{"id":624,"date":"2024-03-05T15:02:18","date_gmt":"2024-03-05T15:02:18","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=624"},"modified":"2024-03-05T15:02:18","modified_gmt":"2024-03-05T15:02:18","slug":"a-couple-has-a-son-and-a-daughter-the-age-of-the-father-is-four-times-that-of-the-son-and-the-age-of-the-daughter-is-one-third-of-her-mother-the-wife-is-six-years-younger-to-her-husband-and-the-sist","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-couple-has-a-son-and-a-daughter-the-age-of-the-father-is-four-times-that-of-the-son-and-the-age-of-the-daughter-is-one-third-of-her-mother-the-wife-is-six-years-younger-to-her-husband-and-the-sist\/","title":{"rendered":"A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of her mother. The wife is six years younger to her husband and the sister in three years older than her brother. The age of the mother is:"},"content":{"rendered":"

[amp_mcq option1=”42 years” option2=”48 years” option3=”54 years” option4=”63 years” correct=”option2″]<\/p>\n

The correct answer is (b), 48 years.<\/p>\n

Let $f$ be the father’s age, $s$ be the son’s age, and $d$ be the daughter’s age. We know that $f = 4s$ and $d = \\frac{1}{3}m$, where $m$ is the mother’s age. We also know that the wife is six years younger than her husband, so $m = f – 6$. Finally, the sister is three years older than her brother, so $d = s + 3$.<\/p>\n

Substituting the first equation into the second equation, we get $d = \\frac{1}{3}(f – 6)$. Substituting the third equation into the second equation, we get $d = s + 3$.<\/p>\n

Equating these two equations, we get $\\frac{1}{3}(f – 6) = s + 3$. Multiplying both sides by 3, we get $f – 6 = 3s + 9$. Adding 6 to both sides, we get $f = 3s + 15$.<\/p>\n

Substituting this into the first equation, we get $3s + 15 = 4s$. Subtracting 3s from both sides, we get $15 = s$. Therefore, the son’s age is 15 years.<\/p>\n

Substituting this into the third equation, we get $d = 15 + 3 = 18$. Therefore, the daughter’s age is 18 years.<\/p>\n

Finally, substituting the first equation into the fourth equation, we get $m = (4)(15) – 6 = 48$. Therefore, the mother’s age is 48 years.<\/p>\n","protected":false},"excerpt":{"rendered":"

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We also know that the wife is six years younger than her husband, so $m = f - 6$. Finally, the sister is three years older than her brother, so $d = s + 3$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-couple-has-a-son-and-a-daughter-the-age-of-the-father-is-four-times-that-of-the-son-and-the-age-of-the-daughter-is-one-third-of-her-mother-the-wife-is-six-years-younger-to-her-husband-and-the-sist\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A couple has a son and a daughter. The age of the father is four times that of the son and the age of the daughter is one-third of her mother. The wife is six years younger to her husband and the sister in three years older than her brother. The age of the mother is:\" \/>\n<meta property=\"og:description\" content=\"Let $f$ be the father's age, $s$ be the son's age, and $d$ be the daughter's age. We know that $f = 4s$ and $d = frac{1}{3}m$, where $m$ is the mother's age. We also know that the wife is six years younger than her husband, so $m = f - 6$. 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