{"id":793330,"date":"2023-10-21T11:12:57","date_gmt":"2023-10-21T05:42:57","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=793330"},"modified":"2023-10-21T11:12:57","modified_gmt":"2023-10-21T05:42:57","slug":"ncert-maths-ratio-and-proportion-class-6-chapter-12-notes","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/","title":{"rendered":"NCERT Maths Ratio and Proportion Class 6 Chapter 12 Notes and Solutions PDF"},"content":{"rendered":"\n<p>In Chapter 12, Ratio and Proportion \u2013 Class 6, students will be introduced to the concepts of Ratio and Proportion. Students will learn the concepts of ratio to determine the relation among the values of two different values. Ratio and Proportion, Class 6 will teach students to solve complex mathematical problems containing a comparison. The chapter Ratio and Proportion of Class 6 teaches students about Ratio, Proportion, and the Unitary Method. Read through for NCERT Maths Ratio and Proportion Class 6 Chapter 12 Notes and Exercise Solutions.<\/p>\n\n\n\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/multiplication-and-division-word-problems\/\"><strong>Multiplication and Division Word Problems<\/strong><\/a><\/p>\n\n\n\n<p class=\"has-text-align-center has-pale-ocean-gradient-background has-background has-medium-font-size\"><a href=\"https:\/\/drive.google.com\/file\/d\/1fOIZblj1cnFHTX8YXJYwaYSr6TcEpT9V\/view?usp=sharing\"><strong>Click here to download NCERT Maths Ratio and Proportion Class 6 Chapter 12 Notes and Solutions PDF<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-maths-ratio-and-proportion-class-6-chapter-12-notes-pdf-available\">NCERT Maths Ratio and Proportion Class 6 Chapter 12 Notes \u2013 PDF Available<\/h2>\n\n\n\n<p>Check the topic-wise notes for NCERT Maths Ratio and Proportion Class 6 Chapter 12 below. You can also download the PDF of the notes and take a printout to study later when you need quick revision before going to the exam hall.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>Top 10 Short Maths Tricks for Complex Problems<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-topic-1-ratio\">Topic 1: Ratio<\/h3>\n\n\n\n<p>When we compare two quantities in terms of \u2018how many times\u2019. This comparison is known as the Ratio. We denote the ratio using the symbol \u2018:\u2019.<\/p>\n\n\n\n<p>However, it must be noted that two quantities can be compared only if they are in the same units of measurement.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Equivalent Ratio: <\/strong>Equivalent ratios can be obtained by multiplying or dividing the numerator and denominator by the same number<\/li>\n<\/ul>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/maths-books-for-competitive-exams\/\"><strong>Maths Books for Competitive Exams<\/strong><\/a><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>BSc Mathematics: Course, Universities and Scope<\/strong><\/p>\n\n\n\n<p>In a simpler manner, Ratios can be explained as \u2013 \u201cWe often make the comparisons between two quantities by taking their difference. However, in certain situations, a more meaningful comparison between quantities is required and that comparison is made by using division, i.e. by seeing how many times one quantity is to the other quantity. This method is known as comparison by ratio.\u201d.<\/p>\n\n\n\n<p>For example, Isha\u2019s weight is 25 kg and her father\u2019s weight is 75 kg. We say that Isha\u2019s father\u2019s weight and Isha\u2019s weight are in the ratio 3 : 1.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-topic-2-proportions\">Topic 2: Proportions<\/h3>\n\n\n\n<p>If two ratios are equal, we say that they are in proportion and use the symbol \u2018<strong>::<\/strong>\u2019 or \u2018=\u2019 to equate the two ratios. If two ratios are not equal, then we say that they are not in proportion.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Respective Terms: <\/strong>In a statement of proportion, the four quantities involved when taken in order are known as respective terms. The first and fourth terms are known as extreme terms. The second and third terms are known as middle terms.<\/li>\n<\/ul>\n\n\n\n<p>For example, in the proportion a:b::c:d, a, b and c, d are the respective terms. And a and d are the extreme terms and c and b are the middle terms.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/maths-for-competitive-exams\/\"><strong>Maths for Competitive Exams<\/strong><\/a><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>Maths Formulas for Class 10<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-topic-3-unitary-method\">Topic 3: Unitary Method<\/h3>\n\n\n\n<p>The method in which first we find the value of one unit and then the value of the required number of units is known as the Unitary Method.<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>SAT Maths Subject Test<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/bodmas-questions\/\"><strong>BODMAS Questions<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-maths-solutions-for-ratio-and-proportion-class-6-chapter-12-free-pdf-download\">NCERT Maths Solutions for Ratio and Proportion Class 6 Chapter 12 \u2013 Free PDF Download\u00a0<\/h2>\n\n\n\n<p>Below we have provided solutions for NCERT Maths Ratio and Proportion Class 6 Chapter 12. Go through for answers to some important questions.\u00a0<\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-exercise-12-1-solutions\">Exercise 12.1 Solutions<\/h3>\n\n\n\n<p><strong>Q 1.\u00a0 There are 20 girls and 15 boys in a class.<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>What is the ratio of number of girls to the number of boys?<\/strong><\/li>\n\n\n\n<li><strong>What is the ratio of number of girls to the total number of students in the class?<\/strong><\/li>\n<\/ol>\n\n\n\n<p><strong>Solutions.<\/strong> The answers are given below.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The ratio of the number of girls to the number of boys can be calculated by dividing both numbers. So, the ratio is given by:<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/wF1Q3bqjXFSkxFTGQfwULsrbsYQQsRNmCXdKwUOJxXs03FgreUY6ONonvjMPhv477TpA2C95PWE_hBfe7cZTG9MdXOm7waLdLQ5s5VmuuY7rIudwltgWGANTAvs5msQcjSMhuY_0l7aOgzT7gum6CRE\" ><\/figure>\n\n\n\n<p>Hence, the ratio of the number of girls to the number of boys is 4:3.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\" start=\"2\">\n<li>Total number of students in the class = number of boys + number of girls = 20 + 15 = 35<\/li>\n<\/ol>\n\n\n\n<p>\u00a0Now, the ratio of number of girls to the total number of students in the class can be calculated by dividing both numbers. So, the ratio is given by:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/5IQuyJxSpQZWd7t-I8SvntBJeCoMBzOJOt1z5V3Q2nzNV_H5JbGlhVR6Jwj_2usFsZDJoFsG_DdIUkiA6YZJMW-RbfWlbvlb_VCq4dCZh2ssBk4Om8s-Ckz1o2LtAWdUPG8k_LWmPw4ALz8vbbTJ71k\" ><\/figure>\n\n\n\n<p>Hence, the ratio of number of girls to the total number of students is 4:7.\u00a0<\/p>\n\n\n\n<p><strong>Q 2. \u00a0 Out of 30 students in a class, 6 like football, 12 like cricket and remaining like tennis. Find the ratio of<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>The number of students liking football to the number of students liking tennis.<\/strong><\/li>\n\n\n\n<li><strong>Number of students liking cricket to the total number of students.<\/strong><\/li>\n<\/ol>\n\n\n\n<p><strong>Solutions.<\/strong> The solutions for each are given below.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Number of students liking tennis = 30 \u2013 (number of students liking football and cricket)<\/li>\n<\/ol>\n\n\n\n<p>So, number of students who like playing tennis = 30 \u2013 (6 + 12) = 12<\/p>\n\n\n\n<p>The ratio of the number of students liking football to the number of students liking tennis can be calculated by:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/54HrIvb0uIO_vjLlSaYe3JYcVvmhYaUZmShgafIhALu2NaXds4ngXv1kXY0D20iIKa-PWI19mgtKTWAY2DMuIjrgM8jcA9DE4CEFHfXs4UliO7WSvUTUsCgE2B29Q4WNzaVLHpD6aDIcJcGHARBpGFk\" ><\/figure>\n\n\n\n<p>So, the ratio is 1:2.<\/p>\n\n\n\n<ol class=\"wp-block-list\" start=\"2\">\n<li>The ratio of students liking cricket to the total number of students can be calculated by:<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/JEhPjAyohoIaC00NAUBo8doLpp-hwn039t8p5zjIO3_lNxtbe8TdlTH5fGnGHWRw---XNHlGesokgOjsMwFGlJIQ8UuPE579i1CpzPI5xkwxFURq5e_xSSV1zwJ8Gi-ajUJyC8RL0303KrJqyhnpSDA\" ><\/figure>\n\n\n\n<p>So, the ratio is 2:5.<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/commerce-with-maths\/\"><strong>Top 20 Career Options in Commerce with Maths Stream<\/strong><\/a><\/p>\n\n\n\n<p><strong>Q 3. Distances travelled by Hamid and Akhtar in an hour are 9 km and 12 km. Find the ratio of the speed of Hamid to the speed of Akhtar.<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>We know, speed is calculated by the distance travelled per unit time. Since Hamid and Akhtar both have travelled for an hour, their speeds can be calculated as follows:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/ovpgbzQyB9qAL7OVpbGgHE701Xq23u-Y1W7pV7qhEuD2CEQMqnFcpeXiqeRoYzcooqtgyCbY5PDeMn7b33BCfNm1aQCi7AdjUlgoiIPE_cvrAaL2MonDpub8xYRU6SoJfEXqVGGpsC1d_S3vi9MvtA0\" ><\/figure>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/ByPqCx6Q65F56OgKsZGU20R-JxjB4woJuqd1fQR1ZrRrHD-SYZHAAlUYSyyxvJzp1BNwbdPYZ9HGt9BdBtG4gMxUVrrR076ZUVSkgjvXDR9Kh26St1ovCpp_Wo6oVYdOEOAaatZinNlwCY7a--hjX8A\" ><\/figure>\n\n\n\n<p>Now that we have calculated Hamid\u2019s and Akhtar\u2019s speeds, we can easily determine their ratio as:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/wCNiHd9a0Cg2ieE_wOpdCjZSMI3SsGde5zjwL0X2YksKaw9U3URS6g2VgaxzsqwDScMZ0gwrzVkJhAQu1lr8jDx19Ct11cMsGWWHd1IwHOPFt6hlU9f-iABIDhLJzg7Uirm4qtT4K3ijre0OweJzW80\" ><\/figure>\n\n\n\n<p>\u2234 The ratio of the speed of Hamid to the speed of Akhtar is 3:4.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/career-options-in-commerce-without-maths\/\"><strong>Courses and Career Options in Commerce Without Maths in 2023<\/strong><\/a><\/p>\n\n\n\n<p><strong>Q 4. Fill in the following blanks:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/jdq757FT-72M-lSBILic5YD7vdGBuON7QMZOuwrSR3aZocXGwhZZp6NPQ7OteA3EsXUxM_3ixif92i_o4tWzamlxtX6O9ifezip2idV0rZQG0E6A6bDc5m-sKd5EdTHZwsHqfNm56004aJzxf6UrFcI\" ><\/figure>\n\n\n\n<p><strong>\u00a0[Are these equivalent ratios?]<\/strong><\/p>\n\n\n\n<p><strong>Solutions.<\/strong> The solutions to each of the given blanks are given below.<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/4niUtULmBcKp-GfySpGkN4v6ixI91GViUpJjXiD4UUOCe-fxlCsxp3nJ2N6PSoTtezlj62yVgFMWd43jPVJnraybcHGp0qNdR3tL1OrpRoXd5fFblRFWooh-vIkFMm2f5rxvDe9sI9MrCF5HtNPgllc\" ><\/figure>\n\n\n\n<p>Now, again:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/09RQeeSNEiNsgjhwOOx0zBI58Ookabus-gi_p-9Ze0virk_W9Z8AIVllpA5wI3mQKcXbs8hZwME7CFsQ6WNiOe4kFb-EV3ae4nYfm5rK_mtYHVM62omlJ1OmmwxqANslnIue3Wk3UMdnkXYk0r5UGSc\" ><\/figure>\n\n\n\n<p>Finally,\u00a0<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/r5U_xlcs-OvAdQbA_YE3yTrH0dkDG36Oyjkkqt290exwFY3DEBNowgQfD32erGo5ZP3IcD97h0SCXc2rs-3yCaTnvYlODqAwf2KdfjHYYLjQzwUOmk3-cXnq1A9bSIRaoe1PFqbZgHcTorJ59Xwj6U8\" ><\/figure>\n\n\n\n<p>\u2234 the value of each \u25fb is 5, 12 and 25. Yes, the given ratios are equivalent ratios.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>NCERT Class 8 Maths Syllabus<\/strong><\/p>\n\n\n\n<p><strong>Q 5. Find the ratio of 30 minutes to 1.5 hours.<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>From the rule of determining the ratios, we know that both quantities must be in the same unit of measurement.\u00a0<\/p>\n\n\n\n<p>We know, 1 hour = 60 minutes\u00a0<\/p>\n\n\n\n<p>\u2234 1.5 hours = 60 + (60 \u00d7 \u00bd) = 90 minutes<\/p>\n\n\n\n<p>Now, the ratio of 30 minutes to 90 minutes is 30\/90 = 1:3<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/ba-mathematics-syllabus\/\"><strong>BA Mathematics Syllabus<\/strong><\/a><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/arts-with-maths\/\"><strong>Pathways for Arts with Maths Students<\/strong><\/a><\/p>\n\n\n\n<p><strong>Q 6. Mother wants to divide \u20b9 36 between her daughters Shreya and Bhoomika in the ratio of their ages. If the age of Shreya is 15 years and the age of Bhoomika is 12 years, find how much Shreya and Bhoomika will get.<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>Let us calculate the ratio between Shreya\u2019s and Bhoomika\u2019s ages:<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/kq34BVLzxCz2TW8WFiptIqOXzJM8NHtFY6iVB0mtu8dKuFRYnpaQRC6InrtHjzk3BjwydRWNT1o3leTG7Fmsrp6fq1J2Ef4tJhj2D0_kzAQLxVdqcSMc3rIpyC5_vV_HgIZdqtqabjr_-jofI_lHr8k\" ><\/figure>\n\n\n\n<p>To divide the money between two sisters by the ratio of their ages, we must calculate the sum of the ratio as:<\/p>\n\n\n\n<p>Ratio = 5:4, Sum of 5 and 4 = 5 + 4 = 9<\/p>\n\n\n\n<p>Hence. Shreya must get 5\/9 of the total money and Bhoomika must get 4\/9 of the total money.<\/p>\n\n\n\n<p>\u2234 money that Shreya must get \u2013\u00a0<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/RTagCd_BuZXFcAXJndYcfS6ZhW1HBQL2kEmjukS9ZTXvdq82bIXxuc50S5nHuBwZnAhSZ93wP6o65tK0gBsj9e3Oaw_Bw0tyN6bjg1_99TFNOwAtn3v9lgFP75Ovh_HmF4bDTZIg6an6YU2KsC8vmh0\" ><\/figure>\n\n\n\n<p>And the money that Bhoomika must get \u2013<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img  decoding=\"async\"  src=\"data:image\/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEUAAP+KeNJXAAAAAXRSTlMAQObYZgAAAAlwSFlzAAAOxAAADsQBlSsOGwAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=\"  alt=\"\"  class=\" pk-lazyload\"  data-pk-sizes=\"auto\"  data-pk-src=\"https:\/\/lh7-us.googleusercontent.com\/xbz3BBvA9j2G9O6-Ac4iP3pnaS1nXJqBMoYtpCk_Z4nH7kmBvoD9hxcwavZ1f2KXrFCZK78i6M3OaGNq4_cjv_VPn6C9Gt6wgfYnLN1vSs9Q0hrwF3VFOkhNTTDkEjWzI7R8xMoj--bd5L9rnTb1aVQ\" ><\/figure>\n\n\n\n<p>So, Shreya must get \u20b9 20 and Bhoomika must get \u20b9 16.<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/career-after-bsc-maths\/\"><strong>Career after BSc Maths<\/strong><\/a><\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-exercise-12-2-solutions\">Exercise 12.2 Solutions<\/h3>\n\n\n\n<p><strong>Q 1. Determine if the following are in proportion.<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>15, 45, 40, 120<\/strong><\/li>\n\n\n\n<li><strong>33, 121, 9,96\u00a0<\/strong><\/li>\n<\/ol>\n\n\n\n<p><strong>Solutions. <\/strong>We know that two ratios are in proportion if they are equal.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>We must simplify the ratio of each of the given respective terms in order to compare their ratios.\u00a0<\/li>\n<\/ol>\n\n\n\n<p>15\/45 = \u2153\u00a0 and\u00a0 40\/120 = \u2153<\/p>\n\n\n\n<p>Hence, 15\/45 = 40\/120, they are in proportion.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\" start=\"2\">\n<li>Now, 33\/121 = 3\/11 and 9\/96 = 3\/32<\/li>\n<\/ol>\n\n\n\n<p>\u2234 33\/121 \u2260 9\/96, they are not in proportion.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>Career in Mathematics and Computing<\/strong><\/p>\n\n\n\n<p><strong>Q 2. Are the following statements true?<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>40 persons : 200 persons = \u20b9 15 : \u20b9 75<\/strong><\/li>\n\n\n\n<li><strong>7.5 litres : 15 litres = 5 kg : 10 kg<\/strong><\/li>\n\n\n\n<li><strong>99 kg : 45 kg = \u20b9 44 : \u20b9 20<\/strong><\/li>\n<\/ol>\n\n\n\n<p><strong>Solutions. <\/strong>Let us see the solution to each of the questions below.\u00a0<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>40\/200 = 1:5\u00a0 and\u00a0 15\/75 = 1:5<\/li>\n<\/ol>\n\n\n\n<p>So, 40\/200 = 15\/75<\/p>\n\n\n\n<p>\u2234 The given proportionality statement is true.<\/p>\n\n\n\n<ol class=\"wp-block-list\" start=\"2\">\n<li>7.5\/15 = 75\/150 = 1:2\u00a0 and\u00a0 5\/10 = 1:2<\/li>\n<\/ol>\n\n\n\n<p>So, 7.5\/15 = 5\/10<\/p>\n\n\n\n<p>\u2234 The given proportionality statement is true.<\/p>\n\n\n\n<ol class=\"wp-block-list\" start=\"3\">\n<li>99\/45 = 11\/5\u00a0 and\u00a0 44\/20 = 11\/5<\/li>\n<\/ol>\n\n\n\n<p>So, 99\/45 = 11\/5<\/p>\n\n\n\n<p>\u2234 The given proportionality statement is true.<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/blog\/courses-after-12th-commerce-without-maths\/\"><strong>Courses after 12th Commerce without Maths<\/strong><\/a><\/p>\n\n\n\n<h3 class=\"wp-block-heading\" id=\"h-exercise-12-3-solutions\">Exercise 12.3 Solutions<\/h3>\n\n\n\n<p><strong>Q 1.\u00a0 If the cost of 7 m of cloth is \u20b9 1470, find the cost of 5 m of cloth.<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>Given, the cost of 7 m of cloth is \u20b9 1470.<\/p>\n\n\n\n<p>Cost of 1 m of cloth = \u20b91470\/7<\/p>\n\n\n\n<p>Now, the cost of 5 m of cloth = 1470\/7 \u00d7 5 = \u20b9 1050<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background has-medium-font-size\"><strong>NCERT Class 11 Maths Solutions & Study Notes<\/strong><\/p>\n\n\n\n<p><strong>Q 2. If it has rained 276 mm in the last 3 days, how many cm of rain will fall in one full week (7 days)? Assume that the rain continues to fall at the same rate.<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>Given that it has rained 276 mm in 3 days. If the rain continues to fall at the same rate then \u2013\u00a0<\/p>\n\n\n\n<p>Amount of rain in 3 days = 276 mm<\/p>\n\n\n\n<p>Amount of rain in 1 day = 276\/3 mm<\/p>\n\n\n\n<p>Now, the amount of rain in 7 days = 276\/3 \u00d7 7 = 644 mm<\/p>\n\n\n\n<p>Now, we know 1 mm = 0.1 cm<\/p>\n\n\n\n<p>\u2234 644 mm = 64.4 cm<\/p>\n\n\n\n<p>So, 64.4 cm of rain will fall in 7 days.\u00a0<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background has-medium-font-size\"><a href=\"https:\/\/leverageedu.com\/explore\/career-counselling\/maths-teacher-interview-questions\/\"><strong>Maths Teacher Interview Questions with Sample Answers<\/strong><\/a><\/p>\n\n\n\n<p><strong>Q 3.\u00a0 Shaina pays \u20b9 15000 as rent for 3 months. How much does she have to pay for a whole year, if the rent per month remains the same?<\/strong><\/p>\n\n\n\n<p><strong>Solutions. <\/strong>Given,<strong> <\/strong>rent for 3 months = \u20b9 15,000<\/p>\n\n\n\n<p>Rent for 1 month = \u20b9 15000\/3<\/p>\n\n\n\n<p>Rent for 12 months = \u20b9 15000\/3 \u00d7 12 = \u20b9 60,000<\/p>\n\n\n\n<p>So, Shaina will have to pay \u20b9 60,000 as rent for 12 months.<\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>CBSE Class 9 Maths Syllabus<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><a href=\"https:\/\/leverageedu.com\/learn\/business-mathematics\/\"><strong>Business Mathematics<\/strong><\/a><\/p>\n\n\n\n<p class=\"has-text-align-center has-light-green-cyan-background-color has-background\"><strong>Applied Mathematics and Computation<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<iframe loading=\"lazy\" title=\"Introduction - Ratio and Proportion - Chapter 12 - Class 6th Maths\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/FEQ3iIDDOSo?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" allowfullscreen><\/iframe>\n<\/div><figcaption class=\"wp-element-caption\">Source: Mathematics Class VI<\/figcaption><\/figure>\n\n\n\n<p>Check out Maths Class 6 Notes and Exercise Solutions for other chapters below.\u00a0<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-regular\" style=\"font-size:18px\"><table class=\"has-background has-fixed-layout\" style=\"background:linear-gradient(135deg,rgb(255,245,203) 0%,rgb(182,227,212) 100%,rgb(51,167,181) 100%)\"><tbody><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-ncert-maths-chapter-1-knowing-our-numbers\/\"><strong>Maths Class 6 Chapter 1 Notes & Solutions<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-ncert-maths-chapter-2-whole-numbers\/\"><strong>Maths Class 6 Chapter 2 Notes & Solutions<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-ncert-maths-chapter-3-playing-with-numbers\/\"><strong>Maths Class 6 Chapter 3 Notes & Solutions<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-ncert-maths-chapter-4-basic-geometrical-ideas\/\"><strong>Maths Class 6 Chapter 4 Notes & Solutions<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-ncert-maths-chapter-5-understanding-elementary-shapes\/\"><strong>Maths Class 6 Chapter 5 Notes & Solutions<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-ncert-class-6-maths-chapter-6-integers\/\"><strong>Maths Class 6 Chapter 6 Notes & Solutions<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-ncert-class-6-maths-chapter-7-fractions\/\"><strong>Maths Class 6 Chapter 7 Notes & Solutions<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-ncert-maths-class-6-chapter-8-decimals\/\"><strong>Maths Class 6 Chapter 8 Notes & Solutions<\/strong><\/a><\/td><\/tr><tr><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-study-material\/cbse-ncert-maths-class-6-chapter-9-notes\/\"><strong>Maths Class 6 Chapter 9 Notes & Solutions<\/strong><\/a><\/td><td><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-ncert-class-6-maths-chapter-10\/\"><strong>Maths Class 6 Chapter 10 Notes & Solutions<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-faqs\">FAQs<\/h2>\n\n\n\n<div class=\"schema-faq wp-block-yoast-faq-block\"><div class=\"schema-faq-section\" id=\"faq-question-1697866695438\"><strong class=\"schema-faq-question\">Q.1. What is Ratio?<\/strong> <p class=\"schema-faq-answer\">Ans: When we compare two quantities in terms of \u2018how many times\u2019. This comparison is known as the Ratio. We denote the ratio using the symbol \u2018:\u2019.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1697866735184\"><strong class=\"schema-faq-question\">Q.2. What is Proportion?<\/strong> <p class=\"schema-faq-answer\">Ans: If two ratios are equal, we say that they are in proportion and use the symbol \u2018::\u2019 or \u2018=\u2019 to equate the two ratios. If two ratios are not equal, then we say that they are not in proportion.<\/p> <\/div> <div class=\"schema-faq-section\" id=\"faq-question-1697866755966\"><strong class=\"schema-faq-question\">Q.3. What is the Unitary Method?<\/strong> <p class=\"schema-faq-answer\">Ans: The method in which first we find the value of one unit and then the value of the required number of units is known as the Unitary Method.<\/p> <\/div> <\/div>\n\n\n\n<p>This was all about NCERT Maths Ratio and Proportion Class 6, Chapter 12 in which we studied the use of variables to make certain rules and that mathematical operations can be operated on variables when we need to determine an unknown value for a given quantity. Download the NCERT Maths Ratio and Proportion Class 6 Notes and Exercise Solutions to ace your exam preparations. Follow the <a href=\"https:\/\/leverageedu.com\/discover\/school-education\/cbse-class-6-maths-notes\/\"><strong>CBSE Class 6 Maths Notes<\/strong><\/a> for more such chapter notes and important questions and answers for preparation for CBSE Class 6 Maths.\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"In Chapter 12, Ratio and Proportion \u2013 Class 6, students will be introduced to the concepts of Ratio&hellip;\n","protected":false},"author":70,"featured_media":793340,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-793330","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.5 (Yoast SEO v27.5) - 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value.","og_url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/","og_site_name":"Leverage Edu Discover","article_published_time":"2023-10-21T05:42:57+00:00","og_image":[{"width":1024,"height":640,"url":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2023\/10\/13134449\/Class-6-Maths-Chapter-12.png","type":"image\/png"}],"author":"Priya Garg","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Priya Garg","Est. reading time":"14 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/#article","isPartOf":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/"},"author":{"name":"Priya 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I am Priya. I provide all sorts of educational content. Working in the content domain for the past seven years, I give the most precise student-friendly information. Professionally, I am a Chemist. So, ethically I don't let any error pass on to the content. All my content is fact-based and contains detailed knowledge of the subject.","sameAs":["http:\/\/www.linkedin.com\/in\/priya-garg-ba4478134"],"url":"https:\/\/leverageedu.com\/discover\/author\/priya\/"},{"@type":"Question","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/#faq-question-1697866695438","position":1,"url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-maths-ratio-and-proportion-class-6-chapter-12-notes\/#faq-question-1697866695438","name":"Q.1. What is Ratio?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"Ans: When we compare two quantities in terms of \u2018how many times\u2019. This comparison is known as the Ratio. 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What is the Unitary Method?","answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"Ans: The method in which first we find the value of one unit and then the value of the required number of units is known as the Unitary Method.","inLanguage":"en-US"},"inLanguage":"en-US"}]}},"acf":[],"_links":{"self":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/793330","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/users\/70"}],"replies":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/comments?post=793330"}],"version-history":[{"count":0,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/793330\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media\/793340"}],"wp:attachment":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media?parent=793330"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/categories?post=793330"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/tags?post=793330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}