{"id":3406,"date":"2022-12-13T04:33:36","date_gmt":"2022-12-13T04:33:36","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=3406"},"modified":"2023-01-31T06:50:52","modified_gmt":"2023-01-31T06:50:52","slug":"ncert-solutions-class-8-maths-chapter-13-direct-and-inverse-proportions-ex-13-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-13-direct-and-inverse-proportions-ex-13-1\/","title":{"rendered":"NCERT Solutions Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 8 Mathematics\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 13 (Direct and Inverse Proportions)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 8 Mathematics <strong>Chapter &#8211; 13 Direct and Inverse Proportions <\/strong>Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<h4><span style=\"color: #000000;\"><strong>Chapter 13: Direct and Indirect Proportions<\/strong><\/span><\/h4>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-maths-chapter-13-direct-and-inverse-proportions-ex-13-2\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 8 Maths Ex &#8211; 13.2<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 13.1<\/span><\/h2>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">1. Following are the car parking charges near a railway station upto:<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">4 hours\u00a0 \u00a0 \u00a0 \u00a0 Rs.60<br \/>\n<\/span><span style=\"color: #000000;\">8 hours\u00a0 \u00a0 \u00a0 \u00a0 Rs.100<br \/>\n<\/span><span style=\"color: #000000;\">12 hours\u00a0 \u00a0 \u00a0 Rs.140<br \/>\n<\/span><span style=\"color: #000000;\">24 hours\u00a0 \u00a0 \u00a0Rs.180<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3429\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-1.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"208\" height=\"109\" \/><br \/>\n<\/span><span style=\"color: #000000;\">Check if the parking charges are in direct proportion to the parking time.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span>Let&#8217;s find the parking charge for 1 hour in all the four cases shown below:<br \/>\n60\/4 = 15\/1 = \u20b915<br \/>\n100\/8 = 25\/2 = \u20b912.5<br \/>\n140\/12 = 35\/3 = \u20b911.67<br \/>\n180\/24 = 15\/2 = \u20b97.50<br \/>\nWe see that the 1 hour charge in all the four cases shown below is not equal.<br \/>\nThus, the parking charges are not in direct proportion.<\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3430\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-2.jpg\" alt=\"NCERT Maths Solutions Class 8\" width=\"374\" height=\"80\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-2.jpg 374w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-2-300x64.jpg 300w\" sizes=\"auto, (max-width: 374px) 100vw, 374px\" \/><br \/>\n<\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">Solution &#8211; <\/span><\/strong><span style=\"color: #000000;\">Let the ratio of parts of red pigment and parts of base be <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a}{b}\" alt=\"\\frac{a}{b}\" align=\"absmiddle\" \/>.<br \/>\n<\/span><span style=\"color: #000000;\"><strong>Case 1:<\/strong> Here, a<sub>1\u00a0<\/sub>= 1, b<sub>1\u00a0<\/sub>= 8<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_1}{b_1}\" alt=\"\\frac{a_1}{b_1}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{8}\" alt=\"\\frac{1}{8}\" align=\"absmiddle\" \/> =\u00a0 <\/span><span style=\"color: #000000;\">k (say)<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case 2:<\/strong> When\u00a0 a<sub>2\u00a0<\/sub>= 4 , b<sub>2\u00a0<\/sub>= ?<br \/>\nk = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_2}{b_2}\" alt=\"\\frac{a_2}{b_2}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">b<sub>2<\/sub> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_2}{k}\" alt=\"\\frac{a_2}{k}\" align=\"absmiddle\" \/>\u00a0 = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{4}{\\frac{1}{8}}\" alt=\"\\frac{4}{\\frac{1}{8}}\" align=\"absmiddle\" \/> = 4 \u00d7 8 = 32<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case 3:<\/strong> When\u00a0 a<sub>3<\/sub>\u00a0= 7 , b<sub>3\u00a0<\/sub>= ?<br \/>\nk = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_3}{b_3}\" alt=\"\\frac{a_3}{b_3}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">b<sub>3\u00a0<\/sub>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_3}{k}\" alt=\"\\frac{a_3}{k}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{7}{\\frac{1}{8}}\" alt=\"\\frac{7}{\\frac{1}{8}}\" align=\"absmiddle\" \/> = 7 \u00d7 8 = 56<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case 4:<\/strong> When a<sub>4\u00a0<\/sub>= 12 , b<sub>4\u00a0<\/sub>=?<br \/>\n<\/span><span style=\"color: #000000;\">k = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_4}{b_4}\" alt=\"\\frac{a_4}{b_4}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">b<sub>4\u00a0<\/sub>= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_4}{k}\" alt=\"\\frac{a_4}{k}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{12}{\\frac{1}{8}}\" alt=\"\\frac{12}{\\frac{1}{8}}\" align=\"absmiddle\" \/> = 12 \u00d7 8 = 96<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case 5:<\/strong> When\u00a0 a<sub>5<\/sub>\u00a0= 20 , b<sub>5\u00a0<\/sub>= ?<br \/>\n<\/span><span style=\"color: #000000;\">k = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_5}{b_5}\" alt=\"\\frac{a_5}{b_5}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">b<sub>5<\/sub> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{a_5}{k}\" alt=\"\\frac{a_5}{k}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{20}{\\frac{1}{8}}\" alt=\"\\frac{20}{\\frac{1}{8}}\" align=\"absmiddle\" \/> = 20 \u00d7 8 = 160<br \/>\n<\/span><span style=\"color: #000000;\">Combine results for all the cases, we have<br \/>\n<\/span><\/p>\n<table border=\"1\">\n<tbody>\n<tr>\n<td><strong>Parts of red pigment<\/strong><\/td>\n<td>1<\/td>\n<td>4<\/td>\n<td>7<\/td>\n<td>12<\/td>\n<td>20<\/td>\n<\/tr>\n<tr>\n<td><strong>Parts of Base<\/strong><\/td>\n<td>8<\/td>\n<td>32<\/td>\n<td>56<\/td>\n<td>96<\/td>\n<td>160<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">3. In Question 2 above, if 1 part of a red pigment requires 75 mL of base, how much red pigment should we mix with 1800 mL of base?<\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong><\/span><span style=\"color: #000000;\">Let the parts of red pigment mix with 1800 mL base be\u00a0 x.<\/span><\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"width: 50.8772%;\"><strong>Parts of red pigment<\/strong><\/td>\n<td style=\"width: 13.7427%; text-align: center;\">1<\/td>\n<td style=\"width: 16.0819%; text-align: center;\">x<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50.8772%;\"><strong>Parts of Base<\/strong><\/td>\n<td style=\"width: 13.7427%; text-align: center;\">75<\/td>\n<td style=\"width: 16.0819%; text-align: center;\">1800<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Since it is in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{75}&amp;space;=&amp;space;\\frac{x}{1800}\" alt=\"\\frac{1}{75} = \\frac{x}{1800}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-10.png\" alt=\"Ncert solutions class 8 chapter 13-10\" \/> 75 \u00d7 x = 1 \u00d7 1800<br \/>\n<\/span><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1\\times&amp;space;1800}{75}\" alt=\"\\frac{1\\times 1800}{75}\" align=\"absmiddle\" \/> = 24<br \/>\n<\/span><span style=\"color: #000000;\">Hence with base 1800 mL, 24 parts red pigment should be mixed.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3431\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-4.jpeg\" alt=\"NCERT Maths Solutions Class 8\" width=\"187\" height=\"82\" \/><br \/>\n<\/strong><\/span><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><span style=\"color: #000000;\">Let the number of bottles filled in five hours be x.<br \/>\n<\/span><span style=\"color: #000000;\">Here ratio of hours and bottles are in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{6}{840}=\\frac{5}{x}\" alt=\"\\frac{6}{840}=\\frac{5}{x}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">6x = 5 \u00d7 840 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5\\times&amp;space;840}{6}\" alt=\"\\frac{5\\times 840}{6}\" align=\"absmiddle\" \/> = 700<br \/>\n<\/span><span style=\"color: #000000;\">Hence machine will fill 700 bottles in five hours.<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the\u00a0<em>actual<\/em> length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?\u00a0<\/span><\/strong><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><span style=\"color: #000000;\">Let enlarged length of bacteria be\u00a0 x .<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3432\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-5.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"284\" height=\"59\" \/><br \/>\n<\/span><span style=\"color: #000000;\">Here length and enlarged length of bacteria are in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5}{50000}&amp;space;=\\frac{x}{20000}\" alt=\"\\frac{5}{50000} =\\frac{x}{20000}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-17.png\" alt=\"Ncert solutions class 8 chapter 13-17\" \/> x \u00d7 50000 = 5 \u00d7 20000<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{5\\times&amp;space;20000}{50000}\" alt=\"\\frac{5\\times 20000}{50000}\" align=\"absmiddle\" \/><br \/>\n<\/span><span style=\"color: #000000;\">x = 2cm<br \/>\n<\/span><span style=\"color: #000000;\">Hence the enlarged length of bacteria is 2 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length if the ship is 28 m, how long is the model ship?<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3433\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Q-6.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"275\" height=\"161\" \/><br \/>\n<\/span><\/strong><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><\/span><span style=\"color: #000000;\">Let the length of model ship be\u00a0 x .<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3434\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Ans-6.jpeg\" alt=\"NCERT Maths Solutions Class 8\" width=\"288\" height=\"78\" \/><br \/>\n<\/span><span style=\"color: #000000;\">Here length of mast and actual length of ship are in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{12}{9}&amp;space;=&amp;space;\\frac{28}{x}\" alt=\"\\frac{12}{9} = \\frac{28}{x}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-23.png\" alt=\"Ncert solutions class 8 chapter 13-23\" \/> x \u00d7 12 = 28 \u00d7 9<\/span><\/p>\n<p>x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{28\\times&amp;space;9}{12}\" alt=\"\\frac{28\\times 9}{12}\" align=\"absmiddle\" \/><br \/>\n<span style=\"color: #000000;\">x = 21 cm<br \/>\n<\/span><span style=\"color: #000000;\">Hence length of the model ship is 21 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">7. Suppose 2 kg of sugar contains 9\u00d710<sup>6<\/sup>\u00a0crystals. How many sugar crystals are there in<br \/>\n<\/span><\/strong><span style=\"color: #000000;\"><strong>(i)<\/strong> 5 kg of sugar?<br \/>\n<strong>(ii)<\/strong> 1.2 kg of sugar?<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong><br \/>\n<\/span><span style=\"color: #000000;\"><strong>(i)\u00a0<\/strong>Let sugar crystals be\u00a0 x.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3435\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Ans-7i.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"300\" height=\"56\" \/><br \/>\n<\/span><span style=\"color: #000000;\">Here, weight of sugar and number of crystals are in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{9\\times&amp;space;10^6}&amp;space;=&amp;space;\\frac{5}{x}\" alt=\"\\frac{2}{9\\times 10^6} = \\frac{5}{x}\" align=\"absmiddle\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-28.png\" alt=\"Ncert solutions class 8 chapter 13-28\" \/> x \u00d7 2 = 5 \u00d7 9 \u00d7 10<sup>6<\/sup><\/span><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-29.png\" alt=\"Ncert solutions class 8 chapter 13-29\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2\\times&amp;space;9&amp;space;\\times&amp;space;10^6}{2}\" alt=\"\\frac{2\\times 9 \\times 10^6}{2}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 22.5 \u00d7 10<sup>6<\/sup><br \/>\n= 2.25 \u00d7 10<sup>7<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Hence the number of sugar crystals is 2.25 \u00d7 10<sup>7\u00a0<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><strong><span style=\"color: #000000;\">(ii)\u00a0 Let sugar crystals be\u00a0 x.<br \/>\n<\/span><\/strong><span style=\"color: #000000;\">Here weight of sugar and number of crystals are in direct proportion.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3436\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Ans-7ii.jpeg\" alt=\"NCERT Maths Solutions Class 8\" width=\"288\" height=\"53\" \/><br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2}{9\\times&amp;space;10^6}&amp;space;=&amp;space;\\frac{1.2}{x}\" alt=\"\\frac{2}{9\\times 10^6} = \\frac{1.2}{x}\" align=\"absmiddle\" \/>\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-35.png\" alt=\"Ncert solutions class 8 chapter 13-35\" \/> x \u00d7 2 = 1.2 \u00d7 9 \u00d7 10<sup>6 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1.2\\times&amp;space;9\\times&amp;space;10^6}{2}\" alt=\"\\frac{1.2\\times 9\\times 10^6}{2}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 0.6 \u00d7 9 \u00d7 10<sup>6 <\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 5.4 \u00d7 10<sup>6<br \/>\n<\/sup><\/span><span style=\"color: #000000;\">Hence the number of sugar crystals is\u00a0 5.4\u00d710<sup>6<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let distance covered in the map be\u00a0 x.<br \/>\n<\/span><\/p>\n<table style=\"border-collapse: collapse; width: 54.1818%; height: 48px;\">\n<tbody>\n<tr style=\"height: 24px;\">\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000;\"><strong>Actual Distance (Km)<\/strong><\/span><\/td>\n<td style=\"width: 10.5454%; height: 24px;\"><span style=\"color: #000000;\">18<\/span><\/td>\n<td style=\"width: 10.303%; height: 24px;\"><span style=\"color: #000000;\">72<\/span><\/td>\n<\/tr>\n<tr style=\"height: 24px;\">\n<td style=\"width: 33.3333%; height: 24px;\"><span style=\"color: #000000;\"><strong>Distance covered in map (cm)<\/strong><\/span><\/td>\n<td style=\"width: 10.5454%; height: 24px;\"><span style=\"color: #000000;\">1<\/span><\/td>\n<td style=\"width: 10.303%; height: 24px;\"><span style=\"color: #000000;\">x<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">Here actual distance and distance covered in the map are in direct proportion. <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{18}{1}&amp;space;=&amp;space;\\frac{72}{x}\" alt=\"\\frac{18}{1} = \\frac{72}{x}\" align=\"absmiddle\" \/><br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-41.png\" alt=\"Ncert solutions class 8 chapter 13-41\" \/> x \u00d7 18 = 72 \u00d7 1 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{72}{18}\" alt=\"\\frac{72}{18}\" align=\"absmiddle\" \/> = 4 cm<br \/>\nHence distance covered in the map is 4 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time<\/strong><br \/>\n<strong>(i)<\/strong> the length of the shadow cast by another pole 10 m 50 cm high<br \/>\n<strong>(ii)<\/strong> the height of a pole which casts a shadow 5 m long.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Here height of the pole and length of the shadow are in direct proportion.<br \/>\n1 m = 100 cm<br \/>\n5 m 60 cm = 5\u00d7100+60 = 560 cm<br \/>\n3 m 20 cm = 3\u00d7100+20 = 320 cm<br \/>\n10 m 50 cm = 10\u00d7100+50 = 1050 cm<br \/>\n5 m = 5\u00d7100 = 500 cm<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i)\u00a0Let the length of the shadow of another pole be\u00a0 x.<br \/>\n<\/strong><\/span><\/p>\n<table style=\"border-collapse: collapse; width: 52.4848%;\">\n<tbody>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\"><strong>Height of pole (cm)<\/strong><\/span><\/td>\n<td style=\"width: 10.7878%;\"><span style=\"color: #000000;\">560<\/span><\/td>\n<td style=\"width: 8.36365%;\"><span style=\"color: #000000;\">1050<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\"><strong>Length of shadow (cm)<\/strong><\/span><\/td>\n<td style=\"width: 10.7878%;\"><span style=\"color: #000000;\">320\u00a0<\/span><\/td>\n<td style=\"width: 8.36365%;\"><span style=\"color: #000000;\">x<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{560}{320}&amp;space;=&amp;space;\\frac{1050}{x}\" alt=\"\\frac{560}{320} = \\frac{1050}{x}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-46.png\" alt=\"Ncert solutions class 8 chapter 13-46\" \/> x \u00d7 560 = 1050 \u00d7 320 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1050\\times&amp;space;320}{560}\" alt=\"\\frac{1050\\times 320}{560}\" align=\"absmiddle\" \/><br \/>\nx= 600 cm = 6m<br \/>\nHence length of the shadow of another pole is 6 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)\u00a0Let the height of the pole be\u00a0 x.<br \/>\n<\/strong><\/span><\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\"><strong>Height of pole (cm)<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\">560<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\">x<\/span><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\"><strong>Length of shadow (cm)<\/strong><\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\">320<\/span><\/td>\n<td style=\"width: 33.3333%;\"><span style=\"color: #000000;\">500<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{560}{320}&amp;space;=&amp;space;\\frac{x}{500}\" alt=\"\\frac{560}{320} = \\frac{x}{500}\" align=\"absmiddle\" \/> \u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-51.png\" alt=\"Ncert solutions class 8 chapter 13-51\" \/> x \u00d7 320 = 560 \u00d7 500<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x =<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{560\\times&amp;space;500}{320}\" alt=\"\\frac{560\\times 500}{320}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">= 875 cm = 8 m 75 cm<br \/>\nHence height of the pole is 8 m 75 cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let distance covered in 5 hours be x km.<br \/>\n1 hour = 60 minutes<br \/>\nTherefore, 5 hours = 5\u00d760 = 300 minutes<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3439\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2022\/12\/NCERT-Solutions-Class-8-Maths-Ex-13.1-Ans-10.png\" alt=\"NCERT Maths Solutions Class 8\" width=\"250\" height=\"55\" \/><br \/>\nHere distance covered and time in direct proportion.<br \/>\n<img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{14}{25}=\\frac{x}{300}\" alt=\"\\frac{14}{25}=\\frac{x}{300}\" align=\"absmiddle\" \/> <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" class=\"\" title=\"NCERT Solution For Class 8 Maths Chapter 13\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/08\/ncert-solutions-class-8-chapter-13-56.png\" alt=\"Ncert solutions class 8 chapter 13-56\" \/> 25 \u00d7 x = 300 \u00d7 14 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">x = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{14\\times&amp;space;300}{25}\" alt=\"\\frac{14\\times 300}{25}\" align=\"absmiddle\" \/><br \/>\nx = 168<br \/>\nTherefore, a truck can travel 168 km in 5 hours.<\/span><\/p>\n<p>&nbsp;<\/p>\n<table style=\"width: 100%; border-collapse: collapse;\" border=\"1\">\n<tbody>\n<tr style=\"height: 32px;\">\n<td style=\"width: 100%; background-color: #f2e079; text-align: center; height: 32px;\"><span style=\"color: #ff0000; font-size: 14pt;\"><strong>NCERT Class 8<sup>th\u00a0<\/sup>Solution\u00a0<\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 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href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"width: 100%; text-align: left; height: 28px;\"><span style=\"font-size: 14pt;\"><strong><span style=\"font-family: georgia, palatino, serif; font-size: 12pt;\"><a class=\"row-title\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-8-social-science\/\" target=\"_blank\" rel=\"noopener\" aria-label=\"\u201cNCERT Solutions Class 6 Social Science\u201d (Edit)\">NCERT Solutions Class 8 Social Science<\/a><\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 8 Mathematics\u00a0 Chapter &#8211; 13 (Direct and Inverse Proportions)\u00a0 The NCERT Solutions in English Language for Class 8 Mathematics Chapter &#8211; 13 Direct and Inverse Proportions Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 13: Direct and Indirect Proportions NCERT Solution Class [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[442],"tags":[468,469,445,5],"class_list":["post-3406","post","type-post","status-publish","format-standard","hentry","category-class-8-maths","tag-ncert-class-8-maths-chapter-13-direct-and-inverse-proportions-in-english","tag-ncert-solutions-class-8-maths-chapter-13-in-english","tag-ncert-solutions-class-8-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 8 Maths 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