Mathematics 2019 Jamb Examination
Mathematics 2019 Jamb Examination
Question 51
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?
Let the assistant work for \( d \) days. Then the bricklayer works for \( d + 10 \) days. \[ 1500(d + 10) + 500d = 95000 \] \[ 1500d + 15000 + 500d = 95000 \] \[ 2000d = 80000 \implies d = 40 \text{ days} \] Assistant's pay: \( 40 \times 500 =\) ₦20,000 .
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Question 52
Find the equation of the locus of a point A(x, y) which is equidistant from B(0, 2) and C(2, 1)
Use the distance formula \( AB^2 = AC^2 \):
\[ (x - 0)^2 + (y - 2)^2 = (x - 2)^2 + (y - 1)^2 \]
\[ x^2 + y^2 - 4y + 4 = x^2 - 4x + 4 + y^2 - 2y + 1 \]
\[ -4y + 4 = -4x - 2y + 5 \]
\[ 4x - 2y = 1 \]
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Question 53
A factory worker earns ₦50,000 per month. He spends 15% on education, ₦13,600 on food, and 3% on electricity. How much does he have left?
Education: \( 15\% \text{ of } 50,000 = 7,500 \)
Electricity: \( 3\% \text{ of } 50,000 = 1,500 \)
Total spent: \( 7,500 + 13,600 + 1,500 = 22,600 \)
Balance: \( 50,000 - 22,600 = 27,400 \).
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Question 54
A binary operation \( \Delta \) is defined by \( a \Delta b = a + 3b + 2 \). Find \( (3 \Delta 2) \Delta 5 \).
First, evaluate \( 3 \Delta 2 = 3 + 3(2) + 2 = 3 + 6 + 2 = 11 \).
Now evaluate \( 11 \Delta 5 = 11 + 3(5) + 2 = 11 + 15 + 2 = 28 \).
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Question 55
If M varies directly as N and inversely as the root of P. Given M = 3, N = 5 and P = 25. Find the value of P when M = 2 and N = 6.
Variation formula: \( M = \frac{kN}{\sqrt{P}} \).
Find \( k \): \( 3 = \frac{k(5)}{\sqrt{25}} \implies 3 = k \).
Find \( P \) when \( M = 2, N = 6 \):
\[ 2 = \frac{3(6)}{\sqrt{P}} \implies 2\sqrt{P} = 18 \implies \sqrt{P} = 9 \]
\[ P = 81 \]
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Question 56
The table below gives the scores of a group of students. Find the mode of the distribution.
\[ \begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Score} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text{Freq} & 4 & 6 & 8 & 4 & 10 & 6 & 2 \\ \hline \end{array} \]
The mode is the score with the highest frequency. The highest frequency is 10, which corresponds to the score 5.
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Question 57
The table below gives the scores of a group of students. Calculate the mean deviation for the distribution. \[ \begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Score (x)} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text{Freq (f)} & 4 & 6 & 8 & 4 & 10 & 6 & 2 \\ \hline \end{array} \]
Find Mean \( \displaystyle (\bar{x}) = \frac{\sum fx}{\sum f} \)
\( \displaystyle = \frac{4+12+24+16+50+36+14}{40}\)
\( \displaystyle = \frac{156}{40} = 3.9 \).
Mean Deviation \( \displaystyle = \frac{\sum f|x - \bar{x}|}{\sum f} \):
\( \displaystyle \sum f|x - 3.9| = 4(2.9) + 6(1.9) + 8(0.9) + 4(0.1) + 10(1.1) + 6(2.1) + 2(3.1) \)
 \( \displaystyle = 11.6 + 11.4 + 7.2 + 0.4 + 11 + 12.6 + 6.2 = 60.4 \)
\( \displaystyle M.D. = \frac{60.4}{40} = 1.51 \).
\( \displaystyle = \frac{4+12+24+16+50+36+14}{40}\)
\( \displaystyle = \frac{156}{40} = 3.9 \).
Mean Deviation \( \displaystyle = \frac{\sum f|x - \bar{x}|}{\sum f} \):
\( \displaystyle \sum f|x - 3.9| = 4(2.9) + 6(1.9) + 8(0.9) + 4(0.1) + 10(1.1) + 6(2.1) + 2(3.1) \)
 \( \displaystyle = 11.6 + 11.4 + 7.2 + 0.4 + 11 + 12.6 + 6.2 = 60.4 \)
\( \displaystyle M.D. = \frac{60.4}{40} = 1.51 \).
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Question 58
Integrate \( \displaystyle \int (4x^{-3} - 7x^2 + 5x - 6) dx \).
Apply the power rule \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
- \( \int 4x^{-3} = \frac{4x^{-2}}{-2} = -2x^{-2} \)
- \( \int -7x^2 = -\frac{7x^3}{3} \)
- \( \int 5x = \frac{5x^2}{2} \)
- \( \int -6 = -6x \)
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Question 59
Find the probability that a number selected at random from 21 to 34 is a multiple of 3
Sample space {21, 22, ..., 34}. Total numbers = \( 34 - 21 + 1 = 14 \).
Multiples of 3 in the range: {21, 24, 27, 30, 33}. Total = 5.
\[ P = \frac{5}{14} \]
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Question 60
If the 3rd and 7th terms of a G.P are 9 and 1/9 respectively. Find the common ratio.
\( T_3 = ar^2 = 9 \)
\( T_7 = ar^6 = 1/9 \)
Divide \( T_7 \) by \( T_3 \):
\[ \frac{ar^6}{ar^2} = \frac{1/9}{9} \implies r^4 = \frac{1}{81} \]
\[ r = \sqrt[4]{\frac{1}{81}} = \frac{1}{3} \]
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