Mathematics 2014 Jamb Examination

To add these binary numbers, we align them by their place values and perform addition from right to left, remembering that in base 2, \(1 + 1 = 10_2\) (write 0, carry 1) and \(1 + 1 + 1 = 11_2\) (write 1, carry 1):

     1 1 0 1 1 1
  + 0 1 0 1 0 0
   ____________
  1 0 0 1 0 1 1
Working right to left: 1+0=1; 1+0=1; 1+1=10 (write 0, carry 1); 0+0+1=1; 1+1=10 (write 0, carry 1); 1+1=10 (write 0, carry 1). The result is 1001011₂.

The selling price at a loss is calculated as: \(\text{Selling Price} = \text{Cost Price} - \text{Loss}\).
Step 1: Calculate the loss amount: 15% of N60,000 = \(0.15 \times 60,000 = 9,000\).
Step 2: Subtract the loss from the cost price: \(60,000 - 9,000 = 51,000\).
Therefore, the woman sold the grinder for N51,000.

Step 1: Multiply the two numbers: \(0.00043 \times 2000 = 0.86\).
Step 2: Convert the result into standard form (scientific notation), which requires one non-zero digit before the decimal point: \(0.86 = 8.6 \times 10^{-1}\).

Let the man's net monthly income be \(I\).
The problem states that 10% of \(I\) is equal to N4,500.\[0.10 \times I = 4,500\]\[I = \frac{4,500}{0.10} = 45,000\]The man's net monthly income is N45,000.

First, express the required logs in terms of \(\log_{10} 7.5\):
\(\log_{10} 75 = \log_{10} (7.5 \times 10) \)
\(= \log_{10} 7.5 + \log_{10} 10 = 0.8751 + 1 = 1.8751\)
\(\log_{10} 750 = \log_{10} (7.5 \times 100) \)
\(= \log_{10} 7.5 + \log_{10} 100 = 0.8751 + 2 = 2.8751\)
Now substitute into the expression:
\(2(1.8751) + 2.8751 = 3.7502 + 2.8751 = 6.6253\).

8x^-2 = 2/25
x^-2 = 2/25 x 1/8
x^-2 = 2/200
x^-2 = 1/100
1/x² = 1/100
x² = 100
x = 10

To simplify an expression with a radical in the denominator, we must multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{2} - \sqrt{3}\) is \(\sqrt{2} + \sqrt{3}\).\[ \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} \times \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \] Expanding the numerator using the FOIL method: \[ (\sqrt{2} + \sqrt{3})(\sqrt{2} + \sqrt{3}) = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2 \] \[ = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6} \] Expanding the denominator using the difference of squares: \[ (\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 \] \[ = 2 - 3 = -1 \] Combining the results: \[ \frac{5 + 2\sqrt{6}}{-1} = -5 - 2\sqrt{6} \] The process of rationalisation removes the irrational component from the divisor, allowing the expression to be written in a linear surd form.

We can solve this using the laws of logarithms or by converting the numbers to powers of 2. First, let's express 8, 16, and 4 as powers of 2: \[ 8 = 2^3, \quad 16 = 2^4, \quad 4 = 2^2 \] Substitute these into the expression: \[ \log_2 2^3 + \log_2 2^4 - \log_2 2^2 \] Using the power law of logarithms, \(\log_b a^n = n \log_b a\): \[ 3\log_2 2 + 4\log_2 2 - 2\log_2 2 \] Since \(\log_a a = 1\): \[ 3(1) + 4(1) - 2(1) = 3 + 4 - 2 \] \[ 7 - 2 = 5 \] Alternatively, using the product and quotient laws: \[ \log_2 \left( \frac{8 \times 16}{4} \right) = \log_2 \left( \frac{128}{4} \right) = \log_2 32 \] \[ 32 = 2^5 \Rightarrow \log_2 2^5 = 5 \]

The union of two sets, P \(\cup\) Q, contains all the elements that are in set P, or in set Q, or in both. We are given: \[ P = \{1, 2, 3, 4, 5\} \] \[ P \cup Q = \{1, 2, 3, 4, 5, 6, 7\} \] To find the elements that must be in Q, we look at the elements in the union that are not present in P. These elements are 6 and 7. While Q could potentially contain some elements that are also in P (like 1, 2, 3, 4, or 5), the most fundamental set Q that satisfies the union given the options provided is the set containing the new elements. Comparing the options, {6, 7} is the only set that accounts for all elements in the union that were missing from P.

To make g the subject of the formula, we need to isolate g on one side of the equation. Starting with the original equation: \[ gt^2 - k - w = 0 \] Step 1: Move the terms without g (i.e., -k and -w) to the other side of the equation by adding them to both sides: \[ gt^2 = k + w \] Step 2: Isolate g by dividing both sides of the equation by \(t^2\): \[ g = \frac{k + w}{t^2} \] This expresses g in terms of k, w, and t, satisfying the requirement to change the subject of the formula.