Mathematics 2017 Jamb Examination

Express all terms as powers of 3: \[ 27^{n+1} = (3^3)^{n+1} = 3^{3n+3} \] \[ 81^n = (3^4)^n = 3^{4n} \] Substitute back into the expression: \[ 3^{n-1} \times \frac{3^{3n+3}}{3^{4n}} = 3^{n-1 + 3n+3 - 4n} \] \[ = 3^{4n+2 - 4n} = 3^2 = 9 \]

The locus of points equidistant from a single straight line consists of a pair of parallel lines on either side of the original line, at the given constant distance.

According to the circle theorem, the angle subtended by an arc at the centre is twice the angle subtended at the circumference. Here, the obtuse angle \( \angle ROT = 2 \times \angle RST \): \[ \angle ROT = 2 \times 60^\circ = 120^\circ \] Since \( x \) is the angle at the centre subtended by the same arc as the circumference angle, \( x = 120^\circ \).

1. Mode: The number that appears most frequently. 10 appears 6 times, 9 appears 4 times. So, Mode = 10. 2. Range: Difference between the highest and lowest values. \[ \text{Highest} = 10, \text{Lowest} = 4 \] \[ \text{Range} = 10 - 4 = 6 \] 3. Sum: \( 10 + 6 = 16 \).

This is a geometric series with first term \( a = \frac{1}{4} \) and common ratio \( r = \frac{1/8}{1/4} = \frac{1}{2} \). The formula for sum to infinity is \( S_\infty = \frac{a}{1 - r} \): \[ S_\infty = \frac{1/4}{1 - 1/2} = \frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2} \]

In any number system, the digits used must be less than the base. The numbers provided consist only of the digits 0 and 1. This is the characteristic of the binary system. Furthermore, performing the addition in base 2 would result in similar strings of 0s and 1s, which is common in such problems. \[ \text{Base } 2 \text{ uses only digits 0 and 1.} \]

Substitute \( x = 2 \) into the expression: \[ 2 + 2(2^2) \] Evaluate the power first: \[ 2^2 = 4 \] Then multiply and add: \[ 2 + 2(4) = 2 + 8 = 10 \]

Let the exterior angle be \( e \). Then the interior angle is \( 2e \). Since the sum of interior and exterior angles is \( 180^\circ \): \[ 2e + e = 180^\circ \implies 3e = 180^\circ \implies e = 60^\circ \] The formula for the exterior angle of a regular polygon with \( n \) sides is \( e = \frac{360^\circ}{n} \): \[ 60^\circ = \frac{360^\circ}{n} \implies n = \frac{360}{60} = 6 \]

The volume \( V \) of a cylinder is \( V = \pi r^2 h \). Diameter = \( 14\text{m} \), so radius \( r = 7\text{m} \). \[ 3080 = \frac{22}{7} \times 7^2 \times h \] \[ 3080 = 22 \times 7 \times h \] \[ 3080 = 154h \implies h = \frac{3080}{154} = 20\text{m} \]

Simplify each surd by finding square factors: - \( 4\sqrt{27} = 4\sqrt{9 \times 3} = 4(3\sqrt{3}) = 12\sqrt{3} \) - \( 5\sqrt{12} = 5\sqrt{4 \times 3} = 5(2\sqrt{3}) = 10\sqrt{3} \) - \( 3\sqrt{75} = 3\sqrt{25 \times 3} = 3(5\sqrt{3}) = 15\sqrt{3} \) Now combine them: \[ 12\sqrt{3} + 10\sqrt{3} - 15\sqrt{3} = (12 + 10 - 15)\sqrt{3} = 7\sqrt{3} \]