2013 - JAMB Mathematics Past Questions and Answers - page 1
1
Convert 2710 to another number in base three
A
10013
B
10103
C
11003
D
10003
correct option: d
\(\begin{array}{c|c}
3 & \text{27 rem 0} \\ \hline
3 & \text{ 9 rem 0} \\ \hline
3 & \text{ 3 rem 0} \\ \hline
3 & \text{ 1 rem 1}\\ \hline
& 0
\end{array}\)
Hence the correct answer is 10003
Users' Answers & Comments3 & \text{27 rem 0} \\ \hline
3 & \text{ 9 rem 0} \\ \hline
3 & \text{ 3 rem 0} \\ \hline
3 & \text{ 1 rem 1}\\ \hline
& 0
\end{array}\)
Hence the correct answer is 10003
2
3 girls share a number of apples in the ration 5:3:2. If the highest share is 40 apples, find the smallest share
A
36
B
24
C
16
D
38
correct option: c
The sum, S of ratio is S = 5 + 3 + 2 = 10.
But highest share = \(\frac{5}{10} \times T\), where T is the total number of apples.
Thus, \(40 = \frac{5}{10} \times T\),
given 40 x 10 = 5T,
\(T = \frac{40 \times 10}{5} = 80\)
Hence the smallest share = \(\frac{2}{10} \times 80\)
= 16 apples
Users' Answers & CommentsBut highest share = \(\frac{5}{10} \times T\), where T is the total number of apples.
Thus, \(40 = \frac{5}{10} \times T\),
given 40 x 10 = 5T,
\(T = \frac{40 \times 10}{5} = 80\)
Hence the smallest share = \(\frac{2}{10} \times 80\)
= 16 apples
3
Evaluate \(\frac{1.25 \times 0.025}{0.05}\), correct to 1 decimal place
A
0.6
B
6.2
C
6.3
D
0.5
correct option: a
\(\frac{1.25 \times 0.025}{0.05}\)
\( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\)
= 125 x 5 x 10-3
= 625 x 0.001
= 0.625
= 0.6 Approx to 1 d.p.
Users' Answers & Comments\( = \frac{125 \times 10^{-2} \times 25 \times 10^{-3}}{5 \times 10^{-2}}\)
= 125 x 5 x 10-3
= 625 x 0.001
= 0.625
= 0.6 Approx to 1 d.p.
4
Calculate the time taken for N3000 to earn N600 if invested at 8% simple interest
A
2\(\frac{1}{2}\) years
B
3 years
C
3\(\frac{1}{2}\) years
D
1\(\frac{1}{2}\) years
correct option: a
Using \(S.I =\frac{P \times T \times R}{100}\)
\(600 =\frac{3000 \times T \times 8}{100}\)
\(T =\frac{600 \times 100}{3000 \times 8}\)
\(\frac{20}{8}\)
= 2\(\frac{1}{2}\) years
Users' Answers & Comments\(600 =\frac{3000 \times T \times 8}{100}\)
\(T =\frac{600 \times 100}{3000 \times 8}\)
\(\frac{20}{8}\)
= 2\(\frac{1}{2}\) years
5
Simplify \(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\)
A
32
B
33
C
35
D
3
correct option: d
\(\frac{3^{-5n}}{9^{1-n}} \times 27^{n + 1}\)
\(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\)
\(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\)
\(3^{-5n - 2(1-n) + 3(n + 1)}\)
\(3^{-5n - 2 + 2n + 3n + 3}\)
\(3^{-5n + 5n + 3 - 2}\)
\(3^{1}\)
= 3
Users' Answers & Comments\(\frac{3^{-5n}}{3^{2(1-n)}} \times 3^{3(n + 1)}\)
\(3^{-5n} \div 3^{2(1-n)} \times 3^{3(n + 1)}\)
\(3^{-5n - 2(1-n) + 3(n + 1)}\)
\(3^{-5n - 2 + 2n + 3n + 3}\)
\(3^{-5n + 5n + 3 - 2}\)
\(3^{1}\)
= 3
6
If log104 = 0.6021, evaluate log1041/3
A
0.3011
B
0.9021
C
1.8063
D
0.2007
7
Simplify \(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\)
A
5
B
\(\frac{1}{5}\)
C
\(\frac{1}{9}\)
D
9
correct option: a
\(\frac{\sqrt{5}(\sqrt{147} - \sqrt{12}}{\sqrt{15}}\)
\(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\)
\(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\)
\(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\)
= 5
Users' Answers & Comments\(\frac{\sqrt{5}(\sqrt{49 \times 3} - \sqrt{4 \times 3}}{\sqrt{5 \times 3}}\)
\(\frac{\sqrt{5}(7\sqrt{3} - 2\sqrt{3}}{\sqrt{5} \times \sqrt{3}}\)
\(\frac{\sqrt{3} (7 - 2}{\sqrt{3}}\)
= 5
8
if P = {x:x is odd, \(-1 < x \leq 20\)} and Q is {y:y is prime, \(-2 < y \leq 25\), find P \(\cap\) Q
A
{3,5,7,11,17,19}
B
{3,5,11,13,17,19}
C
{3,5,7,11,13,17,19}
D
{2,3,5,7,11,13,17,19}
correct option: c
Users' Answers & Comments9
If S = \(\sqrt{t^2 - 4t + 4}\), find t in terms of S
A
S2 - 2
B
S + 2
C
S - 2
D
S2 + 2
correct option: b
S = \(\sqrt{t^2 - 4t + 4}\)
S2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substituting, we have;
Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)
\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)
\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)
\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)
\(t = \frac{2(2 \pm S)}{2}\)
Hence t = 2 + S or t = 2 - S
Users' Answers & CommentsS2 = t2 - 4t + 4
t2 - 4t + 4 - S2 = 0
Using \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Substituting, we have;
Using \(t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4 - S^2)}}{2(1)}\)
\(t = \frac{4 \pm \sqrt{16 - 4(4 - S^2)}}{2}\)
\(t = \frac{4 \pm \sqrt{16 - 16 + 4S^2}}{2}\)
\(t = \frac{4 \pm \sqrt{4S^2}}{2}\)
\(t = \frac{2(2 \pm S)}{2}\)
Hence t = 2 + S or t = 2 - S
10
If x - 4 is a factor of x2 - x - k, then k is
A
4
B
12
C
20
D
2
correct option: b
Let f(x) = x2 - x - k
Then by the factor theorem,
(x - 4): f(4) = (4)2 - (4) - k = 0
16 - 4 - k = 0
12 - k = 0
k = 12
Users' Answers & CommentsThen by the factor theorem,
(x - 4): f(4) = (4)2 - (4) - k = 0
16 - 4 - k = 0
12 - k = 0
k = 12