{"id":5899,"date":"2025-06-09T07:03:10","date_gmt":"2025-06-09T07:03:10","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5899"},"modified":"2025-06-09T07:03:10","modified_gmt":"2025-06-09T07:03:10","slug":"rs-aggarwal-quantitative-aptitude-pdf-download-square-roots-and-roots-cube","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-download-square-roots-and-roots-cube\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free download: SQUARE ROOTS AND CUBE ROOTS"},"content":{"rendered":"<h1 style=\"text-align: center\">SQUARE ROOTS AND CUBE ROOTS<\/h1>\n<h1><strong><u>IMPORTANT FACTS AND FORMULAE<\/u><\/strong><\/h1>\n<p><strong>Square Root:<\/strong> If x<sup>2<\/sup> = <em>y, <\/em>we say that the square root of <em>y <\/em>is x and we write, \u221ay = <em>x.<\/em><\/p>\n<p>Thus, \u221a4 = 2, \u221a9 = 3, \u221a196 = 14.<\/p>\n<p><strong>Cube Root:<\/strong> The cube root of a given number x is the number whose cube is x. We denote the cube root of x by <em><sup>3<\/sup>\u221ax.<\/em><\/p>\n<p>Thus, <sup>3<\/sup>\u221a8\u00a0 = <sup>3<\/sup>\u221a2 x 2 x 2 = 2, <sup>3<\/sup>\u221a343 = <sup>3<\/sup>\u221a7 x 7 x 7 = 7 etc.<\/p>\n<p><strong>Note:<\/strong><\/p>\n<p>1.\u221axy = \u221ax * \u221ay\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2. \u221a(x\/y) = \u221ax \/ \u221ay\u00a0 = (\u221ax \/ \u221ay) * (\u221ay \/ \u221ay) = \u221axy \/ y<\/p>\n<h2><strong>SOLVED EXAMPLES<\/strong><\/h2>\n<p><strong><em>Ex. 1. Evaluate \u221a6084 by factorization method .<\/em><\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0\u00a0 Method:<\/strong> Express the given number as the product of prime factors.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0 6084\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, take the product of these prime factors choosing one out of \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0 3042<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 every pair of the same primes. This product gives the square root\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0\u00a0 1521\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 of the given number.\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\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 3\u00a0\u00a0\u00a0 507<\/p>\n<p>Thus, resolving 6084 into prime factors, we get:\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 13\u00a0\u00a0 169<\/p>\n<p>6084 = 2<sup>2<\/sup> x 3<sup>2<\/sup> x 13<sup>2\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\u00a0\u00a0\u00a0\u00a0 <\/sup>\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 \u00a0\u00a0\u00a0\u00a013\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\ \u221a<em>6084 <\/em>= (2 x 3 x 13) = 78.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p><strong><em>Ex. 2. Find the square root of 1471369.\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\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><\/p>\n<p><strong> Sol.\u00a0\u00a0 \u00a0 Explanation:<\/strong> In the given number, mark off the digits\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00a0 1471369 (1213<\/p>\n<p>in pairs starting from the unit&#8217;s digit. Each pair and\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>the remaining one digit is called a period.\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 22\u00a0\u00a0\u00a0\u00a0 47<\/p>\n<p>Now, 1<sup>2<\/sup> = 1. On subtracting, we get 0 as remainder.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 44<\/p>\n<p>Now, bring down the next period <em>i.e., <\/em>47.\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 241\u00a0\u00a0\u00a0\u00a0\u00a0 313<\/p>\n<p>Now, trial divisor is 1 x 2 = 2 and trial dividend is 47.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 241<\/p>\n<p>So, we take 22 as divisor and put 2 as quotient.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2423\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7269<\/p>\n<p>The remainder is 3.\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\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 7269<\/p>\n<p>Next, we bring down the next period which is 13.\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 x<\/p>\n<p>Now, trial divisor is 12 x 2 = 24 and trial dividend is<\/p>\n<ol start=\"313\">\n<li>So, we take 241 as dividend and 1 as quotient.<\/li>\n<\/ol>\n<p>The remainder is 72. \u00ad<\/p>\n<p>Bring down the next period <em>i.e., <\/em>69.<\/p>\n<p>Now, the trial divisor is 121 x 2 = 242 and the trial<\/p>\n<p>dividend is 7269. So, we take 3as quotient and 2423<\/p>\n<p>as divisor. The remainder is then zero.<\/p>\n<p>Hence, \u221a<em>1471369 <\/em>= 1213.<\/p>\n<p><strong><em>Ex. 3. Evaluate: <\/em><\/strong><strong><em>\u221a248 + <\/em><\/strong><strong><em>\u221a51 + \u221a 169 . <\/em><\/strong><\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0 Given expression = \u221a<em>248 <\/em>+ \u221a<em>51 <\/em>+ 13 = \u221a<em>248 <\/em>+ \u221a64\u00a0\u00a0\u00a0 = \u221a <em>248 <\/em>+ 8 = \u221a256 = 16.<\/p>\n<p>Ex. 4. If a * b * c = \u221a(a + 2)(b + 3) \/ (c + 1), find the value of 6 * 15 * 3.<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 6 * 15 * 3 = \u221a(6 + 2)(15 + 3) \/ (3 + 1) = \u221a8 * 18 \/ 4 = \u221a144 \/ 4 = 12 \/ 4 = 3.<\/p>\n<p><strong>Ex. 5. <em>Find the value of \u221a25\/16.<\/em><\/strong><\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0 \u221a 25 \/ 16\u00a0\u00a0 = \u221a 25 \/ \u221a 16 = 5 \/ 4<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex. 6. <em>What is the square root of 0.0009?<\/em><\/strong><\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a0.0009= \u221a 9 \/ 1000\u00a0 = 3 \/ 100 = 0.03.<\/p>\n<p><strong>Ex. 7. <em>Evaluate \u221a175.2976. <\/em><\/strong><\/p>\n<p><strong>Sol<\/strong>. \u00a0\u00a0\u00a0\u00a0 Method: We make even number of decimal places\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0 175.2976 (13.24\u00a0\u00a0<\/p>\n<p>by affixing a zero, if necessary. Now, we mark off\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1<\/p>\n<p>periods and extract the square root as shown.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 23\u00a0\u00a0\u00a0\u00a0 75\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/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 \u00a0 \u00a0 \u00a0 69<\/p>\n<p>\\\u221a175.2976 = 13.24\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 262\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 629<\/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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0524<\/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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2644\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10576<\/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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010576<\/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\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\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 x<\/p>\n<p><strong>Ex. 8. <em>What will come <\/em>in <em>place of question mark <\/em>in <em>each of the following <\/em>questions? <\/strong><\/p>\n<p><strong><em>(i) \u221a32.4 \/ ?\u00a0 <\/em>= 2<em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) \u221a86.49 <\/em>+ \u221a 5 + ( ? )<sup>2<\/sup> = 12.3.<\/strong><\/p>\n<p><strong> Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 (i) Let \u221a<em>32.4 \/ x <\/em>= 2. Then, 32.4\/x = 4 &lt;=&gt; <em>4x <\/em>= 32.4 &lt;=&gt; <em>x <\/em>= 8.1.<em>\u00a0 \u00a0 \u00a0 \u00a0\u00a0<\/em><\/p>\n<p>(ii) Let \u221a<em>86.49 <\/em>+ \u221a5 + <em>x<sup>2<\/sup> <\/em>= 12.3.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0 Then, <em>9.3 + \u221a5+x<sup>2<\/sup> <\/em>= 12.3 &lt;=&gt; \u221a<em>5+x<sup>2\u00a0 <\/sup><\/em>= 12.3 &#8211; 9.3 = 3<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0 &lt;=&gt; <em>5 + x<sup>2 <\/sup>= 9\u00a0\u00a0 <\/em>&lt;=&gt; x<sup>2 <\/sup>= 9 &#8211; 5= 4\u00a0\u00a0 &lt;=&gt;\u00a0\u00a0 <em>x = \u221a4 = 2. <\/em><strong><em>\u00a0<\/em><\/strong><\/p>\n<p><strong><em>Ex.9. Find the value of \u221a 0.289 \/ 0.00121.<\/em><\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a0.289 \/ 0.00121 = \u221a0.28900\/0.00121 = \u221a28900\/121 = 170 \/ 11.<\/p>\n<p><strong><em>Ex.10. If \u221a1 + (x \/ 144) = 13 \/ 12, the find the value of x.<\/em><\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a<\/strong>1 + (x \/ 144) = 13 \/ 12 \u00de ( 1 + (x \/ 144)) = (13 \/ 12 )<sup>2 <\/sup>= 169 \/ 144<\/p>\n<p>\u00a0 \u00dex \/ 144 = (169 \/ 144) &#8211; 1<\/p>\n<p>\u00a0 \u00dex \/ 144 = 25\/144 \u00de x = 25.<\/p>\n<p><strong><em>Ex. 11. Find the value of \u221a3 up to three places of decimal.<\/em><\/strong><\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0 1\u00a0\u00a0\u00a0 3.000000\u00a0\u00a0 (1.732<\/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 1<\/p>\n<p>27\u00a0\u00a0\u00a0 200<\/p>\n<p>189<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 343\u00a0\u00a0\u00a0\u00a0\u00a0 1100<\/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 1029<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3462\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7100<\/p>\n<p>6924\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 \\\u221a3 = 1.732.<\/p>\n<p><strong>Ex. 12. <em>If \u221a3 <\/em>= 1.732, <em>find the value of \u221a<\/em>192 &#8211; <u>1 <\/u>\u221a48<em> &#8211; \u221a75<\/em> <em>correct to <\/em>3 <em>places<\/em><\/strong><\/p>\n<p><strong><em>\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 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em><\/strong><\/p>\n<p><strong><em>of decimal.\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 <\/em>\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 (S.S.C. 2004)<\/strong><\/p>\n<p><strong> Sol. \u00a0\u00a0\u00a0 <\/strong><em>\u221a<\/em>192 &#8211; (1 \/ 2)\u221a48<em> &#8211; \u221a75 = \u221a64 * 3 &#8211; (1\/2) \u221a 16 * 3\u00a0 &#8211; \u221a 25 * 3<\/em><\/p>\n<p><em>\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 =8\u221a3 &#8211; (1\/2) * 4\u221a3 &#8211; 5\u221a3<\/em><\/p>\n<p><em>\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 =3\u221a3 &#8211; 2\u221a3 = \u221a3 = 1.732<\/em><\/p>\n<p><strong>Ex. 13. Evaluate: \u221a(9.5 * 0.0085 * 18.9) \/ (0.0017 * 1.9 * 0.021)<\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Given exp. = \u221a(9.5 * 0.0085 * 18.9) \/ (0.0017 * 1.9 * 0.021)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, since the sum of decimal places in the numerator and denominator under the \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 radical sign is the same, we remove the decimal.<\/p>\n<p>\\\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Given exp = \u221a(95 * 85 * 18900) \/ (17 * 19 * 21) = \u221a 5 * 5 * 900\u00a0 = 5 * 30 = 150.<\/p>\n<p><strong>Ex. 14. <em>Simplify<\/em>: \u221a <em>[( 12.1 )<sup>2<\/sup> &#8211; (8.<\/em>1)<sup>2<\/sup>] \/ [(0.25)<sup>2<\/sup> + (0.25)(19.95)]<\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Given exp. = \u221a [(12.1 + 8.1)(12.1 &#8211; 8.1)] \/ [(0.25)(0.25 + 19.95)]<\/p>\n<p>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 =\u221a (20.2 * 4) \/( 0.25 * 20.2)\u00a0\u00a0 = \u221a 4 \/ 0.25 = \u221a400 \/ 25 = \u221a16 = 4.<\/p>\n<p><strong>Ex. 15. <em>If x <\/em>= 1 + \u221a2<em> and <\/em>y = 1 &#8211; \u221a2<em>, find the value of (x<sup>2<\/sup> <\/em>+ y<sup>2<\/sup>).<\/strong><\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 <em>x<sup>2<\/sup> <\/em>+ <em>y<sup>2<\/sup> <\/em>= (1 + \u221a2)<sup>2 <\/sup>+ (1 &#8211; \u221a2)<sup>2<\/sup> = 2[(1)<sup>2 <\/sup>+ (\u221a2)<sup>2<\/sup>] = 2 * 3 = 6.<\/p>\n<p><strong>Ex. 16. <em>Evaluate: \u221a0.9 up to 3 places of decimal.<\/em><\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n<p>9\u00a0\u00a0\u00a0 0.900000(0.948<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 81<\/p>\n<p>184\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 900<\/p>\n<p>736<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1888\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16400<\/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 15104\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 \\\u221a0.9 = 0.948<\/p>\n<p><strong><em>Ex.17. If \u221a15 = 3.88, find the value of \u221a (5\/3).<\/em><\/strong><\/p>\n<p><strong>Sol.<\/strong><strong>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>\u221a (5\/3) = \u221a(5 * 3) \/ (3 * 3)\u00a0 = \u221a15 \/ 3 = 3.88 \/ 3 = 1.2933\u2026. = 1.293.<\/p>\n<p>Ex. 18. Find the least square number which is exactly divisible by 10,12,15 and 18.<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 L.C.M. of 10, 12, 15, 18 = 180. Now, 180 = 2 * 2 * 3 * 3 *5 = 2<sup>2<\/sup> * 3<sup>2 <\/sup>* 5.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 To make it a perfect square, it must be multiplied by 5.<\/p>\n<p>\\\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Required number = (2<sup>2<\/sup> * 3<sup>2<\/sup> * 5<sup>2<\/sup>) = 900.<\/p>\n<p>Ex. 19. Find the greatest number of five digits which is a perfect square.<\/p>\n<p>(R.R.B. 1998)<\/p>\n<p><strong>Sol<\/strong>.\u00a0\u00a0\u00a0\u00a0\u00a0 Greatest number of 5 digits is 99999.<\/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\u00a0\u00a0\u00a03\u00a0\u00a0\u00a0 99999(316<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0 9<\/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\u00a0\u00a0 61\u00a0\u00a0\u00a0\u00a0\u00a0 99<\/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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 61<\/p>\n<p>626 \u00a0\u00a0\u00a0\u00a0 3899<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3756<\/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 143<\/p>\n<ul>\n<li>Required number == (99999 &#8211; 143) = 99856.<\/li>\n<\/ul>\n<p><strong><em>Ex. 20. Find the smallest number that must be added to 1780 to make it a perfect<\/em><\/strong><\/p>\n<p><strong><em>square<\/em><\/strong><em>.<\/em><\/p>\n<p><strong>Sol. <\/strong><\/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 4\u00a0\u00a0\u00a0 1780 (42<\/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 16<\/p>\n<p>82\u00a0\u00a0\u00a0\u00a0\u00a0 180\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>164<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0 \\\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Number to be added = (43)<sup>2<\/sup> &#8211; 1780 = 1849 &#8211; 1780 = 69.<\/p>\n<p><em>\u00a0<\/em>Ex. 21. \u221a2 = 1.4142, find the value of \u221a2 \/ (2 + \u221a2).<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 \u221a2 \/ (2 + \u221a2) = \u221a2 \/ (2 + \u221a2) * (2 &#8211; \u221a2) \/ (2 &#8211; \u221a2) = (2\u221a2 \u2013 2) \/ (4 \u2013 2)<\/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 = 2(\u221a2 \u2013 1) \/ 2 = \u221a2 \u2013 1 = 0.4142.<\/p>\n<ol start=\"22\">\n<li><strong><em> If x = (\u221a5 + \u221a3) \/ (\u221a5 &#8211; \u221a3) and y = (\u221a5 &#8211; \u221a3) \/ (\u221a5 + \u221a3), find the value of (x<sup>2 <\/sup>+ y<sup>2<\/sup>).<\/em><\/strong><\/li>\n<\/ol>\n<p><strong>Sol.\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong><\/p>\n<p><em>x <\/em>= [(\u221a5 + \u221a3) \/ (\u221a5 &#8211; \u221a3)] * [(\u221a5 + \u221a3) \/ (\u221a5 + \u221a3)] = (\u221a5 + \u221a3)<sup>2<\/sup> \/ (5 &#8211; 3)<\/p>\n<p>\u00a0\u00a0 =(5 + 3 + 2\u221a15) \/ 2 = 4 + \u221a15.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <em>y <\/em>= [(\u221a5 &#8211; \u221a3) \/ (\u221a5 + \u221a3)] * [(\u221a5 &#8211; \u221a3) \/ (\u221a5 &#8211; \u221a3)] = (\u221a5 &#8211; \u221a3)<sup>2<\/sup> \/ (5 &#8211; 3)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 =(5 + 3 &#8211; 2\u221a15) \/ 2 = 4 &#8211; \u221a15.<\/p>\n<p>\\\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 x<sup>2 <\/sup>+ y<sup>2<\/sup> = (4 + \u221a15)<sup>2 <\/sup>+ (4 &#8211; \u221a15)<sup>2<\/sup> = 2[(4)<sup>2<\/sup> + (\u221a15)<sup>2<\/sup>] = 2 * 31 = 62.<\/p>\n<p>Ex. 23. Find the cube root of 2744.<\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0 Method:<\/strong> Resolve the given number as the product\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0 2744<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 of prime factors and take the product of prime\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0 1372<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 factors, choosing one out of three of the same\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u00a0\u00a0\u00a0\u00a0\u00a0 686\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 prime factors. Resolving 2744 as the product of\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7\u00a0 \u00a0\u00a0\u00a0 343<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 prime factors, we get:\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7\u00a0 \u00ad\u00a0\u00a0\u00a0\u00a0\u00a0 49<\/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\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\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2744 = 2<sup>3<\/sup> x 7<sup>3<\/sup>.\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\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<\/p>\n<ul>\n<li><sup>3<\/sup>\u221a2744= 2 x 7 = 14.<\/li>\n<\/ul>\n<p>Ex. 24. By what least number 4320 be multiplied to obtain a number which is a perfect cube?<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Clearly, 4320 = 2<sup>3<\/sup> * 3<sup>3<\/sup> * 2<sup>2<\/sup> * 5.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 To make it a perfect cube, it must be multiplied by 2 * 5<sup>2 <\/sup>i.e,50.<\/p>\n<p>I understand you&#8217;re seeking a free PDF of the &#8220;Square Roots and Cube Roots&#8221; chapter from R.S. Aggarwal&#8217;s &#8220;Quantitative Aptitude&#8221; book. While I cannot provide direct downloads of copyrighted material, I can guide you to resources that offer practice questions and solutions on this topic.<\/p>\n<p><strong>Available Resources:<\/strong><\/p>\n<ol>\n<li>\n<p><strong>Embibe:<\/strong><\/p>\n<ul>\n<li>Embibe provides practice questions and solutions for the &#8220;Square Roots and Cube Roots&#8221; chapter from R.S. Aggarwal&#8217;s book.<\/li>\n<li>Access the exercises here:<\/li>\n<\/ul>\n<\/li>\n<li>\n<p><strong>Doubtnut:<\/strong><\/p>\n<ul>\n<li>Doubtnut offers solutions to various chapters from R.S. Aggarwal&#8217;s mathematics books, including &#8220;Square Roots and Cube Roots.&#8221;<\/li>\n<li>Explore the solutions here:<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<p><strong>Alternative Study Materials:<\/strong><\/p>\n<p>If you&#8217;re looking for additional study materials on square roots and cube roots, consider the following:<\/p>\n<ol>\n<li>\n<p><strong>NCERT Mathematics Textbooks:<\/strong><\/p>\n<ul>\n<li>The National Council of Educational Research and Training (NCERT) provides free access to mathematics textbooks that cover topics like square roots and cube roots.<\/li>\n<li>Access the Class 8 Mathematics textbook here: <a target=\"_new\" rel=\"noopener\">NCERT Class 8 Maths<\/a><\/li>\n<\/ul>\n<\/li>\n<li>\n<p><strong>Khan Academy:<\/strong><\/p>\n<ul>\n<li>Khan Academy offers free online lessons and practice exercises on square roots and cube roots.<\/li>\n<li>Visit their website:\u00a0<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<p>These resources should help you strengthen your understanding of square roots and cube roots. If you have any other questions or need further assistance, feel free to ask!<\/p>\n<h3>RS Aggarwal Quantitative Aptitude PDF Free download: SQUARE ROOTS AND CUBE ROOTS<\/h3>\n","protected":false},"excerpt":{"rendered":"<p>SQUARE ROOTS AND CUBE ROOTS IMPORTANT FACTS AND FORMULAE Square Root: If x2 = y, we say that the square root of y is x and we write, \u221ay = x. Thus, \u221a4 = 2, \u221a9 = 3, \u221a196 = 14. Cube Root: The cube root of a given number x is the number whose [&hellip;]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[2410,2411,2241,2412,2413,2414,2415,2416,2417,1923,2418,2337,2338,2419,2420,2421,2422,2423,2424,2425,2426,2427,2428],"class_list":["post-5899","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude-pdf","tag-1-to-30-square-and-cube","tag-cubic-root-of-64","tag-quantitative-aptitude-for-competitive-examinations","tag-rs-agarwal-aptitude","tag-rs-agarwal-aptitude-book","tag-rs-aggarwal-aptitude","tag-rs-aggarwal-aptitude-book","tag-rs-aggarwal-aptitude-book-price","tag-rs-aggarwal-competition-book","tag-rs-aggarwal-quantitative-aptitude","tag-rs-aggarwal-quantitative-aptitude-book","tag-rs-aggarwal-quantitative-aptitude-for-competitive-examinations","tag-rs-aggarwal-quantitative-aptitude-price","tag-s-chand-quantitative-aptitude","tag-square-and-cube","tag-square-and-cube-1-to-30","tag-square-and-cube-1-to-50","tag-square-and-cube-roots","tag-square-root-and-cube-root-formula","tag-square-roots-and-cube-roots","tag-squares-and-cubes-till-30","tag-squares-and-cubes-upto-30","tag-the-square-root-of-64-divided-by-the-cube-root-of-64-is"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5899","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=5899"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5899\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5899"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5899"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5899"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}