2024 - JAMB Mathematics Past Questions and Answers - page 7

We are given the equation of a line: \(7x + 5y = 12\), and we need to find the equation of a line passing through the point (5, 7) and parallel to this line. Parallel lines have the same slope. To find the slope of the given line, we rewrite it in slope-intercept form:

\(5y = -7x + 12\)

\(y = -\frac{7}{5}x + \frac{12}{5}\)

The slope is -\(7/5\). Using the point-slope form:

\(y - y_1 = m(x - x_1)\)

Substitute the point (5,7) and slope -7/5:

\(y - 7 = -\frac{7}{5}(x - 5)\)

Expand and simplify:

\(y - 7 = -\frac{7}{5}x + 7\)

\(y = -\frac{7}{5}x + 14\)

Multiply through by 5:

5y = -7x + 70

Simplified: 7x + 5y = 70

μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A = {3, 6, 9, 12, 15, 18}

B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

A ∩ B = {3, 9, 15}

We are given the equation: \(25^{1 - x} \times 5^{x + 2} \div (\frac{1}{125})^{x} = 625^{-1}\). Start by expressing all numbers in terms of base 5:

• \(25 = 5^2\), so \(25^{1 - x} = 5^{2(1 - x)} = 5^{2 - 2x}\)
• \(125 = 5^3\), so \((1/125)^x = (5^{-3})^x = 5^{-3x}\)
• \(625 = 5^4\), so \(625^{-1} = 5^{-4}\)

Now, the equation becomes: \(5^{2 - 2x} \times 5^{x + 2} \div 5^{-3x} = 5^{-4}\).

Combine the exponents: \(5^{2 - 2x + x + 2 + 3x} = 5^{4 + 2x}\)

Set the exponents equal: 4 + 2x = -4, hence 2x = -8 and x = -4.

0.0014 x 0.011 = 0.0000154

In standard form: 1.54 x 10⁻⁵