Mathematics 2011 Jamb Examination

To find the value of q, we convert both numbers from their respective bases to base 10. For 2q3₅: \[ 2(5^2) + q(5^1) + 3(5^0) = 2(25) + 5q + 3(1) = 50 + 5q + 3 = 53 + 5q \] For 77₈: \[ 7(8^1) + 7(8^0) = 7(8) + 7(1) = 56 + 7 = 63 \] Now, equate the two base 10 expressions: \[ 53 + 5q = 63 \] Subtract 53 from both sides: \[ 5q = 10 \] Dividing both sides by 5: \[ q = 2 \] Therefore, the missing digit in the base 5 number is 2.

First, simplify the numerator. Convert the mixed fraction to an improper fraction: \( 3\frac{2}{3} = \frac{11}{3} \). \[ \text{Numerator} = \frac{11}{3} \times \frac{5}{6} \times \frac{2}{3} = \frac{11 \times 5 \times 2}{3 \times 6 \times 3} = \frac{110}{54} = \frac{55}{27} \] Next, simplify the denominator: \[ \text{Denominator} = \frac{11}{15} \times \frac{3}{4} \times \frac{2}{27} = \frac{11 \times 3 \times 2}{15 \times 4 \times 27} = \frac{66}{1620} \] Simplify the denominator fraction by dividing by 6: \[ \frac{66 \div 6}{1620 \div 6} = \frac{11}{270} \] Now, divide the numerator by the denominator: \[ \frac{55}{27} \div \frac{11}{270} = \frac{55}{27} \times \frac{270}{11} \] \[ = \left(\frac{55}{11}\right) \times \left(\frac{270}{27}\right) = 5 \times 10 = 50 \]

Simple Interest (SI) is calculated using the formula: \[ SI = \frac{P \times R \times T}{100} \] Where:
P (Principal) = ₦5,000
R (Rate) = 4%
T (Time in years) = \( \frac{9}{12} \) years = 0.75 years.
Substituting these values into the formula: \[ SI = \frac{5000 \times 4 \times 0.75}{100} \] \[ SI = 50 \times 4 \times 0.75 \] \[ SI = 200 \times 0.75 = 150 \] The simple interest earned over the 9-month period is ₦150.

Given the ratio M:N:Q = 5:4:3, we can assign a common constant \( k \) such that:
\( M = 5k \)
\( N = 4k \)
\( Q = 3k \)
Substitute these values into the required expression: \[ \frac{2N - Q}{M} = \frac{2(4k) - 3k}{5k} \] \[ = \frac{8k - 3k}{5k} \] \[ = \frac{5k}{5k} = 1 \] The ratio simplifies to 1 regardless of the value of \( k \).

First, evaluate the first part of the expression: \[ \left(\frac{16}{81}\right)^{1/4} = \frac{\sqrt[4]{16}}{\sqrt[4]{81}} = \frac{2}{3} \] Next, evaluate the second part. Recall that a negative exponent indicates a reciprocal: \[ \left(\frac{9}{16}\right)^{-1/2} = \left(\frac{16}{9}\right)^{1/2} = \frac{\sqrt{16}}{\sqrt{9}} = \frac{4}{3} \] Finally, perform the division: \[ \frac{2}{3} \div \frac{4}{3} = \frac{2}{3} \times \frac{3}{4} = \frac{2}{4} = \frac{1}{2} \] The expression simplifies to \( \frac{1}{2} \).

Using the laws of logarithms, specifically the addition law \( \log_b A + \log_b C = \log_b (AC) \) and the subtraction law \( \log_b A - \log_b C = \log_b (A/C) \), we can combine the left side of the equation: \[ \log_{3} \left( \frac{18 \times 3}{x} \right) = 3 \] \[ \log_{3} \left( \frac{54}{x} \right) = 3 \] Convert the logarithmic equation into its exponential form: \[ \frac{54}{x} = 3^3 \] \[ \frac{54}{x} = 27 \] Solving for x: \[ x = \frac{54}{27} = 2 \] Therefore, the value of x that satisfies the equation is 2.

To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is \( 3 + \sqrt{5} \): \[ \frac{2 - \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} \] Expand the numerator: \[ (2 \times 3) + (2 \times \sqrt{5}) - (3 \times \sqrt{5}) - (\sqrt{5} \times \sqrt{5}) \] \[ 6 + 2\sqrt{5} - 3\sqrt{5} - 5 = 1 - \sqrt{5} \] Expand the denominator (difference of squares): \[ 3^2 - (\sqrt{5})^2 = 9 - 5 = 4 \] Putting it all together: \[ \frac{1 - \sqrt{5}}{4} \]

This expression follows the difference of two squares identity: \( (a + b)(a - b) = a^2 - b^2 \).
In this case, let \( a = \sqrt{2} \) and \( b = \frac{1}{\sqrt{3}} \). \[ (\sqrt{2})^2 - \left(\dfrac{1}{\sqrt{3}}\right)^2 \] \[ 2 - \frac{1}{3} \] To subtract these values, find a common denominator: \[ \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \] The simplified result of the expression is \( \frac{5}{3} \).

Number of ways to arrange 3 posters out of 7 distinct posters\[ 7 \times 6 \times 5 = 210 \text{ ways}\] This can also be expressed using permutations:\[ ^7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} \]\[=\ \frac{7 \times 6 \times 5 \times 4!}{4!} \]\[= 7 \times 6 \times 5 = 210 \text{ ways}\]

To make R the subject, we isolate it step-by-step.
Start with: \[ T = \frac{KR^2 + M}{3} \] Multiply both sides by 3: \[ 3T = KR^2 + M \] Subtract M from both sides: \[ 3T - M = KR^2 \] Divide both sides by K: \[ \frac{3T - M}{K} = R^2 \] Take the square root of both sides: \[ R = \sqrt{\frac{3T - M}{K}} \] This matches option B.