2006 - JAMB Mathematics Past Questions and Answers - page 1
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)
correct option: a
Users' Answers & Comments2
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
correct option: a
The question will be answered in
\(^5C_4 = \frac{6!}{(6-4)!4!}\=\frac{6!}{2!4!}\=\frac{6\times5\times4!}{2\times1\times4!}\=15\hspace{1mm}ways\)
Users' Answers & Comments\(^5C_4 = \frac{6!}{(6-4)!4!}\=\frac{6!}{2!4!}\=\frac{6\times5\times4!}{2\times1\times4!}\=15\hspace{1mm}ways\)
3
If the mean of five consecutive integers is 30, find the largest of the numbers
A
28
B
30
C
32
D
34
correct option: c
Let the consecutive numbers be a, a+1, a+2, a+3, a+4
Mean = \(\frac{(a+a+1+a+2+a+3+a+4)}{5}\\ 30 = \frac{5a+10}{5}\\ 30 \times 5 = 5a + 10\\ 5a = 150 - 10\\ 5a = 140\\ a = 28 \\ ∴ a + 4 = 28 + 4\\ = 32\)
Users' Answers & CommentsMean = \(\frac{(a+a+1+a+2+a+3+a+4)}{5}\\ 30 = \frac{5a+10}{5}\\ 30 \times 5 = 5a + 10\\ 5a = 150 - 10\\ 5a = 140\\ a = 28 \\ ∴ a + 4 = 28 + 4\\ = 32\)
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
correct option: d
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
Users' Answers & CommentsWhite = 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
correct option: a
\(\sum x = 6x\\
\sum(x-\bar{x})^2 = 2\\
\bar{x} = \frac{\sum x}{n}\\
= \frac{6x}{3}\\
= 2x\\
Variance = \frac{\sum(x-\bar{x})^2}{n}\\
= \frac{2}{3}\)
Users' Answers & Comments6
A
3/20
B
1/4
C
2/5
D
3/4
correct option: c
P(GTN) = 5/20
P(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5
Users' Answers & CommentsP(Qtel) = 3/20
∴P(GTN or Qtel) = (5/20) + (3/20)
= 8/20
= 2/5
7
A
N 28 000
B
N 25 500
C
N15 000
D
N 12 500
correct option: d
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
Users' Answers & CommentsExpenditure 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
correct option: d
\(^{n+1}C_3 = 4(^nC_3)\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\=\frac{n+1}{n-2}=\frac{4}{1}\n+1 = 4(n-2)\n+1 = 4n-8\-3n = -9\\frac{-9}{-3}\n=3\)
Users' Answers & Comments9
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
correct option: b
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
Users' Answers & Commentsy = ∫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
correct option: c
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/x3
Users' Answers & Commentsy = (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/x3