{"id":93423,"date":"2025-06-01T11:50:15","date_gmt":"2025-06-01T11:50:15","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=93423"},"modified":"2025-06-01T11:50:15","modified_gmt":"2025-06-01T11:50:15","slug":"if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/","title":{"rendered":"If $x : y = 3 : 7$, then $\\frac{6x^2 – 2xy + y^2}{xy}$ is equal to"},"content":{"rendered":"

If $x : y = 3 : 7$, then $\\frac{6x^2 – 2xy + y^2}{xy}$ is equal to<\/p>\n

[amp_mcq option1=”$\\frac{62}{21}$” option2=”$\\frac{61}{21}$” option3=”$\\frac{63}{21}$” option4=”$\\frac{64}{21}$” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is $\\frac{61}{21}$.
\n<\/section>\n
\nGiven the ratio $x : y = 3 : 7$. This means that $\\frac{x}{y} = \\frac{3}{7}$.
\nWe can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k.
\nWe need to evaluate the expression $\\frac{6x^2 – 2xy + y^2}{xy}$.
\nSubstitute $x = 3k$ and $y = 7k$ into the expression:
\nNumerator:
\n$6x^2 – 2xy + y^2 = 6(3k)^2 – 2(3k)(7k) + (7k)^2$
\n$= 6(9k^2) – 2(21k^2) + 49k^2$
\n$= 54k^2 – 42k^2 + 49k^2$
\nCombine the terms with $k^2$:
\n$= (54 – 42 + 49)k^2$
\n$= (12 + 49)k^2$
\n$= 61k^2$.
\nDenominator:
\n$xy = (3k)(7k) = 21k^2$.
\nNow, form the ratio:
\n$\\frac{6x^2 – 2xy + y^2}{xy} = \\frac{61k^2}{21k^2}$.
\nSince k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator.
\nThe expression simplifies to $\\frac{61}{21}$.
\n<\/section>\n
\nAlternatively, we can divide the numerator and the denominator of the expression by $y^2$ (assuming y is not zero, which it isn’t since the ratio is 3:7):
\n$\\frac{6x^2 – 2xy + y^2}{xy} = \\frac{\\frac{6x^2}{y^2} – \\frac{2xy}{y^2} + \\frac{y^2}{y^2}}{\\frac{xy}{y^2}} = \\frac{6(\\frac{x}{y})^2 – 2(\\frac{x}{y}) + 1}{(\\frac{x}{y})}$.
\nSubstitute the value of $\\frac{x}{y} = \\frac{3}{7}$:
\n$= \\frac{6(\\frac{3}{7})^2 – 2(\\frac{3}{7}) + 1}{(\\frac{3}{7})} = \\frac{6(\\frac{9}{49}) – \\frac{6}{7} + 1}{\\frac{3}{7}}$
\n$= \\frac{\\frac{54}{49} – \\frac{42}{49} + \\frac{49}{49}}{\\frac{3}{7}} = \\frac{\\frac{54 – 42 + 49}{49}}{\\frac{3}{7}} = \\frac{\\frac{61}{49}}{\\frac{3}{7}}$
\n$= \\frac{61}{49} \\times \\frac{7}{3} = \\frac{61}{7 \\times 7} \\times \\frac{7}{3} = \\frac{61}{7 \\times 3} = \\frac{61}{21}$.
\nBoth methods yield the same result.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

If $x : y = 3 : 7$, then $\\frac{6x^2 – 2xy + y^2}{xy}$ is equal to [amp_mcq option1=”$\\frac{62}{21}$” option2=”$\\frac{61}{21}$” option3=”$\\frac{63}{21}$” option4=”$\\frac{64}{21}$” correct=”option2″] This question was previously asked in UPSC CISF-AC-EXE – 2024 Download PDFAttempt Online The correct answer is $\\frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $\\frac{x}{y} … <\/p>\n

Detailed SolutionIf $x : y = 3 : 7$, then $\\frac{6x^2 – 2xy + y^2}{xy}$ is equal to<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1089],"tags":[1103,1102],"class_list":["post-93423","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1103","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nIf $x : y = 3 : 7$, then $\\frac{6x^2 - 2xy + y^2}{xy}$ is equal to<\/title>\n<meta name=\"description\" content=\"The correct answer is $frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $frac{x}{y} = frac{3}{7}$. We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k. We need to evaluate the expression $frac{6x^2 - 2xy + y^2}{xy}$. Substitute $x = 3k$ and $y = 7k$ into the expression: Numerator: $6x^2 - 2xy + y^2 = 6(3k)^2 - 2(3k)(7k) + (7k)^2$ $= 6(9k^2) - 2(21k^2) + 49k^2$ $= 54k^2 - 42k^2 + 49k^2$ Combine the terms with $k^2$: $= (54 - 42 + 49)k^2$ $= (12 + 49)k^2$ $= 61k^2$. Denominator: $xy = (3k)(7k) = 21k^2$. Now, form the ratio: $frac{6x^2 - 2xy + y^2}{xy} = frac{61k^2}{21k^2}$. Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator. The expression simplifies to $frac{61}{21}$.\" \/>\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\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If $x : y = 3 : 7$, then $\\frac{6x^2 - 2xy + y^2}{xy}$ is equal to\" \/>\n<meta property=\"og:description\" content=\"The correct answer is $frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $frac{x}{y} = frac{3}{7}$. We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k. We need to evaluate the expression $frac{6x^2 - 2xy + y^2}{xy}$. Substitute $x = 3k$ and $y = 7k$ into the expression: Numerator: $6x^2 - 2xy + y^2 = 6(3k)^2 - 2(3k)(7k) + (7k)^2$ $= 6(9k^2) - 2(21k^2) + 49k^2$ $= 54k^2 - 42k^2 + 49k^2$ Combine the terms with $k^2$: $= (54 - 42 + 49)k^2$ $= (12 + 49)k^2$ $= 61k^2$. Denominator: $xy = (3k)(7k) = 21k^2$. Now, form the ratio: $frac{6x^2 - 2xy + y^2}{xy} = frac{61k^2}{21k^2}$. Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator. The expression simplifies to $frac{61}{21}$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:50:15+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"If $x : y = 3 : 7$, then $\\frac{6x^2 - 2xy + y^2}{xy}$ is equal to","description":"The correct answer is $frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $frac{x}{y} = frac{3}{7}$. We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k. We need to evaluate the expression $frac{6x^2 - 2xy + y^2}{xy}$. Substitute $x = 3k$ and $y = 7k$ into the expression: Numerator: $6x^2 - 2xy + y^2 = 6(3k)^2 - 2(3k)(7k) + (7k)^2$ $= 6(9k^2) - 2(21k^2) + 49k^2$ $= 54k^2 - 42k^2 + 49k^2$ Combine the terms with $k^2$: $= (54 - 42 + 49)k^2$ $= (12 + 49)k^2$ $= 61k^2$. Denominator: $xy = (3k)(7k) = 21k^2$. Now, form the ratio: $frac{6x^2 - 2xy + y^2}{xy} = frac{61k^2}{21k^2}$. Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator. The expression simplifies to $frac{61}{21}$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/","og_locale":"en_US","og_type":"article","og_title":"If $x : y = 3 : 7$, then $\\frac{6x^2 - 2xy + y^2}{xy}$ is equal to","og_description":"The correct answer is $frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $frac{x}{y} = frac{3}{7}$. We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k. We need to evaluate the expression $frac{6x^2 - 2xy + y^2}{xy}$. Substitute $x = 3k$ and $y = 7k$ into the expression: Numerator: $6x^2 - 2xy + y^2 = 6(3k)^2 - 2(3k)(7k) + (7k)^2$ $= 6(9k^2) - 2(21k^2) + 49k^2$ $= 54k^2 - 42k^2 + 49k^2$ Combine the terms with $k^2$: $= (54 - 42 + 49)k^2$ $= (12 + 49)k^2$ $= 61k^2$. Denominator: $xy = (3k)(7k) = 21k^2$. Now, form the ratio: $frac{6x^2 - 2xy + y^2}{xy} = frac{61k^2}{21k^2}$. Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator. The expression simplifies to $frac{61}{21}$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:50:15+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/","url":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/","name":"If $x : y = 3 : 7$, then $\\frac{6x^2 - 2xy + y^2}{xy}$ is equal to","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:50:15+00:00","dateModified":"2025-06-01T11:50:15+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is $\\frac{61}{21}$. Given the ratio $x : y = 3 : 7$. This means that $\\frac{x}{y} = \\frac{3}{7}$. We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k. We need to evaluate the expression $\\frac{6x^2 - 2xy + y^2}{xy}$. Substitute $x = 3k$ and $y = 7k$ into the expression: Numerator: $6x^2 - 2xy + y^2 = 6(3k)^2 - 2(3k)(7k) + (7k)^2$ $= 6(9k^2) - 2(21k^2) + 49k^2$ $= 54k^2 - 42k^2 + 49k^2$ Combine the terms with $k^2$: $= (54 - 42 + 49)k^2$ $= (12 + 49)k^2$ $= 61k^2$. Denominator: $xy = (3k)(7k) = 21k^2$. Now, form the ratio: $\\frac{6x^2 - 2xy + y^2}{xy} = \\frac{61k^2}{21k^2}$. Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator. The expression simplifies to $\\frac{61}{21}$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/if-x-y-3-7-then-frac6x2-2xy-y2xy-is-equal-to\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CISF-AC-EXE","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cisf-ac-exe\/"},{"@type":"ListItem","position":3,"name":"If $x : y = 3 : 7$, then $\\frac{6x^2 – 2xy + y^2}{xy}$ is equal to"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/93423","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=93423"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/93423\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=93423"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=93423"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=93423"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}