2007 - JAMB Mathematics Past Questions and Answers - page 1
1
A particle p moves between points S and T such that angle SPT is always constant. Find the locus of P.
A
it is a straight line perpendicular to ST.
B
it is a quadrant of a circle with ST as diameter
C
it is a perpendicular bisector of ST
D
it is a semi-circle with ST as diameter
2
The area of a square is 444 sqcm. Find the length of the diagonal.
A
12\(\sqrt{2}\) cm
B
11\(\sqrt{3}\)cm
C
13 cm
D
12 cm
correct option: a
Area S2 = 144sqcm
s = \(\sqrt{144}\) = 12 cm
Length of diagonal = \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144}\) + 144
= \(\sqrt{228}\)
= 12\(\sqrt{2}\)cm
Users' Answers & Commentss = \(\sqrt{144}\) = 12 cm
Length of diagonal = \(\sqrt{12^2 + 12^2}\)
= \(\sqrt{144}\) + 144
= \(\sqrt{228}\)
= 12\(\sqrt{2}\)cm
3
Find the value of \(\frac{tan60^o - tan30^o}{tan60^o + tan30^o}\)
A
1
B
\(\frac{1}{2}\)
C
\(\frac{3}{\sqrt{3}}\)
D
\(\frac{2}{\sqrt{3}}\)
correct option: b
\(\frac{tan60^o - tan30^o}{tan60^o + tan30^o}\) = \(\frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{\sqrt{3} - \frac{1}{\sqrt{3}}}\)
= \(\frac{3 - 1}{3 + 1}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
Users' Answers & Comments= \(\frac{3 - 1}{3 + 1}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
4
Calculate the length of an arc of a circle of diameter 14cm, which subtends an angle of 90o at the centre of the circle
A
14\(\pi\) cm
B
7\(\pi\) cm
C
5\(\pi\) cm
D
4\(\pi\) cm
correct option: d
Length of an arc \(\frac{\theta}{360^o}\pi d\)
= \(\frac{90^o}{360^o}\) x \(\pi\) x 4cm
= 4\(\pi\) cm
Users' Answers & Comments= \(\frac{90^o}{360^o}\) x \(\pi\) x 4cm
= 4\(\pi\) cm
5
What is the value of k if the mid-point of the line joining (1 - k, -4) and (2, k - 1) is (-k, k)?
A
-4
B
-1
C
-2
D
-3
correct option: d
The midpoint (-k, k) = (\(\frac{1 - k + 2}{2}, \frac{-4 + k + 1}{2}\))
\(\Rightarrow\) -k = \(\frac{3 - k}{2}\)
-2k = 3 - k
-2k + k = 3
k = -3
Users' Answers & Comments\(\Rightarrow\) -k = \(\frac{3 - k}{2}\)
-2k = 3 - k
-2k + k = 3
k = -3
6
\(\begin{array}{c|c} \text{Age in years} & 10 & 11 & 12\\
\hline \text{No. of pupils} & 6 & 27 & 7\end{array}\)
the table above shows the number of pupil in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
the table above shows the number of pupil in each age group in a class. What is the probability that a pupil chosen at random is at least 11 years old?
A
\(\frac{27}{40}\)
B
\(\frac{3}{30}\)
C
\(\frac{17}{20}\)
D
\(\frac{33}{40}\)
correct option: c
Total number of pupils = 6 + 27 + 7 = 40
Number of pupils that are at least 11 years old = 27 + 7
= 34
therefore prob. (at least 11 years) = \(\frac{34}{17}\)
= \(\frac{17}{20}\)
Users' Answers & CommentsNumber of pupils that are at least 11 years old = 27 + 7
= 34
therefore prob. (at least 11 years) = \(\frac{34}{17}\)
= \(\frac{17}{20}\)
7
If 5, 8, 6 and 2 occur with frequencies 3, 2, 4 and 1 respectively, find the product of the modal and the median number
A
40
B
30
C
48
D
36
correct option: d
The mode = 6
therefore the array is 21, 53, 64, 82
The median = 6
The product = 6 x 6 = 36
Users' Answers & Commentstherefore the array is 21, 53, 64, 82
The median = 6
The product = 6 x 6 = 36
8
In how many ways can 6 subjects be selected from 10 subjects for an examination?
A
218
B
210
C
215
D
216
correct option: b
No. of ways \(^{10}C_{6}\) = \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\)
= 210
Users' Answers & Comments= 210
9
in a basket, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that the fruit is either a grape or banana?
A
\(\frac{5}{30}\)
B
\(\frac{6}{30}\)
C
\(\frac{11}{30}\)
D
\(\frac{17}{30}\)
correct option: d
Total fruits = 6 + 11 + 13 = 30
prob. (Grape) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
prob. (Banana) = \(\frac{11}{30}\)
prob. (a grape or a banana) = \(\frac{1}{5}\) + \(\frac{11}{30}\)
= \(\frac{17}{30}\)
Users' Answers & Commentsprob. (Grape) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
prob. (Banana) = \(\frac{11}{30}\)
prob. (a grape or a banana) = \(\frac{1}{5}\) + \(\frac{11}{30}\)
= \(\frac{17}{30}\)
10
a senatorial candidate had planned to visit seven cities prior to a primary election. However, he could only visit four of the cities. How many different itineraries could be considered?
A
840
B
720
C
640
D
520