{"id":5900,"date":"2025-06-09T07:09:19","date_gmt":"2025-06-09T07:09:19","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5900"},"modified":"2025-06-09T07:09:19","modified_gmt":"2025-06-09T07:09:19","slug":"rs-aggarwal-quantitative-aptitude-pdf-average-download","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-average-download\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free download: AVERAGE"},"content":{"rendered":"<h1 style=\"text-align: center\"><strong>6.AVERAGE<\/strong><\/h1>\n<p><strong>Ex.1:Find the average of all prime numbers between 30 and 50?<\/strong><\/p>\n<p><strong>Sol:<\/strong> there are five prime numbers between 30 and 50.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 They are 31,37,41,43 and 47.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Therefore the required average=(31+37+41+43+47)\/5\u00a0 \u00f3199\/5\u00a0 \u00f3 39.8.<\/p>\n<p><strong>Ex.2. find the average of first 40 natural numbers?<\/strong><\/p>\n<p><strong>Sol:<\/strong>\u00a0\u00a0\u00a0 sum\u00a0 of first n natural numbers=n(n+1)\/2;<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So,sum of 40 natural numbers=(40*41)\/2\u00a0 \u00f3820.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Therefore\u00a0 the required average=(820\/40) \u00f320.5.<\/p>\n<p><strong>Ex.3. find the average of first 20 multiples of 7?<\/strong><\/p>\n<p><strong>Sol:<\/strong>\u00a0\u00a0\u00a0 Required average =7(1+2+3+\u2026\u2026.+20)\/20\u00a0\u00a0 \u00f3(7*20*21)\/(20*2) \u00f3(147\/2)=73.5.<\/p>\n<p><strong>Ex.4. the average of four consecutive even numbers is 27. find the largest of these<\/strong> <strong>numbers?<\/strong><\/p>\n<p><strong>Sol:<\/strong>\u00a0\u00a0\u00a0 let the numbers be x,x+2,x+4 andx+6. then,<\/p>\n<p>(x+(x+2)+(x+4)+(x+6))\/4) = 27\u00a0<\/p>\n<p>\u00f3(4x+12)\/4 = 27<\/p>\n<p>\u00f3x+3=27\u00a0\u00a0\u00a0\u00a0 \u00a0\u00f3 x=24.<\/p>\n<p>Therefore the largest number=(x+6)=24+6=30.<\/p>\n<p>Ex.5. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?<\/p>\n<p><strong>Sol:<\/strong>\u00a0\u00a0 total weight of(36+44) students=(36*40+44*35)kg\u00a0 =2980kg.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Therefore weight of the total class=(2980\/80)kg\u00a0 =37.25kg.<\/p>\n<p>Ex:6.nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?<\/p>\n<p><strong>Sol:<\/strong> Let the average expenditure of all nine be Rs.x<\/p>\n<p>Then 12*8+(x+8)=9x or 8x=104 or x=13.<\/p>\n<p>Total money spent = 9x=Rs.(9*13)=Rs.117.<\/p>\n<p>Ex.7: Of the three numbers, second is twice the first and is also thrice the third.If the average of the three numbers is 44.Find the largest number.<\/p>\n<p><strong>Sol<\/strong>: Let the third number be x.<\/p>\n<p>Then second number = 3x.<\/p>\n<p>First number=3x\/2.<\/p>\n<p>Therefore x+3x+(3x\/2)=(44*3) or x=24<\/p>\n<p>So largest number= 2<sup>nd<\/sup> number=3x=72.<\/p>\n<p><strong>Ex.8:The average of\u00a0 25 result is 18.The average of 1<sup>st<\/sup> 12 of them is 14 &amp; that of last 12 is 17.Find the 13<sup>th<\/sup> result.<\/strong><\/p>\n<p><strong>Sol<\/strong>: Clearly 13<sup>th<\/sup> result=(sum of 25 results)-(sum of 24 results)<\/p>\n<p>=(18*25)-(14*12)+(17*12)<\/p>\n<p>=450-(168+204)<\/p>\n<p>=450-372<\/p>\n<p>=78.<\/p>\n<p><strong>Ex.9:The Average of 11 results is 16, if the average of the 1<sup>st<\/sup> 6 results is 58 &amp; that of the last 63. Find the 6<sup>th<\/sup> result.<\/strong><\/p>\n<p><strong>Sol<\/strong>: 6<sup>th<\/sup> result = (58*6+63*6-60*11)=66<\/p>\n<p>Ex.10:The average waight of A,B,C is 45 Kg. The avg wgt of A &amp; B be 40Kg &amp; that of B,C be 43Kg. Find the wgt of B.<\/p>\n<p><strong>Sol. <\/strong>Let A,B,c represent their individual wgts.<\/p>\n<p>Then,<\/p>\n<p>A+B+C=(45*3)Kg=135Kg<\/p>\n<p>A+B=(40*2)Kg=80Kg &amp; B+C=(43*2)Kg=86Kg<\/p>\n<p>B=(A+B)+(B+C)-(A+B+C)<\/p>\n<p>=(80+86-135)Kg<\/p>\n<p>=31Kg.<\/p>\n<p>Ex. 11. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.<\/p>\n<p><strong>Sol.<\/strong> Total age of 39 persons = (39 x 15) years<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 585 years.<\/p>\n<p>Average age of 40 persons= 15 yrs 3 months<\/p>\n<p>= 61\/4 years.<\/p>\n<p>Total age of 40 persons = <u>(_(61\/4 )<\/u>x 40) years= 610 years.<\/p>\n<p>:. Age of the teacher = (610 &#8211; 585) years=25 years.<\/p>\n<p><strong>Ex. 12. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new<\/strong><\/p>\n<p><strong> man.<\/strong><\/p>\n<p><strong>Sol.<\/strong> Total weight increased =(1.8 x 10) kg =18 kg.<\/p>\n<p>:. Weight of the new man =(53 + 18) kg =71 kg.<\/p>\n<p><strong>Ex. 13. There were 35 students in a hostel. Due to the admission of 7 new students, ;he expenses of the mess were increased by Rs. 42 per day while the average expenditure per head diminished by Rs 1. Wbat was the original expenditure of the mess?<\/strong><\/p>\n<p><strong>Sol.<\/strong> Let the original average expenditure be Rs. x. Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 42 (x &#8211; 1) &#8211; <em>35x=<\/em>42\u00a0 \u00a0 \u00f3<em> 7x= <\/em>84\u00a0\u00a0 \u00f3\u00a0\u00a0 x =12.<\/p>\n<p>Original expenditure = Rs. (35 x 12) =Rs. 420. .<\/p>\n<ol start=\"14\">\n<li><strong> A batsman makes a score of 87 runs in the 17th inning and thus increases his avg by 3. Find his average after 17th inning.<\/strong><\/li>\n<\/ol>\n<p><strong>Sol.<\/strong> Let the average after 17th inning = x.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0 Then, average after 16th inning = <em>(x <\/em>&#8211; 3).<\/p>\n<p>:. 16 (x &#8211; 3) + 87 = <em>17x <\/em>or <em>x <\/em>= (87 &#8211; 48) = 39.<\/p>\n<p><strong>Ex.15. Distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km perhour. Find the average speed of the train during the whole journey.<\/strong><\/p>\n<p><strong>Sol.<\/strong> Required average speed = ((2xy)\/(x+y)) km \/ hr<\/p>\n<p>=(2 x 84 x 56)\/(84+56)km\/hr<\/p>\n<p>\u00a0= (2*84*56)\/140 km\/hr<\/p>\n<p>=67.2 km\/hr.<\/p>\n<p data-start=\"0\" data-end=\"284\">Accessing copyrighted materials, such as the &#8220;Quantitative Aptitude&#8221; book by R.S. Aggarwal, through unauthorized free downloads is both illegal and unethical. To obtain this resource legally and ensure you have the most accurate and up-to-date content, consider the following options:<\/p>\n<p data-start=\"286\" data-end=\"506\"><strong data-start=\"286\" data-end=\"308\">Official Purchase:<\/strong> You can purchase the latest edition of &#8220;Quantitative Aptitude&#8221; by R.S. Aggarwal from reputable retailers. This ensures you receive a legitimate copy with all the necessary content for your studies.<\/p>\n<p data-start=\"508\" data-end=\"692\"><strong data-start=\"508\" data-end=\"527\">Free Resources:<\/strong> While the complete book may not be legally available for free, there are authorized resources that offer practice questions and explanations on the &#8220;Average&#8221; topic:<\/p>\n<ul data-start=\"694\" data-end=\"941\">\n<li data-start=\"694\" data-end=\"941\"><strong data-start=\"696\" data-end=\"716\">The Company Boy:<\/strong> This platform provides practice questions and explanations on &#8220;Average&#8221; based on R.S. Aggarwal&#8217;s content. It&#8217;s a valuable resource for understanding the concepts and practicing problems.<\/li>\n<\/ul>\n<p data-start=\"943\" data-end=\"1177\">Investing in official and authorized study materials not only ensures you receive high-quality and accurate content but also supports the authors and publishers who have dedicated their efforts to creating these educational resources.<\/p>\n<h3 data-start=\"943\" data-end=\"1177\"><a href=\"https:\/\/eltsindia.com\/StudyMaterialFiles\/ea007acc-bc55-4e17-8470-0d9e85313252quantitative-aptitude-for-competitive-examinations-by-rs-aggarwal-reprint-2017.pdf\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Quantitative Aptitude PDF Free download: AVERAGE<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>6.AVERAGE Ex.1:Find the average of all prime numbers between 30 and 50? Sol: there are five prime numbers between 30 and 50. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 They are 31,37,41,43 and 47. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Therefore the required average=(31+37+41+43+47)\/5\u00a0 \u00f3199\/5\u00a0 \u00f3 39.8. Ex.2. find the average of first 40 natural numbers? Sol:\u00a0\u00a0\u00a0 sum\u00a0 of first n natural numbers=n(n+1)\/2; \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 So,sum of [&hellip;]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2443],"tags":[2429,2430,2431,2432,2433,2434,2435,2436,2437,2438,2439,2440,2441,2442,1923,2444],"class_list":["post-5900","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude-pdf-free-download","tag-arithmetic-mean","tag-average","tag-average-calculator","tag-average-female-height","tag-average-formula","tag-average-height","tag-average-height-for-men","tag-average-height-for-women","tag-average-male-height","tag-cumulative-gpa-calculator","tag-gpa-meaning","tag-mean-median-mode-calculator","tag-mean-median-mode-range","tag-median-calculator","tag-rs-aggarwal-quantitative-aptitude","tag-weighted-grade-calculator"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5900","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=5900"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5900\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5900"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5900"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5900"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}