Mathematics 2010 Jamb Examination

To find the value of \( r \), we must convert both numbers from their respective bases (base 8 and base 9) into base 10 and then equate them.
Expanding \( 6r7_{8} \) in base 10: \[ 6(8^2) + r(8^1) + 7(8^0) = 6(64) + 8r + 7(1) \] \[ = 384 + 8r + 7 = 391 + 8r \] Expanding \( 511_{9} \) in base 10: \[ 5(9^2) + 1(9^1) + 1(9^0) = 5(81) + 9 + 1 \] \[ = 405 + 9 + 1 = 415 \] Now, set the two expressions equal to each other: \[ 391 + 8r = 415 \] \[ 8r = 415 - 391 \] \[ 8r = 24 \] \[ r = \frac{24}{8} \] \[ r = 3 \] By converting both sides to a common decimal base, we determine that the missing digit in the base 8 number must be 3.

We follow the BODMAS rule, starting with 'of' and then proceeding to division. First, let's convert the mixed fractions to improper fractions: \[ 9\frac{1}{2} = \frac{19}{2} \quad \text{and} \quad 1\frac{5}{19} = \frac{24}{19} \] Step 1: Solve the 'of' part: \[ \frac{3}{4} \times \frac{4}{9} = \frac{3}{9} = \frac{1}{3} \] Step 2: Perform the first division inside the bracket: \[ \frac{1}{3} \div \frac{19}{2} = \frac{1}{3} \times \frac{2}{19} = \frac{2}{57} \] Step 3: Perform the final division outside the bracket: \[ \frac{2}{57} \div \frac{24}{19} = \frac{2}{57} \times \frac{19}{24} \] Simplifying the multiplication: \[ \frac{2}{24} \times \frac{19}{57} = \frac{1}{12} \times \frac{1}{3} = \frac{1}{36} \] The expression simplifies step-by-step to 1/36.

Percentage error is calculated by finding the absolute difference between the measured value and the actual value, dividing that by the actual value, and then multiplying by 100. The formula is: \[ \text{Percentage Error} = \frac{|\text{Measured Value} - \text{Actual Value}|}{\text{Actual Value}} \times 100 \] Given: \[ \text{Measured Value} = 1.26\text{m} \] \[ \text{Actual Value} = 1.25\text{m} \] Step 1: Find the error: \[ \text{Error} = 1.26 - 1.25 = 0.01\text{m} \] Step 2: Calculate percentage error: \[ \text{Percentage Error} = \frac{0.01}{1.25} \times 100 \] \[ \text{Percentage Error} = \frac{1}{1.25} = 0.8\% \] Thus, the inaccuracy in the measurement relative to the true length is 0.80%.

The formula for Simple Interest (I) is given by: \[ I = \frac{P \times R \times T}{100} \] Where: \( I = 24 \) (the interest), \( P = 400 \) (the principal), \( T = 3 \) (the time in years), \( R = ? \) (the rate we need to find). Rearranging the formula to solve for R: \[ R = \frac{I \times 100}{P \times T} \] Substituting the values: \[ R = \frac{24 \times 100}{400 \times 3} \] \[ R = \frac{2400}{1200} \] \[ R = 2 \] The annual interest rate required for the principal of \text{₦}400 to generate \text{₦}24 over 3 years is 2%.

If p : q = 23 : 56; then the sum S₁ = 23 + 56 = 79.
If q : r = 34 : 12; then the sum S₂ = 34 + 12 = 46.
Let p + q = T₁.
Then\[ q = \left(\frac{56}{79}\right) T_1.\] Again, let q + r = T₂.

Then\[ q = \left(\frac{34}{46}\right) T_2.\] Since\; q = q,\[ \frac{56}{79} T_1 = \frac{34}{46} T_2.\] \[ \frac{T_1}{T_2} = \frac{34 \times 79}{56 \times 46}.\] \[ \frac{T_1}{T_2} = \frac{2686}{2576}.\] Taking proportional values,\[ T_1 = 27,\quad T_2 = 25.\] Now,\[ p = \left(\frac{23}{79}\right) T_1
= \left(\frac{23}{79}\right) 27
= 12,\] \[ q = \left(\frac{56}{79}\right) T_1
= \left(\frac{56}{79}\right) 27
= 15,\] \[ r = \left(\frac{12}{46}\right) T_2
= \left(\frac{12}{46}\right) 25
= 10.\] Hence,\[ p : q : r = 12 : 15 : 10.\]

The formula for the \( n \)-th term of an Arithmetic Progression (A.P.) is: \[ T_n = a + (n - 1)d \] Given: 1) \( T_3 = a + 2d = -9 \) 2) \( T_7 = a + 6d = -29 \) Subtract equation (1) from equation (2): \[ (a + 6d) - (a + 2d) = -29 - (-9) \] \[ 4d = -20 \] \[ d = -5 \] Substitute \( d = -5 \) into equation (1): \[ a + 2(-5) = -9 \] \[ a - 10 = -9 \] \[ a = 1 \] Now, find the 10th term (\( T_{10} \)): \[ T_{10} = a + 9d \] \[ T_{10} = 1 + 9(-5) \] \[ T_{10} = 1 - 45 \] \[ T_{10} = -44 \] The 10th term of the sequence is -44.

To find the value of \( x \) where the function attains its minimum (or maximum) value, we differentiate the function and set it equal to zero (the stationary point). Given: \[ y = x^2 - 2x - 3 \] Differentiating with respect to \( x \): \[ \frac{dy}{dx} = 2x - 2 \] Set the derivative to zero: \[ 2x - 2 = 0 \] \[ 2x = 2 \] \[ x = 1 \] Since the coefficient of \( x^2 \) is positive (\( +1 \)), the parabola opens upwards, confirming that this stationary point is a minimum. Therefore, the function reaches its lowest point when \( x = 1 \).

Two lines are parallel if they have the same gradient (slope). The given equation is in the form \( y = mx + c \), where \( m = -4 \). Therefore, the new line must also have a gradient of -4. Using the point-slope form with point (2, 3) and gradient \( m = -4 \): \[ y - y_1 = m(x - x_1) \] \[ y - 3 = -4(x - 2) \] \[ y - 3 = -4x + 8 \] To arrange this into the general form \( ax + by + c = 0 \): \[ y + 4x - 3 - 8 = 0 \] \[ y + 4x - 11 = 0 \] This equation represents the line that is parallel to the original and passes through the specified coordinates.

To make \( Q \) the subject, we need to isolate it on one side of the equation. Start with: \[ P = \frac{M}{5}(X + Q) + 1 \] Step 1: Subtract 1 from both sides: \[ P - 1 = \frac{M}{5}(X + Q) \] Step 2: Multiply both sides by 5 to remove the denominator: \[ 5(P - 1) = M(X + Q) \] \[ 5P - 5 = M(X + Q) \] Step 3: Divide both sides by M: \[ \frac{5P - 5}{M} = X + Q \] Step 4: Subtract X from both sides to isolate Q: \[ Q = \frac{5P - 5}{M} - X \] To express this as a single fraction, put \( X \) over the common denominator \( M \): \[ Q = \frac{5P - 5 - MX}{M} \] Rearranging the terms in the numerator: \[ Q = \frac{5P - MX - 5}{M} \] This matches option B.

The expression \( 27x^3 - 8y^3 \) is a 'difference of two cubes'. The general algebraic identity for the difference of two cubes is: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] In this problem: \[ 27x^3 = (3x)^3 \Rightarrow a = 3x \] \[ 8y^3 = (2y)^3 \Rightarrow b = 2y \] Applying the identity: \[ 27x^3 - 8y^3 = (3x - 2y)((3x)^2 + (3x)(2y) + (2y)^2) \] \[ 27x^3 - 8y^3 = (3x - 2y)(9x^2 + 6xy + 4y^2) \] Since the question identifies \( 9x^2 + 6xy + 4y^2 \) as one of the factors, the other factor must be the first term in the product, which is \( 3x - 2y \).