{"id":90968,"date":"2025-06-01T10:42:11","date_gmt":"2025-06-01T10:42:11","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90968"},"modified":"2025-06-01T10:42:11","modified_gmt":"2025-06-01T10:42:11","slug":"which-one-of-the-following-is-the-average-of-first-five-multiples-of-e","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/","title":{"rendered":"Which one of the following is the average of first five multiples of e"},"content":{"rendered":"

Which one of the following is the average of first five multiples of each of the numbers from 11, 12, 13, …, 20 ?<\/p>\n

[amp_mcq option1=”40.5″ option2=”42.5″ option3=”44.5″ option4=”46.5″ correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is D) 46.5.
\n<\/section>\n
\nWe need to find the average of the first five multiples for each number from 11 to 20.
\nFor any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$.
\nThe sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$.
\nThe average of these five multiples is $\\frac{15N}{5} = 3N$.
\nSo, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$.
\nWe need to find the average of the values $3 \\times 11, 3 \\times 12, …, 3 \\times 20$.
\n<\/section>\n
\nThe numbers are $33, 36, 39, …, 60$. There are $20 – 11 + 1 = 10$ such numbers.
\nThe average of these 10 numbers is $\\frac{(3 \\times 11) + (3 \\times 12) + … + (3 \\times 20)}{10}$.
\nWe can factor out 3 from the numerator: $\\frac{3 \\times (11 + 12 + … + 20)}{10}$.
\nThe numbers 11, 12, …, 20 form an arithmetic progression. The sum of an arithmetic progression is $\\frac{\\text{number of terms}}{2} \\times (\\text{first term} + \\text{last term})$.
\nSum of $11 + … + 20 = \\frac{10}{2} \\times (11 + 20) = 5 \\times 31 = 155$.
\nThe required average is $\\frac{3 \\times 155}{10} = \\frac{465}{10} = 46.5$.
\nAlternatively, the average of $3N$ for $N=11, …, 20$ is $3 \\times (\\text{Average of } 11, …, 20)$. The average of $11, …, 20$ is $\\frac{11+20}{2} = 15.5$. The required average is $3 \\times 15.5 = 46.5$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Which one of the following is the average of first five multiples of each of the numbers from 11, 12, 13, …, 20 ? [amp_mcq option1=”40.5″ option2=”42.5″ option3=”44.5″ option4=”46.5″ correct=”option4″] This question was previously asked in UPSC CAPF – 2024 Download PDFAttempt Online The correct answer is D) 46.5. We need to find the average … <\/p>\n

Detailed SolutionWhich one of the following is the average of first five multiples of e<\/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":[1103,1102],"class_list":["post-90968","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1103","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nWhich one of the following is the average of first five multiples of e<\/title>\n<meta name=\"description\" content=\"The correct answer is D) 46.5. We need to find the average of the first five multiples for each number from 11 to 20. For any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$. The sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$. The average of these five multiples is $frac{15N}{5} = 3N$. So, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$. We need to find the average of the values $3 times 11, 3 times 12, ..., 3 times 20$.\" \/>\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\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Which one of the following is the average of first five multiples of e\" \/>\n<meta property=\"og:description\" content=\"The correct answer is D) 46.5. We need to find the average of the first five multiples for each number from 11 to 20. For any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$. The sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$. The average of these five multiples is $frac{15N}{5} = 3N$. So, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$. We need to find the average of the values $3 times 11, 3 times 12, ..., 3 times 20$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:42:11+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":"Which one of the following is the average of first five multiples of e","description":"The correct answer is D) 46.5. We need to find the average of the first five multiples for each number from 11 to 20. For any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$. The sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$. The average of these five multiples is $frac{15N}{5} = 3N$. So, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$. We need to find the average of the values $3 times 11, 3 times 12, ..., 3 times 20$.","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\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/","og_locale":"en_US","og_type":"article","og_title":"Which one of the following is the average of first five multiples of e","og_description":"The correct answer is D) 46.5. We need to find the average of the first five multiples for each number from 11 to 20. For any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$. The sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$. The average of these five multiples is $frac{15N}{5} = 3N$. So, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$. We need to find the average of the values $3 times 11, 3 times 12, ..., 3 times 20$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:42:11+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\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/","url":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/","name":"Which one of the following is the average of first five multiples of e","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:42:11+00:00","dateModified":"2025-06-01T10:42:11+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is D) 46.5. We need to find the average of the first five multiples for each number from 11 to 20. For any number $N$, the first five multiples are $N, 2N, 3N, 4N, 5N$. The sum of these multiples is $N + 2N + 3N + 4N + 5N = (1+2+3+4+5)N = 15N$. The average of these five multiples is $\\frac{15N}{5} = 3N$. So, for each number $N$ from 11 to 20, the average of its first five multiples is $3N$. We need to find the average of the values $3 \\times 11, 3 \\times 12, ..., 3 \\times 20$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/which-one-of-the-following-is-the-average-of-first-five-multiples-of-e\/#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":"Which one of the following is the average of first five multiples of e"}]},{"@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\/90968","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=90968"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90968\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90968"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90968"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}