{"id":31745,"date":"2026-08-07T12:35:33","date_gmt":"2026-08-07T07:05:33","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31745"},"modified":"2026-08-07T13:24:21","modified_gmt":"2026-08-07T07:54:21","slug":"fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/","title":{"rendered":"Fractions In Disguise &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)"},"content":{"rendered":"\n<p><strong><strong>Fractions In Disguise<\/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>Fractions In Disguise<\/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: Shambhavi owns a stationery shop. She procures 200 page notebooks at \u20b9 36 per book. She sells them with a profit margin of 20%. Find the selling price.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cost Price {tex}({CP})={\/tex} \u20b9 36 per book<\/li>\n\n\n\n<li>Profit margin {tex}=20 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: <\/strong>Calculate the profit amount Profit {tex}=20 \\%{\/tex} of CP<br>Profit {tex}=(\\frac {20 }{ 100}) \\times 36{\/tex}<br>Profit {tex}= 0.20 \\times 36{\/tex}<br>Profit {tex}={\/tex} \u20b9 7.20<br><strong>Step 2:<\/strong> Calculate Selling Price<br>Selling Price {tex}(S P)=C P+{\/tex} Profit<br>{tex}S P=36+7.20 {\/tex}<br>SP = \u20b9 43.20<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.2: A utensil store is offering a 35% discount on the cooker with an MRP \u20b9 1800. What is the selling price? If the cost price was \u20b9 900, what is the percentage profit made after the sale?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Part 1: Finding Selling Price<\/strong><br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{MRP}=\u20b9 1800{\/tex}<\/li>\n\n\n\n<li>Discount {tex}=35 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate discount amount Discount amount {tex}=35 \\%{\/tex} of 1800<br>Discount amount {tex}=(35 \/ 100) \\times 1800{\/tex}<br>Discount amount {tex}=0.35 \\times 1800{\/tex}<br>Discount amount {tex}=\u20b9 630{\/tex}<br><strong>Step 2:<\/strong> Calculate Selling Price<br>SP = MRP &#8211; Discount<br>SP = 1800 &#8211; 630<br>SP = \u20b9 1170<br><strong>Part 2: Finding Profit Percentage<\/strong><br><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cost Price (CP) = \u20b9 900<\/li>\n\n\n\n<li>Selling Price (SP) = \u20b9 1170<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate profit Profit = SP &#8211; CP Profit = 1170 &#8211; 900 Profit = \u20b9 270<br><strong>Step 2:<\/strong> Calculate profit percentage Profit percentage {tex}=({\/tex}Profit {tex}\/ {CP}) \\times 100 \\%{\/tex}<br>Profit percentage {tex}=(270 \/ 900) \\times 100 \\%{\/tex}<br>Profit percentage {tex}=0.30 \\times 100 \\%{\/tex}<br>Profit percentage = 30%<br>Therefore, the selling price is \u20b9 1170, and the shopkeeper makes a profit of {tex}30 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.3: Surya wants to use a deep orange colour to capture the sunset. He mixes some red paint and yellow paint to make this colour. The red paint makes up {tex}\\frac 34{\/tex} of this mixture. What percentage of the colour is made with red?&nbsp;<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}\\frac{3}{4}{\/tex} is 3 out of every 4.<br>That is, 6 out of every 8 (equivalent fraction).<br>That is, 30 out of every 40.<br>That is, 75 out of every 100.<br>{tex}\\frac{3}{4}=\\frac{6}{8}=\\frac{30}{40}=\\frac{75}{100}{\/tex}<br>This means {tex}75 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.4: Surya won some prize money in a contest. He wants to save {tex}\\frac{2}{5}{\/tex} of the money to purchase a new canvas. Express this quantity as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Try to understand the different methods for solving this problem, as shown below.<br><strong>Method 1:<\/strong><br>{tex} \\frac{2}{5}=\\frac{20}{50}=\\frac{40}{100} {\/tex}<br>{tex} =40 \\% . {\/tex}<br><strong>Method 2:<\/strong><br>{tex} \\frac{2}{5}=\\frac{x}{100} {\/tex}<br>{tex} x=\\frac{2}{5} \\times 100=40 {\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.5: Given a percentage, can you express it as a fraction? For example, express 24% as a fraction.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Since a percentage is a fraction, {tex}24 \\%{\/tex} is the same as {tex}\\frac{24}{100}{\/tex}.<br>We can find other equivalent forms of {tex}\\frac{24}{100}=\\frac{12}{50}=\\frac{6}{25}=\\frac{48}{200}{\/tex}.<br>In general, we can say that a percentage, {tex}z \\%{\/tex}, can be expressed by any of the fractions that are equivalent to {tex}\\frac{z}{100}{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.6: Express the {tex}\\frac 35{\/tex}&nbsp;fraction as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>To convert a fraction to percentage, multiply it by 100.<br>{tex}\\frac 35=(\\frac 35) \\times 100 \\% {\/tex}<br>{tex} =\\frac {(3 \\times 100) }{ 5 \\% }{\/tex}<br>{tex} =\\frac {300 }{ 5 \\% }{\/tex}<br>{tex} =60 \\% {\/tex}<br>Therefore, {tex}\\frac 3 5=60 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.7: Express the {tex}\\frac {7}{14}{\/tex}&nbsp;fraction as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>First, let&#8217;s simplify the fraction:<br>{tex}\\frac { 7}{ 14}= \\frac 1 2 {\/tex}<br>Now, converting to percentage:<br>{tex}\\frac 1 2=(\\frac 12) \\times 100 \\% {\/tex}<br>{tex} =\\frac {100 }{ 2 \\% }{\/tex}<br>{tex} =50 \\% {\/tex}<br>Therefore, {tex}\\frac {7 }{ 14}=50 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.8: Express the {tex}\\frac {9}{20}{\/tex}&nbsp;fraction as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Converting to percentage:<br>{tex}\\frac { 9}{20}=(\\frac {9 }{ 20}) \\times 100 \\% {\/tex}<br>{tex} =\\frac {(9 \\times 100) }{ 20 \\% }{\/tex}<br>{tex} =\\frac {900 }{ 20 \\%} {\/tex}<br>{tex} =45 \\% {\/tex}<br>Therefore, {tex}\\frac {9 }{ 20}=45 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.9: Express the {tex}\\frac {72}{150}{\/tex}&nbsp;fraction as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Converting to percentage:<br>{tex}\\frac { 72 }{ 150}=(\\frac {72 }{ 150}) \\times 100 \\% {\/tex}<br>{tex} =\\frac {(72 \\times 100) }{ 150 \\% }{\/tex}<br>{tex} =\\frac {7200 }{ 150 \\% }{\/tex}<br>{tex} =48 \\% {\/tex}<br>Therefore, {tex}\\frac {72 }{ 150}=48 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.10: Express the {tex}\\frac 13{\/tex}&nbsp;fraction as&nbsp;a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Converting to percentage:<br>{tex}\\frac 1 3=(\\frac 1 3) \\times 100 \\% {\/tex}<br>{tex} =\\frac {100 }{ 3 \\%} {\/tex}<br>{tex} =33.33 \\% \\text { (or } 33 \\frac{1}{3} \\% \\text { ) } {\/tex}<br>Therefore, {tex}\\frac 13{\/tex}&nbsp;= 33.33%<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.11: Express the {tex}\\frac {5}{11}{\/tex}&nbsp;fraction as a percentage.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Converting to percentage:<br>{tex}\\frac {5 }{11}=(\\frac {5 }{ 11}) \\times 100 \\% {\/tex}<br>{tex} =\\frac {500 }{ 11 \\%} {\/tex}<br>{tex}=45.45 \\% \\text { (approx.)}{\/tex}<br>Therefore,&nbsp;{tex}\\frac {5}{11}\\approx 45.45 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.12: Nandini has 25 marbles, of which 15 are white. What percentage of her marbles are white?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>10%<\/li>\n\n\n\n<li>15%<\/li>\n\n\n\n<li>25%<\/li>\n\n\n\n<li>60%<\/li>\n\n\n\n<li>40%<\/li>\n\n\n\n<li>None of these<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total marbles {tex}=25{\/tex}<br>White marbles = 15<br>Fraction of white marbles {tex}=\\frac {15 }{ 25}=\\frac 3 5{\/tex}<br>Converting to percentage:<br>{tex}(\\frac 3 5) \\times 100 \\%=\\frac {(3 \\times 100) }{ 5 \\%}{\/tex}&nbsp;{tex}=\\frac {300 }{ 5 \\%}=60 \\% {\/tex}<br>Therefore, the correct answer is (iv) 60%.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.13: In a school, 15 of the 80 students come to school by walking. What percentage of the students come by walking?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total students {tex}=80{\/tex}<br>Students who walk {tex}=15{\/tex}<br>Fraction of students who walk =&nbsp;{tex}\\frac {15}{80}{\/tex}<br>Simplifying: {tex}\\frac {15}{80}{\/tex} {tex}=\\frac {3 }{ 16}{\/tex}<br>Converting to percentage:<br>{tex}(\\frac {3 }{ 16}) \\times 100 \\%=\\frac {(3 \\times 100) }{ 16 \\%}{\/tex}&nbsp;{tex}=\\frac {300 }{ 16 \\%}=18.75 \\% {\/tex}<br>Therefore, {tex}18.75 \\%{\/tex} of students come to school by walking.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.14: A group of friends is participating in a long-distance run. The positions of each of them after 15 minutes are shown in the following picture. Match (among the given options) what percentage of the race each of them has approximately completed.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768556235-nxtngw.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Based on the positions shown in the diagram:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Person A<\/strong> is very close to the start {tex}\\rightarrow{\/tex} approximately {tex}{2 0} \\boldsymbol{\\%}{\/tex} completed<\/li>\n\n\n\n<li><strong>Person B<\/strong> is about one-third through the race {tex}\\rightarrow{\/tex} approximately {tex}{3 8 \\%}{\/tex} completed<\/li>\n\n\n\n<li><strong>Person C<\/strong> is more than halfway {tex}\\rightarrow{\/tex} approximately {tex}{5 5 \\%}{\/tex} completed<\/li>\n\n\n\n<li><strong>Person D<\/strong> is close to the finish {tex}\\rightarrow{\/tex} approximately {tex}{7 2 \\%}{\/tex} completed<\/li>\n<\/ul>\n\n\n\n<p><strong>Matching:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}{A} \\rightarrow 20 \\%{\/tex}<\/li>\n\n\n\n<li>{tex}{B} \\rightarrow 38 \\%{\/tex}<\/li>\n\n\n\n<li>{tex}{C} \\rightarrow 55 \\%{\/tex}<\/li>\n\n\n\n<li>{tex}{D} \\rightarrow 72 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.15: Pairs of quantities are shown below. Identify and write appropriate symbols \u2018&gt;\u2019, \u2018&lt;\u2019, \u2018=\u2019 in the blanks. Try to do it without calculations.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>50% ________ 5%<\/li>\n\n\n\n<li>{tex}\\frac{5}{10} {\/tex}&nbsp;________ 50%<\/li>\n\n\n\n<li>{tex}\\frac{3}{11}{\/tex}&nbsp;________ 61%<\/li>\n\n\n\n<li>{tex}30 \\%{\/tex} ________ {tex}\\frac{1}{3}{\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}50 \\%{\/tex} ________ 5%<br>50% &gt; 5%<br>({tex}50 \\%{\/tex} means 50 out of 100, while {tex}5 \\%{\/tex} means 5 out of 100. Clearly 50 is greater than 5.)<\/li>\n\n\n\n<li>{tex}\\frac {5}{1 0}{\/tex} ________&nbsp;50%<br>{tex}\\frac {5}{1 0}{\/tex}&nbsp;= 50%<br>({tex}\\frac {5}{1 0}{\/tex} {tex}=\\frac 1 2=50 \\%{\/tex})<\/li>\n\n\n\n<li>{tex}\\frac{{3}}{{1 1}}{\/tex} ________ 61%<br>{tex}\\frac{{3}}{{1 1}}{\/tex}&nbsp;&lt; 61%<br>{tex}(\\frac {3}{11} \\approx 27.27 \\%{\/tex}, which is less than {tex}61 \\%){\/tex}<\/li>\n\n\n\n<li>30% ________ {tex}\\frac 13{\/tex}<br>30% &lt; {tex}\\frac 13{\/tex}<br>{tex}(30 \\%=\\frac {30 }{ 100}=0.3{\/tex}, while {tex}\\frac 1 3 \\approx 0.333{\/tex} or {tex}33.33 \\%){\/tex}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.16: Madhu and Madhav each ate biscuits of a different variety. Madhu\u2019s biscuits had 25% sugar, while Madhav\u2019s had 35% sugar. Can you tell who ate more sugar?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>As we just saw, percentages represent fractional quantities or proportions. It would be inappropriate to compare just the percentages when they are referring to different quantities or wholes. That is, if they both had 100 g of biscuits, then clearly Madhav ate more sugar- 35 g (35% of 100 g is 35 g per 100 g) vs. Madhu\u2019s 25 g (25% of 100 g is 25 g per 100 g)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.17: We can find {tex}50 \\%{\/tex} of a value by multiplying {tex}\\frac{1}{2}{\/tex} with the value. Will multiplying the value by 0.5 also give the answer for {tex}50 \\%{\/tex} of the value?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, since {tex}\\frac{1}{2}=0.5{\/tex}.<br>{tex} 50 \\%=\\frac{50}{100}=\\frac{1}{2}=\\frac{0.5}{1}=0.5 . {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.18: The maximum marks in a test are 75. If students score 80% or above in the test, they get an A grade. How much should Zubin score at least to get an A grade?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>We can find 80% of 75 in different ways, using our understanding of fraction and decimal multiplication, as well as of proportionality.<br>Fraction Multiplication {tex}\\rightarrow \\frac{80}{100} \\times 75{\/tex}&nbsp;{tex} =\\frac{4}{5} \\times 75=60 . {\/tex}<br>Decimal Multiplication {tex}\\rightarrow 0.8 \\times 75=60{\/tex}.<br>Proportional Reasoning {tex}\\rightarrow{\/tex}&nbsp;Out of 100, the minimum mark is 80.<br>Out of 75, it is&nbsp;{tex} \\frac{75 \\times 80}{100}=60 \\text {.}{\/tex}<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768557430-9pck7x.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.19: To prepare a particular millet kanji (porridge), suppose the ratio of millet to water to be mixed for boiling is 2:7. What percentage does the millet constitute in this mixture? If 500 ml of the mixture is to be made, how much millet should be used?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This situation can be modelled as shown in the bar model on the right side.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768557618-y2etmx.jpg\"><br>The ratio of millet to the volume of the mixture is 2:9. In other words, in one unit of the mixture, millet occupies {tex}\\frac{2}{9}{\/tex} units and water occupies {tex}\\frac{7}{9}{\/tex} units.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.20: A cyclist cycles from Delhi to Agra and completes 40% of the journey. If he has covered 92 km, how many more kilometres does he have to travel to reach Agra?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let us first try to model this situation by a bar model.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768557864-dh5cyx.jpg\"><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.21: Kishanlal recently opened a garment shop. He aims to achieve a daily sales of at least \u20b9 5000. The sales on the first 2 days were \u20b9 2000 and \u20b9 3500. What percentage of his target did he achieve?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The percentage target achieved is visualised below.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768557967-wnbyww.jpg\"><br>% of target:<br>{tex} \\frac{2000}{5000} \\times 100=40 \\% {\/tex}<br>{tex} 40 \\%=\\frac{40}{100}=\\frac{2}{5}=0.4 {\/tex}<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768558012-c49s5w.jpg\"><br>% of target:<br>{tex} \\frac{3500}{5000} \\times 100=70 \\% {\/tex}<br>{tex} 70 \\%=\\frac{70}{100}=\\frac{7}{10}=0.7 {\/tex}<br>It is 40% on Day 1 and 70% on Day 2.<br>Another way of saying it is-&nbsp;he was 60% short of his target on Day 1 and 30% short of his target on Day 2.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.22: A farmer harvested 260 kg of wheat last year. This year, they harvested 650 kg of wheat. What percentage of last year\u2019s harvest is this year\u2019s harvest?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This year&#8217;s harvest {tex}=\\frac{650}{260} \\times 100=250 \\%{\/tex} of last year&#8217;s harvest.<br>{tex}250 \\%{\/tex} indicates that it is 2.5 times the original value.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.23: Find the missing numbers.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768559253-b9ypx8.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: {tex}20 \\%{\/tex} of a number {tex}=75{\/tex}<br>Let the number be x.<br>{tex} 20 \\% \\text { of } x=75 {\/tex}<br>{tex} (\\frac {20 }{ 100}) \\times x=75 {\/tex}<br>{tex} x=75 \\times(\\frac {100 }{ 20}) {\/tex}<br>{tex} x=75 \\times 5 {\/tex}<br>{tex} x=375 {\/tex}<br>Therefore, {tex}100 \\%{\/tex} of the number {tex}={3 7 5}{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.24: Find the missing numbers.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768559444-mqsuk3.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: 60% of a number {tex}=90{\/tex}<br>Let the number be x.<br>{tex} 60 \\% \\text { of } x=90 {\/tex}<br>{tex} (\\frac {60 }{100}) \\times x=90 {\/tex}<br>{tex} x=90 \\times(\\frac {100 }{ 60}) {\/tex}<br>{tex} x=90 \\times(\\frac {10 }{ 6}) {\/tex}<br>{tex} x=\\frac {900 }{ 6 }{\/tex}<br>{tex} x=150 {\/tex}<br>Therefore, {tex}100 \\%{\/tex} of the number {tex}=150{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.25: Find the missing numbers<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768559635-3b74x7.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The question needs to specify what value is obtained. Assuming we need to find {tex}100 \\%{\/tex} of 140:<br>{tex} 100 \\% \\text { of } 140={1 4 0} {\/tex}<br>If a different percentage is implied, please provide the complete information.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.26: Find the value of 25% of 160 and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}25 \\%{\/tex} of {tex}160=(\\frac {25 }{ 100}) \\times 160{\/tex}<br>{tex} =(\\frac 1 4) \\times 160 {\/tex}<br>{tex} =\\frac {160 }{ 4} {\/tex}<br>{tex} =40 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.27: Find the value of 16% of 250 and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}16 \\% \\text { of } 250=(\\frac {16 }{ 100}) \\times 250 {\/tex}<br>{tex} =\\frac {(16 \\times 250) }{ 100 }{\/tex}<br>{tex} =\\frac {4000 }{ 100 }{\/tex}<br>{tex} =40 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.28: Find the value of 62% of 360 and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}62 \\%{\/tex} of {tex}360=(\\frac {62 }{ 100}) \\times 360{\/tex}<br>{tex} =\\frac {(62 \\times 360) }{ 100 }{\/tex}<br>{tex} =\\frac {22,320 }{ 100 }{\/tex}<br>{tex} =223.2 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.29: Find the value of 140% of 40 and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}140 \\%{\/tex} of {tex}40=(\\frac {140 }{ 100}) \\times 40{\/tex}<br>{tex} =\\frac {(140 \\times 40) }{ 100 }{\/tex}<br>{tex} =\\frac {5600 }{ 100 }{\/tex}<br>{tex} =56 {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.30: Find the value of 1% of 1 hour and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>1 hour {tex}=60{\/tex} minutes<br>{tex} 1 \\% \\text { of } 60 \\text { minutes }=(\\frac {1 }{ 100}) \\times 60 {\/tex}<br>{tex} =\\frac {60 }{100 }{\/tex}<br>{tex} =0.6 \\text { minutes } {\/tex}<br>{tex} =0.6 \\times 60 \\text { seconds } {\/tex}<br>{tex} =36 \\text { seconds}{\/tex}<br>Therefore, {tex}1 \\%{\/tex} of 1 hour = {tex}{0 . 6}{\/tex} minutes or {tex}{3 6}{\/tex} seconds.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.31: Find the value of 7% of 10 kg and also draw their bar models.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>{tex}7 \\%{\/tex} of {tex}10 {~kg}=(\\frac {7 }{ 100}) \\times 10{\/tex}<br>{tex} =\\frac {70 }{ 100 }{\/tex}<br>{tex} =0.7 {~kg} {\/tex}<br>{tex} =700 \\text { grams}{\/tex}<br>Therefore, 7% of 10 kg = {tex}{0 . 7 ~ k g}{\/tex} or {tex}{7 0 0}{\/tex} grams.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.32: Surya made 60 ml of deep orange paint, how much red paint did he use if red paint made up {tex}\\frac{3}{4}{\/tex} of the deep orange paint?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total deep orange paint {tex}=60 {ml}{\/tex}<br>Red paint {tex}={\/tex} {tex}\\frac 34{\/tex} of total paint<br>Red paint {tex}=(\\frac 3 4) \\times 60 {ml}{\/tex}<br>{tex} =\\frac {(3 \\times 60) }{ 4} {ml} {\/tex}<br>{tex} =\\frac {180 }{ 4 }{ml} {\/tex}<br>{tex} =45 \\ {ml} {\/tex}<br>Therefore, Surya used {tex}{4 5 ~ m l}{\/tex} of red paint.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.33: Pairs of quantities are shown below. Identify and write appropriate symbols \u2018&gt;\u2019, \u2018&lt;\u2019, \u2018=\u2019 in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}50 \\% \\text { of } 510 \\ \\square \\ 50 \\% \\text { of } 515 {\/tex}<\/li>\n\n\n\n<li>{tex}37 \\% \\text { of } 148\\ \\square \\ 73 \\% \\text { of } 148 {\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}50 \\%{\/tex} of {tex}510=(\\frac {50 }{ 100}) \\times 510=255{\/tex}<br>{tex} 50 \\% \\text { of } 515=(\\frac {50 }{ 100}) \\times 515=257.5 {\/tex}<br>{tex} 255&lt;257.5 {\/tex}<br>Therefore, 50% of 510 &lt; 50% of 515<\/li>\n\n\n\n<li>Since the base value (148) is the same, we can directly compare the percentages.<br>{tex} 37 \\%&lt;73 \\% {\/tex}<br>Therefore, 37% of {tex}148&lt;73 \\%{\/tex} of 148<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.34: Pairs of quantities are shown below. Identify and write appropriate symbols \u2018&gt;\u2019, \u2018&lt;\u2019, \u2018=\u2019 in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}29 \\% \\text { of } 43 \\ \\square \\ 92 \\% \\text { of } 110 {\/tex}<\/li>\n\n\n\n<li>{tex}30 \\% \\text { of } 40 \\ \\square \\ 40 \\% \\text { of } 50 {\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}29 \\%{\/tex} of {tex}43 \\approx 0.29 \\times 43 \\approx 12.47{\/tex}<br>{tex}92 \\%{\/tex} of {tex}110=0.92 \\times 110=101.2{\/tex}<br>12.47 &lt; 101.2<br>Therefore, 29% of {tex}43&lt;92 \\%{\/tex} of 110<\/li>\n\n\n\n<li>30% of 40 {tex}=(\\frac {30 }{ 100}) \\times 40=12{\/tex}<br>{tex}40 \\%{\/tex} of {tex}50=(\\frac {40 }{ 100}) \\times 50=20{\/tex}<br>12 &lt; 20<br>Therefore, {tex}30 \\%{\/tex} of {tex}40&lt;40 \\%{\/tex} of 50<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.35: Pairs of quantities are shown below. Identify and write appropriate symbols \u2018&gt;\u2019, \u2018&lt;\u2019, \u2018=\u2019 in the boxes. Visualising or estimating can help. Compute only if necessary or for verification.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}45 \\%{\/tex} of 200 {tex}\\square{\/tex} 10% of 490<\/li>\n\n\n\n<li>{tex}30 \\%{\/tex} of 80 {tex}\\square{\/tex}&nbsp;{tex}24 \\%{\/tex} of 64<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>45% of {tex}200=(\\frac {45 }{ 100}) \\times 200=90{\/tex}<br>{tex} 10 \\% \\text { of } 490=(\\frac {10 }{ 100}) \\times 490=49 {\/tex}<br>{tex} 90&gt;49 {\/tex}<br>Therefore, 45% of {tex}200&gt;10 \\%{\/tex} of 490<\/li>\n\n\n\n<li>{tex}30 \\%{\/tex} of {tex}80=(\\frac {30 }{ 100}) \\times 80=24{\/tex}<br>{tex} 24 \\% \\text { of } 64=(\\frac {24 }{ 100}) \\times 64=15.36 {\/tex}<br>{tex} 24&gt;15.36 {\/tex}<br>Therefore, {tex}30 \\%{\/tex} of {tex}80&gt;24 \\%{\/tex} of 64<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.36: 30% of k is 70, 60% of k is ________, 90% of k is ________, 120% of k is ________.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: 30% of {tex}{k}=70{\/tex}<br>First, let&#8217;s find k:<br>{tex} (\\frac {30 }{ 100}) \\times k=70 {\/tex}<br>{tex} k=70 \\times(\\frac {100 }{ 30}) {\/tex}<br>{tex} k=\\frac {7000 }{ 30 }{\/tex}<br>{tex} k=\\frac {700 }{ 3}=233.33 {\/tex}<br>Now finding the required values:<br><strong>60% of <\/strong>{tex}{k}=60 \\%{\/tex} is double of 30%<br>So, {tex}60 \\%{\/tex} of {tex}{k}=2 \\times 70=140{\/tex}<br>{tex}{9 0 \\%}{\/tex} of {tex}{k}=90 \\%{\/tex} is three times of {tex}30 \\%{\/tex}<br>So, {tex}90 \\%{\/tex} of {tex}{k}=3 \\times 70=210{\/tex}<br><strong>120% of<\/strong> {tex}{k}=120 \\%{\/tex} is four times of 30%<br>So, {tex}120 \\%{\/tex} of {tex}{k}=4 \\times 70={2 8 0}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.37: 100% of m is 215, 10% of m is ________, 1% of m is ________, 6% of m is ________.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: 100% of m = 215<br>This means {tex}{m}=215{\/tex}<br>{tex} {1 0 \\%} \\text { of } {m}=(\\frac {10 }{ 100}) \\times 215=\\frac {215 }{ 10}={2 1 . 5} {\/tex}<br>{tex} {1 \\%} \\text { of } {m}=(\\frac {1 }{ 100}) \\times 215=\\frac {215 }{ 100}={2 . 1 5} {\/tex}<br>{tex} {6 \\%} \\text { of } {m}=6 \\times(1 \\% \\text { of } m)=6 \\times 2.15={1 2 . 9} {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.38: 90% of n is 270, 9% of n is ________, 18% of n is ________, 100% of n is ________.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: {tex}90 \\%{\/tex} of {tex}n=270{\/tex}<br>9% of {tex}{n}={9 \\%}{\/tex} is one-tenth of 90%<br>So, {tex}9 \\%{\/tex} of {tex}n=\\frac {270 }{ 10}=30{\/tex}<br>18% of {tex}{n}=18 \\%{\/tex} is double of 9%<br>So, {tex}18 \\%{\/tex} of {tex}n=2 \\times 30=60{\/tex}<br>100% of {tex}n={\/tex} First find {tex}n{\/tex}:<br>{tex}(\\frac {90 }{ 100}) \\times n=270{\/tex}<br>{tex}n=270 \\times(\\frac {100 }{ 90}){\/tex}<br>{tex}{n}=\\frac {27000 }{ 90}{\/tex}<br>{tex}n=300{\/tex}<br>Therefore, 100% of {tex}n={3 0 0}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.39: 3 is ________% of 300.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the percentage be {tex}{x} \\%{\/tex}.<br>{tex} x \\% \\text { of } 300=3 {\/tex}<br>{tex} (\\frac {x }{ 100}) \\times 300=3 {\/tex}<br>{tex} 3 x=3 {\/tex}<br>{tex} x=\\frac 33=1 {\/tex}<br>Therefore, 3 is {tex}{1 \\%}{\/tex} of 300.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.40: ________ is 40% of 4.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the number be y.<br>{tex} y=40 \\% \\text { of } 4 {\/tex}<br>{tex} y=(\\frac {40 }{ 100}) \\times 4 {\/tex}<br>{tex} y=\\frac {160 }{ 100} {\/tex}<br>{tex} y=1.6 {\/tex}<br>Therefore, {tex}{1 . 6}{\/tex} is {tex}40 \\%{\/tex} of 4.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.41: 40 is 80% of ________.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the number be {tex}z{\/tex}.<br>{tex} 40=80 \\% \\text { of } z {\/tex}<br>{tex} 40=(\\frac {80 }{ 100}) \\times z {\/tex}<br>{tex} 40=(\\frac 4 5) \\times z {\/tex}<br>{tex} z=40 \\times(\\frac 5 4) {\/tex}<br>{tex} z=\\frac {200 }{ 4 }{\/tex}<br>{tex} z=50 {\/tex}<br>Therefore, 40 is {tex}80 \\%{\/tex} of 50.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.42: Is 10% of a day longer than 1% of a week?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let&#8217;s calculate both:<br>{tex}{1 0 \\%}{\/tex} <strong>of a day:<\/strong> 1 day {tex}=24{\/tex} hours {tex}10 \\%{\/tex} of 24 hours {tex}=(\\frac {10 }{ 100}) \\times 24=2.4{\/tex} hours<br><strong>1% of a week:<\/strong> 1 week {tex}=7{\/tex} days {tex}=7 \\times 24=168{\/tex} hours 1% of 168 hours {tex}=(\\frac {1 }{ 100}) \\times{\/tex} 168 = 1.68 hours<br>Comparing: 2.4 hours &gt; 1.68 hours<br>Therefore, Yes, 10% of a day (2.4 hours) is longer than 1% of a week (1.68 hours).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.43: Mariam\u2019s farm has a peculiar bull. One day she gave the bull 2 units of fodder and the bull ate 1 unit. The next day, she gave the bull 3 units of fodder and the bull ate 2 units. The day after, she gave the bull 4 units and the bull ate 3 units. This continued, and on the 99th day she gave the bull 100 units and the bull ate 99 units. Represent these quantities as percentages. This task can be distributed among the class. What do you observe?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let&#8217;s calculate the percentage eaten each day:<br><strong>Day 1:<\/strong> Ate 1 out of {tex}2=(\\frac 1 2) \\times 100 \\%=50 \\%{\/tex}<br><strong>Day 2:<\/strong> Ate 2 out of {tex}3=(\\frac 2 3) \\times 100 \\%=66.67 \\%{\/tex}<br><strong>Day 3:<\/strong> Ate 3 out of {tex}4=(\\frac 3 4) \\times 100 \\%=75 \\%{\/tex}<br><strong>Day 4:<\/strong> Ate 4 out of {tex}5=(\\frac 4 5) \\times 100 \\%=80 \\%{\/tex}<br><strong>Day 5:<\/strong> Ate 5 out of {tex}6=(\\frac 5 6) \\times 100 \\%=83.33 \\%{\/tex}<br>&#8230;continuing this pattern&#8230;<br><strong>Day 10:<\/strong> Ate 10 out of {tex}11=(\\frac {10 }{ 11}) \\times 100 \\%=90.91 \\%{\/tex}<br><strong>Day 20:<\/strong> Ate 20 out of {tex}21=(\\frac {20 }{ 21}) \\times 100 \\%=95.24 \\%{\/tex}<br><strong>Day 50:<\/strong> Ate 50 out of {tex}51=(\\frac {50 }{ 51}) \\times 100 \\%=98.04 \\%{\/tex}<br><strong>Day 99:<\/strong> Ate 99 out of {tex}100=(\\frac {99 }{ 100}) \\times 100 \\%=99 \\%{\/tex}<br><strong>Observation:&nbsp;<\/strong>The percentage of fodder eaten by the bull increases each day and approaches {tex}100 \\%{\/tex} as the days progress. The bull&#8217;s eating percentage follows the pattern: {tex}(\\frac {n }{(n+1)}) \\times 100 \\%{\/tex}, where {tex}n{\/tex} is the day number. As {tex}n{\/tex} increases, the percentage gets closer and closer to {tex}100 \\%{\/tex}, but never quite reaches it.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.44: Workers in a coffee plantation take 18 days to pick coffee berries in 20% of the plantation. How many days will they take to complete the picking work for the entire plantation, assuming the rate of work stays the same? Why is this assumption necessary?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Time taken for {tex}20 \\%{\/tex} of plantation {tex}=18{\/tex} days<\/li>\n\n\n\n<li>Rate of work remains constant<\/li>\n<\/ul>\n\n\n\n<p><strong>Method 1: Using Proportion<\/strong><br>If 20% takes 18 days, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}20 \\% \\rightarrow 18{\/tex} days<\/li>\n\n\n\n<li>{tex}100 \\% \\rightarrow{\/tex} ?<\/li>\n<\/ul>\n\n\n\n<p>Using proportion: {tex}20: 18:: 100: x{\/tex}<br>{tex} x=\\frac {(100 \\times 18) }{ 20 }{\/tex}<br>{tex} x=\\frac {1800 }{ 20 }{\/tex}<br>{tex} x=90 \\text { days}{\/tex}<br><strong>Method 2: Using Logic<\/strong><br>{tex}20 \\%{\/tex} of work takes 18 days {tex}100 \\%=5 \\times 20 \\%{\/tex}<br>So, {tex}100 \\%{\/tex} will take {tex}=5 \\times 18=90{\/tex} days<br><strong>Why is the assumption necessary?<\/strong><br>The assumption that the rate of work stays the same is necessary because:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Workers might get tired over time, reducing their efficiency<\/li>\n\n\n\n<li>Different parts of the plantation might have varying difficulty levels<\/li>\n\n\n\n<li>Weather conditions might change<\/li>\n\n\n\n<li>Some workers might take breaks or leave<\/li>\n<\/ol>\n\n\n\n<p>Without this assumption, we cannot accurately predict the total time needed, as the rate could vary significantly.<br>Therefore, workers will take {tex}{9 0}{\/tex} <strong>days<\/strong> to complete the entire plantation.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.45: The badminton coach has planned the training sessions such that the ratio of warm up : play : cool down is 10% : 80% : 10%. If he wants to conduct a training of 90 minutes. How long should each activity be done?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total training time {tex}=90{\/tex} minutes<br>The ratio is: Warm up : Play : Cool down {tex}=10 \\%: 80 \\%: 10 \\%{\/tex}<br><strong>Warm up time:<\/strong> {tex}10 \\%{\/tex} of 90 minutes {tex}=(\\frac {10 }{ 100}) \\times 90=9{\/tex} minutes<br><strong>Play time:<\/strong> 80% of 90 minutes {tex}=(\\frac {80 }{100}) \\times 90=72{\/tex} minutes<br><strong>Cool down time:<\/strong> {tex}10 \\%{\/tex} of 90 minutes {tex}=(\\frac {10 }{ 100}) \\times 90=9{\/tex} minutes<br><strong>Verification:<\/strong> {tex}9+72+9=90{\/tex} minutes<br>Therefore:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Warm up {tex}=9{\/tex} minutes<\/strong><\/li>\n\n\n\n<li><strong>Play {tex}=72{\/tex} minutes<\/strong><\/li>\n\n\n\n<li><strong>Cool down {tex}=9{\/tex} minutes<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.46: An estimated 90% of the world\u2019s population lives in the Northern Hemisphere. Find the (approximate) number of people living in the Northern Hemisphere based on this year\u2019s worldwide population.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Current world population {tex}(2024) \\approx 8{\/tex} billion {tex}(8,000,000,000){\/tex}<br>Population in Northern Hemisphere {tex}=90 \\%{\/tex} of world population<br>{tex} =(\\frac {90 }{ 100}) \\times 8,000,000,000 {\/tex}<br>{tex} =(\\frac {9 }{ 10}) \\times 8,000,000,000 {\/tex}<br>{tex} =\\frac {72,000,000,000 }{ 10} {\/tex}<br>{tex} =7,200,000,000 {\/tex}<br>{tex} =7.2 \\text { billion}{\/tex}<br>Therefore, approximately {tex}{7 . 2}{\/tex} billion people (or {tex}{7 , 2 0 0}{\/tex} million people) live in the Northern Hemisphere.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.47: A recipe for the dish, halwa, for 4 people has the following ingredients in the given proportions-\u2009Rava: 40%, Sugar: 40%, and Ghee: 20%.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>If you want to make halwa for 8 people, what is the proportion of each of the above ingredients?<\/li>\n\n\n\n<li>If the total weight of the ingredients is 2 kg, how much rava, sugar and ghee are present?<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The proportions remain the same regardless of the number of people.<br>When we increase the quantity from 4 people to 8 people, we double the amounts, but the proportions (percentages) stay constant.<br>Therefore:<ul><li><strong>Rava: 40%<\/strong><\/li><li><strong>Sugar: 40%<\/strong><\/li><li><strong>Ghee: 20%<\/strong><\/li><\/ul>The proportions do not change when scaling the recipe.<\/li>\n\n\n\n<li>Total weight {tex}=2 {~kg}=2000{\/tex} grams<br><strong>Rava:<\/strong> {tex}40 \\%{\/tex} of {tex}2000 {~g}=(\\frac {40 }{ 100}) \\times 2000=800{\/tex} grams {tex}=0.8 {~kg}{\/tex}<br><strong>Sugar:<\/strong> 40% of {tex}2000 {~g}=(\\frac {40 }{ 100}) \\times 2000=800{\/tex} grams {tex}=0.8 {~kg}{\/tex}<br><strong>Ghee:<\/strong> 20% of {tex}2000 {~g}=(\\frac {20 }{ 100}) \\times 2000=400{\/tex} grams {tex}=0.4 {~kg}{\/tex}<br><strong>Verification: <\/strong>{tex}800+800+400=2000{\/tex} grams<br>Therefore:\n<ul class=\"wp-block-list\">\n<li>Rava {tex}=800{\/tex} grams {tex}(0.8 {~kg}){\/tex}<\/li>\n\n\n\n<li>Sugar {tex}=800{\/tex} grams (0.8 kg)<\/li>\n\n\n\n<li>Ghee {tex}=400{\/tex} grams {tex}(0.4 {~kg}){\/tex}<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.48: Eesha scored 42 marks out of 50 on an English test and 70 marks out of 80 in a Science test. Since she lost only 8 marks in English but 10 marks in Science, she thinks she has done better at English. Reema does not agree! She argues that since Eesha has scored more marks in Science, she has done better at Science. Vishu thinks we cannot compare the scores because the maximum marks are different. Who do you think is correct?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>If the maximum marks are the same, the comparison becomes easier, isn&#8217;t it? For these kinds of comparisons, we need to convert both values to percentages.<br>English score as a percentage {tex}=\\frac{42}{50} \\times 100=84 \\%{\/tex}<br>Science score as a percentage {tex}=\\frac{70}{80} \\times 100=87.5 \\%{\/tex}.<br>The Science score (as a percentage) is higher than the English score (as a percentage). So, we can conclude that Eesha has scored better on the Science test.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.49: Madhu and Madhav recently learnt about the importance of reading labels on processed food before purchase. They are at a shop to buy badam drink mix. They are looking at two products and wondering which has a larger share of badam. Can you figure it out? Which product uses a smaller proportion of food chemicals?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>It is easier to compare the proportions of the ingredients if we convert them into percentages. For example,<br>DEF\u2019s sugar content as a percentage of total weight&nbsp;{tex}=\\frac{99}{150} \\times 100=66 \\%{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.50: Do the following two statements mean the same thing?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The population of this state in 1991 is 165% of that in 1961.<\/li>\n\n\n\n<li>The population of this state has increased by 65% from 1961 to 1991.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Yes, both mean the same. Suppose p is the population of the state in 1961 and q is the population of the state in 1991.<br><strong>Statement A implies,<\/strong><br>{tex} q=165 \\% \\text { of } p {\/tex}<br>{tex} q=\\frac{165}{100} \\times p=1.65 p {\/tex}<br><strong>Statement B implies,<\/strong><br>{tex} q=p+65 \\% \\text { of } p {\/tex}<br>{tex} q=p+0.65 \\times p=1.65 p {\/tex}<br>In other words, the population of the state in 1991 is 1.65 times that in 1961.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.51: Find out the percentage profit Kishanlal made on this sweater.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768566457-6nz6em.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>We shall consider the cost price to be 100% to find out the percentage profit made with reference to the cost price. The following rough diagram describes this situation.<br>The profit amount is \u20b9 130.<br>The percentage profit is&nbsp;{tex}\\frac{130}{300} \\times 100=43.3 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.52: The rice stock in Raghu\u2019s provision store is getting old. He had purchased the rice at \u20b9 35 per kg. To clear his stock, he sells 10 kg rice for \u20b9 300. Find out the percentage loss.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The amount Raghu had paid towards buying the 10 kg rice is \u20b9 350. He sold it for \u20b9 300.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768566774-y5d7fq.jpg\"><br>The loss is \u20b9 350 &#8211; \u20b9 300 = \u20b9 50.<br>The percentage loss is&nbsp;{tex}\\frac{50}{350} \\times 100=14.28 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.53: Shyamala had procured decorative vases at \u20b9 2650 per piece. One of the pieces was slightly damaged. She decides to sell it at a loss of 18%. How much will she get by selling this piece?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Two methods of solving this are shown.<br><strong>Method 1:<\/strong><br>With respect to the buying price being 100%, the selling price is {tex}18 \\%{\/tex} less than the buying price. That is, the selling price would be {tex}82 \\%{\/tex}.<br>{tex} 82 \\% \\text { of } 2650=0.82 \\times 2650=2173 . {\/tex}<br><strong>Method 2:<\/strong><br>The loss amount is 18% of 2650.<br>That is, {tex}\\frac {18}{100}\\times 2650{\/tex}&nbsp;= 477.<br>Reducing this from the buying price, 2650 &#8211; 477 = 2173.<br>The sale amount of the damaged vase would be \u20b9 2173.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.54: If one deposits \u20b9 6000 in the bank, what is the amount after 3 years?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>That depends on the choice of FD. There are two possibilities:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Option 1:<\/strong> The interest is paid out regularly (for example, every year).<br>The principal amount is returned after the maturity period.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768818292-a7g82n.jpg\"><\/li>\n\n\n\n<li><strong>Option 2:<\/strong> The interest gained every time (say after each year) is added back to the FD, thus increasing the principal amount for the subsequent period. After the maturity period, the entire amount is returned. This phenomenon is called <strong>compounding<\/strong>.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768818502-qazysa.jpg\"><br>We can see that with compounding, the final amount is more.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.55: What percent is the total amount received with respect to the amount deposited in both the options?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>This can be calculated by finding {tex}\\frac{\\text { total amount received }}{\\text { amount deposited }} \\times 100{\/tex}.<br><strong>Without Compounding<\/strong><br>{tex} \\frac{7800}{6000} \\times 100=130 \\%=1.3 \\text {.}{\/tex}<br>In other words, the total amount<br>{tex} \\text { received }= 6000 \\times(1+0.1+0.1+0.1) {\/tex}<br>{tex} = 6000 \\times 1.3 {\/tex}<br>The percentage gain over 3 years is {tex}30 \\%{\/tex}.<br><strong>With Compounding<\/strong><br>{tex} \\frac{7986}{6000} \\times 100=133.1 \\%=1.331 {\/tex}<br>In other words, the total amount received<br>{tex} =6000 \\times 1.1 \\times 1.1 \\times 1.1 {\/tex}<br>{tex} =6000 \\times 1.331 {\/tex}<br>The percentage gain over 3 years is {tex}33.1 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.56: What is the amount we get back if we invest \u20b9 6000 at an interest rate of 10% p.a. for \u2018t\u2019 years?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>No Compounding<\/strong><br>Here, the interest gained every term is paid back. Therefore, the principal amount for every term shall remain the same, and as a result, the interest gained every term also shall be the same.<br>The interest gained in 1 term is {tex}6000 \\times 0.1{\/tex}<br>The interest gained in 3 terms is&nbsp;{tex} 6000 \\times 0.1 \\times 3 {\/tex}<br>The total amount at the end of an FD of 3 years is&nbsp;{tex} 6000+(6000 \\times 0.1 \\times 3) . {\/tex}<br>The interest gained in 1 term is {tex}p \\times r{\/tex} ({tex}p{\/tex} is the principal, {tex}r{\/tex} is the rate of interest in percentage)<br>The interest gained in {tex}t{\/tex} terms is&nbsp;{tex} p \\times r \\times t {\/tex}<br>The total amount at the end of an FD of {tex}t{\/tex} years is&nbsp;{tex} p+(p \\times r \\times t)=p+p r t {\/tex}<br>{tex} =p(1+r t) . {\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.57: A TV is bought at a price of \u20b9 21,000. After 1 year, the value of the TV depreciates by 5%. Find the value of the TV after one year.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The amount of reduction in the&nbsp;{tex}\\text {value is } 5 \\% \\text { of } {\/tex}21,000<br>{tex} =0.05 \\times 21,000 {\/tex}&nbsp;{tex} =1050 {\/tex}.<br>The current value is {tex}21,000-1050{\/tex}<br>= 19,950.<br>The value of the TV after 1 year will be {tex}95 \\%{\/tex} of the current value<br>{tex} =95 \\% \\text { of } 21,000 =0.95 \\times 21,000 {\/tex}<br>{tex} =19,950 . {\/tex}<br>The value of the TV after 1 year will be \u20b9 19,950.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.58: The population of a village was observed to be reducing by about 10% every decade. If the current population is 1250, what is the expected population after 3 decades?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>The population 1 decade later will be 0.9 times the population of the current decade.<br>Therefore, the population after 1 decade will be {tex}1250 \\times 0.9{\/tex}.<br>The population after 2 decades will be {tex}1250 \\times 0.9 \\times 0.9{\/tex}.<br>The population after 3 decades will be {tex}1250 \\times 0.9 \\times 0.9 \\times 0.9 =911.25{\/tex}.<br>First&nbsp;{tex}\\text {decade&#8217;s decrease } =0.1 \\times 1250 {\/tex}&nbsp;{tex} =125 {\/tex}<br>Population after 1 decade&nbsp;{tex} =1250-125=1125 . {\/tex}<br>Second decade&#8217;s decrease&nbsp;{tex} =0.1 \\times 1125=112.5 \\cong 112 . {\/tex}<br>Population after 2 decades&nbsp;{tex} \\text { = } 1125-112=1013 {\/tex}<br>Third decade&#8217;s population decrease&nbsp;{tex} =0.1 \\times 1013=101.3 \\cong 101 . {\/tex}<br>Population after 3 decades&nbsp;{tex} \\text { = } 1013-101 \\text { = } 912 . {\/tex}<br>Rounding off, we can say that the expected population after 3 decades will be around 910.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.59: A bakery called Cakely is offering a 30% + 20% discount on all cakes. Another bakery called Cakify is offering a 50% discount on all cakes. Would you rather choose Cakely or Cakify if you want the cheaper cost?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>It seems that both the options should give the same benefit. Although mathematically {tex}30 \\%+20 \\%{\/tex} is the same as {tex}50 \\%{\/tex}, the usage of {tex}30 \\%+20 \\%{\/tex} in shopping means compounding.<br>Suppose you want to buy a cake worth \u20b9 200.<br>Cakely&#8217;s {tex}30 \\%+20 \\% \\rightarrow{\/tex}<br>Applying the {tex}30 \\%{\/tex} discount {tex}\\rightarrow{\/tex} the price of cake is \u20b9 200 &#8211; \u20b9 60 = \u20b9 140.<br>Applying the {tex}20 \\%{\/tex} discount on \u20b9 140 {tex}\\rightarrow{\/tex}&nbsp;the price of cake is \u20b9 140 &#8211; \u20b9 28 = \u20b9 112.<br>Cakify&#8217;s 50% {tex}\\rightarrow{\/tex} The 50% discount makes the price of the cake \u20b9 100.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.60: After Surbhi bought cookware from the wholesaler, she kept a profit margin of 50% on all the products. To clear off the remaining stock, she thought she would offer a 50% discount and come out without any loss.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Do you think she didn\u2019t make any loss?<\/li>\n\n\n\n<li>If she had sold goods (originally) for \u20b9 12,000 after discount, how much loss did she incur? What is the percentage loss?<\/li>\n\n\n\n<li>What should have been the percentage discount offered so that she sold the goods at the price she had bought (i.e., no profit or loss)?<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let us model the situation first.<br>Suppose the worth of the products she bought from the wholesaler is {tex}x{\/tex}.<br>The worth corresponding to the selling price (with a {tex}50 \\%{\/tex} margin) is {tex}1.5 x{\/tex}.<br>A {tex}50 \\%{\/tex} discount on this price will make the worth {tex}0.75 x{\/tex}.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769073067-eab5su.jpg\"><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>This means the selling price is {tex}\\frac{3}{4}{\/tex} of the price the goods were bought at, i.e., a {tex}25 \\%{\/tex} loss.<\/li>\n\n\n\n<li>If she had sold goods worth \u20b9 12,000,<br>{tex} 0.75 x=12,000 {\/tex}<br>{tex} x=16,000 . {\/tex}<br>She lost \u20b9 4000.<\/li>\n\n\n\n<li>To sell the goods at the same price, the discount offered should be<br>{tex} 1.5 x-d \\times(1.5 x)=x {\/tex}<br>{tex} d=\\frac{1}{3}=0.33 {\/tex}<br>The discount offered should have been {tex}33.33 \\%{\/tex}.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.61: If a shopkeeper buys a geometry box for \u20b9 75 and sells it for \u20b9 110, what is his profit margin with respect to the cost?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cost Price (CP) {tex}={\/tex} \u20b9 75<\/li>\n\n\n\n<li>Selling Price (SP) = \u20b9 110<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate profit Profit {tex}=S P-C P{\/tex}<br>Profit {tex}=110-75{\/tex}<br>Profit {tex}=\u20b9 35{\/tex}<br><strong>Step 2:<\/strong> Calculate profit percentage with respect to cost Profit percentage = {tex}\\frac {\\text {Profit}}{\\text {CP}}{\/tex}{tex}\\times 100 \\%{\/tex} Profit percentage {tex}=(\\frac {35 }{ 75}) \\times 100 \\%{\/tex}<br>Profit percentage {tex}=0.4667 \\times 100 \\%{\/tex}<br>Profit percentage {tex}=46.67 \\%{\/tex} (approximately {tex}47 \\%{\/tex})<br>Therefore, the shopkeeper&#8217;s profit margin is {tex}46.67 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.62: I am a carpenter and I make chairs. The cost of materials for a chair is \u20b9 475 and I want to have a profit margin of 50%. At what price should I sell a chair?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Cost Price {tex}({CP})={\/tex} \u20b9 475<\/li>\n\n\n\n<li>Profit margin {tex}=50 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate profit amount Profit {tex}=50 \\%{\/tex} of CP Profit {tex}=(\\frac {50 }{ 100}) \\times 475{\/tex} Profit {tex}= 0.50 \\times 475{\/tex} Profit {tex}=\u20b9 237.50{\/tex}<br><strong>Step 2:<\/strong> Calculate Selling Price {tex}S P=C P+{\/tex} Profit<br>{tex}S P=475+237.50 {\/tex}<br>SP = \u20b9 712.50<br><strong>Alternative Method:<\/strong> {tex}{SP}={CP} \\times(1+{\/tex} profit percentage)<br>{tex} {SP}=475 \\times(1+0.50){\/tex}<br>{tex}SP= 475 \\times 1.50 {\/tex}<br>{tex}{SP}=\u20b9 \\ 712.50{\/tex}<br>Therefore, the carpenter should sell each chair at \u20b9 712.50.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.63: The total sales of a company (also called revenue) was \u20b9 2.5 crore last year. They had a healthy profit margin of 25%. What was the total expenditure (costs) of the company last year?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Revenue (Total Sales) = \u20b9 2.5 crore<\/li>\n\n\n\n<li>Profit margin {tex}=25 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Understanding:<\/strong> Profit margin of {tex}25 \\%{\/tex} means profit is {tex}25 \\%{\/tex} of the revenue.<br><strong>Step 1:<\/strong> Calculate profit amount Profit {tex}=25 \\%{\/tex} of Revenue<br>Profit {tex}=(25 \/ 100) \\times 2.5{\/tex} crore<br>Profit {tex}=0.25 \\times 2.5{\/tex} crore<br>Profit {tex}=\u20b9 0.625{\/tex} crore<br><strong>Step 2:<\/strong> Calculate total expenditure Revenue {tex}={\/tex} Expenditure + Profit {tex}2.5={\/tex} Expenditure + 0.625<br>Expenditure {tex}=2.5-0.625{\/tex}<br>Expenditure {tex}=\u20b9 1.875{\/tex} crore<br>Converting to easier format: \u20b9 1.875 crore {tex}={\/tex} \u20b9 1,87,50,000<br>Therefore, the total expenditure of the company last year was \u20b9 1.875 crore or \u20b9 1,87,50,000.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.64: A clothing shop offers a 25% discount on all shirts. If the original price of a shirt is \u20b9 300, how much will Anwar have to pay to buy this shirt?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original Price (Marked Price) = \u20b9 300<\/li>\n\n\n\n<li>Discount = 25%<\/li>\n<\/ul>\n\n\n\n<p><strong>Method 1:<\/strong><br><strong>Step 1:<\/strong> Calculate discount amount<br>Discount {tex}=25 \\%{\/tex} of 300<br>Discount {tex}=(25 \/ 100) \\times{\/tex} 300<br>Discount {tex}=0.25 \\times 300{\/tex}<br>Discount {tex}=\u20b9 75{\/tex}<br><strong>Step 2:<\/strong> Calculate selling price Selling Price {tex}={\/tex} Original Price &#8211; Discount SP {tex}=300- 75{\/tex}<br>SP = \u20b9 225<br><strong>Method 2:<\/strong> {tex}S P={\/tex} Original Price {tex}\\times(1-{\/tex} discount percentage)<br>{tex} S P=300 \\times(1-0.25) {\/tex}<br>{tex}S P= 300 \\times 0.75 {\/tex}<br>SP = \u20b9 225<br>Therefore, Anwar will have to pay \u20b9 225 for the shirt.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.65: The petrol price in 2015 was \u20b9 60 and \u20b9 100 in 2025. What is the percentage increase in the price of petrol?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>50%<\/li>\n\n\n\n<li>40%<\/li>\n\n\n\n<li>60%<\/li>\n\n\n\n<li>66.66%<\/li>\n\n\n\n<li>140%<\/li>\n\n\n\n<li>160.66%<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Price in 2015 = \u20b9 60<\/li>\n\n\n\n<li>Price in {tex}2025=\u20b9 \\ 100{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate increase in price Increase = Price in 2025 &#8211; Price in 2015<br>Increase = 100 &#8211; 60 Increase = \u20b9 40<br><strong>Step 2:<\/strong> Calculate percentage increase Percentage increase = (Increase\/Original Price{tex}) \\times 100 \\%{\/tex}<br>Percentage increase {tex}=(40 \/ 60) \\times 100 \\%{\/tex}<br>Percentage increase {tex}=(2 \/ 3) \\times 100 \\%{\/tex}<br>Percentage increase {tex}=66.67 \\%{\/tex} (approximately {tex}66.66 \\%{\/tex})<br>Therefore, the correct answer is <strong>(iv) 66.66%.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.66: Samson bought a car for \u20b9 4,40,000 after getting a 15% discount from the car dealer. What was the original price of the car?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Price paid after discount {tex}({SP})=\u20b9 4,40,000{\/tex}<\/li>\n\n\n\n<li>Discount {tex}=15 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Understanding: <\/strong>If there&#8217;s a {tex}15 \\%{\/tex} discount, Samson paid {tex}85 \\%{\/tex} of the original price.<br><strong>Step 1:<\/strong> Set up the equation {tex}85 \\%{\/tex} of Original Price<br>{tex}=\u20b9 4,40,000(85 \/ 100) \\times{\/tex} Original Price<br>{tex}=4,40,0000.85 \\times{\/tex} Original Price<br>{tex}=4,40,000{\/tex}<br><strong>Step 2:<\/strong> Calculate original price<br>Original Price {tex}=4,40,000 \/ 0.85{\/tex}<br>Original Price {tex}= 4,40,000 \\div 0.85{\/tex}<br>Original Price {tex}={\/tex} \u20b9 5,17,647.06 (approximately \u20b9 5,17,647)<br><strong>Verification:&nbsp;<\/strong>Discount {tex}=15 \\%{\/tex} of 5,17,647 {tex}=0.15 \\times 5,17,647=\u20b9 77,647{\/tex}<br>Price after discount {tex}=5,17,647-77,647=\u20b9 4,40,000 \\checkmark{\/tex}<br>Therefore, the original price of the car was approximately \u20b9 5,17,647.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.67: 1600 people voted in an election and the winner got 500 votes. What percent of the total votes did the winner get? Can you guess the minimum number of candidates who stood for the election?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Part 1: Percentage of votes won<\/strong><br>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total votes {tex}=1600{\/tex}<\/li>\n\n\n\n<li>Winner&#8217;s votes {tex}=500{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Percentage of votes {tex}=({\/tex}Winner&#8217;s votes\/Total votes{tex}) \\times 100 \\%{\/tex}<br>Percentage of votes {tex}= (500 \/ 1600) \\times 100 \\%{\/tex}<br>Percentage of votes {tex}=(5 \/ 16) \\times 100 \\%{\/tex}<br>Percentage of votes {tex}={\/tex} 31.25%<br><strong>Part 2:<\/strong> <strong>Minimum number of candidates<\/strong><br>The winner got 500 votes out of 1600. Remaining votes {tex}=1600-500=1100{\/tex} votes<br>For the winner to win with 500 votes, no other candidate should have more than 500 votes.<br>If there was only 1 other candidate, that candidate would have 1100 votes (more than the winner) &#8211; so the winner wouldn&#8217;t win.<br>If there were 2 other candidates sharing 1100 votes equally: Each would get 550 votes (still more than 500) &#8211; winner wouldn&#8217;t win.<br>If there were 3 other candidates: Maximum any could get (if votes distributed unevenly) could be more than 500.<br>For the winner to definitely win with 500 votes, we need at least 3 candidates total (winner +2 others), where the other 1100 votes are distributed such that no single candidate gets more than 500.<br>Therefore, the winner got {tex}31.25 \\%{\/tex} of total votes, and the minimum number of candidates who stood for the election is {tex}{3}{\/tex} <strong>candidates<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.68: The price of 1 kg of rice was \u20b9 38 in 2024. It is \u20b9 42 in 2025. What is the rate of inflation? (Inflation is the percentage increase in prices.)<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Price in 2024 = \u20b9 38 per kg<\/li>\n\n\n\n<li>Price in 2025 = \u20b9 42 per kg<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: <\/strong>Calculate increase in price Increase = Price in 2025 &#8211; Price in 2024 Increase = 42 &#8211; 38 Increase = \u20b9 4<br><strong>Step 2:<\/strong> Calculate rate of inflation<br>Rate of inflation {tex}=({\/tex}Increase\/Original Price{tex}) \\times 100 \\%{\/tex}<br>Rate of inflation {tex}=(4 \/ 38) \\times 100 \\%{\/tex}<br>Rate of inflation {tex}=0.1053 \\times 100 \\%{\/tex}<br>Rate of inflation = 10.53% (approximately)<br>Therefore, the rate of inflation is approximately {tex}10.53 \\%{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.69: A number increased by 20% becomes 90. What is the number?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A number increased by {tex}20 \\%=90{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Let the original number be x.<br><strong>Step 1:<\/strong> Set up the equation Number {tex}+20 \\%{\/tex} of Number {tex}=90 {x}+(20 \/ 100) \\times {x}{\/tex}&nbsp;<br>{tex}=90 {x}+ 0.20 x=90{\/tex}<br>{tex}1.20 x=90{\/tex}<br><strong>Step 2:<\/strong> Solve for x<br>{tex}{x}=90 \/ 1.20{\/tex}<br>{tex} {x}=90 \\div 1.20 {\/tex}<br>{tex}{x}=75{\/tex}<br><strong>Verification:&nbsp;<\/strong>20% of 75 {tex}=(20 \/ 100) \\times 75=15{\/tex}<br>{tex}75+15=90{\/tex}<br>Therefore, the number is 75.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.70: A milkman sold two buffaloes for \u20b9 80,000 each. On one of them, he made a profit of 5% and on the other a loss of 10%. Find his overall profit or loss.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Selling price of each buffalo = \u20b9 80,000<\/li>\n\n\n\n<li><strong>Buffalo 1:<\/strong> Profit of {tex}5 \\%{\/tex}<\/li>\n\n\n\n<li><strong>Buffalo 2:<\/strong> Loss of {tex}10 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>For Buffalo 1 (5% profit):<\/strong><br>{tex} S P=C P+5 \\% \\text { of } C P \\ 80,000{\/tex}&nbsp;{tex}=C P \\times(1+0.05) \\ 80,000{\/tex}&nbsp;<br>{tex}=C P \\times 1.05 \\ C P_1{\/tex}&nbsp;{tex}=\\frac {80,000 }{ 1.05} {\/tex}<br>{tex} C P_1=\u20b9 \\ 76,190.48 {\/tex}<br><strong>For Buffalo 2 (10% loss):<\/strong><br>{tex} S P=C P-10 \\% \\text { of } C P{\/tex}&nbsp;<br>80,000&nbsp;{tex}=C P \\times(1-0.10) {\/tex}<br>80,000&nbsp;{tex}=C P \\times 0.90 C P_2\\ =\\frac {80,000 }{ 0.90} {\/tex}<br>{tex} C P_2=\u20b9 \\ 88,888.89 {\/tex}<br><strong>Overall calculation:<\/strong><br>Total Cost Price {tex}={CP}_1+{CP}_2{\/tex}<br>Total {tex}{CP}=76,190.48+88,888.89{\/tex}<br>Total {tex}{CP}={\/tex} \u20b9 1,65,079.37<br>Total Selling Price {tex}=80,000+80,000{\/tex}<br>Total SP {tex}=\u20b9 \\ 1,60,000{\/tex}<br>Overall Loss {tex}={\/tex} Total CP &#8211; Total SP<br>Overall Loss {tex}=1,65,079.37-1,60,000{\/tex}<br>Overall Loss = \u20b9 5,079.37<br>Percentage Loss {tex}=(5,079.37 \/ 1,65,079.37) \\times 100 \\%{\/tex}<br>Percentage Loss {tex}=3.08 \\%{\/tex}<br>Therefore, the milkman incurred an overall loss of approximately \u20b9 5,079 or 3.08%.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.71: The population of elephants in a national park increased by 5% in the last decade. If the population of the elephants last decade is p, the population now is<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>p {tex}\\times{\/tex} 0.5<\/li>\n\n\n\n<li>p {tex}\\times{\/tex} 0.05<\/li>\n\n\n\n<li>p {tex}\\times{\/tex} 1.5<\/li>\n\n\n\n<li>p {tex}\\times{\/tex} 1.05<\/li>\n\n\n\n<li>p + 1.50<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Original population {tex}={p}{\/tex}<\/li>\n\n\n\n<li>Increase {tex}=5 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1:<\/strong> Calculate new population<br>New population {tex}={\/tex} Original {tex}+5 \\%{\/tex} of Original<br>New population {tex}=p+(5 \/ 100) \\times p{\/tex}<br>New population {tex}=p+0.05 p{\/tex}<br>New population {tex}=1.05 p{\/tex}<br>New population {tex}={p} \\times 1.05{\/tex}<br>Therefore, the correct answer is (iv) {tex}{p} \\times {1 . 0 5}{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.72: Which of the following statement(s) mean the same as &#8211; &#8220;The demand for cameras has fallen by 85% in the last decade&#8221;?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The demand now is 85% of the demand a decade ago.<\/li>\n\n\n\n<li>The demand a decade ago was 85% of the demand now.<\/li>\n\n\n\n<li>The demand now is 15% of the demand a decade ago.<\/li>\n\n\n\n<li>The demand a decade ago was 15% of the demand now.<\/li>\n\n\n\n<li>The demand a decade ago was 185% of the demand now.<\/li>\n\n\n\n<li>The demand now is 185% of the demand a decade ago.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let&#8217;s analyze what &#8220;demand has fallen by {tex}85 \\%{\/tex}&#8221; means:<br>If original demand (a decade ago) {tex}=100 \\%{\/tex}<br>Fallen by {tex}85 \\%{\/tex} means the demand decreased by {tex}85 \\%{\/tex}<br>Current demand {tex}=100 \\%-85 \\%=15 \\%{\/tex}<br>So, current demand is {tex}15 \\%{\/tex} of what it was a decade ago.<br><strong>Checking each statement:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The demand now is {tex}85 \\%{\/tex} of the demand a decade ago.\n<ul class=\"wp-block-list\">\n<li>This would mean demand fell by only {tex}15 \\%{\/tex}. <strong>Incorrect<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The demand a decade ago was {tex}85 \\%{\/tex} of the demand now.\n<ul class=\"wp-block-list\">\n<li>This would mean demand increased. <strong>Incorrect<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The demand now is {tex}15 \\%{\/tex} of the demand a decade ago.\n<ul class=\"wp-block-list\">\n<li>If demand fell by {tex}85 \\%{\/tex}, then {tex}100 \\%-85 \\%=15 \\%{\/tex} remains. <strong>Correct \u2713<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The demand a decade ago was {tex}15 \\%{\/tex} of the demand now.\n<ul class=\"wp-block-list\">\n<li>This would mean demand increased significantly. <strong>Incorrect<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>The demand a decade ago was {tex}185 \\%{\/tex} of the demand now.<ul><li>Let&#8217;s check: If current demand {tex}=15 \\%{\/tex}, then original {tex}=100 \\%{\/tex}<\/li><li>{tex}100 \\%=(100 \/ 15) \\times 15 \\%=6.67 \\times 15 \\% \\approx 667 \\%{\/tex} of current<\/li><li>We can also think: If now {tex}=15{\/tex}, then ago {tex}=100{\/tex}<\/li><li>So ago {tex}=(100 \/ 15) \\times{\/tex} now {tex}\\approx 6.67 \\times{\/tex} now<\/li><li>Wait, let&#8217;s recalculate: If now is {tex}15 \\%{\/tex} of ago, then ago {tex}=(100 \/ 15) \\times{\/tex} now = 6.67&nbsp;{tex}\\times{\/tex}&nbsp;now<\/li><li>This doesn&#8217;t match {tex}185 \\%{\/tex}. <strong>Incorrect<\/strong><\/li><\/ul>Actually, let me recalculate ({tex}v{\/tex}): If current demand {tex}=x{\/tex}, and it is {tex}15 \\%{\/tex} of original demand {tex}({d}){\/tex} Then {tex}{x}=0.15 {~d}{\/tex} So {tex}{d}={x} \/ 0.15=(100 \/ 15) {x}=6.67 {x}{\/tex}<br>This means original demand was 667% of current demand, not 185%. <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>The demand now is {tex}185 \\%{\/tex} of the demand a decade ago.<ul><li>This would mean demand increased by {tex}85 \\%{\/tex}. <strong>Incorrect<\/strong><\/li><\/ul>Therefore, only statement (iii) is correct.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.73: Bank of Yahapur offers an interest of 10% p.a. Compare how much one gets if they deposit \u20b9 20,000 for a period of 2 years with compounding and without compounding annually.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Principal {tex}(P)=\u20b9\\ 20,000{\/tex}<\/li>\n\n\n\n<li>Rate of interest {tex}(r)=10 \\%{\/tex} p.a. {tex}=0.10{\/tex}<\/li>\n\n\n\n<li>Time period {tex}({t})=2{\/tex} years<\/li>\n<\/ul>\n\n\n\n<p><strong>Without Compounding (Simple Interest):<\/strong><br>Interest for Year 1 {tex}={P} \\times {r}=20,000 \\times 0.10={\/tex} \u20b9 2,000<br>Interest for Year 2 {tex}={P} \\times {r}= 20,000 \\times 0.10=\u20b9 2,000{\/tex}<br>Total Interest {tex}=2,000+2,000=\u20b9\\ 4,000{\/tex}<br>Total Amount {tex}={\/tex} Principal + Total Interest<br>Total Amount {tex}=20,000+4,000=\u20b9 24,000{\/tex}<br><strong>Alternative Formula:&nbsp;<\/strong>Amount {tex}={P}(1+{rt}){\/tex}<br>Amount {tex}=20,000(1+0.10 \\times 2){\/tex}<br>Amount {tex}= 20,000(1+0.20){\/tex}<br>Amount {tex}=20,000 \\times 1.20{\/tex}<br>Amount {tex}=\u20b9 24,000{\/tex}<br><strong>With Compounding:<\/strong><br>Year 1: Amount {tex}=P \\times(1+r)=20,000 \\times 1.10=\u20b9 22,000{\/tex}<br>Year 2: Amount {tex}=22,000 \\times(1 +{r}){\/tex}&nbsp;{tex}=22,000 \\times 1.10=\u20b9 24,200{\/tex}<br><strong>Comparison:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Without compounding: \u20b9 24,000<\/li>\n\n\n\n<li>With compounding: \u20b9 24,200<\/li>\n\n\n\n<li>Difference: \u20b9 24,200 &#8211; \u20b9 24,000 = \u20b9 200<\/li>\n<\/ul>\n\n\n\n<p>Therefore, with compounding, one gets \u20b9 200 more than without compounding.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.74: Bank of Wahapur offers an interest of 5% p.a. Compare how much one gets if one deposits \u20b9 20,000 for a period of 4 years with compounding and without compounding annually.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Principal {tex}(P)=\u20b9 20,000{\/tex}<\/li>\n\n\n\n<li>Rate of interest {tex}(r)=5 \\%{\/tex} p.a. {tex}=0.05{\/tex}<\/li>\n\n\n\n<li>Time {tex}\\operatorname{period}({t})=4{\/tex} years<\/li>\n<\/ul>\n\n\n\n<p><strong>Without Compounding:<\/strong><br>Interest per year {tex}={P} \\times {r}=20,000 \\times 0.05=\u20b9 1,000{\/tex}<br>Total Interest for 4 years {tex}=1,000 \\times{\/tex} 4 = \u20b9 4,000<br>Total Amount {tex}=P+{\/tex} Total Interest<br>Amount {tex}=20,000+4,000=\u20b9 \\ 24,000{\/tex}<br><strong>Using Formula:&nbsp;<\/strong>Amount {tex}={P}(1+{rt}){\/tex}<br>Amount {tex}=20,000(1+0.05 \\times 4){\/tex}<br>Amount {tex}= 20,000(1+0.20){\/tex}<br>Amount {tex}=20,000 \\times 1.20{\/tex}<br>Amount {tex}=\u20b9 24,000{\/tex}<br><strong>With Compounding:<\/strong><br>Year 1: {tex}20,000 \\times 1.05=\u20b9 21,000{\/tex}<br>Year 2: {tex}21,000 \\times 1.05=\u20b9 22,050{\/tex}<br>Year 3: {tex}22,050 \\times 1.05=\u20b9 23,152.50{\/tex}<br>Year {tex}4: 23,152.50 \\times 1.05{\/tex}&nbsp;{tex}=\u20b9 24,310.125 \\approx \u20b9 24,310.13{\/tex}<br><strong>Using Formula:&nbsp;<\/strong>Amount {tex}={P}(1+{r})^ {t}{\/tex}<br>Amount {tex}=20,000(1.05)^4{\/tex}<br>Amount {tex}=20,000 \\times{\/tex} 1.21550625<br>Amount {tex}={\/tex} \u20b9 {tex}24,310.125 \\approx{\/tex} \u20b9 {tex}24,310.13{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.75: Jasmine invests amount \u2018p\u2019 for 4 years at an interest of 6% p.a. Which of the following expression(s) describe the total amount she will get after 4 years when compounding is not done?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}p \\times 6 \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p \\times 0.6 \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p \\times \\frac{0.6}{100} \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p \\times \\frac{0.06}{100} \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p \\times 1.6 \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p \\times 1.06 \\times 4{\/tex}<\/li>\n\n\n\n<li>{tex}p+(p \\times 0.06 \\times 4){\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>When compounding is not done, we use simple interest.<br>Formula: Amount {tex}={\/tex} Principal + Simple Interest<br>Amount {tex}={P}+({P} \\times {r} \\times {t}){\/tex}<br>Amount {tex}={P} +({P} \\times 0.06 \\times 4){\/tex}<br>Amount {tex}={P}(1+0.06 \\times 4){\/tex}<br>Amount {tex}={P}(1+0.24){\/tex}<br>Amount {tex}={P} \\times 1.24{\/tex}<br>Now let&#8217;s check each option:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}{p} \\times 6 \\times 4=24 {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}{p} \\times 0.6 \\times 4=2.4 {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}{p} \\times 0.6 \/ 100 \\times 4={p} \\times 0.006 \\times 4=0.024 {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}{p} \\times 0.06 \/ 100 \\times 4={p} \\times 0.0006 \\times 4=0.0024 {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}{p} \\times 1.6 \\times 4=6.4 {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}{p} \\times 1.06 \\times 4=4.24 {p}{\/tex} <strong>Incorrect <\/strong>(This would be for compound interest formula applied incorrectly)<\/li>\n\n\n\n<li>{tex}p+(p \\times 0.06 \\times 4)=p+0.24 p=1.24 p{\/tex} <strong>Correct<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Therefore, only option <strong>(vii) <\/strong>{tex}{p}+({p} \\times {0 . 0 6} \\times {4}){\/tex} is correct.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.76: The post office offers an interest of 7% p.a. How much interest would one get if one invests \u20b9 50,000 for 3 years without compounding? How much more would one get if it was compounded?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Principal {tex}(P)=\u20b9 \\ 50,000{\/tex}<\/li>\n\n\n\n<li>Rate {tex}(r)=7 \\%{\/tex} p.a. {tex}=0.07{\/tex}<\/li>\n\n\n\n<li>Time {tex}({t})=3{\/tex} years<\/li>\n<\/ul>\n\n\n\n<p><strong>Without Compounding (Simple Interest):<\/strong><br>Simple Interest {tex}={P} \\times {r} \\times {t}\\ {\/tex}<br>{tex}{\\text {SI}}=50,000 \\times 0.07 \\times 3\\ {\/tex}&nbsp;<br>{tex} {\\text {SI}}=50,000 \\times 0.21{\/tex}<br>{tex}{\\text {SI}}=\u20b9 \\ 10,500{\/tex}<br>Total Amount {tex}={P}+{SI}=50,000+10,500=\u20b9 \\ 60,500{\/tex}<br><strong>With Compounding:<\/strong><br>Amount {tex}=P(1+r)^ t{\/tex} Amount {tex}=50,000(1.07)^3{\/tex}<br>Amount {tex}=50,000 \\times 1.225043{\/tex}<br>Amount = \u20b9 61,252.15<br>Compound Interest {tex}={\/tex} Amount &#8211; Principal<br>{tex}{CI}=61,252.15-50,000{\/tex}<br>{tex} {CI}=\u20b9 11,252.15{\/tex}<br><strong>Difference:&nbsp;<\/strong>Extra interest with compounding {tex}={CI}-{SI}{\/tex}<br>Extra interest {tex}=11,252.15{\/tex}&nbsp;&#8211; 10,500<br>Extra interest = \u20b9 752.15<br>Therefore, without compounding, one would get \u20b9 10,500 as interest. With compounding, one would get \u20b9 752.15 more, making the total interest \u20b9 11,252.15.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.77: Giridhar borrows a loan of \u20b9 12,500 at 12% per annum for 3 years without compounding and Raghava borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Principal for both {tex}=\u20b9 12,500{\/tex}<\/li>\n\n\n\n<li>Time period for both {tex}=3{\/tex} years<\/li>\n<\/ul>\n\n\n\n<p><strong>Giridhar (12% without compounding):<\/strong><br>Simple Interest {tex}={P} \\times {r} \\times {t}{\/tex}<br>{tex} {SI}=12,500 \\times 0.12 \\times 3 {\/tex}&nbsp;<br>{tex}{SI}=12,500 \\times 0.36{\/tex}<br>{tex}{SI}=\u20b9\\ 4,500{\/tex}<br><strong>Raghava (10% with compounding):<\/strong><br>Amount {tex}=P(1+r)^ t{\/tex}<br>Amount {tex}=12,500(1.10)^3{\/tex}<br>Amount {tex}=12,500 \\times 1.331{\/tex}<br>Amount {tex}={\/tex} \u20b9 16,637.50<br>Compound Interest {tex}={\/tex} Amount &#8211; Principal<br>{tex}{CI}=16,637.50-12,500 {\/tex}<br>{tex}{CI}=\u20b9 4,137.50{\/tex}<br><strong>Comparison:&nbsp;<\/strong>Giridhar&#8217;s interest {tex}={\/tex} \u20b9 4,500<br>Raghava&#8217;s interest {tex}={\/tex} \u20b9 4,137.50<br>Difference {tex}=4,500-4,137.50=\u20b9 362.50{\/tex}<br>Therefore, Giridhar pays more interest by \u20b9 362.50.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.78: Consider an amount \u20b9 1000. If this grows at 10% p.a., how long will it take to double when compounding is done vs. when compounding is not done? Is compounding an example of exponential growth and not-compounding an example of linear growth?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Principal {tex}(P)=\u20b9 1,000{\/tex}<\/li>\n\n\n\n<li>Rate (r) = 10% p.a. = 0.10<\/li>\n\n\n\n<li>Target amount = \u20b9 2,000 (double)<\/li>\n<\/ul>\n\n\n\n<p><strong>Without Compounding (Simple Interest):<\/strong><br>Amount {tex}=P(1+r t)\\ 2,000{\/tex}&nbsp;<br>{tex}=1,000(1+0.10 \\times t) 2=1+0.10 t{\/tex}<br>{tex}t=1 \/ 0.10 t=10{\/tex} years<br><strong>With Compounding:<\/strong><br>Amount {tex}=P(1+r)^t\\ 2,000=1,000(1.10)^t \\ 2=(1.10)^t{\/tex}<br>Taking logarithm on both sides: {tex}\\log (2)={t} \\times \\log (1.10){\/tex}<br>{tex} {t}=\\log (2) \/ \\log (1.10){\/tex}&nbsp;<br>{tex}t= 0.3010 \/ 0.0414 t \\approx 7.27{\/tex} years<br>So it takes between 7 and 8 years, approximately 7.27 years.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.79: The population of a city is rising by about 3% every year. If the current population is 1.5 crore, what is the expected population after 3 years?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Current population {tex}({P})=1.5{\/tex} crore {tex}=1,50,00,000{\/tex}<\/li>\n\n\n\n<li>Growth rate {tex}(r)=3 \\%{\/tex} per year {tex}=0.03{\/tex}<\/li>\n\n\n\n<li>Time period {tex}({t})=3{\/tex} years<\/li>\n<\/ul>\n\n\n\n<p>Population growth follows compounding (exponential growth).<br><strong>Formula: <\/strong>Population after t years {tex}={P}(1+{r})^{{t}}{\/tex}<br><strong>Calculation:<\/strong> Population after 3 years {tex}=1.5{\/tex} crore {tex}\\times(1.03)^3=1.5{\/tex} crore {tex}\\times 1.092727{\/tex} = 1.639 crore (approximately)<br><strong>Detailed calculation:<\/strong> After Year 1: {tex}1.5 \\times 1.03=1.545{\/tex} crore<br>After Year 2: {tex}1.545 \\times 1.03=1.59135{\/tex} crore<br>After Year 3: {tex}1.59135 \\times 1.03=1.6391{\/tex} crore<br>Converting: 1.6391 crore {tex}=1,63,91,000{\/tex}<br>Therefore, the expected population after 3 years is approximately 1.64 crore or 1,63,91,000.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.80: In a laboratory, the number of bacteria in a certain experiment increases at the rate of 2.5% per hour. Find the number of bacteria at the end of 2 hours if the initial count is 5,06,000.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial count {tex}(P)=5,06,000{\/tex}<\/li>\n\n\n\n<li>Growth rate {tex}(r)=2.5 \\%{\/tex} per hour {tex}=0.025{\/tex}<\/li>\n\n\n\n<li>Time period {tex}({t})=2{\/tex} hours<\/li>\n<\/ul>\n\n\n\n<p>Bacterial growth follows compounding (exponential growth).<br><strong>Formula:&nbsp;<\/strong>Number of bacteria after {tex}t{\/tex} hours {tex}=P(1+r)^t{\/tex}<br><strong>Calculation:&nbsp;<\/strong>Number after 2 hours {tex}=5,06,000 \\times(1.025)^2=5,06,000 \\times 1.050625={\/tex} 5,31,616.25<br>Since we cannot have fractional bacteria, we round to the nearest whole number.<br><strong>Detailed calculation:&nbsp;<\/strong>After 1 hour: 5,06,000 {tex}\\times 1.025=5,18,650{\/tex}<br>After 2 hours: {tex}5,18,650 \\times 1.025=5,31,616.25{\/tex}<br>Therefore, the number of bacteria at the end of 2 hours is approximately {tex}5,31,616{\/tex}.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.81: The population of Bengaluru in 2025 is about 250% of its population in 2000. If the population in 2000 was 50 lakhs, what is the population in 2025?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Population in {tex}2000=50{\/tex} lakhs<\/li>\n\n\n\n<li>Population in {tex}2025=250 \\%{\/tex} of population in 2000<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculation:&nbsp;<\/strong>Population in 2025 {tex}=250 \\%{\/tex} of 50 lakhs {tex}=(\\frac {250 }{ 100}) \\times 50{\/tex} lakhs {tex}=2.5 \\times{\/tex} 50 lakhs = 125 lakhs<br>Converting: 125 lakhs {tex}=1.25{\/tex} crore {tex}=1,25,00,000{\/tex}<br><strong>Understanding:<\/strong> 250% means 2.5 times the original value.<br>Therefore, the population of Bengaluru in 2025 is 125 lakhs or 1.25 crore.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.82: The population of the world in 2025 is about 8.2 billion. The populations of some countries in 2025 are given. Match them with their approximate percentage share of the worldwide population<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769075907-6gt2jr.jpg\"><br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1769075952-kpjv8p.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>World population {tex}=8.2{\/tex} billion {tex}=8,200{\/tex} million<br><strong>For Germany (83 million):&nbsp;<\/strong>Percentage {tex}=(83 \/ 8200) \\times 100 \\%{\/tex}&nbsp;{tex}=0.0101 \\times 100 \\%= 1.01 \\% \\approx 1 \\%{\/tex}<br><strong>For India ({tex}{1 . 4 6}{\/tex} billion {tex}{= 1 , 4 6 0}{\/tex} million):<\/strong> Percentage {tex}=(1,460 \/ 8,200) \\times 100 \\%{\/tex}&nbsp;{tex}= 0.178 \\times 100 \\%{\/tex}&nbsp;{tex}=17.8 \\% \\approx 18 \\%{\/tex}<br><strong>For Bangladesh (175 million):<\/strong> Percentage {tex}=(175 \/ 8,200) \\times 100 \\%{\/tex}&nbsp;{tex}=0.0213 \\times 100 \\% =2.13 \\% \\approx 2 \\%{\/tex}<br><strong>For USA (347 million):&nbsp;<\/strong>Percentage {tex}=(347 \/ 8,200) \\times 100 \\%{\/tex}&nbsp;{tex}=0.0423 \\times 100 \\%={\/tex} 4.23% {tex}\\approx{\/tex} <strong>Approximately 4<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.83: The price of a mobile phone is \u20b9 8,250. A GST of 18% is added to the price. Which of the following gives the final price of the phone including the GST?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}8250+18{\/tex}<\/li>\n\n\n\n<li>{tex}8250+1800{\/tex}<\/li>\n\n\n\n<li>{tex}8250+\\frac{18}{100}{\/tex}<\/li>\n\n\n\n<li>{tex}8250 \\times 18{\/tex}<\/li>\n\n\n\n<li>{tex}8250 \\times 1.18{\/tex}<\/li>\n\n\n\n<li>{tex}8250+8250 \\times 0.18{\/tex}<\/li>\n\n\n\n<li>{tex}1.8 \\times 8250{\/tex}<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Price = \u20b9 8,250<\/li>\n\n\n\n<li>{tex}{GST}=18 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculating final price:&nbsp;<\/strong>GST amount {tex}=18 \\%{\/tex} of 8,250<br>GST amount {tex}=(18 \/ 100) \\times{\/tex} 8,250<br>GST amount {tex}=0.18 \\times 8,250{\/tex}<br>GST amount {tex}=\u20b9 1,485{\/tex}<br>Final price {tex}={\/tex} Price + GST amount<br>Final price {tex}=8,250+1,485{\/tex}<br>Final price {tex}=\u20b9 9,735{\/tex}<br><strong>Alternative method:&nbsp;<\/strong>Final price {tex}={\/tex} Price {tex}\\times(1+{\/tex} GST rate)<br>Final price {tex}=8,250 \\times(1+{\/tex} 0.18 )<br>Final price {tex}=8,250 \\times 1.18{\/tex}<br>Final price {tex}=\u20b9 9,735{\/tex}<br><strong>Now checking each option:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}8250+18=8,268{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}8250+1800=10,050{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}8250+18 \/ 100=8250+0.18=8,250.18{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}8250 \\times 18=1,48,500{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>{tex}8250 \\times 1.18=9,735{\/tex} <strong>Correct<\/strong><\/li>\n\n\n\n<li>{tex}8250+8250 \\times 0.18=8250+1485=9,735{\/tex} <strong>Correct<\/strong><\/li>\n\n\n\n<li>{tex}1.8 \\times 8250=14,850{\/tex} <strong>Incorrect<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Therefore, options (v) {tex}{8 2 5 0} \\times {1 . 1 8}{\/tex} and (vi) {tex}{8 2 5 0}+{8 2 5 0} \\times {0 . 1 8}{\/tex} are correct.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.84: The monthly percentage change in population (compared to the previous month) of mice in a lab is given: Month 1 change was +5%, Month 2 change was -2%, and Month 3 change was -3%. Which of the following statement(s) are true? The initial population is p.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>The population after three months was p {tex}\\times{\/tex} 0.05 {tex}\\times{\/tex} 0.02 {tex}\\times{\/tex} 0.03.<\/li>\n\n\n\n<li>The population after three months was p {tex}\\times{\/tex} 1.05 {tex}\\times{\/tex} 0.98 {tex}\\times{\/tex} 0.97.<\/li>\n\n\n\n<li>The population after three months was p + 0.05 &#8211; 0.02 &#8211; 0.03.<\/li>\n\n\n\n<li>The population after three months was p.<\/li>\n\n\n\n<li>The population after three months was more than p.<\/li>\n\n\n\n<li>The population after three months was less than p.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial population {tex}={p}{\/tex}<\/li>\n\n\n\n<li>Month {tex}1:+5 \\%{\/tex} change<\/li>\n\n\n\n<li>Month 2: -2% change<\/li>\n\n\n\n<li>Month 3: {tex}-3 \\%{\/tex} change<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculating population after each month:<\/strong><br>After Month 1: Population increases by {tex}5 \\%={p} \\times(1+0.05)={p} \\times 1.05{\/tex}<br>After Month 2: Population decreases by {tex}2 \\%={p} \\times 1.05 \\times(1-0.02){\/tex}&nbsp;{tex}={p} \\times 1.05 \\times 0.98{\/tex}<br>After Month 3: Population decreases by {tex}3 \\%={p} \\times 1.05 \\times 0.98 \\times(1-0.03){\/tex}&nbsp;{tex}={p} \\times 1.05 \\times 0.98 \\times 0.97{\/tex}<br><strong>Calculating the final value:<\/strong> {tex}={p} \\times 1.05 \\times 0.98 \\times 0.97={p} \\times 0.998{\/tex} (approximately) {tex}={\/tex} 0.998p<br>Since {tex}0.998&lt;1{\/tex}, the final population is less than {tex}p{\/tex}.<br><strong>Checking each statement:<\/strong><\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>{tex}{p} \\times 0.05 \\times 0.02 \\times 0.03={p} \\times 0.00003{\/tex} <strong>Incorrect <\/strong>&#8211; This doesn&#8217;t represent percentage changes correctly<\/li>\n\n\n\n<li>{tex}{p} \\times 1.05 \\times 0.98 \\times 0.97=0.998 {p}{\/tex} <strong>Correct<\/strong><\/li>\n\n\n\n<li>{tex}{p}+0.05-0.02-0.03={p}{\/tex} <strong>Incorrect <\/strong>&#8211; Percentage changes multiply, not add<\/li>\n\n\n\n<li>The population after three months was {tex}{p} .0 .998 {p} \\neq {p}{\/tex} <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>The population after three months was more than p. 0.998 p &lt; p <strong>Incorrect<\/strong><\/li>\n\n\n\n<li>The population after three months was less than {tex}{p} .0 .998 {p}&lt;{p}{\/tex} <strong>Correct<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Therefore, statements (ii) and (vi) are true.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.85: A shopkeeper initially set the price of a product with a 35% profit margin. Due to poor sales, he decided to offer a 30% discount on the selling price. Will he make a profit or a loss? Give reasons for your answer.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the cost price of the product be \u20b9 100 (for easy calculation).<br><strong>Step 1: Calculate initial selling price with {tex}{3 5 \\%}{\/tex} profit<\/strong><br>Selling Price {tex}={CP} \\times(1+{\/tex} profit percentage)<br>{tex} {SP}=100 \\times(1+0.35) {\/tex}<br>{tex}{SP}=100 \\times 1.35 {\/tex}<br>S.P = \u20b9 135<br><strong>Step 2:<\/strong> Calculate selling price after 30% discount<br>Discount {tex}=30 \\%{\/tex} of SP<br>Discount {tex}=0.30 \\times 135=\u20b9 40.50{\/tex}<br>Final Selling Price {tex}={SP}-{\/tex} Discount<br>Final {tex}{SP}=135-40.50={\/tex} \u20b9 94.50<br><strong>Alternative calculation:&nbsp;<\/strong>Final SP {tex}={SP} \\times{\/tex} (1- discount percentage)<br>Final SP {tex}=135 \\times{\/tex} (1-0.30)<br>Final SP=135 {tex}\\times{\/tex}&nbsp;0.70<br>Final SP = \u20b9 94.50<br><strong>Step 3: Compare with cost price<\/strong><br>Cost Price {tex}={\/tex} \u20b9 100<br>Final Selling Price {tex}={\/tex} \u20b9 94.50<br>Since Final SP &lt; CP, the shopkeeper makes a loss.<br>Loss {tex}={CP}-{\/tex} FinalSP<br>{tex}=100-94.50=\u20b9 5.50{\/tex}<br>Loss Percentage {tex}=({\/tex}Loss{tex}\/{\/tex}CP{tex}) \\times 100 \\%{\/tex}<br>Loss Percentage {tex}=(5.50 \/ 100) \\times 100 \\%{\/tex}<br>Loss Percentage {tex}=5.5 \\%{\/tex}<br><strong>Verification using general formula:&nbsp;<\/strong>If {tex}C P=x{\/tex}, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>SP with {tex}35 \\%{\/tex} profit {tex}=1.35 {x}{\/tex}<\/li>\n\n\n\n<li>After {tex}30 \\%{\/tex} discount {tex}=1.35 x \\times 0.70=0.945 x{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Since {tex}0.945&lt;1{\/tex}, there&#8217;s a loss of {tex}(1-0.945) \\times 100 \\%=5.5 \\%{\/tex}<br>Therefore, the shopkeeper will make a <strong>loss of<\/strong> {tex}{5 . 5} \\boldsymbol{\\%}{\/tex}.<br><strong>Reason:<\/strong> Although he initially added a {tex}35 \\%{\/tex} profit margin, the {tex}30 \\%{\/tex} discount is calculated on the increased selling price (not the cost price), which results in a larger absolute discount amount that exceeds the original profit.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.86: What percentage of area is occupied by the region marked \u2018E\u2019 in the figure?<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768884457-jssd76.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Without seeing the exact figure, I&#8217;ll explain the general approach:<br><strong>Step 1:<\/strong> Identify the total area<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The total figure represents 100%<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2:<\/strong> Determine what fraction region E occupies<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Count the number of equal parts in the figure<\/li>\n\n\n\n<li>Count how many parts region E occupies<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3:<\/strong> Calculate percentage<br>Percentage {tex}=({\/tex}Area of {tex}{E} \/{\/tex} Total Area{tex}) \\times 100 \\%{\/tex}<br><strong>Example:<\/strong> If the figure is divided into equal parts:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If total parts {tex}=10{\/tex} and E occupies 2 parts<\/li>\n\n\n\n<li>Percentage of {tex}E=(2 \/ 10) \\times 100 \\%=20 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.87: What is 5% of 40? What is 40% of 5? What is 25% of 12? What is 12% of 25? What is 15% of 60? What is 60% of 15? What do you notice? Can you make a general statement and justify it using algebra, comparing x% of y and y% of x?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p><strong>Calculating each:<\/strong><br>{tex}5 \\%{\/tex} of {tex}40=(5 \/ 100) \\times 40=0.05 \\times 40=2{\/tex}<br>{tex}40 \\%{\/tex} of {tex}5=(40 \/ 100) \\times 5=0.40 \\times 5=2{\/tex}<br>{tex}25 \\%{\/tex} of {tex}12=(25 \/ 100) \\times 12=0.25 \\times 12=3{\/tex}<br>{tex}12 \\%{\/tex} of {tex}25=(12 \/ 100) \\times 25=0.12 \\times 25=3{\/tex}<br>{tex}15 \\%{\/tex} of {tex}60=(15 \/ 100) \\times 60=0.15 \\times 60=9{\/tex}<br>{tex}60 \\%{\/tex} of {tex}15=(60 \/ 100) \\times 15=0.60 \\times 15=9{\/tex}<br><strong>Observation:&nbsp;<\/strong>In each pair, the results are equal!<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>{tex}5 \\%{\/tex} of {tex}40=40 \\%{\/tex} of 5<\/li>\n\n\n\n<li>{tex}25 \\%{\/tex} of {tex}12=12 \\%{\/tex} of 25<\/li>\n\n\n\n<li>{tex}15 \\%{\/tex} of {tex}60=60 \\%{\/tex} of 15<\/li>\n<\/ul>\n\n\n\n<p><strong>General Statement: {tex}{x \\%}{\/tex} of {tex}{y}={y \\%}{\/tex} of <\/strong>{tex}{x}{\/tex} for any values of x and y.<br><strong>Algebraic Justification:<\/strong><br>Let&#8217;s prove that {tex}{x} \\%{\/tex} of y equals {tex}{y} \\%{\/tex} of x.<br>{tex}x \\%{\/tex} of {tex}y=(x \/ 100) \\times y=x y \/ 100{\/tex}<br>{tex}y \\%{\/tex} of {tex}x=(y \/ 100) \\times x=y x \/ 100{\/tex}<br>Since multiplication is commutative {tex}(x y=y x){\/tex}:<br>{tex}x y \/ 100=y x \/ 100{\/tex}<br>Therefore, {tex}{x} \\%{\/tex} of {tex}{y}={y} \\%{\/tex} of {tex}{x}{\/tex}<br>This is a beautiful mathematical property that shows the symmetry in percentage calculations!<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.88: A school is organising an excursion for its students. 40% of them are Grade 8 students and the rest are Grade 9 students. Among these Grade 8 students, 60% are girls.<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>What percentage of the students going to the excursion are Grade 8 girls?<\/li>\n\n\n\n<li>If the total number of students going to the excursion is 160, how many of them are Grade 8 girls?<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li><strong>Finding percentage of Grade 8 girls:<\/strong><br>Grade 8 students {tex}=40 \\%{\/tex} of total students<br>Among Grade 8 students, girls {tex}=60 \\%{\/tex}<br>Grade 8 girls {tex}=60 \\%{\/tex} of (40% of total) {tex}=60 \\%{\/tex} of 40% {tex}=(60 \/ 100) \\times(40 \/ 100){\/tex} of total {tex}= 0.60 \\times 0.40{\/tex} of total {tex}=0.24{\/tex} of total {tex}=24 \\%{\/tex} of total<br><strong>Alternative method:&nbsp;<\/strong>If total students {tex}=100{\/tex}<br>Grade 8 students {tex}=40 \\%{\/tex} of {tex}100=40{\/tex}<br>Grade 8 girls {tex}=60 \\%{\/tex} of {tex}40=24{\/tex}<br>Percentage {tex}=(24 \/ 100) \\times 100 \\%=24 \\%{\/tex}<br>Therefore, {tex}{2 4 \\%}{\/tex} of the students going to the excursion are Grade {tex}{8}{\/tex} girls.<\/li>\n\n\n\n<li><strong>Finding number of Grade 8 girls when total <\/strong>{tex}=160{\/tex}:<br>Grade 8 students = 40% of 160&nbsp;{tex}=0.40 \\times 160=64{\/tex}<br>Grade 8 girls = 60% of 64&nbsp;{tex}=0.60 \\times 64=38.4{\/tex}&nbsp;<br>The answer 38.4 suggests there might be rounding involved, but mathematically it works out. In practical terms, this would be approximately {tex}{3 8}{\/tex} or {tex}{3 9}{\/tex} students.<br>However, using exact calculation: {tex}{3 8 . 4}{\/tex}, which we can round to {tex}{3 8}{\/tex} students (assuming we round down) or the problem expects us to work with the exact percentage giving us the decimal answer.<br>Therefore, there are {tex}{3 8}{\/tex} Grade {tex}{8}{\/tex} girls (or precisely 38.4 as per calculation).<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.89: A shopkeeper sells pencils at a price such that the selling price of 3 pencils is equal to the cost of 5 pencils. Does he make a profit or a loss? What is his profit or loss percentage?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Given: Selling price of 3 pencils {tex}={\/tex} Cost price of 5 pencils<br>Let the cost price of one pencil be \u20b9 x.<br><strong>Step 1: Set up the relationship<\/strong><br>Cost price of 5 pencils {tex}=5 x{\/tex}<br>Selling price of 3 pencils {tex}=5 x{\/tex}<br>Therefore, selling price of 1 pencil {tex}=5 x \/ 3{\/tex}<br><strong>Step 2: Compare cost and selling price for 1 pencil<\/strong><br>Cost price of 1 pencil {tex}=x{\/tex}<br>Selling price of 1 pencil {tex}=5 x \/ 3{\/tex}<br>Since {tex}5 x \/ 3&gt;x{\/tex} (because {tex}5 \/ 3&gt;1{\/tex}), the shopkeeper makes a profit.<br><strong>Step 3: Calculate profit<\/strong><br>Profit per pencil {tex}=S P-C P=5 x \/ 3-x{\/tex}&nbsp;{tex}=5 x \/ 3-3 x \/ 3=2 x \/ 3{\/tex}<br><strong>Step 4: Calculate profit percentage<\/strong><br>Profit% {tex}=({\/tex}Profit{tex}\/{\/tex}CP){tex} \\times 100 \\%=(2 x \/ 3 \\div x) \\times 100 \\%{\/tex}&nbsp;{tex}=(2 x \/ 3 \\times 1 \/ x) \\times 100 \\%{\/tex}&nbsp;{tex}=(2 \/ 3) \\times 100 \\%=66.67 \\%{\/tex} (or {tex}66 \\frac{2}{3} \\%{\/tex})<br><strong>Alternative approach:<\/strong> If CP of 1 pencil {tex}=\u20b9 1{\/tex}, then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>CP of 5 pencils = \u20b9 5<\/li>\n\n\n\n<li>SP of 3 pencils = \u20b9 5<\/li>\n\n\n\n<li>SP of 1 pencil {tex}={\/tex} \u20b9 5\/3 = \u20b9 1.67<\/li>\n<\/ul>\n\n\n\n<p>Profit {tex}=1.67-1=\u20b9 0.67{\/tex}<br>Profit {tex}\\%=(0.67 \/ 1) \\times 100 \\%=66.67 \\%{\/tex}<br>Therefore, the shopkeeper makes a profit of {tex}{6 6 . 6 7 \\%}{\/tex} (or {tex}{6 6} \\frac{{2}}{{3}} \\boldsymbol{\\%}{\/tex}).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.90: The bus fares were increased by 3% last year and by 4% this year. What is the overall percentage price increase in the last 2 years?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let the original bus fare be {tex}\u20b9 100{\/tex}.<br><strong>Step 1: Calculate fare after first year (3% increase)<\/strong><br>Fare after Year {tex}1={\/tex} Original fare {tex}\\times(1+0.03)=100 \\times 1.03=\u20b9 103{\/tex}<br><strong>Step 2: Calculate fare after second year (4% increase on Year 1 fare)<\/strong><br>Fare after Year {tex}2={\/tex} Fare after Year {tex}1 \\times(1+0.04)=103 \\times 1.04=\u20b9 107.12{\/tex}<br><strong>Step 3: Calculate overall percentage increase<\/strong><br>Overall increase {tex}={\/tex} Final fare &#8211; Original fare {tex}=107.12-100={\/tex} \u20b9 7.12<br>Overall percentage increase {tex}=(7.12 \/ 100) \\times 100 \\%=7.12 \\%{\/tex}<br><strong>Alternative formula approach:<\/strong> Overall multiplier {tex}=1.03 \\times 1.04=1.0712{\/tex}<br>Overall increase {tex}=(1.0712-1) \\times 100 \\%=7.12 \\%{\/tex}<br>Therefore, the overall percentage price increase in the last 2 years is 7.12%.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.91: If the length of a rectangle is increased by 10% and the area is unchanged, by what percentage (exactly) does the breadth decrease by?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Let original length {tex}={\/tex} L<br>Let original breadth {tex}={\/tex} B<br>Original area {tex}={L} \\times {B}{\/tex}<br><strong>Step 1: Calculate new length<\/strong><br>New length {tex}=L+10 \\%{\/tex} of L<br>{tex}=L \\times(1+0.10)=1.1 L{\/tex}<br><strong>Step 2: Find new breadth (area remains same)<\/strong><br>Original area {tex}={\/tex} New area {tex}{L} \\times {B}=1.1 {\/tex}&nbsp;<br>{tex}{L} \\times{\/tex}&nbsp;(New breadth)<br>{tex} {B}=1.1 \\times({\/tex}New breadth)<br>New breadth = B\/1.1<br>New breadth = (10\/11)B<br><strong>Step 3: Calculate decrease in breadth<\/strong><br>Decrease = Original breadth &#8211; New breadth = B &#8211; (10\/11)B = (11\/11)B &#8211; (10\/11)B = (1\/11)B<br><strong>Step 4:<\/strong> <strong>Calculate percentage decrease<\/strong><br>Percentage decrease = (Decrease\/Original breadth){tex} \\times 100 \\%=[(1 \/ 11) {B} \\div {B}] \\times 100 \\%{\/tex}&nbsp;{tex}=(1 \/ 11) \\times 100 \\%=9.09 \\%{\/tex} (or {tex}91 \/ 11 \\%{\/tex})<br>Exact answer: 100\/11% = {tex}91 \/ 11 \\%{\/tex}<br>Therefore, the breadth decreases by exactly {tex}{9 1 \/ 1 1 \\%}{\/tex} <strong>or 9.09%<\/strong> (approximately).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.92: The percentage of ingredients in a 65 g chips packet is shown in the picture. Find out the weight each ingredient makes up in this packet.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768886536-r7rmjm.jpg\"><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total weight of packet {tex}=65 {~g}{\/tex}<br>Assuming the percentages shown in the picture are: (Please refer to your textbook for exact percentages)<br><strong>General formula: <\/strong>Weight of ingredient {tex}=({\/tex} Percentage {tex}\/ 100) \\times{\/tex} Total weight<br><strong>Example calculation<\/strong> (using hypothetical percentages):<br>If the ingredients are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Potato: {tex}60 \\%{\/tex}<\/li>\n\n\n\n<li>Oil: 30%<\/li>\n\n\n\n<li>Salt and spices: {tex}10 \\%{\/tex}<\/li>\n<\/ul>\n\n\n\n<p>Then:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Weight of Potato {tex}=(60 \/ 100) \\times 65{\/tex}&nbsp;{tex}=0.60 \\times 65=39 {~g}{\/tex}<\/li>\n\n\n\n<li>Weight of Oil {tex}=(30 \/ 100) \\times 65{\/tex}&nbsp;{tex}=0.30 \\times 65=19.5 {~g}{\/tex}<\/li>\n\n\n\n<li>Weight of Salt and spices {tex}=(10 \/ 100) \\times 65{\/tex}&nbsp;{tex}=0.10 \\times 65=6.5 {~g}{\/tex}<\/li>\n<\/ul>\n\n\n\n<p><strong>Verification:<\/strong> {tex}39+19.5+6.5=65 {~g}{\/tex}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.93: Three shops sell the same items at the same price. The shops offer deals as follows:<br>Shop A: &#8220;Buy 1 and get 1 free&#8221;<br>Shop B: &#8220;Buy 2 and get 1 free&#8221;<br>Shop C: &#8220;Buy 3 and get 1 free&#8221;<br>Answer the following:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>If the price of one item is \u20b9 100, what is the effective price per item in each shop? Arrange the shops from cheapest to costliest.<\/li>\n\n\n\n<li>For each shop, calculate the percentage discount on the items.<br>[<strong>Hint:<\/strong> Compare the free items to the total items you receive.]<\/li>\n\n\n\n<li>Suppose you need 4 items. Which shop would you choose? Why?<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>Price of one item {tex}=\u20b9 100{\/tex}<br><strong>Shop A: <\/strong>Buy 1 get 1 free<ul><li>Pay for 1 item, get 2 items total<\/li><li>Amount paid = \u20b9 100<\/li><li>Items received {tex}=2{\/tex}<\/li><li>Effective price per item {tex}=100 \/ 2=\u20b9 50{\/tex}<\/li><\/ul><strong>Shop B: <\/strong>Buy 2 get 1 free<ul><li>Pay for 2 items, get 3 items total<\/li><li>Amount paid = \u20b9 200<\/li><li>Items received = 3<\/li><li>Effective price per item = {tex}\\frac {200 }{ 3}{\/tex}&nbsp;= \u20b9 66.67<\/li><\/ul><strong>Shop C: Buy {tex}{3}{\/tex} get {tex}{1}{\/tex} free<\/strong><ul><li>Pay for 3 items, get 4 items total<\/li><li>Amount paid = \u20b9 300<\/li><li>Items received {tex}=4{\/tex}<\/li><li>Effective price per item {tex}\\frac {300}{4}{\/tex}&nbsp;= \u20b9 75<\/li><\/ul><strong>Arrangement from cheapest to costliest: Shop A (\u20b9 50) &lt; Shop B (\u20b9 66.67) &lt; Shop C (\u20b975)<\/strong><\/li>\n\n\n\n<li><strong>Shop A:&nbsp;<\/strong>Original price for 2 items = \u20b9 200 Amount paid = \u20b9 100<br>Discount {tex}=200-100=\u20b9 100{\/tex}<br>Percentage discount {tex}=(100 \/ 200) \\times 100 \\%=50 \\%{\/tex}<br><strong>Shop B:<\/strong> Original price for 3 items = \u20b9 300<br>Amount paid =&nbsp;\u20b9 200<br>Discount {tex}=300-200=\u20b9 100{\/tex}<br>Percentage discount {tex}=(100 \/ 300) \\times 100 \\%=33.33 \\%{\/tex}<br><strong>Shop C:<\/strong> Original price for 4 items {tex}={\/tex} \u20b9 400<br>Amount paid {tex}={\/tex} \u20b9 300<br>Discount {tex}=400-300=\u20b9 100{\/tex}<br>Percentage discount {tex}=(100 \/ 400) \\times 100 \\%=25 \\%{\/tex}<br><strong>Summary:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Shop A: 50% discount<\/li>\n\n\n\n<li>Shop B: 33.33% discount<\/li>\n\n\n\n<li>Shop C: 25% discount<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Shop A:<\/strong><br>Buy 1 get {tex}1 \\rightarrow{\/tex} To get 4 items, pay for 2<br>Cost {tex}=2 \\times 100=2002 \\times 100=\u20b9 200{\/tex}<br><strong>Shop B:<\/strong><br>Buy 2 get {tex}1 \\rightarrow{\/tex} For 4 items, pay for 3<br>Cost {tex}=3 \\times 100=3003 \\times 100=\u20b9 300{\/tex}<br><strong>Shop C:<\/strong><br>Buy 3 get {tex}1 \\rightarrow{\/tex} Get 4 items by paying for 3<br>Cost {tex}=3 \\times 100=3003 \\times 100=\u20b9 300{\/tex}<br>Best choice: Shop A because it gives 4 items for the lowest cost of \u20b9 200.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.94: In a room of 100 people, 99% are left-handed. How many left-handed people have to leave the room to bring that percentage down to 98%?<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>Total people = 100<br>Left-handed people = 99<br>Right-handed people = 1<br>Let x left-handed people leave the room.<br>Then,<br>Left-handed remaining {tex}=99-{x}{\/tex}<br>Total people remaining {tex}=100-{x}{\/tex}<br>We want left-handed people to be {tex}98 \\%{\/tex} of the remaining people:<br>{tex} \\frac{99-x}{100-x} =\\frac{98}{100} {\/tex}<br>{tex} 100(99-x) =98(100-x) {\/tex}<br>{tex} 9900-100 x =9800-98 x {\/tex}<br>{tex} 100 =2 x {\/tex}<br>{tex} x =50 {\/tex}<br>Therefore, 50 left-handed people must leave the room.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>Q.95: Look at the following graph.<br><img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768888565-ws4qej.jpg\"><br>Based on the graph, which of the following statement(s) are valid?<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>People in their twenties are the most computer-literate among all age groups.<\/li>\n\n\n\n<li>Women lag behind in the ability to use computers across age groups.<\/li>\n\n\n\n<li>There are more people in their twenties than teenagers.<\/li>\n\n\n\n<li>More than a quarter of people in their thirties can use computers.<\/li>\n\n\n\n<li>Less than 1 in 10 aged 60 and above can use computers.<\/li>\n\n\n\n<li>Half of the people in their twenties can use computers.<\/li>\n<\/ol>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<ol start=\"1\" class=\"wp-block-list\">\n<li>People in their twenties are the most computer-literate among all age groups.<br>Teenagers {tex}=24 \\%+29 \\%=53 \\%{\/tex}<br>Twenties {tex}=26 \\%+37 \\%=63 \\%{\/tex} (highest)<br>This statement is true.<\/li>\n\n\n\n<li>Women lag behind in the ability to use computers across age groups.<br>In every age group, the percentage for females is lower than that for males.<br>This statement is true.<\/li>\n\n\n\n<li>There are more people in their twenties than teenagers.<br>The graph shows computer usage, not population size.<br>This statement is false.<\/li>\n\n\n\n<li>More than a quarter of people in their thirties can use computers.<br>Thirties {tex}=14 \\%+25 \\%=39 \\%{\/tex}, which is more than {tex}25 \\%{\/tex}.<br>This statement is true.<\/li>\n\n\n\n<li>Less than 1 in 10 aged 60 and above can use computers.<br>Seniors {tex}=2 \\%+4 \\%=6 \\%{\/tex}, which is less than {tex}10 \\%{\/tex}.<br>This statement is true.<\/li>\n\n\n\n<li>Half of the people in their twenties can use computers.<br>Twenties {tex}=63 \\%{\/tex}, not exactly 50%.<br>This statement is false.<\/li>\n<\/ol>\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>Fractions In Disguise &#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 Fractions In Disguise \u2013 NCERT Solutions Q.1: Shambhavi owns a stationery shop. She procures 200 page notebooks at \u20b9 36 per book. She sells them &#8230; <a title=\"Fractions In Disguise &#8211; NCERT Solutions Class 8 Maths (Ganita Prakash)\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/\" aria-label=\"More on Fractions In Disguise &#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-31745","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>Fractions In Disguise - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide<\/title>\n<meta name=\"description\" content=\"Fractions In Disguise - 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\/fractions-in-disguise-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=\"Fractions In Disguise - NCERT Solutions Class 8 Maths (Ganita Prakash) | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Fractions In Disguise - NCERT Solutions Class 8 Maths (Ganita Prakash) includes all the questions with solutions\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-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=\"2026-08-07T07:05:33+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2026-08-07T07:54:21+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1768556235-nxtngw.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=\"55 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/mycbseguide.com\/blog\/fractions-in-disguise-ncert-solutions-class-8-maths-ganita-prakash\/\"},\"author\":{\"name\":\"myCBSEguide\",\"@id\":\"https:\/\/mycbseguide.com\/blog\/#\/schema\/person\/10b8c7820ff29025ab8524da7c025f65\"},\"headline\":\"Fractions In Disguise &#8211; 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