(a) An open rectangular tank is made of a steel plate of area 1440\(m^{2}\). Its length is twice its width . If the depth of the tank is 4m less than its width, find its length.
(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.
(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.
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Correct Answer: Option n
Explanation:
(a) Width = x m; Length = 2x m; Height = (x - 4) m.
Total surface area = 2Lh + WL + 2Wh
= \(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\)
= \(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\)
= \(8x^{2} - 24x = 1440\)
= \(x^{2} - 3x - 180 = 0\)
= \((x - 15)(x + 12) = 0\)
\(x = 15m\)
Length = 2(15m) = 30m.
(b) \(I = \frac{PRT}{100}\)
\(I_{P} + I_{Q} = I_{T}\)
\(\frac{3000 \times x \times 1}{100} + \frac{2000 \times y \times 1}{100} = 640\)
\(30x + 20y = 640 \implies 3x + 2y = 64 .... (1)\)
\(\frac{2000 \times x \times 1}{100} + \frac{3000 \times y \times 1}{100} = 660\)
\(20x + 30y = 660 \implies 2x + 3y = 66 .... (2)\)
To eliminate x , multiply (1) by 2 and (2) by 3.
\(6x + 4y = 128 ... (1)\)
\(6x + 9y = 198 .... (2)\)
(2) - (1) : \(9y - 4y = 198 - 128 \implies 5y = 70\)
\(y = 14\)
\(2x + 3y = 66 \implies 2x + 3(14) = 66\)
\(2x + 42 = 66 \implies 2x = 24\)
\(x = 12\)
\(\therefore x = 12 ; y = 14\).
(a) Width = x m; Length = 2x m; Height = (x - 4) m.
Total surface area = 2Lh + WL + 2Wh
= \(2(2x)(x - 4) + x(2x) + 2(x)(x - 4) = 1440 m^{2}\)
= \(4x^{2} - 16x + 2x^{2} + 2x^{2} - 8x = 1440\)
= \(8x^{2} - 24x = 1440\)
= \(x^{2} - 3x - 180 = 0\)
= \((x - 15)(x + 12) = 0\)
\(x = 15m\)
Length = 2(15m) = 30m.
(b) \(I = \frac{PRT}{100}\)
\(I_{P} + I_{Q} = I_{T}\)
\(\frac{3000 \times x \times 1}{100} + \frac{2000 \times y \times 1}{100} = 640\)
\(30x + 20y = 640 \implies 3x + 2y = 64 .... (1)\)
\(\frac{2000 \times x \times 1}{100} + \frac{3000 \times y \times 1}{100} = 660\)
\(20x + 30y = 660 \implies 2x + 3y = 66 .... (2)\)
To eliminate x , multiply (1) by 2 and (2) by 3.
\(6x + 4y = 128 ... (1)\)
\(6x + 9y = 198 .... (2)\)
(2) - (1) : \(9y - 4y = 198 - 128 \implies 5y = 70\)
\(y = 14\)
\(2x + 3y = 66 \implies 2x + 3(14) = 66\)
\(2x + 42 = 66 \implies 2x = 24\)
\(x = 12\)
\(\therefore x = 12 ; y = 14\).