2019 - JAMB Mathematics Past Questions and Answers - page 6
A bricklayer charges ₦1,500 per day for himself and ₦500 per day for his assistant. If a two bedroom flat was built for ₦95,000 and the bricklayer worked 10 days more than his assistant, how much did the assistant receive?
N20,000
N28,000
N31,200
N41,000
Let the number of days worked by the assistant = d
\(\therefore\) The bricklayer worked (d + 10) days.
1500(d + 10) + 500(d) = N 95,000
1500d + 15,000 + 500d= N 95,000
2000d = N 95,000 - N 15,000
2000t = N 80,000
d = 40 days
\(\therefore\), the assistant worked for 40 days and received N (500 x 40)
= N 20,000
Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)
4x + 2y = 3
4x - 3y = 1
4x - 2y = 1
4x + 2y = -1
Given that A(x, y) is the point of equidistance between B and C,
AB = AC
(AB)\(^2\) = (AC)\(^2\)
Hence,
(x - 0)\(^2\) + (y - 2)\(^2\) = (x - 2)\(^2\) + (y - 1)\(^2\)
x\(^2\) + y\(^2\) - 4y + 4 = x\(^2\) - 4x + 4 + y\(^2\) - 2y + 1
x\(^2\) - x\(^2\) + y\(^2\) - y\(^2\) + 4x - 4y + 2y = 5 - 4
4x - 2y = 1
A factory worker earns ₦50,000 per month out of which he spends 15% on his children's education, ₦13,600 on Food, 3% on electricity and uses the rest for his personal purpose. How much does he have left?
N21,850
N18,780
N27,400
N32,500
Education => \(\frac{15}{100} \times N 50,000\) = N 7,500
Food => N 13,600
Electricity => \(\frac{3}{100} \times N 50,000\) = N 1,500
Leftover => N (50,000 - (7,500 + 13,600 + 1,500))
= N 27,400
A binary operation Δ is defined by a Δ b = a + 3b + 2.
Find (3 Δ 2) Δ 5
a Δ b = a + 3b + 2
(3 Δ 2) Δ 5 = (3 + 3(2) + 2) Δ 5
= 11 Δ 5
= 11 + 3(5) + 2
= 28
If M varies directly as N and inversely as the root of P. Given that M = 3, N = 5 and P = 25. Find the value of P when M = 2 and N = 6.
\(M \propto N \) ; \(M \propto \frac{1}{\sqrt{P}}\).
\(\therefore M \propto \frac{N}{\sqrt{P}}\)
\(M = \frac{k N}{\sqrt{P}}\)
when M = 3, N = 5 and P = 25;
\(3 = \frac{5k}{\sqrt{25}}\)
\(k = 3\)
\(M = \frac{3N}{\sqrt{P}}\)
when M = 2 and N = 6,
\(2 = \frac{3(6)}{\sqrt{P}} \implies \sqrt{P} = \frac{18}{2}\)
\(\sqrt{P} = 9 \implies P = 9^2\)
P = 81
This table below gives the scores of a group of students in a Further Mathematics Test.
| Score | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Frequency | 4 | 6 | 8 | 4 | 10 | 6 | 2 |
Find the mode of the distribution.
7
10
5
4
This table shown gives the scores of a group of students in a Further Mathematics Test.
| Score | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Frequency | 4 | 6 | 8 | 4 | 10 | 6 | 2 |
Calculate the mean deviation for the distribution
4.32
2.81
1.51
3.90
| Score(x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | Total |
| Frequency (f) | 4 | 6 | 8 | 4 | 10 | 6 | 2 | 40 |
| fx | 4 | 12 | 24 | 16 | 50 | 36 | 14 | 156 |
| x - \(\bar{x}\) | -2.9 | -1.9 | -0.9 | 0.1 | 1.1 | 2.1 | 3.1 | |
| |x - \(\bar{x}\)| | 2.9 | 1.9 | 0.9 | 0.1 | 1.1 | 2.1 | 3.1 | |
| f|x - \(\bar{x}\)| | 11.6 | 11.4 | 7.2 | 0.4 | 11 | 12.6 | 6.2 | 60.4 |
the mean = \(\frac{\sum fx}{\sum f}\)
= \(\frac{156}{40}\)
= 3.9
M.D = \(\frac{\sum f|x - \bar{x}|}{\sum f}\)
= \(\frac{60.4}{40}\)
= 1.51
Integrate \(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\).
\(\int (4x^{-3} - 7x^2 + 5x - 6) \mathrm d x\)
= \(\frac{4x^{-3 + 1}}{-3 + 1} - \frac{7x^{2 + 1}}{2 + 1} + \frac{5x^{1 + 1}}{1 + 1} - 6x\)
= \(-2x^{-2} - \frac{7}{3} x^3 + \frac{5}{2} x^2 - 6x\)
Find the probability that a number selected at random from 21 to 34 is a multiple of 3
\(\frac{3}{11}\)
\(\frac{2}{9}\)
\(\frac{5}{14}\)
\(\frac{5}{13}\)
Set, S = {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34}
n(S) = 14
multiples of 3 = {21, 24, 27, 30, 33}
n(multiples of 3) = 5
Probability of picking a multiple of 3 = 5/14
If the 3rd and 7th terms of a G.P are 9 and 1/9 respectively. Find the common ratio.
\(T_n = ar^{n - 1}\) (terms of a G.P)
\(T_3 = ar^2 = 9\) ... (i)
\(T_7 = ar^6 = \frac{1}{9}\) ... (ii)
Divide (i) by (ii);
\(\frac{ar^6}{ar^2} = \frac{\frac{1}{9}}{9}\)
\(r^4 = \frac{1}{81}\)
\(r^4 = (\frac{1}{3})^4\)
\(r = \frac{1}{3}\)