{"id":92630,"date":"2025-06-01T11:29:29","date_gmt":"2025-06-01T11:29:29","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=92630"},"modified":"2025-06-01T11:29:29","modified_gmt":"2025-06-01T11:29:29","slug":"a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/","title":{"rendered":"A painter wants to paint a picture (rectangular portrait) occupying 72"},"content":{"rendered":"

A painter wants to paint a picture (rectangular portrait) occupying 72 square inches on a canvas allowing a margin of 4 inches on the top and at the bottom and 2 inches on each side. What will be the smallest dimension of the canvas?<\/p>\n

[amp_mcq option1=”12\u2033 x 17\u2033” option2=”7\u2033 x 31\u2033” option3=”13\u2033 x 16\u2033” option4=”10\u2033 x 20\u2033” correct=”option4″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CISF-AC-EXE – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w \\times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side.
\n<\/section>\n
\n– The dimensions of the canvas will be:
\n – Canvas Width = Picture Width + 2 * Side Margin = $w + 2 \\times 2 = w + 4$
\n – Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 \\times 4 = h + 8$
\n– The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) \\times (h+8)$ such that $w \\times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options.
\n– The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$.
\n– Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$.
\n– To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$.
\n– Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 \\implies 2w^2 = 72 \\implies w^2 = 36$.
\n– Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$.
\n– Check if $2w=h$: $2 \\times 6 = 12$, which is true.
\n– The picture dimensions that minimize the canvas area are 6 inches by 12 inches.
\n– The corresponding canvas dimensions are:
\n – Canvas Width = $w + 4 = 6 + 4 = 10$ inches
\n – Canvas Height = $h + 8 = 12 + 8 = 20$ inches
\n– The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D).
\n– Let’s check if the other options correspond to valid picture dimensions and their canvas areas:
\n – Option A: Canvas 12×17. Picture width = 12-4=8. Picture height = 17-8=9. 8×9=72. Valid. Canvas area = 12×17 = 204.
\n – Option B: Canvas 7×31. Picture width = 7-4=3. Picture height = 31-8=23. 3×23=69. Not 72. Invalid option.
\n – Option C: Canvas 13×16. Picture width = 13-4=9. Picture height = 16-8=8. 9×8=72. Valid. Canvas area = 13×16 = 208.
\n – Option D: Canvas 10×20. Picture width = 10-4=6. Picture height = 20-8=12. 6×12=72. Valid. Canvas area = 10×20 = 200.
\n– Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10″x20″ option).
\n<\/section>\n
\nThe problem implicitly asks for the canvas dimensions that result in the minimum possible canvas area while accommodating the picture with the specified margins. The method of minimizing the area $104 + 8w + 4h$ subject to $wh = 72$ using calculus (finding the derivative with respect to $w$ after substituting $h=72\/w$) or AM-GM inequality ($8w + 4h \\ge 2\\sqrt{32wh} = 2\\sqrt{32 \\times 72} = 2\\sqrt{2304} = 2 \\times 48 = 96$, minimum when $8w=4h$) both confirm that the minimum area occurs when $w=6$ and $h=12$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A painter wants to paint a picture (rectangular portrait) occupying 72 square inches on a canvas allowing a margin of 4 inches on the top and at the bottom and 2 inches on each side. What will be the smallest dimension of the canvas? [amp_mcq option1=”12\u2033 x 17\u2033” option2=”7\u2033 x 31\u2033” option3=”13\u2033 x 16\u2033” option4=”10\u2033 … <\/p>\n

Detailed SolutionA painter wants to paint a picture (rectangular portrait) occupying 72<\/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":[1119,1102],"class_list":["post-92630","post","type-post","status-publish","format-standard","hentry","category-upsc-cisf-ac-exe","tag-1119","tag-quantitative-aptitude-and-reasoning","no-featured-image-padding"],"yoast_head":"\nA painter wants to paint a picture (rectangular portrait) occupying 72<\/title>\n<meta name=\"description\" content=\"The correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side. - The dimensions of the canvas will be: - Canvas Width = Picture Width + 2 * Side Margin = $w + 2 times 2 = w + 4$ - Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 times 4 = h + 8$ - The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) times (h+8)$ such that $w times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options. - The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$. - Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$. - To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$. - Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 implies 2w^2 = 72 implies w^2 = 36$. - Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$. - Check if $2w=h$: $2 times 6 = 12$, which is true. - The picture dimensions that minimize the canvas area are 6 inches by 12 inches. - The corresponding canvas dimensions are: - Canvas Width = $w + 4 = 6 + 4 = 10$ inches - Canvas Height = $h + 8 = 12 + 8 = 20$ inches - The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D). - Let's check if the other options correspond to valid picture dimensions and their canvas areas: - Option A: Canvas 12x17. Picture width = 12-4=8. Picture height = 17-8=9. 8x9=72. Valid. Canvas area = 12x17 = 204. - Option B: Canvas 7x31. Picture width = 7-4=3. Picture height = 31-8=23. 3x23=69. Not 72. Invalid option. - Option C: Canvas 13x16. Picture width = 13-4=9. Picture height = 16-8=8. 9x8=72. Valid. Canvas area = 13x16 = 208. - Option D: Canvas 10x20. Picture width = 10-4=6. Picture height = 20-8=12. 6x12=72. Valid. Canvas area = 10x20 = 200. - Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10"x20" option).\" \/>\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-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A painter wants to paint a picture (rectangular portrait) occupying 72\" \/>\n<meta property=\"og:description\" content=\"The correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side. - The dimensions of the canvas will be: - Canvas Width = Picture Width + 2 * Side Margin = $w + 2 times 2 = w + 4$ - Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 times 4 = h + 8$ - The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) times (h+8)$ such that $w times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options. - The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$. - Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$. - To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$. - Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 implies 2w^2 = 72 implies w^2 = 36$. - Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$. - Check if $2w=h$: $2 times 6 = 12$, which is true. - The picture dimensions that minimize the canvas area are 6 inches by 12 inches. - The corresponding canvas dimensions are: - Canvas Width = $w + 4 = 6 + 4 = 10$ inches - Canvas Height = $h + 8 = 12 + 8 = 20$ inches - The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D). - Let's check if the other options correspond to valid picture dimensions and their canvas areas: - Option A: Canvas 12x17. Picture width = 12-4=8. Picture height = 17-8=9. 8x9=72. Valid. Canvas area = 12x17 = 204. - Option B: Canvas 7x31. Picture width = 7-4=3. Picture height = 31-8=23. 3x23=69. Not 72. Invalid option. - Option C: Canvas 13x16. Picture width = 13-4=9. Picture height = 16-8=8. 9x8=72. Valid. Canvas area = 13x16 = 208. - Option D: Canvas 10x20. Picture width = 10-4=6. Picture height = 20-8=12. 6x12=72. Valid. Canvas area = 10x20 = 200. - Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10"x20" option).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T11:29:29+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=\"3 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"A painter wants to paint a picture (rectangular portrait) occupying 72","description":"The correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side. - The dimensions of the canvas will be: - Canvas Width = Picture Width + 2 * Side Margin = $w + 2 times 2 = w + 4$ - Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 times 4 = h + 8$ - The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) times (h+8)$ such that $w times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options. - The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$. - Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$. - To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$. - Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 implies 2w^2 = 72 implies w^2 = 36$. - Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$. - Check if $2w=h$: $2 times 6 = 12$, which is true. - The picture dimensions that minimize the canvas area are 6 inches by 12 inches. - The corresponding canvas dimensions are: - Canvas Width = $w + 4 = 6 + 4 = 10$ inches - Canvas Height = $h + 8 = 12 + 8 = 20$ inches - The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D). - Let's check if the other options correspond to valid picture dimensions and their canvas areas: - Option A: Canvas 12x17. Picture width = 12-4=8. Picture height = 17-8=9. 8x9=72. Valid. Canvas area = 12x17 = 204. - Option B: Canvas 7x31. Picture width = 7-4=3. Picture height = 31-8=23. 3x23=69. Not 72. Invalid option. - Option C: Canvas 13x16. Picture width = 13-4=9. Picture height = 16-8=8. 9x8=72. Valid. Canvas area = 13x16 = 208. - Option D: Canvas 10x20. Picture width = 10-4=6. Picture height = 20-8=12. 6x12=72. Valid. Canvas area = 10x20 = 200. - Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10\"x20\" option).","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\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/","og_locale":"en_US","og_type":"article","og_title":"A painter wants to paint a picture (rectangular portrait) occupying 72","og_description":"The correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side. - The dimensions of the canvas will be: - Canvas Width = Picture Width + 2 * Side Margin = $w + 2 times 2 = w + 4$ - Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 times 4 = h + 8$ - The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) times (h+8)$ such that $w times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options. - The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$. - Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$. - To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$. - Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 implies 2w^2 = 72 implies w^2 = 36$. - Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$. - Check if $2w=h$: $2 times 6 = 12$, which is true. - The picture dimensions that minimize the canvas area are 6 inches by 12 inches. - The corresponding canvas dimensions are: - Canvas Width = $w + 4 = 6 + 4 = 10$ inches - Canvas Height = $h + 8 = 12 + 8 = 20$ inches - The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D). - Let's check if the other options correspond to valid picture dimensions and their canvas areas: - Option A: Canvas 12x17. Picture width = 12-4=8. Picture height = 17-8=9. 8x9=72. Valid. Canvas area = 12x17 = 204. - Option B: Canvas 7x31. Picture width = 7-4=3. Picture height = 31-8=23. 3x23=69. Not 72. Invalid option. - Option C: Canvas 13x16. Picture width = 13-4=9. Picture height = 16-8=8. 9x8=72. Valid. Canvas area = 13x16 = 208. - Option D: Canvas 10x20. Picture width = 10-4=6. Picture height = 20-8=12. 6x12=72. Valid. Canvas area = 10x20 = 200. - Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10\"x20\" option).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T11:29:29+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/","name":"A painter wants to paint a picture (rectangular portrait) occupying 72","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T11:29:29+00:00","dateModified":"2025-06-01T11:29:29+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is (D) 10\u2033 x 20\u2033. We are given that the rectangular picture has an area of 72 square inches. Let the dimensions of the picture be width $w$ and height $h$, so $w \\times h = 72$. The canvas has a margin of 4 inches on the top and bottom and 2 inches on each side. - The dimensions of the canvas will be: - Canvas Width = Picture Width + 2 * Side Margin = $w + 2 \\times 2 = w + 4$ - Canvas Height = Picture Height + 2 * Top\/Bottom Margin = $h + 2 \\times 4 = h + 8$ - The question asks for the smallest dimension of the canvas. Among the given options, we need to find the canvas dimensions $(w+4) \\times (h+8)$ such that $w \\times h = 72$. We are looking for the pair $(w+4, h+8)$ that represents the smallest overall canvas size (minimum area) or has the smallest minimum dimension among the options. - The canvas area is $A = (w+4)(h+8) = wh + 8w + 4h + 32$. - Since $wh = 72$, $A = 72 + 8w + 4h + 32 = 104 + 8w + 4h$. - To minimize the canvas area, we need to minimize $8w + 4h$ subject to $wh = 72$. This occurs when $8w = 4h$, i.e., $2w = h$. - Substituting $h = 2w$ into $wh = 72$: $w(2w) = 72 \\implies 2w^2 = 72 \\implies w^2 = 36$. - Since dimensions must be positive, $w = 6$. Then $h = 72\/6 = 12$. - Check if $2w=h$: $2 \\times 6 = 12$, which is true. - The picture dimensions that minimize the canvas area are 6 inches by 12 inches. - The corresponding canvas dimensions are: - Canvas Width = $w + 4 = 6 + 4 = 10$ inches - Canvas Height = $h + 8 = 12 + 8 = 20$ inches - The calculated minimal canvas dimensions are 10\u2033 x 20\u2033, which is option (D). - Let's check if the other options correspond to valid picture dimensions and their canvas areas: - Option A: Canvas 12x17. Picture width = 12-4=8. Picture height = 17-8=9. 8x9=72. Valid. Canvas area = 12x17 = 204. - Option B: Canvas 7x31. Picture width = 7-4=3. Picture height = 31-8=23. 3x23=69. Not 72. Invalid option. - Option C: Canvas 13x16. Picture width = 13-4=9. Picture height = 16-8=8. 9x8=72. Valid. Canvas area = 13x16 = 208. - Option D: Canvas 10x20. Picture width = 10-4=6. Picture height = 20-8=12. 6x12=72. Valid. Canvas area = 10x20 = 200. - Comparing the valid canvas areas (204, 208, 200), the minimum area is 200 sq inches, corresponding to the 10\u2033 x 20\u2033 canvas. The smallest dimension among the options provided that satisfy the conditions and result in the minimum canvas size is 10 inches (from the 10\"x20\" option).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-painter-wants-to-paint-a-picture-rectangular-portrait-occupying-72\/#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":"A painter wants to paint a picture (rectangular portrait) occupying 72"}]},{"@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\/92630","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=92630"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/92630\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=92630"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=92630"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=92630"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}