{"id":415,"date":"2022-04-13T10:55:50","date_gmt":"2022-04-13T10:55:50","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=415"},"modified":"2025-03-20T07:15:53","modified_gmt":"2025-03-20T07:15:53","slug":"v-3-weighted-averages-word-problems","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/v-3-weighted-averages-word-problems\/","title":{"rendered":"V.3 Weighted averages: word problems"},"content":{"rendered":"\n

Weighted averages: word problems<\/strong><\/h2>\n\n\n\n

Key Notes :<\/strong><\/p>\n\n\n\n

\ud83c\udfaf What is a Weighted Average?<\/strong><\/h3>\n\n\n\n
    \n
  • A weighted average<\/strong> gives different values more importance (weight) based on how frequently or significantly they occur.<\/li>\n<\/ul>\n\n\n\n

    The formula is:<\/strong><\/p>\n\n\n\n

    Weighted Average = \u2211(value\u00d7weight) \/ \u2211weights<\/p>\n\n\n\n

    \u2705 Example:<\/strong> If a test counts for 60% of your grade and a project counts for 40%, the average isn’t a simple mean \u2014 the test carries more “weight.”<\/p>\n\n\n\n

    \n\n\n\n

    \ud83d\udd0d Understanding Weights<\/strong><\/h3>\n\n\n\n
      \n
    • Weights<\/strong> are percentages, frequencies, or any numbers representing importance.<\/li>\n\n\n\n
    • Total weight<\/strong> should add up to 100% (if dealing with percentages) or the total number of items.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      \ud83d\udee0\ufe0fTypes of Weighted Average Problems<\/strong><\/h3>\n\n\n\n
        \n
      • Grades or scores:<\/strong> Different assignments\/tests have different weights.<\/li>\n\n\n\n
      • Mixture problems:<\/strong> Two or more substances with different prices or concentrations are mixed.<\/li>\n\n\n\n
      • Speed or rate problems:<\/strong> Traveling different distances at different speeds.<\/li>\n\n\n\n
      • Financial averages:<\/strong> Calculating average prices, investments, or returns with different amounts invested.<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        \ud83e\udde0 Solving Word Problems Step-by-Step<\/strong><\/h3>\n\n\n\n
          \n
        • Identify the values and weights<\/strong> \u2014 What\u2019s being averaged? What are the weights?<\/li>\n\n\n\n
        • Set up the equation<\/strong> \u2014 Multiply each value by its weight, then divide the total by the sum of the weights.<\/li>\n\n\n\n
        • Solve the equation<\/strong> \u2014 Simplify carefully to avoid mistakes.<\/li>\n\n\n\n
        • Check the answer<\/strong> \u2014 Does the result make sense?<\/li>\n<\/ul>\n\n\n\n

          Learn with an example<\/strong><\/p>\n\n\n\n

          \n
          \n

          \ud83d\uddfc Daniel, a chemist, has 2 litres of a solution that contains 15% hydrochloric acid (HCl), plus unlimited 25% HCl solution. For a particular experiment, however, he needs to prepare some that contains 24% HCl. How much 25% HCl solution should be added to the 15% HCl solution to get the 24% HCl solution needed for the experiment?<\/strong><\/p>\n\n\n\n

          \ud83d\uddfc Write your answer as a whole number or as a decimal rounded to the nearest tenth.<\/em>______ litres<\/p>\n<\/div><\/div>\n\n\n\n

          Step 1: Define a variable and make a table.<\/strong><\/p>\n\n\n\n

          Let x<\/em> represent the quantity of 15% HCl solution. Now make a table.<\/p>\n\n\n\n

          Start by filling in the first two columns with what you know.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of HCl solution (litres)<\/strong><\/td>Amount of HCl (litres)<\/strong><\/td><\/tr>
          15% HCl solution<\/strong><\/td>0.15<\/td>2<\/td><\/td><\/tr>
          25% HCl solution<\/strong><\/td>0.25<\/td>x<\/em><\/td><\/td><\/tr>
          24% HCl solution<\/strong><\/td>0.24<\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Next, finish filling in the second column. Since there are 2 litres of 15% HCl solution and x<\/em> litres of 25% HCl solution, there are a total of 2 + x<\/em> litres of 24% HCl solution. Fill in 2 + x<\/em>.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of HCl solution (litres)<\/strong><\/td>Amount of HCl (litres)<\/strong><\/td><\/tr>
          15% HCl solution<\/strong><\/td>0.15<\/td>2<\/td><\/td><\/tr>
          25% HCl solution<\/strong><\/td>0.25<\/td>x<\/em><\/td><\/td><\/tr>
          24% HCl solution<\/strong><\/td>0.24<\/td>2 + x<\/em><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Finally, multiply across to fill in the third column.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of HCl solution (litres)<\/strong><\/td>Amount of HCl (litres)<\/strong><\/td><\/tr>
          15% HCl solution<\/strong><\/td>0.15<\/td>2<\/td>0.15(2)<\/td><\/tr>
          25% HCl solution<\/strong><\/td>0.25<\/td>x<\/em><\/td>0.25x<\/em><\/td><\/tr>
          24% HCl solution<\/strong><\/td>0.24<\/td>2 + x<\/em><\/td>0.24(2 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Step 2: Write an equation and solve.<\/strong><\/p>\n\n\n\n

          Using the information in the table, write an equation.<\/p>\n\n\n\n

          amount of HCl in 15% HCl solution<\/strong><\/td>plus<\/td>amount of HCl in 25% HCl solution<\/strong><\/td>equals<\/td>amount of HCl in 24% HCl solution<\/strong><\/td><\/tr>
           <\/td><\/td><\/td><\/td><\/td><\/tr>
          0.15(2)<\/td>+<\/td>0.25x<\/em><\/td>=<\/td>0.24(2 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Solve the equation.<\/p>\n\n\n\n

          0.15(2) + 0.25x<\/em><\/td> = <\/td>0.24(2 + x<\/em>)<\/td><\/td><\/tr>
          <\/td><\/tr>
          0.3 + 0.25x<\/em><\/td> = <\/td>0.48 + 0.24x<\/em><\/td>Multiply<\/em><\/td><\/tr>
          <\/td><\/tr>
          0.3 + 0.01x<\/em><\/td> = <\/td>0.48<\/td>Subtract 0.24x from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          0.01x<\/em><\/td> = <\/td>0.18<\/td>Subtract 0.3 from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          x<\/em><\/td> = <\/td>18<\/td>Divide both sides by 0.01<\/em><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          To get the 24% HCl solution needed for the experiment, Daniel should add 18 litres of the 25% HCl solution.<\/p>\n<\/div><\/div>\n\n\n\n

          \n
          \n

          \ud83d\uddfc The manager at a health foods store mixes a unique superfruit juice cocktail that costs \u20b91,063.10 per litre to make. The cocktail includes mixed fruit juice and a\u00e7ai berry juice, which cost \u20b9344.15 per litre and \u20b92,740.65 per litre, respectively. The manager has already opened 50 litres of the mixed fruit juice. How many litres of the a\u00e7ai berry juice does he need to add?<\/strong><\/p>\n\n\n\n

          \ud83d\uddfc Write your answer as a whole number or as a decimal rounded to the nearest tenth.<\/em>______ litres<\/p>\n<\/div><\/div>\n\n\n\n

          Step 1: Define a variable and make a table.<\/strong><\/p>\n\n\n\n

          Let x<\/em> represent the quantity of mixed fruit juice. Now make a table.<\/p>\n\n\n\n

          Start by filling in the first two columns with what you know.<\/p>\n\n\n\n

          <\/td>Price per litre<\/strong><\/td>Number of litres<\/strong><\/td>Total price<\/strong><\/td><\/tr>
          Mixed fruit juice<\/strong><\/td>344.15<\/td>50<\/td><\/td><\/tr>
          A\u00e7ai berry juice<\/strong><\/td>2,740.65<\/td>x<\/em><\/td><\/td><\/tr>
          Superfruit juice cocktail<\/strong><\/td>1,063.1<\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Next, finish filling in the second column. Since there are 50 litres of mixed fruit juice and x<\/em> litres of a\u00e7ai berry juice, there are a total of 50 + x<\/em> litres of superfruit juice cocktail. Fill in 50 + x<\/em>.<\/p>\n\n\n\n

          <\/td>Price per litre<\/strong><\/td>Number of litres<\/strong><\/td>Total price<\/strong><\/td><\/tr>
          Mixed fruit juice<\/strong><\/td>344.15<\/td>50<\/td><\/td><\/tr>
          A\u00e7ai berry juice<\/strong><\/td>2,740.65<\/td>x<\/em><\/td><\/td><\/tr>
          Superfruit juice cocktail<\/strong><\/td>1,063.1<\/td>50 + x<\/em><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Finally, multiply across to fill in the third column.<\/p>\n\n\n\n

          <\/td>Price per litre<\/strong><\/td>Number of litres<\/strong><\/td>Total price<\/strong><\/td><\/tr>
          Mixed fruit juice<\/strong><\/td>344.15<\/td>50<\/td>344.15(50)<\/td><\/tr>
          A\u00e7ai berry juice<\/strong><\/td>2,740.65<\/td>x<\/em><\/td>2,740.65x<\/em><\/td><\/tr>
          Superfruit juice cocktail<\/strong><\/td>1,063.1<\/td>50 + x<\/em><\/td>1,063.1(50 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Step 2: Write an equation and solve.<\/strong><\/p>\n\n\n\n

          Using the information in the table, write an equation.<\/p>\n\n\n\n

          total price of mixed fruit juice<\/strong><\/td>plus<\/td>total price of a\u00e7ai berry juice<\/strong><\/td>equals<\/td>total price of superfruit juice cocktail<\/strong><\/td><\/tr>
           <\/td><\/td><\/td><\/td><\/td><\/tr>
          344.15(50)<\/td>+<\/td>2,740.65x<\/em><\/td>=<\/td>1,063.1(50 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Solve the equation.<\/p>\n\n\n\n

          344.15(50) + 2,740.65x<\/em><\/td> = <\/td>1,063.1(50 + x<\/em>)<\/td><\/td><\/tr>
          <\/td><\/tr>
          17,207.5 + 2,740.65x<\/em><\/td> = <\/td>53,155 + 1,063.1x<\/em><\/td>Multiply<\/em><\/td><\/tr>
          <\/td><\/tr>
          17,207.5 + 1,677.55x<\/em><\/td> = <\/td>53,155<\/td>Subtract 1,063.1x from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          1,677.55x<\/em><\/td> = <\/td>35,947.5<\/td>Subtract 17,207.5 from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          x<\/em><\/td> \u2248 <\/td>21.4<\/td>Divide both sides by 1,677.55<\/em><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          The manager needs to add about 21.4 litres of the a\u00e7ai berry juice.<\/p>\n<\/div><\/div>\n\n\n\n

          \n
          \n

          \ud83d\uddfc For a big cleaning job, Ayana wants to mix some ammonia-based cleaners to get a certain concentration of ammonia. She has 8 litres of cleaner that is 15% ammonia. She also has a large supply of cleaner that contains 20% ammonia. How many litres of the 20%-ammonia cleaner does Ayana need to add to the 15%-ammonia cleaner to obtain a batch that contains 18% ammonia?<\/strong><\/p>\n\n\n\n

          \ud83d\uddfc Write your answer as a whole number or as a decimal rounded to the nearest tenth.<\/em>_____ litres<\/p>\n<\/div><\/div>\n\n\n\n

          Step 1: Define a variable and make a table.<\/strong><\/p>\n\n\n\n

          Let x<\/em> represent the quantity of 15% cleaner. Now make a table.<\/p>\n\n\n\n

          Start by filling in the first two columns with what you know.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of cleaner (litres)<\/strong><\/td>Amount of ammonia (litres)<\/strong><\/td><\/tr>
          15% cleaner<\/strong><\/td>0.15<\/td>8<\/td><\/td><\/tr>
          20% cleaner<\/strong><\/td>0.2<\/td>x<\/em><\/td><\/td><\/tr>
          18% cleaner<\/strong><\/td>0.18<\/td><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Next, finish filling in the second column. Since there are 8 litres of 15% cleaner and x<\/em> litres of 20% cleaner, there are a total of 8 + x<\/em> litres of 18% cleaner. Fill in 8 + x<\/em>.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of cleaner (litres)<\/strong><\/td>Amount of ammonia (litres)<\/strong><\/td><\/tr>
          15% cleaner<\/strong><\/td>0.15<\/td>8<\/td><\/td><\/tr>
          20% cleaner<\/strong><\/td>0.2<\/td>x<\/em><\/td><\/td><\/tr>
          18% cleaner<\/strong><\/td>0.18<\/td>8 + x<\/em><\/td><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Finally, multiply across to fill in the third column.<\/p>\n\n\n\n

          <\/td>Concentration<\/strong><\/td>Amount of cleaner (litres)<\/strong><\/td>Amount of ammonia (litres)<\/strong><\/td><\/tr>
          15% cleaner<\/strong><\/td>0.15<\/td>8<\/td>0.15(8)<\/td><\/tr>
          20% cleaner<\/strong><\/td>0.2<\/td>x<\/em><\/td>0.2x<\/em><\/td><\/tr>
          18% cleaner<\/strong><\/td>0.18<\/td>8 + x<\/em><\/td>0.18(8 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Step 2: Write an equation and solve.<\/strong><\/p>\n\n\n\n

          Using the information in the table, write an equation.<\/p>\n\n\n\n

          amount of ammonia in 15% cleaner<\/strong><\/td>plus<\/td>amount of ammonia in 20% cleaner<\/strong><\/td>equals<\/td>amount of ammonia in 18% cleaner<\/strong><\/td><\/tr>
           <\/td><\/td><\/td><\/td><\/td><\/tr>
          0.15(8)<\/td>+<\/td>0.2x<\/em><\/td>=<\/td>0.18(8 + x<\/em>)<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Solve the equation.<\/p>\n\n\n\n

          0.15(8) + 0.2x<\/em><\/td> = <\/td>0.18(8 + x<\/em>)<\/td><\/td><\/tr>
          <\/td><\/tr>
          1.2 + 0.2x<\/em><\/td> = <\/td>1.44 + 0.18x<\/em><\/td>Multiply<\/em><\/td><\/tr>
          <\/td><\/tr>
          1.2 + 0.02x<\/em><\/td> = <\/td>1.44<\/td>Subtract 0.18x from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          0.02x<\/em><\/td> = <\/td>0.24<\/td>Subtract 1.2 from both sides<\/em><\/td><\/tr>
          <\/td><\/tr>
          x<\/em><\/td> = <\/td>12<\/td>Divide both sides by 0.02<\/em><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

          Ayana needs to add 12 litres of the 20%-ammonia cleaner.\"\"<\/p>\n<\/div><\/div>\n\n\n\n

          Let’s practice!\ud83d\udd8a\ufe0f<\/p>\n\n\n\n

          \n
          \n
          \"\"<\/a><\/figure>\n<\/div>\n\n\n\n
          \n
          \"\"<\/a><\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

          Weighted averages: word problems Key Notes : \ud83c\udfaf What is a Weighted Average? The formula is: Weighted Average = \u2211(value\u00d7weight) \/ \u2211weights \u2705 Example: If a test counts for 60% of your grade and a project counts for 40%, the average isn’t a simple mean \u2014 the test carries more “weight.” \ud83d\udd0d Understanding Weights \ud83d\udee0\ufe0fTypes ofContinue reading “V.3 Weighted averages: word problems”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-415","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/415","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=415"}],"version-history":[{"count":14,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/415\/revisions"}],"predecessor-version":[{"id":17652,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/415\/revisions\/17652"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=415"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}