(a) Simplify : \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7 : 2. Find their present ages.
(b) The sum of the ages of a woman and her daughter is 46 years. In 4 years' time, the ratio of their ages will be 7 : 2. Find their present ages.
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Correct Answer: Option n
Explanation:
(a) \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
\(3^{-5n} \times (3^{2})^{n - 1} \times (3^{3})^{n + 1}\)
\(3^{-5n} \times 3^{2n - 2} \times 3^{3n + 3}\)
\(3^{-5n + 2n - 2 + 3n + 3}\)
= \(3^{1} = 3\)
(b) Let the daughter's age be c and the woman's age be d.
\(c + d = 46 .... (1)\)
In 4 years time, the daughter's age = c + 4
The woman's age = d + 4
\(\frac{d + 4}{c + 4} = \frac{7}{2}\)
\(2(d + 4) = 7(c + 4) \implies 2d + 8 = 7c + 28\)
\(2d - 7c = 28 - 8 \implies 2d - 7c = 20 ... (2)\)
\(c + d = 46 \implies d = 46 - c\)
\(\therefore 2(46 - c) - 7c = 20\)
\(92 - 2c - 7c = 20 \implies 92 - 20 = 9c\)
\(72 = 9c \implies c = 8\)
\(d = 46 - c \implies d = 46 - 8 = 38\)
Therefore, the daughter is 8 years old and the woman is 38 years.
(a) \(\frac{1}{3^{5n}} \times 9^{n - 1} \times 27^{n + 1}\)
\(3^{-5n} \times (3^{2})^{n - 1} \times (3^{3})^{n + 1}\)
\(3^{-5n} \times 3^{2n - 2} \times 3^{3n + 3}\)
\(3^{-5n + 2n - 2 + 3n + 3}\)
= \(3^{1} = 3\)
(b) Let the daughter's age be c and the woman's age be d.
\(c + d = 46 .... (1)\)
In 4 years time, the daughter's age = c + 4
The woman's age = d + 4
\(\frac{d + 4}{c + 4} = \frac{7}{2}\)
\(2(d + 4) = 7(c + 4) \implies 2d + 8 = 7c + 28\)
\(2d - 7c = 28 - 8 \implies 2d - 7c = 20 ... (2)\)
\(c + d = 46 \implies d = 46 - c\)
\(\therefore 2(46 - c) - 7c = 20\)
\(92 - 2c - 7c = 20 \implies 92 - 20 = 9c\)
\(72 = 9c \implies c = 8\)
\(d = 46 - c \implies d = 46 - 8 = 38\)
Therefore, the daughter is 8 years old and the woman is 38 years.