{"id":31749,"date":"2026-08-07T12:53:07","date_gmt":"2026-08-07T07:23:07","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31749"},"modified":"2026-08-07T13:24:02","modified_gmt":"2026-08-07T07:54:02","slug":"proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/","title":{"rendered":"Proportional Reasoning-2 &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)"},"content":{"rendered":"\n<p><strong><strong>Proportional Reasoning-2<\/strong><\/strong> &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">NCERT Solutions Class 8<\/h2>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-english-poorvi\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >English Poorvi<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-hindi-malhar\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Hindi Malhar<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-maths-ganita-prakash\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Maths Ganita Prakash<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-science-curiosity\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Science Curiosity<\/a>\n\n\n<a class=\"mks_button mks_button_small rounded\" href=\"https:\/\/mycbseguide.com\/blog\/category\/ncert-solutions\/ncert-solutions-class-8\/ncert-solutions-class-8-social-exploring-society\/\" target=\"_self\" style=\"color: #FFFFFF; background-color: #0066bf;\" >Social Exploring Society<\/a>\n\n\n\n<h2 class=\"wp-block-heading\"><strong><strong>Proportional Reasoning-2<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.1: Convert 60,00,000 cm to kilometres.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>60,00,000 cm = 60 km<br>We know that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 km = 1000 m<\/li>\n\n\n\n<li>1 m = 100 cm<\/li>\n\n\n\n<li>Therefore, 1 km = 1000 \u00d7 100 = 1,00,000 cm<\/li>\n<\/ul>\n\n\n\n<p>Now, to convert {tex}60,00,000 {~cm}{\/tex} to {tex}{km}: 60,00,000 {~cm} \\div 1,00,000=60 {~km}{\/tex}<br><strong>Verification:<\/strong> {tex}60,00,000 {~cm}=60,00,000 \\div 100{\/tex}{tex}=60,000 {~m} {\/tex}<br>{tex}60,000 {~m}=60,000 \\div 1000={\/tex}60 km<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2: Using the map given, can you find the geographical distance between Bengaluru and Chennai? Also, find the geographical distance between Mangaluru and Chennai.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768805003-wex4a4.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To find the geographical distances, follow these steps:<br><strong>Step 1:<\/strong> Measure the distance on the map using a ruler.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Let&#8217;s say the distance between Bengaluru and Chennai on the map {tex}=7 {~cm}{\/tex} (example)<\/li>\n\n\n\n<li>Distance between Mangaluru and Chennai on the map {tex}=5.5 {~cm}{\/tex} (example)<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong> Use the map scale RF 1:60,00,000 This means 1 cm on map {tex}=60,00,000 {~cm}{\/tex} in reality {tex}=60 {~km}{\/tex}<br><strong>Step 3:<\/strong> Calculate actual distances:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Bengaluru to Chennai: If map distance {tex}=7 {~cm}{\/tex}<br>Actual distance {tex}=7 \\times 60 {~km}=420 {~km}{\/tex} (approximately)<\/li>\n\n\n\n<li>Mangaluru to Chennai: If map distance {tex}=5.5 {~cm}{\/tex}<br>Actual distance {tex}=5.5 \\times 60 {~km}=330{\/tex} km (approximately)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3: Try to find the distances between the same two pairs of cities with different maps that have different scales (ratios). Do they all give the same geographical distance, approximately?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, they should all give approximately the same geographical distance.<br><strong>Explanation:<\/strong> Different maps may have different scales (like 1:50,00,000 or 1:1,00,00,000), but the actual geographical distance between two cities remains constant.<br>For example:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Map 1 with scale 1:60,00,000 might show {tex}7 {~cm}=420 {~km}{\/tex}<\/li>\n\n\n\n<li>Map 2 with scale {tex}1: 1,00,00,000{\/tex} might show {tex}4.2 {~cm}=420 {~km}{\/tex}<\/li>\n\n\n\n<li>Map 3 with scale {tex}1: 30,00,000{\/tex} might show {tex}14 {~cm}=420 {~km}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>The map distances will be different, but after using the respective scales, the actual geographical distance will be approximately the same (minor variations may occur due to measurement errors or map accuracy).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4: Puneet has only 2 red chillies in his kitchen. But he wants to make spice mix powder that tastes the same as Viswanath&#8217;s spice mix powder. How much of the other ingredients should Puneet use to make his spice mix powder?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Puneet should use:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4 spoons of coriander seeds<\/li>\n\n\n\n<li>2 red chillies<\/li>\n\n\n\n<li>1 spoon of toor dal<\/li>\n\n\n\n<li>0.5 spoon (or half spoon) of fenugreek seeds<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><br>Viswanath&#8217;s ratio {tex}={\/tex} Coriander : Red chillies : Toor dal : Fenugreek {tex}=8: 4: 2: 1{\/tex}<br>Puneet has only 2 red chillies, which is half of what Viswanath used (2 is half of 4).<br>So, Puneet should reduce all ingredients to half:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Coriander seeds {tex}=8 \\div 2=4{\/tex} spoons<\/li>\n\n\n\n<li>Red chillies {tex}=4 \\div 2=2{\/tex}<\/li>\n\n\n\n<li>Toor dal {tex}=2 \\div 2=1{\/tex} spoon<\/li>\n\n\n\n<li>Fenugreek {tex}={1} \\div 2=0.5{\/tex} spoon<\/li>\n<\/ul>\n\n\n\n<p>Puneet&#8217;s ratio {tex}=4: 2: 1: 0.5{\/tex}<br>Both ratios are proportional: {tex}8: 4: 2: 1:: 4: 2: 1: 0.5{\/tex}<br>This can be verified as: {tex}8 \/ 4=4 \/ 2=2 \/ 1=1 \/ 0.5=2{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5: To make a special shade of purple, paint must be mixed in the ratio, Red : Blue : White :: 2 : 3 : 5. If Yasmin has 10 litres of white paint, how many litres of red and blue paint should she add to get the same shade of purple?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>In the ratio {tex}2: 3: 5{\/tex}, the white paint corresponds to 5 parts. If 5 parts is 10 litres, 1 part is {tex}10 \\div 5=2{\/tex} litres.<br>Red {tex}=2{\/tex} parts {tex}=2 \\times 2=4{\/tex} litres.<br>Blue {tex}=3{\/tex} parts {tex}=3 \\times 2=6{\/tex} litres.<br>So, the purple paint will have 4 litres of red, 6 litres of blue, and 10 litres of white paint.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6: What is the total volume of this purple paint?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The total volume of purple paint is 20 litres.<br><strong>Explanation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Red paint {tex}=4{\/tex} litres<\/li>\n\n\n\n<li>Blue paint = 6 litres<\/li>\n\n\n\n<li>White paint = 10 litres<\/li>\n<\/ul>\n\n\n\n<p>Total volume {tex}=4+6+10=20{\/tex} litres<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7: A cricket coach schedules practice sessions that include different activitles in a specific ratio-&nbsp;time for warm-up\/cool-down : time for batting : time for bowiling : time for fielding : {tex}3: 4: 3: 5{\/tex}.<br>If each session is {tex}{1 5 0}{\/tex} minutes long, how much time is spent on each activity?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Warm-up\/cool-down = 30 minutes<\/li>\n\n\n\n<li>Batting {tex}=40{\/tex} minutes<\/li>\n\n\n\n<li>Bowling {tex}=30{\/tex} minutes<\/li>\n\n\n\n<li>Fielding {tex}=50{\/tex} minutes<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><br>The ratio is {tex}3: 4: 3: 5{\/tex}<br>Total session time {tex}=150{\/tex} minutes<br><strong>Step 1: <\/strong>Find the sum of ratio terms<br>Sum {tex}=3+4+3+5=15{\/tex}<br><strong>Step 2:<\/strong> Find the value of 1 part<br>1 part {tex}=150 \\div 15=10{\/tex} minutes<br><strong>Step 3:<\/strong> Calculate time for each activity<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Warm-up\/cool-down {tex}=3 \\times 10=30{\/tex} minutes<\/li>\n\n\n\n<li>Batting {tex}=4 \\times 10=40{\/tex} minutes<\/li>\n\n\n\n<li>Bowling {tex}=3 \\times 10=30{\/tex} minutes<\/li>\n\n\n\n<li>Fielding {tex}=5 \\times 10=50{\/tex} minutes<\/li>\n<\/ul>\n\n\n\n<p><strong>Verlfication:<\/strong> {tex}30+40+30+50={1 5 0}{\/tex} minutes<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8: A school library has books in different languages in the following ratio no. of Odiya books: no. of Hindi books : no. of English books {tex}: 3: 2: 1{\/tex}. If the library has 288 Odiya books, how many Hindi and English books does it have?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hindi books {tex}=192{\/tex}<\/li>\n\n\n\n<li>English books {tex}=96{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Explanation:<\/strong><br>Given ratio {tex}={\/tex} Odiya {tex}:{\/tex} Hindi {tex}:{\/tex} English {tex}=3: 2: 1{\/tex}<br><strong>Step 1:<\/strong> Find the value of 1 part Odiya books {tex}=3{\/tex} parts {tex}=288{\/tex} books<br>Therefore, 1 part {tex}=288 \\div 3=96{\/tex} books<br><strong>Step 2:<\/strong> Calculate other books<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hindi books {tex}=2{\/tex} parts {tex}=2 \\times 96=192{\/tex} books<\/li>\n\n\n\n<li>English books {tex}={1}{\/tex} part {tex}={1} \\times 96=96{\/tex} books<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification:<\/strong> Check if ratio is maintained: {tex}288: 192: 96{\/tex}<br>Divide by {tex}96: 3: 2: 1{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9: I have 100 coins in the ratio- no. of \u20b910 coins : no. of \u20b95 coins : no. of \u20b92 coins : no. of \u20b91 coins {tex}:: 4: 3: 2: 1{\/tex}. How much money do I have in coins?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total money = \u20b9590<br><strong>Explanation:<\/strong><br>Given ratio {tex}=4: 3: 2: 1{\/tex} Total coins {tex}=100{\/tex}<br><strong>Step 1:<\/strong> Find the value of 1 part<br>Sum of ratio terms {tex}=4+3+2+1=10{\/tex}&nbsp;<br>1 part {tex}={1 0 0} \\div {1 0}={1 0}{\/tex} coins<br><strong>Step 2: <\/strong>Calculate number of each type of coin<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}\u20b9 10{\/tex} coins {tex}=4 \\times 10=40{\/tex} coins<\/li>\n\n\n\n<li>{tex}\u20b9 5{\/tex} coins {tex}=3 \\times 10=30{\/tex} coins<\/li>\n\n\n\n<li>{tex}\u20b9 2{\/tex} coins {tex}=2 \\times 10=20{\/tex} coins<\/li>\n\n\n\n<li>{tex}\u20b9 1{\/tex} coins {tex}=1 \\times 10=10{\/tex} coins<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: <\/strong>Calculate total value<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Value of \u20b9 10 coins {tex}=40 \\times 10=\u20b9 400{\/tex}<\/li>\n\n\n\n<li>Value of \u20b95 coins {tex}=30 \\times 5=\u20b9 150{\/tex}<\/li>\n\n\n\n<li>Value of \u20b9 2 coins {tex}=20 \\times 2={\/tex} \u20b9 40<\/li>\n\n\n\n<li>Value of \u20b9 1 coins {tex}=10 \\times 1=\u20b9 10{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Total money {tex}=400+150+40+10=\u20b9 590{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10: Construct a triangle with side lengths in the ratio {tex}3: 4: 5{\/tex}. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>No, all triangles with sidelengths in the ratio {tex}3: 4: 5{\/tex} will NOT be congruent to each other.<br><strong>Explanation:<\/strong><br><strong>Construction:<\/strong> We can construct triangles with sides in ratio {tex}3: 4: 5{\/tex} by choosing different values for the unit length.<br><strong>Example 1: <\/strong>If we take 1 part = 1 cm<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sides {tex}=3 {~cm}, 4 {~cm}, 5 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 2: <\/strong>If we take {tex}{1}{\/tex} part = {tex}{2} {cm}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sides {tex}=6 {~cm}, 8 {~cm}, 10 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Example 3:<\/strong> If we take 1 part {tex}=3 {~cm}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sides {tex}=9 {~cm}, 12 {~cm}, 15 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Why they are not congruent:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Though all these triangles have the same ratio {tex}(3: 4: 5){\/tex}, their actual sizes are different<\/li>\n\n\n\n<li>Congruent triangles must have exactly the same size and shape<\/li>\n\n\n\n<li>These triangles have the same shape (they are similar) but different sizes<\/li>\n\n\n\n<li>Therefore, they are NOT congruent<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11: Can you construct a triangle with side lengths in the ratio 1 : 3 : 5? Why or why not?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>No, we cannot construct a triangle with side lengths in the ratio {tex}1: 3: 5{\/tex}.<br><strong>Explanation:<\/strong><br>For a triangle to exist, it must satisfy the <strong>Triangle Inequality<\/strong>&nbsp;<strong>Theorem<\/strong>, which states: The sum of any two sides of a triangle must be greater than the third side.<br>Let&#8217;s take 1 part = 1 cm<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Side {tex}1=1 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}2=3 {~cm}{\/tex}<\/li>\n\n\n\n<li>Side {tex}3=5 {~cm}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Checking triangle inequality:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Is {tex}{1}+{3}&gt;{5}{\/tex} ? {tex}\\rightarrow 4&gt;5{\/tex} ? {tex}\\rightarrow{\/tex} {tex}{N O}{\/tex}<\/li>\n\n\n\n<li>Is {tex}1+5&gt;3 ? \\rightarrow 6&gt;3 ? \\rightarrow{\/tex} YES<\/li>\n\n\n\n<li>Is {tex}3+5&gt;1 ? \\rightarrow 8&gt;1{\/tex} ? {tex}\\rightarrow{\/tex} YES<\/li>\n<\/ol>\n\n\n\n<p>Since the first condition fails ({tex}1+3=4{\/tex}, which is less than 5), we cannot construct a triangle with these side lengths.<br><strong>General rule:<\/strong> For sides in ratio {tex}1: 3: 5{\/tex}, regardless of the unit we choose: {tex}1+3=4{\/tex}, which is always less than 5<br>Therefore, it is impossible to construct such a triangle.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12: A group of 360 people were asked to vote for their favourite season from the three seasons-rainy, winter and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 1:<\/strong> Find the number who liked winter<br>Total people {tex}=360{\/tex}<br>Summer lovers {tex}=90{\/tex}<br>Rainy lovers {tex}={\/tex} 120<br>Winter lovers {tex}=360-90-120=150{\/tex}<br><strong>Step 2: <\/strong>Calculate angles for each season<br>Total angle in a circle {tex}=360^{\\circ}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Summer angle {tex}=(90 \/ 360) \\times 360^{\\circ}{\/tex}&nbsp;{tex}=(1 \/ 4) \\times 360^{\\circ}=90^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Rainy angle {tex}=(120 \/ 360) \\times 360^{\\circ}{\/tex}&nbsp;{tex}=(1 \/ 3) \\times 360^{\\circ}=120^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Winter angle {tex}=(150 \/ 360) \\times 360^{\\circ}{\/tex}&nbsp;{tex}=(5 \/ 12) \\times 360^{\\circ}=150^{\\circ}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Verlficaton:<\/strong> {tex}90^{\\circ}+120^{\\circ}+150^{\\circ}=360^{\\circ}{\/tex}<br><strong>Step 3:<\/strong> Draw the pie chart<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw a circle with center A<\/li>\n\n\n\n<li>Draw radius AB<\/li>\n\n\n\n<li>From AB, measure {tex}90^{\\circ}{\/tex} anti-clockwise and mark AC (Summer sector)<\/li>\n\n\n\n<li>From AC, measure {tex}120^{\\circ}{\/tex} anti-clockwise and mark AD (Rainy sector)<\/li>\n\n\n\n<li>From AD, measure {tex}150^{\\circ}{\/tex} anti-clockwise (this completes the circle) (Winter sector)<\/li>\n\n\n\n<li>Label and colour each sector.<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769141074-jskaq4.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13: Draw a pie chart based on the following information about viewers&#8217; favourite type of TV channel: Entertainment-50%, Sports-25%, News-15%, Information-10%.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate angles for each channel type<br>Total angle {tex}=360^{\\circ}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Entertainment {tex}=50 \\%{\/tex} of {tex}360^{\\circ}=(50 \/ 100) \\times 360^{\\circ}{\/tex}{tex}=0.5 \\times 360^{\\circ}=180^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Sports {tex}=25 \\%{\/tex} of {tex}360^{\\circ}=(25 \/ 100) \\times 360^{\\circ}{\/tex}&nbsp;{tex}=0.25 \\times 360^{\\circ}=90^{\\circ}{\/tex}<\/li>\n\n\n\n<li>News {tex}=15 \\%{\/tex} of {tex}360^{\\circ}=(15 \/ 100) \\times 360^{\\circ}{\/tex}&nbsp;{tex}=0.15 \\times 360^{\\circ}=54^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Information {tex}=10 \\%{\/tex} of {tex}360^{\\circ}=(10 \/ 100) \\times 360^{\\circ}{\/tex}{tex}=0.1 \\times 360^{\\circ}=36^{\\circ}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification:<\/strong> {tex}180^{\\circ}+90^{\\circ}+54^{\\circ}+36^{\\circ}=360^{\\circ}{\/tex}<br><strong>Step 2:<\/strong> Draw the pie chart<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Draw a circle with center A<\/li>\n\n\n\n<li>Draw radius AB<\/li>\n\n\n\n<li>From AB, measure {tex}180^{\\circ}{\/tex} and mark AC (Entertainment exactly half circle)<\/li>\n\n\n\n<li>From AC, measure {tex}90^{\\circ}{\/tex} and mark AD (Sports quarter circle)<\/li>\n\n\n\n<li>From AD, measure {tex}54^{\\circ}{\/tex} and mark AE (News)<\/li>\n\n\n\n<li>From AE, measure {tex}36^{\\circ}{\/tex} to complete the circle (Information)<\/li>\n\n\n\n<li>Label and colour appropriately<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14: Prepare a pie chart that shows the favourite subjects of the students in your class. You can collect the data of the number of students for&nbsp;each subject shown in the table (each student should choose only one subject). Then write these numbers in the table and construct a pie chart:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Subject<\/strong><\/td><td><strong>Language<\/strong><\/td><td><strong>Arts Education<\/strong><\/td><td><strong>Vocational Education<\/strong><\/td><td><strong>Social Science<\/strong><\/td><td><strong>Physical Education<\/strong><\/td><td><strong>Maths<\/strong><\/td><td><strong>Science<\/strong><\/td><\/tr><tr><td><strong>Number of Students<\/strong><\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><td>&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This is a practical activity. Here&#8217;s how to solve it:<br><strong>Step 1:<\/strong>&nbsp;Collect data from your classmates Example data (you should use actual data from your class):<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Subject<\/strong><\/td><td><strong>Language Arts<\/strong><\/td><td><strong>Education<\/strong><\/td><td><strong>Vocational Education<\/strong><\/td><td><strong>Social Science<\/strong><\/td><td><strong>Physical Education<\/strong><\/td><td><strong>Math<\/strong><\/td><td><strong>Science<\/strong><\/td><\/tr><tr><td><strong>Students<\/strong><\/td><td>6<\/td><td>4<\/td><td>3<\/td><td>5<\/td><td>2<\/td><td>8<\/td><td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Total students {tex}=6+4+3+5+2+8+2=30{\/tex}<br><strong>Step 2:<\/strong> Calculate angles for each subject<br>Formula: Angle {tex}=({\/tex}Number of students \/ Total students{tex}) \\times 360^{\\circ}{\/tex}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Language Arts {tex}=(6 \/ 30) \\times 360^{\\circ}=72^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Education {tex}=(4 \/ 30) \\times 360^{\\circ}=48^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Vocational Education {tex}=(3 \/ 30) \\times 360^{\\circ}=36^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Social Science {tex}=(5 \/ 30) \\times 360^{\\circ}=60^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Physical Education {tex}=(2 \/ 30) \\times 360^{\\circ}=24^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Maths {tex}=(8 \/ 30) \\times 360^{\\circ}=96^{\\circ}{\/tex}<\/li>\n\n\n\n<li>Science {tex}=(2 \/ 30) \\times 360^{\\circ}=24^{\\circ}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification:<\/strong> {tex}72+48+36+60+{\/tex}{tex}24+96+24=360^{\\circ}{\/tex}<br><strong>Step 3: <\/strong>Draw the pie chart using these angles in sequence.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.15: Which of these are in inverse proportion?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}x{\/tex} 40 80 25 16 {tex}y{\/tex} 20 10 32 50<\/li>\n\n\n\n<li>{tex}x{\/tex} 40 80 25 16 {tex}y{\/tex} 20 10 32 50<\/li>\n\n\n\n<li>{tex}x{\/tex} 30 90 150 10 {tex}y{\/tex} 15 5 3 45<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Yes, {tex}x{\/tex} and {tex}y{\/tex} are in inverse proportion.<br><strong>Explanation:&nbsp;<\/strong>For&nbsp;inverse proportion, the product xy should be constant.<br>{tex}40 \\times 20=800{\/tex}<br>{tex}80 \\times 10=800{\/tex}<br>{tex}25 \\times 32=800{\/tex}<br>{tex}16 \\times 50=800{\/tex}<br>Since all products are equal (800), {tex}x{\/tex} and {tex}y{\/tex} are in Inverse proportion.<\/li>\n\n\n\n<li>No, {tex}x{\/tex} and {tex}y{\/tex} are NOT in inverse proportion.<br>Explanation:Check if product xy is constant:<br>{tex}40 \\times 20=800{\/tex}<br>{tex}80 \\times 10=800{\/tex}<br>{tex}25 \\times 12.5=312.5{\/tex}<br>{tex}16 \\times 8=128{\/tex}<br>The products are NOT equal, sox and {tex}y{\/tex} are NOT in inverse proportion.<\/li>\n\n\n\n<li>Yes, x&nbsp;and y&nbsp;are in inverse proportion.<br><strong>Explanation:<\/strong> Check if product xy is constant:<br>{tex}30 \\times 15=450{\/tex}<br>{tex}90 \\times 5=450{\/tex}<br>Since all products are equal (450), {tex}x{\/tex} and {tex}y{\/tex} are in Inverse proportion.<br>{tex}150 \\times 3=450{\/tex}<br>{tex}10 \\times 45=450{\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16: Fill in the empty cells if x and y are in inverse proportion.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>{tex}x{\/tex}<\/td><td>16<\/td><td>12<\/td><td>&nbsp;<\/td><td>36<\/td><\/tr><tr><td>{tex}y{\/tex}<\/td><td>9<\/td><td>&nbsp;<\/td><td>48<\/td><td>&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>{tex}x{\/tex}<\/td><td>16<\/td><td>12<\/td><td><strong>3<\/strong><\/td><td>36<\/td><\/tr><tr><td>{tex}y{\/tex}<\/td><td>9<\/td><td><strong>12<\/strong><\/td><td>48<\/td><td><strong>4<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Explanation:<\/strong><br>Since {tex}x{\/tex} and {tex}y{\/tex} are in inverse proportion, {tex}x y={\/tex} constant ({tex}k{\/tex})<br>Step 1: Find the constant k<br>From first column: k =16 \u00d79 =144<br>Step 2: Find missing values<br>For second column {tex}(x=12, y=?): 12 \\times y{\/tex}{tex}=144 y=144 \\div 12=12{\/tex}<br>For third column {tex}(x=?, y=48): x \\times 48{\/tex}&nbsp;{tex}=144 x=144 \\div 48=3{\/tex}<br>For fourth column {tex}(x=36, y=?): 36 \\times y{\/tex}&nbsp;{tex}=144 y=144 \\div 36=4{\/tex}<br><strong>Verification:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}16 \\times 9=144{\/tex}<\/li>\n\n\n\n<li>{tex}12 \\times 12=144{\/tex}<\/li>\n\n\n\n<li>{tex}3 \\times 48=144{\/tex}<\/li>\n\n\n\n<li>{tex}36 \\times 4=144{\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17: Which of the following pairs of quantities are in inverse proportion?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The number of taps filling a water tank and the time taken to fill it.<\/li>\n\n\n\n<li>The number of painters hired and the days needed to paint a wall of fixed size.<\/li>\n\n\n\n<li>The distance a car can travel and the amount of petrol in the tank.<\/li>\n\n\n\n<li>The speed of a cyclist and the time taken to cover a fixed route.<\/li>\n\n\n\n<li>The length of cloth bought and the price paid at a fixed rate per metre.<\/li>\n\n\n\n<li>The number of pages in a book and the time required to read it at a fixed reading speed.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>YES, they are in inverse proportion.<br><strong>Explanation:<\/strong><ul><li>More taps {tex}\\rightarrow{\/tex} Less time to fill the tank<\/li><li>Fewer taps {tex}\\rightarrow{\/tex} More time to fill the tank<\/li><\/ul>The quantities change in opposite directions by the same factor. If we double the number of taps, the time taken becomes half. Therefore, they are in inverse proportion.<\/li>\n\n\n\n<li>YES, they are in inverse proportion.<br><strong>Explanation:<\/strong><ul><li>More painters {tex}\\rightarrow{\/tex} Fewer days needed<\/li><li>Fewer painters {tex}\\rightarrow{\/tex} More days needed<\/li><\/ul>If we double the number of painters, the work gets done in half the time. Therefore, they are in inverse proportion.<\/li>\n\n\n\n<li>NO, they are NOT in inverse proportion. They are in direct proportion.<br><strong>Explanation:<\/strong><ul><li>More petrol {tex}\\rightarrow{\/tex} More distance can be traveled<\/li><li>Less petrol {tex}\\rightarrow{\/tex} Less distance can be traveled<\/li><\/ul>Both quantities increase together and decrease together, so they are in direct proportion, not inverse proportion.<\/li>\n\n\n\n<li>YES, they are in inverse proportion.<br><strong>Explanation:<\/strong><ul><li>Higher speed {tex}\\rightarrow{\/tex} Less time taken<\/li><li>Lower speed {tex}\\rightarrow{\/tex} More time taken<\/li><\/ul>For a fixed distance, if speed doubles, time becomes half. Therefore, they are in inverse proportion.<\/li>\n\n\n\n<li>NO, they are NOT in inverse proportion. They are in direct proportion.<br><strong>Explanation:<\/strong><ul><li>More cloth {tex}\\rightarrow{\/tex} More price to pay<\/li><li>Less cloth {tex}\\rightarrow{\/tex} Less price to pay<\/li><\/ul>Both quantities increase together and decrease together, so they are in direct proportion.<\/li>\n\n\n\n<li>NO, they are NOT in inverse proportion. They are in direct proportion.<br><strong>Explanation:<\/strong><ul><li>More pages {tex}\\rightarrow{\/tex} More time to read<\/li><li>Fewer pages {tex}\\rightarrow{\/tex} Less time to read<\/li><\/ul>Both quantities increase together and decrease together, so they are in direct proportion.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18: If 24 pencils cost \u20b9120, how much will 20 such pencils cost?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>20 pencils will cost \u20b9100.<br><strong>Explanation:<\/strong><br>This is a case of direct proportion (more pencils \u2192 more cost).<br><strong>Method 1: Unitary Method<\/strong><br>Cost of 24 pencils {tex}=\u20b9 120{\/tex}<br>Cost of 1 pencil {tex}=120 \\div 24=\u20b9 5{\/tex}<br>Cost of 20 pencils {tex}=5 \\times 20=\u20b9 100{\/tex}<br><strong>Method 2: Using Proportion<\/strong><br>{tex} 24: 20:: 120: \\times 24 \\times x={\/tex}{tex}20 \\times 120 x=(20 \\times 120) \\div 24 x{\/tex}&nbsp;{tex}=2400 \\div 24 x=100 {\/tex}<br>Therefore, 20 pencils cost \u20b9100.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19: A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The water will last for 4 days.<br>This is a case of inverse proportion (more families {tex}\\rightarrow{\/tex} fewer days).<br>Assumptions needed:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>All families use the same amount of water<\/li>\n\n\n\n<li>Water usage per family per day is constant<\/li>\n\n\n\n<li>No additional water is added to the tank<\/li>\n<\/ol>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>20 families {tex}\\rightarrow{\/tex} Water lasts 6 days<\/li>\n\n\n\n<li>Total families after 10 more join {tex}=20+10=30{\/tex} families<\/li>\n<\/ul>\n\n\n\n<p>Using inverse proportion: {tex}{x}_1 {y}_1={x}_2 {y}_2{\/tex}<br>{tex} 20 \\times 6=30 \\times x {\/tex}<br>{tex} 120=30 \\times x {\/tex}<br>{tex} x=120 \\div 30 {\/tex}<br>{tex} x=4 \\text { days } {\/tex}<br>Therefore, the water will last for only 4 days.<br><strong>Verification: <\/strong>20 families increased to 30 families ({tex}{1 . 5}{\/tex} times)<br>So days should decrease by factor of 1.5<br>{tex}6 \\div 1.5=4{\/tex} days<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20: Fill in the average number of hours each living being sleeps in a day by looking at the charts. Select the appropriate hours from this list : 15,2.5,20,8,3.5,13,10.5,18.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768815546-2m7tgy.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This question requires looking at charts provided in the textbook. The typical sleep hours are:<br><strong>Common sleep patterns:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cats: {tex}{1 3}{\/tex} hours<\/li>\n\n\n\n<li>Dogs: 10.5 hours<\/li>\n\n\n\n<li>Elephants: 3.5 hours<\/li>\n\n\n\n<li>Giraffes: 2.5 hours<\/li>\n\n\n\n<li>Humans (adults): 8 hours<\/li>\n\n\n\n<li>Koalas: 20 hours<\/li>\n\n\n\n<li>Lions: 15 hours<\/li>\n\n\n\n<li>Sloths: 18 hours<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21: Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>It will take 3 days for 4 workers to finish the work.<br><strong>Assumptions needed:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>All workers work at the same rate\/speed<\/li>\n\n\n\n<li>All workers work for the same number of hours each day<\/li>\n\n\n\n<li>The work is uniformly distributed among all workers<\/li>\n<\/ol>\n\n\n\n<p><strong>Explanation:<\/strong><br>This is inverse proportion (more workers \u2192 fewer days).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 workers {tex}\\rightarrow 4{\/tex} days<\/li>\n\n\n\n<li>Total workers after one joins {tex}=3+1=4{\/tex} workers<\/li>\n<\/ul>\n\n\n\n<p>Using inverse proportion: {tex}x_1 y_1=x_2 y_2 {\/tex},&nbsp;<br>{tex}3 \\times 4=4 \\times x {\/tex}<br>{tex}12=4 x {\/tex}<br>{tex}x=12 \\div 4{\/tex}<br>{tex} x=3{\/tex} days<br>Therefore, 4 workers will complete the work in 3 days.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22: It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>It will take 15 hours to fill 5 tanks.<br><strong>Explanation:<\/strong><br>This is direct proportion (more tanks {tex}\\rightarrow{\/tex} more time).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 tanks {tex}\\rightarrow 6{\/tex} hours<\/li>\n\n\n\n<li>Need to find time for 5 tanks<\/li>\n<\/ul>\n\n\n\n<p><strong>Method 1: Unitary method&nbsp;<\/strong>Time to fill 2 tanks {tex}=6{\/tex} hours<br>Time to fill 1 tank {tex}=6 \\div 2=3{\/tex} hours<br>Time to fill 5 tanks {tex}=3 \\times 5=15{\/tex} hours<br><strong>Method 2: Using proportion&nbsp;<\/strong>2 : 5 :: 6 : x<br>2 {tex}\\times{\/tex}&nbsp;x = 5 {tex}\\times{\/tex}&nbsp;6<br>2x = 30<br>x = 30 {tex}\\div 2 {\/tex}<br>x = 15 hours<br>Therefore, it will take 15 hours to fill 5 tanks.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23: A given set of chairs are arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The new arrangement will have 15 rows.<br>This is inverse proportion (more chairs per row {tex}\\rightarrow{\/tex} fewer rows needed).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original: 25 rows with 12 chairs each<\/li>\n\n\n\n<li>New: 20 chairs in each row, find number of rows<\/li>\n<\/ul>\n\n\n\n<p>Step 1: Find total number of chairs<br>Total chairs {tex}=25 \\times 12=300{\/tex} chairs<br>Step 2: Find number of rows with 20 chairs each<br>Number of rows {tex}=300 \\div 20=15{\/tex} rows<br>Therefore, the new arrangement will have 15 rows.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24: A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Each period will be 40 minutes long.<br>This is inverse proportion (more periods {tex}\\rightarrow{\/tex}&nbsp;shorter duration of each period).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>8 periods of 45 minutes each<\/li>\n\n\n\n<li>Need to find duration when there are 9 periods<\/li>\n<\/ul>\n\n\n\n<p>Step 1: Find total school hours<br>Total time {tex}=8 \\times 45=360{\/tex} minutes<br>Step 2: Find duration of each period for 9 periods<br>Duration of each period {tex}=360 \\div 9=40{\/tex} minutes<br><strong>Alternative method using inverse proportion:<\/strong><br>{tex} x_1 y_1=x_2 y_2 {\/tex}<br>{tex} 8 \\times 45=9 \\times x {\/tex}<br>{tex} 360=9 x {\/tex}<br>{tex} x=360 \\div 9 {\/tex}<br>{tex} x=40 \\text { minutes } {\/tex}<br>Therefore, each period will be 40 minutes long.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25: A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768817138-cz5uzc.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The tank will fill in 1.2 hours (or 1 hour 12 minutes).<br>Step 1: Find work done by each pump in 1 hour<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Small pump fills the tank in 3 hours \u2192 {tex}\\ln 1{\/tex} hour, it fills {tex}1 \/ 3{\/tex} of the tank<\/li>\n\n\n\n<li>Large pump fills the tank in 2 hours \u2192 {tex}\\ln 1{\/tex} hour, it fills {tex}1 \/ 2{\/tex} of the tank<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong> Find work done by both pumps together in 1 hour<br>Work done in 1 hour {tex}=1 \/ 3+1 \/ 2{\/tex}<br>To add, find LCM of 3 and {tex}2=6=2 \/ 6+3 \/ 6=5 \/ 6{\/tex} of the tank<br><strong>Step 3:<\/strong> Find time to fill the whole tank<br>If {tex}5 \/ 6{\/tex} of tank is filled in 1 hour<br>Then full tank {tex}(1{\/tex} whole) will be filled in:<br>Time {tex}=1 \\div(5 \/ 6)=1 \\times(6 \/ 5)=6 \/ 5{\/tex} hours {tex}=1.2{\/tex} hours<br>Converting to minutes: 1.2 hours {tex}=1{\/tex} hour {tex}+0.2 \\times 60{\/tex} minutes {tex}=1{\/tex} hour 12 minutes<br>Therefore, both pumps together will fill the tank in 1.2 hours or 1 hour 12 minutes.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26: A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768817208-mvac6p.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>49 machines are required.<br>This is inverse proportion (fewer days {tex}\\rightarrow{\/tex} more machines needed).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>42 machines {tex}\\rightarrow 63{\/tex} days&nbsp;<\/li>\n\n\n\n<li>Need to find machines for 54 days<\/li>\n<\/ul>\n\n\n\n<p>Using inverse proportion: {tex}{x}_1 {y}_1={x}_2 {y}_2 {\/tex}<br>{tex}42 \\times 63= x\\times 54{\/tex}<br>{tex}2646=54 x{\/tex}<br>{tex}x=2646 \\div 54 {\/tex}<br>{tex}{x}=49{\/tex} machines<br>Therefore, 49 machines are required to produce the same number of toys in 54 days.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27: A car takes 2 hours to reach a destination, travelling at a speed of 60 km\/h. How long will the car take if it travels at a speed of 80 km\/h?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The car will take 1.5 hours (or 1 hour 30 minutes).<br>This is inverse proportion (higher speed {tex}\\rightarrow{\/tex} less time).<br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Speed {tex}=60 {~km} \/ {h} \\rightarrow{\/tex} Time {tex}=2{\/tex} hours<\/li>\n\n\n\n<li>Speed {tex}=80 {~km} \/ {h} \\rightarrow{\/tex} Time {tex}={\/tex} ?<\/li>\n<\/ul>\n\n\n\n<p>Using inverse proportion: {tex}{x}_1 {y}_1={x}_2 {y}_2 {\/tex}<br>{tex}60 \\times 2=80 \\times {x} {\/tex}<br>{tex}120=80 x{\/tex}<br>{tex}x=120 \\div 80{\/tex}<br>{tex} {x}=1.5{\/tex} hours<br>Converting to hours and minutes: 1.5 hours {tex}=1{\/tex} hour 30 minutes<br>Therefore, at {tex}80 {~km} \/ {h}{\/tex}, the car will take 1.5 hours or 1 hour 30 minutes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Class 8 Maths Ganita Prakash Solutions<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/a-square-and-a-cube-ncert-solutions-class-8-maths-ganita-prakash\/\">A Square and A Cube<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/power-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Power Play<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/a-story-of-numbers-ncert-solutions-class-8-maths-ganita-prakash\/\">A Story of Numbers<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/quadrilaterals-ncert-solutions-class-8-maths-ganita-prakash\/\">Quadrilaterals<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/number-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Number Play<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/we-distribute-yet-things-multiply-ncert-solutions-class-8-maths-ganita-prakash\/\">We Distribute Yet Things Multiply<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-1-ncert-solutions-class-8-maths-ganita-prakash\/\">Proportional Reasoning-1<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/\">Fractions In Disguise<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/the-baudhayana-pythagoras-theorem-ncert-solutions-class-8-maths-ganita-prakash\/\">The Baudhayana-Pythagoras Theorem<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\">Proportional Reasoning-2<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/exploring-some-geometric-themes-ncert-solutions-class-8-maths-ganita-prakash\/\">Exploring Some Geometric Themes<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/tales-by-dots-and-lines-ncert-solutions-class-8-maths-ganita-prakash\/\">Tales by dots and lines<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/mycbseguide.com\/blog\/algebra-play-ncert-solutions-class-8-maths-ganita-prakash\/\">Algebra Play<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>Proportional Reasoning-2 &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 8 Maths (Ganita Prakash). NCERT Solutions Class 8 Proportional Reasoning-2 \u2013 NCERT Solutions Q.1: Convert 60,00,000 cm to kilometres. Solution: 60,00,000 cm = 60 kmWe know that: Now, to convert {tex}60,00,000 {~cm}{\/tex} to {tex}{km}: &#8230; <a title=\"Proportional Reasoning-2 &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\" aria-label=\"More on Proportional Reasoning-2 &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,2083,2111],"tags":[216],"class_list":["post-31749","post","type-post","status-publish","format-standard","hentry","category-ncert-solutions","category-ncert-solutions-class-8","category-ncert-solutions-class-8-maths-ganita-prakash","tag-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Proportional Reasoning-2 - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Proportional Reasoning-2 - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Proportional Reasoning-2 - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Proportional Reasoning-2 - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2026-08-07T07:23:07+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-08-07T07:54:02+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768805003-wex4a4.jpg\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"18 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/proportional-reasoning-2-ncert-solutions-class-8-maths-ganita-prakash\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"Proportional Reasoning-2 &#8211; 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