2022 - JAMB Mathematics Past Questions and Answers - page 1
Solve for k in the equation \(\frac{1}{8}^{k+2}\) = 1
2
-4
-2
4
\((8^{-1})^{k+2}\) = \((8^{0})\)
cancel out base 8:
-1(k+2) = 0
-k -2 = 0
: k = -2
Evaluate \((101_{two})^3\)
\(111101_{two}\)
\(11111101_{two}\)
\(1111101_{two}\)
\(11001_{two}\)
\((101_{two})^3\) = \((5_{ten})^3\)
5\(^3\) = 125
125 = \(1111101_{two}\)
The shaded portion in the Venn diagram above represents...?
F - (E n F) - (G n F)
E' n F n G'
E u F u G
F
- we can express the diagram as E' n F n G'
- sets E and G are present in the Universal Set but are not included in the expression
- the shaded portion in the diagram is set F
If a fair coin is tossed twice, what is the probability of obtaining at least one head?
0.25
0.75
0.5
0.33
S=HH, HT, TH, TT
We can get the result by subtracting the probability of getting no heads with 1.
That is, the probability of getting at least one head = 1 minus the probability of getting no heads:
P=1−1/4
= 3/4
= 0.75
Give that the mean of 2, 5, (x+1), (x+2), 7 and 9 is 6, find the median.
5.5
5
6.5
6
To find the value of x,
6 = \(\frac{2 + 5 + x+1 + x+2 + 7 + 9}{6}\)
6 * 6 = 2 + 5 + x+1 + x+2 + 7 + 9
36 = 2x + 26
36 - 26 = 2x
10 = 2x
x = 5
median = \(\frac{7+6}{2}\)
= 6.5
If sin θ = - \(\frac{3}{5}\) and θ lies in the third quadrant, find cos θ
\(\frac{4}{5}\)
- \(\frac{5}{4}\)
\(\frac{5}{4}\)
- \(\frac{4}{5}\)
sin θ = \(\frac{opp}{hyp}\) → \(\frac{-3}{5}\)
opp = -3, hyp = 5
using Pythagoras' formula:
hyp\(^2\) = adj\(^2\) + opp\(^2\)
adj\(^2\) = hyp\(^2\) - opp\(^2\)
adj\(^2\) = 5\(^2\) - 3\(^2\) → 25 - 9
adj\(^2\) = 16
adj = 4
cos θ = \(\frac{adj}{hyp}\) → \(\frac{4}{5}\)
In the third quadrant, cos θ is negative
=> - \(\frac{4}{5}\)
Simplify \(\frac{1}{3-√2}\) in the form of p + q√2
\(\frac{7}{3}\) - \(\frac{1}{7√2}\)
\(\frac{7}{3}\) + \(\frac{1}{7√2}\)
\(\frac{3}{7}\) - \(\frac{1}{7√2}\)
\(\frac{3}{7}\) + \(\frac{√2}{7}\)
Rationalizing the denominator using conjugates with the numerator an integer, \({3+√2}\)
\(\frac{1}{3-√2}\)
=> \(\frac{1 * [3+√2]}{[3-√2][3+√2]}\)
= \(\frac{3+√2}{9 -3√2 + 3√2 + √4}\)
= \(\frac{3+√2}{9 - 2}\) → \(\frac{3+√2}{7}\)
= \(\frac{3}{7}\) + \(\frac{√2}{7}\)
Find the length of a chord 3cm from the centre of a circle of radius 5cm.
8cm
5.6cm
7cm
6.5cm
Using the Pythagoras theorem,
Hyp\(^2\) = adj\(^2\) + opp\(^2\)
5\(^2\) = opp\(^2\) + 3\(^2\)
5\(^2\) - 3\(^2\) = adj\(^2\)
4 = adj
Hence, length of the chord = 2 * 4 = 8cm
Mr Adu spends his annual salary on food(f), rent(r), car maintenance, gifts(g), savings(s) and some miscellaneous (m) as indicated in the table below:
| F | R | C | G | S | M |
| 28% | 15% | 20% | 14% | 10% | 13% |
If the information on the table is represented on a pie chart, what angle represents his spending on food?
108.5
100.8
98.8
120.5
A rectangular pyramid has an area of 24cm\(^2\) and a height of 7.5cm. Calculate the volume of the pyramid.
65.0cm\(^3\)
70.5cm\(^3\)
56.5cm\(^3\)
60.0cm\(^3\)
Volume of a rectangular pyramid, v = \(\frac{length * width * height}{3}\) or \(\frac{area * height}{3}\)
v = \(\frac{24 * 7.5}{3}\) → \(\frac{180}{3}\)
= 60cm\(^3\)