(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
(b)
The diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\(cm^{2}\), find the values of x.
(b)
The diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\(cm^{2}\), find the values of x.
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Correct Answer: Option n
Explanation:
(a) \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
Multiply through by the LCM of 4 and 8, to clear fractions :
\(8 \times 4 + 8 \times \frac{3}{4}(x + 2) \leq 8 \times \frac{3}{8}x + 8 \)
\(32 + 6(x + 2) \leq 3x + 8 \)
\(32 + 6x + 12 \leq 3x + 8 \)
\(44 + 6x \leq 3x + 8\)
\(6x - 3x \leq 8 - 44 \implies 3x \leq -36\)
\(x \leq - 12\).
(b)
Area \((A_{1})\), of rectangle PQRS = \(20(20 + x) cm^{2}\)
Area \((A_{2})\), of square = \(x \times x = x^{2} cm^{2}\)
\(\implies 20(20 + x) - x^{2} = 484\)
\(400 + 20x - x^{2} = 484\)
\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\)
\(x^{2} - 14x - 6x + 84 = 0\)
\(x( x - 14) - 6(x - 14) = 0 \)
\((x - 14)(x - 6) = 0\)
\(\therefore \text{x = 6 cm or 14 cm}\)
(a) \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
Multiply through by the LCM of 4 and 8, to clear fractions :
\(8 \times 4 + 8 \times \frac{3}{4}(x + 2) \leq 8 \times \frac{3}{8}x + 8 \)
\(32 + 6(x + 2) \leq 3x + 8 \)
\(32 + 6x + 12 \leq 3x + 8 \)
\(44 + 6x \leq 3x + 8\)
\(6x - 3x \leq 8 - 44 \implies 3x \leq -36\)
\(x \leq - 12\).
(b)
Area \((A_{1})\), of rectangle PQRS = \(20(20 + x) cm^{2}\)
Area \((A_{2})\), of square = \(x \times x = x^{2} cm^{2}\)
\(\implies 20(20 + x) - x^{2} = 484\)
\(400 + 20x - x^{2} = 484\)
\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\)
\(x^{2} - 14x - 6x + 84 = 0\)
\(x( x - 14) - 6(x - 14) = 0 \)
\((x - 14)(x - 6) = 0\)
\(\therefore \text{x = 6 cm or 14 cm}\)