A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is : [#1127]

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Q1. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :
Q1. A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is :

(A) 60 years
(A) 60 years

(B) 51 years
(B) 51 years

(D) 45 years
(D) 45 years

Answer: (D) 45 years
Answer: (D) 45 years

The son becomes 15 years old after 3 years. Hence son's present age = 15 years - 3 years = 12 years Man is four times as old as his son. Hence Man's present age = 12 years * 4 = 48 years Man is 3 years older than his wife. Hence present age of his wife = 48 years - 3 years = 45 years.

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@1127

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2024-05-15

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Related MCQ Quizzes

Q1. The value of $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ is :
Q1. The value of $\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}$ is :

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\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}

\sqrt{0.02*0.02} + \sqrt{1.2*1.2} + \sqrt{1.3*1.3} + \sqrt{0.03*0.03}

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\sqrt{0.04} + \sqrt{1.44} + \sqrt{1.69} + \sqrt{0.0009}

\sqrt{0.02*0.02} + \sqrt{1.2*1.2} + \sqrt{1.3*1.3} + \sqrt{0.03*0.03}

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Q2. Find the value of $\sqrt{\frac{225}{729}} - \sqrt{\frac{25}{144}} - \sqrt{\frac{16}{81}}$
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(A)

\frac{- 5}{9}

(A)

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\sqrt{\frac{225}{729}} - \sqrt{\frac{25}{144}} - \sqrt{\frac{16}{81}}

\sqrt{\frac{15 * 15}{27 * 27}} - \sqrt{\frac{5 * 5}{12 * 12}} - \sqrt{\frac{4 * 4}{9 * 9}}

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\frac{98}{2}

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A man is 3 years older than his wife and four times as old as his son. If the son becomes 15 years old after 3 years, the present age of his wife is : [#1127]

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