2020 - JAMB Mathematics Past Questions and Answers - page 4
The chord ST of a circle is equal to the radius r of the circle. Find the length of arc ST
\(\frac{\pi r}{3}\)
\(\frac{\pi r}{2}\)
\(\frac{\pi r}{12}\)
\(\frac{\pi r}{6}\)
\(\frac{ \frac{r}{2}}{r}\) Sin \(\theta\) = \(\frac{1}{2}\)
\(\theta\) = sin\(^{-1}\) (\(\frac{1}{2}\)) = 30\(^o\) = 60\(^o\)
Length of the minor arc:
ST = \(\frac{\theta}{360}\) x 2\(\pi r\)
\(\frac{60}{360} \times 2 \pi \times r = \frac{\pi}{3}\)
A sector of circle of radius 7.2cm which substends an angle of 300\(^o\) at the centre is used to form a cone. What s the radius of the base of the cone?
8cm
9cm
6cm
7cm
\(\frac{Q}{360}\) = \(\frac{r}{L}\)
\(\frac{300}{360}\) = \(\frac{r}{7.2}\)
r = \(\frac{300 \times 7.2}{360}\)
= 8cm
A cylindrical tank has a capacity of 3080m\(^3\). What is the depth of the tank if the diameter of its base is 14m?
25m
23m
22m
20m
***
V = 2080cm\(^3\), h = ?
r = 7cm
Volume:
v= \(\pi r^2h\)
h = \(\frac{V}{\pi r^2} = \frac{3080}{\frac{22}{7} \times 49}\)
\(\frac{3080}{54}\) = 20cm
h = 20cm
The acres for rice, pineapple, cassava, cocoa and palm oil in a certain district are given respectively as 2, 5, 3, 11 and 9. What is the angle of the sector of cassava in a pie chart?
180\(^o\)
36\(^o\)
60\(^o\)
108\(^o\)
Total number of acres = 2 + 5 + 3 + 11 + 9 = 30
The angle of acres = 2 + 5 + 3 + 11 + 9 = 30
The angle of the cassava sector = \(\frac{3}{30} \times 360^o = 36^o\)
Three consecutive terms of a geometric progression are give as n - 2, n and n + 3. Find the common ratio
\(\frac{3}{2}\)
\(\frac{2}{3}\)
\(\frac{1}{2}\)
\(\frac{1}{4}\)
\(\frac{h}{n - 2} = \frac{n + 3}{n}\)
n\(^2\) = (n + 3) (n - 2)
n\(^2\) = n\(^2\) + n - 6
n\(^2\) + n - 6 - n\(^2\) = 0
n - 6 = 0
n = 6
The common ratio => \(\frac{n}{n - 2} = \frac{6}{6 - 2} = \frac{6}{4}\) = \(\frac{3}{2}\)
In a class of 40 students, 32 offer mathematics, 24 offer physics and 4 offer neither mathematics nor physics. How many offer both mathematics and physics?
4
8
16
20
The sum of the interior angle of pentagon is 6x + 6y. Find y in terms of x.
y = 6 - x
y = 90 - x
y = 120 - x
y = 150 - x
Sum of the interior angles = (2n - 4) 90\(^o\)
A pentagon has 5 sides(n), hence, n = 5
Therefore, the sum of interior angles = 6 x 90\(^o\) = 540\(^o\)
6x + 6y = 540\(^o\)
6(x + y) = 540\(^o\)
x + y = \(\frac{540^o}{6}\) = 90\(^o\)
y = 90\(^o\)
y = 90 - x
The mean age group of some students is 15years. When the age of a teacher, 45 years old, is added to the ages of the students, the mean of their ages become 18 years. Find the number of students in the group.
7
9
15
42
x \(\frac{\sum x}{N}\)
15 = \(\frac{\sum x }{N}\)
\(\sum x\) = 15N........(i)
y = \(\frac{\sum y}{Ny} = \frac{\sum x + 45}{N + 1}\)
\(\frac{18}{1} = \frac{15N + 45}{N + 1}\)
18(N + 1) = 15N + 45
18N + 18 = 15N + 45
18N - 15N = 45 - 18
3N = 27
N = \(\frac{27}{3}\)
= 9
A surveyor walks 500m up a hill which slopes at an angle of 30\(^o\). Calculate the vertical height through which he rises
252m
500m
250m
255m
\(\frac{h}{500}\) = sin 30\(^o\)
= 500 sin 30\(^o\)
= 500 x \(\frac{1}{2}\)
= 250m
h = 250m
Find the non-zero positive value of x which satisfies the equation
\(\begin{bmatrix} x & 1 & 0 \ 1 & x & 1 \ 0 & 1 & x \end{bmatrix}\) = 0
2
\(\sqrt{3}\)
\(\sqrt{2}\)
1