{"id":5976,"date":"2025-06-09T06:36:54","date_gmt":"2025-06-09T06:36:54","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5976"},"modified":"2025-06-09T06:36:54","modified_gmt":"2025-06-09T06:36:54","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-ratio-proportion-and","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-ratio-proportion-and\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free Download: RATIO AND PROPORTION"},"content":{"rendered":"

RATIO AND PROPORTION<\/strong><\/h1>\n

IMPORTANT FACTS <\/u><\/strong>AND FORMULAE<\/u><\/strong><\/h2>\n

\u00a0<\/strong>RATIO:<\/strong> The ratio of two quantities a and b in the same units, is the fraction a\/b and we write it as a:b.<\/p>\n

In the ratio a:b, we call a as the first term or antecedent<\/strong> and b, the second\u00a0 term or consequent.<\/strong><\/p>\n

Ex.<\/strong> The ratio 5: 9 represents 5\/9 with antecedent = 5, consequent = 9.<\/p>\n

Rule:<\/strong> The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.<\/p>\n

Ex.\u00a0 <\/strong>4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.<\/p>\n

    \n
  1. \n

    PROPORTION: The equality of two ratios is called proportion.<\/em><\/strong><\/h2>\n<\/li>\n<\/ol>\n

    If a:<\/em> b = c: d, we write, a:<\/em> b:: c : d and we say that a, <\/em>b, c, d are in proportion . Here a <\/em>and d are called extremes, while b and c are called mean terms.<\/p>\n

    \u00a0Product of means = Product of extremes.<\/p>\n

    Thus, a: b:: c : d <=> (b x c) = (a x d).<\/p>\n

      \n
    1. (i) Fourth Proportional: If a <\/em>: b = c: d, then d is called the fourth proportional<\/li>\n<\/ol>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 to a, <\/em>b, c.<\/p>\n

      \u00a0\u00a0 (ii) <\/em>Third Proportional: If a<\/em>: b = b: c, then c is called the third proportional to<\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a and b.<\/p>\n

      \u00a0\u00a0 (iii) Mean Proportional: Mean proportional between a <\/em>and b <\/em>is square root of ab<\/em><\/p>\n

        \n
      1. (i) COMPARISON OF RATIOS:<\/li>\n<\/ol>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 We say that (a<\/em>: b) <\/em>> (c: d) <\/em><=>\u00a0 (a\/b)>(c \/d).<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0 (ii) COMPOUNDED RATIO:<\/p>\n

        \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The compounded ratio of the ratios (a: b), <\/em>(c: d), <\/em>(e : f) is (ace<\/em>: bdf)<\/p>\n

          \n
        1. (i) Duplicate ratio <\/em>of (a <\/em>: b) is (a2<\/sup> <\/em>: b2<\/sup>).<\/li>\n<\/ol>\n

          \u00a0\u00a0\u00a0\u00a0\u00a0 (ii) Sub-duplicate ratio <\/em>of (a : b) <\/em>is (<\/em>\u221aa <\/em>: \u221ab).<\/em><\/p>\n

          \u00a0\u00a0\u00a0\u00a0 (iii)Triplicate ratio <\/em>of (a : b) <\/em>is (a3<\/sup> <\/em>: b3<\/sup>).<\/p>\n

          \u00a0\u00a0\u00a0\u00a0 (iv) <\/em>Sub-triplicate ratio of (a : b) <\/em>is (a<\/em> \u2153 : b<\/em> \u2153 ).<\/p>\n

          \u00a0\u00a0\u00a0\u00a0 (v) <\/em>If (a\/b)=(c\/d), then\u00a0 ((a+b)\/(a-b))=((c+d)\/(c-d))\u00a0\u00a0\u00a0 (Componendo and dividendo)<\/strong><\/p>\n

            \n
          1. VARIATION:<\/li>\n<\/ol>\n

            (i) We say that x is directly proportional to y, <\/em>if x = ky\u00a0 <\/em>for some constant k and<\/p>\n

            \u00a0\u00a0\u00a0\u00a0 we write, x \u00b5 y.<\/em><\/p>\n

            (ii) <\/em>We say that x is inversely proportional to y, <\/em>if xy = k <\/em>for some constant k and<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 we write, x\u221e(1\/y)<\/p>\n

            SOLVED PROBLEMS<\/h1>\n

            Ex. 1. If <\/em>a : b <\/em>= 5 : 9 and <\/em>b : c = 4: 7, find <\/em>a : b : c.<\/strong><\/p>\n

            Sol.<\/strong>\u00a0 a:b=5:9 and b:c=4:7= (4X9\/4): (7×9\/4) = 9:63\/4<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a:b:c = 5:9:63\/4 =20:36:63.<\/em><\/p>\n

            Ex. 2. Find:<\/em><\/strong><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (i) the fourth proportional <\/em>to 4, 9, 12;<\/strong><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 (ii) the third proportional <\/em>to 16 and <\/em>36;<\/strong><\/p>\n

            \u00a0\u00a0\u00a0\u00a0 iii) the <\/em><\/strong>mean proportional between 0.08 and 0.18.<\/em><\/strong><\/p>\n

            \u00a0<\/em><\/strong>Sol.<\/strong><\/p>\n

            \u00a0 \u00a0 \u00a0 <\/em>i) <\/em>Let the fourth proportional to 4, 9, 12 be x.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, 4 : 9 : : 12 : x\u00a0 \u00f34 x x=9×12 \u00f3 X=(9 x 12)\/14=27;<\/p>\n

            \u00a0\u00a0 Fourth proportional to 4, 9, 12 is 27.<\/p>\n

            \u00a0(ii) <\/em>Let the third proportional to 16 and 36 be x.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 Then, 16 : 36 : : 36 : x\u00a0 \u00f316 x x <\/em>= 36 x 36 \u00f3 x=(36 x 36)\/16 =81<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 Third proportional to 16 and 36 is 81.<\/p>\n

            (iii) <\/em>Mean proportional between 0.08 and 0.18<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 \u00d60.08 x 0.18 =\u00d68\/100 x 18\/100= \u00d6144\/(100 x 100)=12\/100=0.12<\/p>\n

            Ex. 3. If <\/em>x : y <\/em>= 3 : 4, find <\/em>(4x + 5y) <\/em>: (5x – 2y).<\/em><\/strong><\/p>\n

            \u00a0<\/em><\/strong>Sol.<\/strong>\u00a0 X\/Y=3\/4 \u00f3 (4x+5y)\/(5x+2y)= (4( x\/y)+5)\/(5 (x\/y)-2) =(4(3\/4)+5)\/(5(3\/4)-2)<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =(3+5)\/(7\/4)=32\/7<\/p>\n

            \u00a0<\/em>Ex. 4. Divide Rs. <\/em>672 in the ratio <\/em>5 : 3<\/strong>.<\/p>\n

            Sol.<\/strong> Sum of ratio terms = (5 + 3) = 8.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 First part = Rs. (672 x (5\/8)) = Rs. 420; Second part = Rs. (672 x (3\/8)) = Rs. 252.<\/p>\n

            Ex. 5. Divide Rs. <\/em>1162 among A, B, <\/em>C in the ratio <\/em>35 : 28 : 20.<\/em><\/strong><\/p>\n

            Sol.<\/strong> Sum of ratio terms = (35 + 28 + 20) = 83.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A’s share = Rs. (1162 x (35\/83))= Rs. 490; B’s share = Rs. (1162 x(28\/83))= Rs. 392;<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C’s share = Rs. (1162 x (20\/83))= Rs. 280.<\/p>\n

            Ex. 6. A bag contains <\/em>50 p, <\/em>25 P and <\/em>10 p coins in the ratio <\/em>5: 9: 4, amounting <\/em>to<\/strong><\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Rs. 206. Find the <\/em><\/strong>number of coins <\/em>of each type.<\/em><\/strong><\/p>\n

            \u00a0<\/em><\/strong>\u00a0Sol.<\/strong> Let the number of 50 p, 25 P and 10 p coins be 5x, 9x and 4x respectively.<\/p>\n

            \u00a0\u00a0 \u00a0\u00a0\u00a0 (5x\/2)+( 9x\/ 4)+(4x\/10)=206<\/em>\u00f3<\/em> 50x + 45x <\/em>+ 8x = 4120\u00f31O3x = 4120 \u00f3x=40.<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 Number of 50 p coins = (5 x 40) = 200; Number of 25 p coins = (9 x 40) = 360;<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0 Number of 10 p coins = (4 x 40) = 160.<\/p>\n

            \u00a0Ex. 7. A <\/em>mixture contains alcohol <\/em>and water <\/em>in the ratio <\/em>4 : 3. If <\/em>5 litres of water is <\/em>added to the\u00a0\u00a0\u00a0\u00a0 mixture, the ratio becomes <\/em>4: 5. Find the quantity of alcohol <\/em>in the given mixture<\/strong><\/p>\n

            Sol.<\/strong> Let the quantity of alcohol and water be 4x litres and 3x litres respectively<\/p>\n

            \u00a0\u00a0\u00a0 4x\/(3x+5)=4\/5 \u00f320x=4(3x+5)\u00f38x=20 \u00f3x=2.5<\/p>\n

            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Quantity of alcohol = (4 x 2.5) litres = 10 litres.<\/p>\n

            “Quantitative Aptitude” by R.S. Aggarwal is a comprehensive resource for various mathematical topics, including “Ratio and Proportion.” While accessing the complete book in PDF format for free may infringe upon copyright laws, there are legitimate ways to study this specific chapter:<\/p>\n

              \n
            1. \n

              Online Practice Questions<\/strong>:<\/p>\n

                \n
              • Websites like provide practice questions specifically on “Ratio and Proportion” from R.S. Aggarwal’s book. This allows you to practice relevant problems directly online.<\/li>\n<\/ul>\n<\/li>\n
              • \n

                Educational Platforms<\/strong>:<\/p>\n

                  \n
                • Platforms such as \u00a0offer solutions and explanations for exercises related to “Ratio and Proportion.” This can help you understand the methodology behind solving these problems.<\/li>\n<\/ul>\n<\/li>\n
                • \n

                  Purchase the Book<\/strong>:<\/p>\n

                    \n
                  • For comprehensive coverage and practice, consider purchasing the book through authorized sellers. This ensures you have access to all topics and practice exercises.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n

                    Please be cautious of unauthorized websites offering free downloads of the complete book, as they may violate copyright laws and could pose security risks. Utilizing authorized resources not only supports the creators but also guarantees the accuracy and quality of the study materials you use in your preparation.<\/p>\n

                    RS Aggarwal Quantitative Aptitude PDF Free Download: RATIO AND PROPORTION<\/h3>\n","protected":false},"excerpt":{"rendered":"

                    RATIO AND PROPORTION IMPORTANT FACTS AND FORMULAE \u00a0RATIO: The ratio of two quantities a and b in the same units, is the fraction a\/b and we write it as a:b. In the ratio a:b, we call a as the first term or antecedent and b, the second\u00a0 term or consequent. Ex. The ratio 5: 9 […]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[126,127],"tags":[],"class_list":["post-5976","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude","category-rs-aggarwal-quantitative-aptitude-pdf"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5976","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=5976"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5976\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5976"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5976"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5976"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}