 32/36 and 48/36
Highest fraction = 54/36
Lowest fraction = 6/36
Range = Highest - Lowest
(54/36) - (6/36)
= 48/36
= 4/3
40.

Triangle SPT is the solution of the linear inequalities

A. 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0, -2 ≤ x ≤ - 1

B. -2 ≤ x ≤ 2, y ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

C. 2y - x - 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0

D. 2y - x - 2 ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

Detailed Solution

Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1
31.

If y = x sinx, find dy/dx when x = π/2.

A. -π/2

B. -1

C. 1

D. π/2

Detailed Solution

y = x sin x
dy/dx = 1 x sinx + x cosx
= sinx + x cosx
At x = π/2, = sin (π/2) + (π/2) cos (π/2)
= 1 + (π/2) x (0) = 1
32.

Find the dimensions of a rectangle of greatest area which has a fixed perimeter p.

A. square of sides p

B. square of sides 2p

C. square of sides (p/2)

D. square of sides (p/4)

Detailed Solution

Let the rectangle be a square of sides p/4.

So that perimeter of square = 4p
4 x (p/4) = p.
33.

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. Find the square of the mode.

A. 49

B. 121

C. 25

D. 64

Detailed Solution

Mode = score with highest frequency = 11.
Square of 11 = 121
34.

Given the scores: 4, 7, 8, 11, 13, 8 with corresponding frequencies: 3, 5, 2, 7, 2, 1 respectively. The mean score is

A. 7.0

B. 8.7

C. 9.5

D. 11.0

Detailed Solution

Mean = ∑fx/∑f
Mean = (4x3) + (7x5) + ... + (8x1) all divided by 20
Mean = 174/20 = 8.7
35.

Teams P and Q are involved in a game of football. What is the probability that the game ends in a draw?

A. 2/3

B. 1/2

C. 1/3

D. 1/4

Detailed Solution

P (games end in draw)
=> Team P wins and Q wins
P(P wins) = 1/2
P(Q wins) = 1/2
Therefore P (games ends in draw) = 1/2 x 1/2 = 1/4
36.

If 6Pr = 6, find the value of 6Pr+1

A. 30

B. 33

C. 35

D. 15

Detailed Solution

6Pr = 6 Thus r = 1

N.B 6P1 = \(\frac{\text{6!}}{\text{(6 - 1)!}}\)

= \(\frac{\text{6!}}{\text{5!}}\)

= \(\frac{6 \times \text{5!}}{\text{5!}}\) = 6

6Pr + 1 = 6P2 = \(\frac{\text{6!}}{(6 - 2)\text{!}} = \frac{6\text{!}}{\text{4!}}\)

= \(\frac{6 \times 5 \times \text{4!}}{\text{4!}}\)

= 30
37.

Find the variance of 2, 6, 8, 6, 2 and 6

A. 6

B. 5

C. √6

D. √5

Detailed Solution

mean of X = x = (2+6+8+6+2+6)/6 = 30/6 = 5
X → 2,6,8,6,2,6
X - x = -3, 1, 3, 1, 3, 1
(X - x)2 = 9, 1, 9, 1, 9, 1
∑(X-x) = 9+1+9+1+9+1 = 30
Variance = ∑(X-x)2/n = 30/6 = 5
38.

Find the number of ways of selecting 8 subjects from 12 subjects for an examination

A. 490

B. 495

C. 496

D. 498

Detailed Solution

Combination is the number (n) of ways of selecting a number of (m) of objects from n
\(^{12}C_{8}\frac{12!}{8!(12-8)!}\\=\frac{12!}{8!4!}\\\frac{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\\After\hspace{1mm} cancelling \hspace{1mm}out \hspace{1mm}we\hspace{1mm} have\\11\times 5\times 9 = 495 \)
39.

Find the range of 1/6, 1/3, 3/2, 2/3, 8/9 and 4/3

A. 3/4

B. 5/6

C. 7/6

D. 4/3

Detailed Solution

1/6, 1/3, 3/2, 2/3, 8/9 and 4/3
L.C.M of the denominators = 36
The fraction now becomes
6/36, 12/36, 54/36, 24/36, 32/36 and 48/36
Highest fraction = 54/36
Lowest fraction = 6/36
Range = Highest - Lowest
(54/36) - (6/36)
= 48/36
= 4/3
40.

Triangle SPT is the solution of the linear inequalities

A. 2y - x - 2 ≤ 0, y + 2x + 2 ≥ 0, -2 ≤ x ≤ - 1

B. -2 ≤ x ≤ 2, y ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

C. 2y - x - 2 ≤ 0, y + 2x + 2 ≤ 0, y ≥ 0, x ≤ 0

D. 2y - x - 2 ≥ 0, y + 2x + 2 ≤ 0, x ≤ 0

Detailed Solution

Δ SPT is the solution of the inequalities
2y - x - 2 ≤ 0
2y ≤ 2 + x
y ≤ 2/2 + x/2
y ≤ 1 + 1/2x
y + 2x + 2 ≥ 0,
y ≥ -2 -2x
∴ the solution of the inequalities
2y - x - 2 ≤ 0,
y + 2x + 2 ≥ 0
= -2 ≤ x ≤ -1