{"id":90159,"date":"2025-06-01T10:21:58","date_gmt":"2025-06-01T10:21:58","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90159"},"modified":"2025-06-01T10:21:58","modified_gmt":"2025-06-01T10:21:58","slug":"the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/","title":{"rendered":"The length of a rectangle is increased by 60%. By what per cent would"},"content":{"rendered":"

The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area ?<\/p>\n

[amp_mcq option1=”37.5 %” option2=”60%” option3=”75%” option4=”120%” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2017<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe width would have to be decreased by 37.5% to maintain the same area.
\n<\/section>\n
\n– Let the original length be $L$ and original width be $W$. The original area is $A = L \\times W$.
\n– The length is increased by 60%, so the new length $L’$ is $L + 0.60L = 1.60L$.
\n– Let the new width be $W’$. The new area $A’ = L’ \\times W’ = 1.60L \\times W’$.
\n– To maintain the same area, $A’ = A$, so $1.60L \\times W’ = L \\times W$.
\n– We can solve for $W’$: $W’ = \\frac{L \\times W}{1.60L} = \\frac{W}{1.60} = \\frac{W}{8\/5} = \\frac{5}{8}W$.
\n– The decrease in width is $W – W’ = W – \\frac{5}{8}W = \\frac{3}{8}W$.
\n– The percentage decrease in width is $\\frac{\\text{Decrease in width}}{\\text{Original width}} \\times 100\\% = \\frac{(3\/8)W}{W} \\times 100\\% = \\frac{3}{8} \\times 100\\%$.
\n– $\\frac{3}{8} = 0.375$, so the percentage decrease is $0.375 \\times 100\\% = 37.5\\%$.
\n<\/section>\n
\nThis problem demonstrates the inverse relationship between dimensions when the area is kept constant. If one dimension is increased, the other must be decreased proportionally to maintain the same area. The percentage change in one dimension results in a different percentage change in the other dimension for a constant area, especially when the changes are expressed relative to the original values.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The length of a rectangle is increased by 60%. By what per cent would the width have to be decreased to maintain the same area ? [amp_mcq option1=”37.5 %” option2=”60%” option3=”75%” option4=”120%” correct=”option1″] This question was previously asked in UPSC CAPF – 2017 Download PDFAttempt Online The width would have to be decreased by 37.5% … <\/p>\n

Detailed SolutionThe length of a rectangle is increased by 60%. By what per cent would<\/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":[1085],"tags":[1101,1102],"class_list":["post-90159","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1101","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nThe length of a rectangle is increased by 60%. By what per cent would<\/title>\n<meta name=\"description\" content=\"The width would have to be decreased by 37.5% to maintain the same area. - Let the original length be $L$ and original width be $W$. The original area is $A = L times W$. - The length is increased by 60%, so the new length $L'$ is $L + 0.60L = 1.60L$. - Let the new width be $W'$. The new area $A' = L' times W' = 1.60L times W'$. - To maintain the same area, $A' = A$, so $1.60L times W' = L times W$. - We can solve for $W'$: $W' = frac{L times W}{1.60L} = frac{W}{1.60} = frac{W}{8\/5} = frac{5}{8}W$. - The decrease in width is $W - W' = W - frac{5}{8}W = frac{3}{8}W$. - The percentage decrease in width is $frac{text{Decrease in width}}{text{Original width}} times 100% = frac{(3\/8)W}{W} times 100% = frac{3}{8} times 100%$. - $frac{3}{8} = 0.375$, so the percentage decrease is $0.375 times 100% = 37.5%$.\" \/>\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\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The length of a rectangle is increased by 60%. By what per cent would\" \/>\n<meta property=\"og:description\" content=\"The width would have to be decreased by 37.5% to maintain the same area. - Let the original length be $L$ and original width be $W$. The original area is $A = L times W$. - The length is increased by 60%, so the new length $L'$ is $L + 0.60L = 1.60L$. - Let the new width be $W'$. The new area $A' = L' times W' = 1.60L times W'$. - To maintain the same area, $A' = A$, so $1.60L times W' = L times W$. - We can solve for $W'$: $W' = frac{L times W}{1.60L} = frac{W}{1.60} = frac{W}{8\/5} = frac{5}{8}W$. - The decrease in width is $W - W' = W - frac{5}{8}W = frac{3}{8}W$. - The percentage decrease in width is $frac{text{Decrease in width}}{text{Original width}} times 100% = frac{(3\/8)W}{W} times 100% = frac{3}{8} times 100%$. - $frac{3}{8} = 0.375$, so the percentage decrease is $0.375 times 100% = 37.5%$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:21:58+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":"The length of a rectangle is increased by 60%. By what per cent would","description":"The width would have to be decreased by 37.5% to maintain the same area. - Let the original length be $L$ and original width be $W$. The original area is $A = L times W$. - The length is increased by 60%, so the new length $L'$ is $L + 0.60L = 1.60L$. - Let the new width be $W'$. The new area $A' = L' times W' = 1.60L times W'$. - To maintain the same area, $A' = A$, so $1.60L times W' = L times W$. - We can solve for $W'$: $W' = frac{L times W}{1.60L} = frac{W}{1.60} = frac{W}{8\/5} = frac{5}{8}W$. - The decrease in width is $W - W' = W - frac{5}{8}W = frac{3}{8}W$. - The percentage decrease in width is $frac{text{Decrease in width}}{text{Original width}} times 100% = frac{(3\/8)W}{W} times 100% = frac{3}{8} times 100%$. - $frac{3}{8} = 0.375$, so the percentage decrease is $0.375 times 100% = 37.5%$.","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\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/","og_locale":"en_US","og_type":"article","og_title":"The length of a rectangle is increased by 60%. By what per cent would","og_description":"The width would have to be decreased by 37.5% to maintain the same area. - Let the original length be $L$ and original width be $W$. The original area is $A = L times W$. - The length is increased by 60%, so the new length $L'$ is $L + 0.60L = 1.60L$. - Let the new width be $W'$. The new area $A' = L' times W' = 1.60L times W'$. - To maintain the same area, $A' = A$, so $1.60L times W' = L times W$. - We can solve for $W'$: $W' = frac{L times W}{1.60L} = frac{W}{1.60} = frac{W}{8\/5} = frac{5}{8}W$. - The decrease in width is $W - W' = W - frac{5}{8}W = frac{3}{8}W$. - The percentage decrease in width is $frac{text{Decrease in width}}{text{Original width}} times 100% = frac{(3\/8)W}{W} times 100% = frac{3}{8} times 100%$. - $frac{3}{8} = 0.375$, so the percentage decrease is $0.375 times 100% = 37.5%$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:21:58+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\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/","name":"The length of a rectangle is increased by 60%. By what per cent would","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:21:58+00:00","dateModified":"2025-06-01T10:21:58+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The width would have to be decreased by 37.5% to maintain the same area. - Let the original length be $L$ and original width be $W$. The original area is $A = L \\times W$. - The length is increased by 60%, so the new length $L'$ is $L + 0.60L = 1.60L$. - Let the new width be $W'$. The new area $A' = L' \\times W' = 1.60L \\times W'$. - To maintain the same area, $A' = A$, so $1.60L \\times W' = L \\times W$. - We can solve for $W'$: $W' = \\frac{L \\times W}{1.60L} = \\frac{W}{1.60} = \\frac{W}{8\/5} = \\frac{5}{8}W$. - The decrease in width is $W - W' = W - \\frac{5}{8}W = \\frac{3}{8}W$. - The percentage decrease in width is $\\frac{\\text{Decrease in width}}{\\text{Original width}} \\times 100\\% = \\frac{(3\/8)W}{W} \\times 100\\% = \\frac{3}{8} \\times 100\\%$. - $\\frac{3}{8} = 0.375$, so the percentage decrease is $0.375 \\times 100\\% = 37.5\\%$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-length-of-a-rectangle-is-increased-by-60-by-what-per-cent-would\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"The length of a rectangle is increased by 60%. 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