Mathematics 2017 Jamb Examination
Mathematics 2017 Jamb Examination
Question 21
Evaluate \( \dfrac{0.00000231}{0.007} \) and leave the answer in standard form
Convert both numbers to standard form first:
\[ 0.00000231 = 2.31 \times 10^{-6} \]
\[ 0.007 = 7 \times 10^{-3} \]
Now divide:
\[ \frac{2.31 \times 10^{-6}}{7 \times 10^{-3}} = \frac{2.31}{7} \times 10^{-6 - (-3)} \]
\[ = 0.33 \times 10^{-3} = 3.3 \times 10^{-4} \]
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Question 22
If a rod 10cm in length was measured as 10.5cm, calculate the percentage error
Percentage error is calculated as:
\[ \frac{\text{Error}}{\text{Actual Value}} \times 100 \]
\[ \text{Error} = |10.5 - 10| = 0.5 \]
\[ \text{Percentage Error} = \frac{0.5}{10} \times 100 = 5\% \]
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Question 23
Find the principal which amounts to ₦ 5,500 at a simple interest in 5 years at 2% per annum
Amount \( (A) = Principal (P) + Interest (I) \). Simple Interest \( I = \frac{PRT}{100} \). \[ A = P + \frac{PRT}{100} = P(1 + \frac{RT}{100}) \] Given \( A = 5500, R = 2, T = 5 \): \[ 5500 = P(1 + \frac{2 \times 5}{100}) = P(1 + 0.1) \] \[ 5500 = 1.1P \implies P = \frac{5500}{1.1} = 5000 \]
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Question 24

The pie chart shows the allocation of money to each sector in a farm. The total amount allocated to the farm is ₦ 80 000. Find the amount allocated to fertilizer
From the pie chart, the angle for fertilizer is \( 90^\circ \) (indicated by the right-angle symbol). \[ \text{Amount} = \frac{\text{Angle}}{360^\circ} \times \text{Total Amount} \] \[ = \frac{90}{360} \times 80,000 = \frac{1}{4} \times 80,000 = 20,000 \]
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Question 25
In how many ways can the word MATHEMATICS be arranged?
The word MATHEMATICS has 11 letters. We identify the repeating letters:
- M appears 2 times
- A appears 2 times
- T appears 2 times
Using the formula for permutations of multi-sets:
\[ \text{Total arrangements} = \frac{11!}{2!2!2!} \]
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Question 26
In how many ways can the word MACICITA be arranged?
The word MACICITA consists of 8 letters. We check for repeated letters:
- A appears 2 times
- C appears 2 times
- I appears 2 times
The number of arrangements is calculated using the formula for permutations of identical objects:
\[ \frac{8!}{2! 2! 2!} \]
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Question 27
\( y \) is inversely proportional to \( x \) and \( y \) is 6 when \( x = 7 \). Find the constant of the variation
For inverse proportion, the relationship is \( y = \frac{k}{x} \), where \( k \) is the constant of variation.
Rearranging for \( k \):
\[ k = y \times x \]
\[ k = 6 \times 7 = 42 \]
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Question 28
Find the equation of the locus of a point \( p(x, y) \) such that \( pv = pw \), where \( v = (1, 1) \) and \( w = (3, 5) \)
The locus of a point equidistant from two fixed points is the perpendicular bisector of the line joining those points. We set the squared distances equal: \[ (x - 1)^2 + (y - 1)^2 = (x - 3)^2 + (y - 5)^2 \] \[ x^2 - 2x + 1 + y^2 - 2y + 1 \]\[= x^2 - 6x + 9 + y^2 - 10y + 25 \] \[ -2x - 2y + 2 = -6x - 10y + 34 \] \[ 4x + 8y = 32 \] Divide through by 4: \[ x + 2y = 8 \]
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Question 29
Find \( \int(x^2 + 3x - 5)dx \)
Use the power rule for integration \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
\[ \int x^2 dx = \frac{x^3}{3} \]
\[ \int 3x dx = \frac{3x^2}{2} \]
\[ \int -5 dx = -5x \]
Adding the constant of integration \( k \), we get:
\[ \frac{x^3}{3} + \frac{3x^2}{2} - 5x + k \]
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Question 30

In the diagram above MN is a chord of a circle KMN centre O and radius 10cm. If \( \angle MON = 140^\circ \), find, to the nearest cm, the length of the chord MN.
To find the chord length \( L \), use the formula \( L = 2r \sin(\frac{\theta}{2}) \), where \( r = 10 \) and \( \theta = 140^\circ \): \[ MN = 2(10) \sin(70^\circ) \] \[ MN = 20 \times 0.9397 \approx 18.79 \] To the nearest cm, the length is 19cm.
