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

Paper 1 | Objectives

1 - 10 of 50 Questions

# Question Ans
1.

The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?

A. (33, 4)

B. (22, 4)

C. (33, 12)

D. (12, 4)

Detailed Solution

The mode (m) = 4.
No of students who scored 4 and above = 12 + 10 + 6 + 4 + 1 = 33.
\(\therefore\) (n, m) = (33, 4)
2.

A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?

A. 15

B. 20

C. 30

D. 45

Detailed Solution

The question will be answered in
\(^5C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)
3.

If the mean of five consecutive integers is 30, find the largest of the numbers

A. 28

B. 30

C. 32

D. 34

Detailed Solution

Let the consecutive numbers be a, a+1, a+2, a+3, a+4
Mean = \(\frac{(a+a+1+a+2+a+3+a+4)}{5}\\
30 = \frac{5a+10}{5}\\
30 \times 5 = 5a + 10\\
5a = 150 - 10\\
5a = 140\\
a = 28 \\
∴ a + 4 = 28 + 4\\
= 32\)
4.

A bag contains 5 black, 4 white and x red marbles. If the probability of picking a red marble is 2/3, find the value of x

A. 8

B. 10

C. 4

D. 6

Detailed Solution

Black = 5
White = 4
Red = x
Total = 9+x
P(red) = 2/3
x / (9+x) = 2/5
5x = 2(9+x)
5x = 18+2x
5x-2 = 18
3x = 18
x = 6
5.

Find the variance of 2x, 2x-1 and 2x+1

A. 2/3

B. 2

C. √(2/3)

D. 1

Detailed Solution

\(\sum x = 6x\\
\sum(x-\bar{x})^2 = 2\\
\bar{x} = \frac{\sum x}{n}\\
= \frac{6x}{3}\\
= 2x\\
Variance = \frac{\sum(x-\bar{x})^2}{n}\\
= \frac{2}{3}\)
6.

The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?

A. 3/20

B. 1/4

C. 2/5

D. 3/4

Detailed Solution

P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5
7.

The pie chart above shows the expenditure of a family whose income is N30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?

A. N 28 000

B. N 25 500

C. N15 000

D. N 12 500

Detailed Solution

Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90o
Expenditure on transport = 45o
x + 2x + 90 + 45 = 360o
3x + 135 = 360o
3x = 360 - 135
3x = 225
x = 225/3 = 75o
∴∠ for food = 2x = 2 * 75
= 150
360o = N30,000
1o = ?
1o = 30,000/360
150o = (30000/360) * (150/1)
= N12500
8.

For what of n is n+1C3 = 4(nC3)?

A. 6

B. 5

C. 4

D. 3

Detailed Solution

\(^{n+1}C_3 = 4(^nC_3)\\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\\=\frac{n+1}{n-2}=\frac{4}{1}\\n+1 = 4(n-2)\\n+1 = 4n-8\\-3n = -9\\\frac{-9}{-3}\\n=3\)
9.

The gradient of a curve is 2x + 7 and the curve passes through point (2, 0). find the equation of the curve.

A. y = x2 + 7x + 9

B. y = x2 + 7x - 18

C. y = x2 + 7x + 18

D. y = x2 + 14x + 11

Detailed Solution

dy/dx = 2x + 7
y = ∫2x + 7
y = x2 + 7x + C at (2,0)
0 = 22 + 7(2) + C
0 = 4 + 14 + C
0 = 18 + C
C = -18
∴ The equation is y = x2 + 7x - 18
10.

Differentiate (x2 - 1/x)2 with respect to x

A. 4x2 - 4x - 2/x

B. 4x2 - 2 + 2/x3

C. 4x2 - 2 - 2/x3

D. 4x2 - 3x + 2/x

Detailed Solution

y = (x2 - 1/x)2
y = (x2 - 1/x)(x2 - 1/x)
y = x4 - x - x + 1/x2
y = x4 - 2x + 1/x2
y= x4 - 2x + x-2
dy/dx = 4x2 - 2 - 2x-3
= 4x2 - 2 - 2
1.

The table above shows the scores of a group of students in a physics test. If the mode is m and the number of students who scored 4 or more is n, what is (n, m)?

A. (33, 4)

B. (22, 4)

C. (33, 12)

D. (12, 4)

Detailed Solution

The mode (m) = 4.
No of students who scored 4 and above = 12 + 10 + 6 + 4 + 1 = 33.
\(\therefore\) (n, m) = (33, 4)
2.

A final examination requires that a student answer any 4 out of 6 questions. In how many ways can this be done?

A. 15

B. 20

C. 30

D. 45

Detailed Solution

The question will be answered in
\(^5C_4 = \frac{6!}{(6-4)!4!}\\=\frac{6!}{2!4!}\\=\frac{6\times5\times4!}{2\times1\times4!}\\=15\hspace{1mm}ways\)
3.

If the mean of five consecutive integers is 30, find the largest of the numbers

A. 28

B. 30

C. 32

D. 34

Detailed Solution

Let the consecutive numbers be a, a+1, a+2, a+3, a+4
Mean = \(\frac{(a+a+1+a+2+a+3+a+4)}{5}\\
30 = \frac{5a+10}{5}\\
30 \times 5 = 5a + 10\\
5a = 150 - 10\\
5a = 140\\
a = 28 \\
∴ a + 4 = 28 + 4\\
= 32\)
4.

A bag contains 5 black, 4 white and x red marbles. If the probability of picking a red marble is 2/3, find the value of x

A. 8

B. 10

C. 4

D. 6

Detailed Solution

Black = 5
White = 4
Red = x
Total = 9+x
P(red) = 2/3
x / (9+x) = 2/5
5x = 2(9+x)
5x = 18+2x
5x-2 = 18
3x = 18
x = 6
5.

Find the variance of 2x, 2x-1 and 2x+1

A. 2/3

B. 2

C. √(2/3)

D. 1

Detailed Solution

\(\sum x = 6x\\
\sum(x-\bar{x})^2 = 2\\
\bar{x} = \frac{\sum x}{n}\\
= \frac{6x}{3}\\
= 2x\\
Variance = \frac{\sum(x-\bar{x})^2}{n}\\
= \frac{2}{3}\)
6.

The table above shows the distribution of recharge cards of four major GSM operators. What is the probability that a recharge card selected at random will be GTN or Qtel?

A. 3/20

B. 1/4

C. 2/5

D. 3/4

Detailed Solution

P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5
7.

The pie chart above shows the expenditure of a family whose income is N30,000. If the expenditure on food is twice that on housing and that on school fees is twice that on transport, how much does the family spend on food?

A. N 28 000

B. N 25 500

C. N15 000

D. N 12 500

Detailed Solution

Expenditure on housing = x
Expenditure on food = 2x
Expenditure on school fees = 90o
Expenditure on transport = 45o
x + 2x + 90 + 45 = 360o
3x + 135 = 360o
3x = 360 - 135
3x = 225
x = 225/3 = 75o
∴∠ for food = 2x = 2 * 75
= 150
360o = N30,000
1o = ?
1o = 30,000/360
150o = (30000/360) * (150/1)
= N12500
8.

For what of n is n+1C3 = 4(nC3)?

A. 6

B. 5

C. 4

D. 3

Detailed Solution

\(^{n+1}C_3 = 4(^nC_3)\\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\\=\frac{n+1}{n-2}=\frac{4}{1}\\n+1 = 4(n-2)\\n+1 = 4n-8\\-3n = -9\\\frac{-9}{-3}\\n=3\)
9.

The gradient of a curve is 2x + 7 and the curve passes through point (2, 0). find the equation of the curve.

A. y = x2 + 7x + 9

B. y = x2 + 7x - 18

C. y = x2 + 7x + 18

D. y = x2 + 14x + 11

Detailed Solution

dy/dx = 2x + 7
y = ∫2x + 7
y = x2 + 7x + C at (2,0)
0 = 22 + 7(2) + C
0 = 4 + 14 + C
0 = 18 + C
C = -18
∴ The equation is y = x2 + 7x - 18
10.

Differentiate (x2 - 1/x)2 with respect to x

A. 4x2 - 4x - 2/x

B. 4x2 - 2 + 2/x3

C. 4x2 - 2 - 2/x3

D. 4x2 - 3x + 2/x

Detailed Solution

y = (x2 - 1/x)2
y = (x2 - 1/x)(x2 - 1/x)
y = x4 - x - x + 1/x2
y = x4 - 2x + 1/x2
y= x4 - 2x + x-2
dy/dx = 4x2 - 2 - 2x-3
= 4x2 - 2 - 2