2007 JAMB Mathematics Past Questions & Answers - page 1

JAMB CBT App

Study offline with EduPadi JAMB CBT app that has so many features, including thousands of past questions, JAMB syllabus, novels, etc.

Install The App

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


Correct Option: D


Solution
The locus of P 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


Solution
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

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


Solution
\(\frac{tan60^o - tan30^o}{tan60^o + tan30^o}\) = \(\frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{\sqrt{3} - \frac{1}{\sqrt{3}}}\)

= \(\frac{3 - 1}{3 + 1}\)

= \(\frac{2}{4}\)

= \(\frac{1}{2}\)

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


Solution
Length of an arc \(\frac{\theta}{360^o}\pi d\)

= \(\frac{90^o}{360^o}\) 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


Solution
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

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?
A \(\frac{27}{40}\)
B \(\frac{3}{30}\)
C \(\frac{17}{20}\)
D \(\frac{33}{40}\)


Correct Option: C


Solution
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) = \(\frac{34}{17}\)

= \(\frac{17}{20}\)

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


Solution
The mode = 6

therefore 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


Solution
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

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


Solution
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) = \(\frac{1}{5}\) + \(\frac{11}{30}\)

= \(\frac{17}{30}\)

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


Correct Option: A


Solution
No of ways = \(^{7}P_{4}\)

= 7 x 6 x 5 x 4

= 840

JAMB CBT APP

EduPadi JAMB CBT app that has thousands of past questions and answers, JAMB syllabus, novels, etc. And works offline!


Install App