{"id":31872,"date":"2026-09-17T17:07:28","date_gmt":"2026-09-17T11:37:28","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31872"},"modified":"2026-09-17T17:08:06","modified_gmt":"2026-09-17T11:38:06","slug":"working-with-fractions-ncert-solutions-class-7-maths-ganita-prakash","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/working-with-fractions-ncert-solutions-class-7-maths-ganita-prakash\/","title":{"rendered":"Working With Fractions – NCERT Solutions Class 7 Maths (Ganita Prakash)"},"content":{"rendered":"\n

Working With Fractions<\/strong><\/strong> – NCERT Solutions Class 7 Maths (Ganita Prakash) includes all the questions with solutions given in the NCERT Class 7 Maths (Ganita Prakash).<\/p>\n\n\n\n

NCERT Solutions Class 7<\/h2>\n\n\nEnglish Poorvi<\/a>\n\n\nHindi Malhar<\/a>\n\n\nMaths Ganita Prakash<\/a>\n\n\nScience Curiosity<\/a>\n\n\nSocial Exploring Society<\/a>\n\n\n\n

Working With Fractions<\/strong><\/strong> \u2013 NCERT Solutions<\/h2>\n\n\n\n
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Q.1: Tenzin drinks {tex}\\frac{1}{2}{\/tex} glass of milk every day. How many glasses of milk does he drink in a week? How many glasses of milk did he drink in the month of January?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Glass of milk drank every day {tex}=\\frac{1}{2}{\/tex}
Glasses of milk drank in a week {tex}=\\frac{1}{2} \\times 7=\\frac{7}{2}{\/tex}
Days in month of January = 31
Glasses of milk drank in month of January {tex}=31 \\times \\frac{1}{2}=\\frac{31}{2}=15 \\frac{1}{2}{\/tex}.<\/p>\n\n\n\n

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Q.2: A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make ________ km of the water canal. If they work 5 days a week, they can make ________ km of the water canal in a week.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

The length of canal made by workers in 8 days {tex}=1 {~km}{\/tex}
The length of canal made by workers in 1 day {tex}=\\frac{1}{8} {~km}{\/tex}
By working 5 days in a week, the length of canal made by workers {tex}=5 \\times \\frac{1}{8}=\\frac{5}{8} {~km}{\/tex}.<\/p>\n\n\n\n

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Q.3: Manju and two of her neighbours buy 5 litres of oil every week and share it equally among the 3 families. How much oil does each family get in a week? How much oil will one family get in 4 weeks?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

Amount of oil shared by 3 families in a week = 5 litres
Amount of oil one family got in a week {tex}=\\frac{5}{3}{\/tex} litres
Amount of oil one family got in 4 weeks {tex}=4 \\times \\frac{5}{3}=\\frac{20}{3}=6 \\frac{2}{3}{\/tex} litres.<\/p>\n\n\n\n

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Q.4: Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets {tex}\\frac{5}{6}{\/tex} hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

No. of days from Monday to Thursday = 3
Delay in moon setting time per day {tex}=\\frac{5}{6}{\/tex} hour
Delay in moon setting time over 3 days {tex}=3 \\times \\frac{5}{6}{\/tex} hour {tex}=\\frac{5}{2}=2.5{\/tex} hours
Thus, the Moon will set 2.5 hours or 2 hours 30 minutes after 10 PM on Thursday.<\/p>\n\n\n\n

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Q.5: Multiply and then convert it into a mixed fraction:<\/p>\n\n\n\n

    \n
  1. {tex} 7 \\times \\frac{3}{5} {\/tex}<\/li>\n\n\n\n
  2. {tex} 4 \\times \\frac{1}{3} {\/tex}<\/li>\n\n\n\n
  3. {tex} \\frac{9}{7} \\times 6 {\/tex}<\/li>\n\n\n\n
  4. {tex} \\frac{13}{11} \\times 6 {\/tex}<\/li>\n<\/ol>\n\n\n\n

    Solution:<\/p>\n\n\n\n

      \n
    1. {tex}7 \\times \\frac{3}{5}=\\frac{21}{5}=4 \\frac{1}{5}{\/tex}.<\/li>\n\n\n\n
    2. {tex}4 \\times \\frac{1}{3}=\\frac{4}{3}{\/tex}.<\/li>\n\n\n\n
    3. {tex}\\frac{9}{7} \\times 6=\\frac{54}{7}{\/tex}.<\/li>\n\n\n\n
    4. {tex}\\frac{13}{11} \\times 6=\\frac{78}{11}{\/tex}.<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Q.6: Find the following products. Use a unit square as a whole for representing the fractions:<\/p>\n\n\n\n

        \n
      1. {tex} \\frac{1}{3} \\times \\frac{1}{5} {\/tex}<\/li>\n\n\n\n
      2. {tex} \\frac{1}{4} \\times \\frac{1}{3} {\/tex}<\/li>\n\n\n\n
      3. {tex} \\frac{1}{5} \\times \\frac{1}{2} {\/tex}<\/li>\n\n\n\n
      4. {tex} \\frac{1}{6} \\times \\frac{1}{5} {\/tex}<\/li>\n<\/ol>\n\n\n\n

        Solution:<\/p>\n\n\n\n

          \n
        1. {tex} \\frac{1}{3} \\times \\frac{1}{5} {\/tex}
          {tex}\\frac{1}{3}{\/tex} (multiplier) {tex}\\times \\frac{1}{5}{\/tex} (multiplicand)
          Number of rows {tex}={\/tex} Denominator of the multiplicand {tex}=5{\/tex}
          Number of columns {tex}={\/tex} Denominator of the multiplier {tex}=3{\/tex}
          Thus, the whole is divided into {tex}(5 \\times 3)=15{\/tex} equal parts
          So, {tex}\\frac{1}{3} \\times \\frac{1}{5}=\\frac{1}{3 \\times 5}=\\frac{1}{15}{\/tex}.
          <\/li>\n\n\n\n
        2. {tex}\\frac{1}{4} \\times \\frac{1}{3}{\/tex}
          {tex}\\frac{1}{4}{\/tex} (multiplier) {tex}\\times \\frac{1}{3}{\/tex} (multiplicand)
          Number of rows {tex}={\/tex} Denominator of the multiplicand {tex}=3{\/tex}
          Number of columns {tex}={\/tex} Denominator of the multiplier {tex}=4{\/tex}
          Thus, the whole is divided into {tex}(3 \\times 4)=12{\/tex} equal parts
          So, {tex}\\frac{1}{4} \\times \\frac{1}{3}=\\frac{1}{4 \\times 3}=\\frac{1}{12}{\/tex}.
          <\/li>\n\n\n\n
        3. {tex}\\frac{1}{5} \\times \\frac{1}{2}{\/tex}
          {tex}\\frac{1}{5}({\/tex}multiplier{tex}) \\times \\frac{1}{2}({\/tex}multiplicand{tex}){\/tex}
          Number of rows {tex}={\/tex} Denominator of the multiplicand {tex}=2{\/tex}
          Number of columns {tex}={\/tex} Denominator of the multiplier {tex}=5{\/tex}
          Thus, the whole is divided into {tex}(2 \\times 5)=10{\/tex} equal parts
          So, {tex}\\frac{1}{5} \\times \\frac{1}{2}=\\frac{1}{5 \\times 2}=\\frac{1}{10}{\/tex}.
          <\/li>\n\n\n\n
        4. {tex}\\frac{1}{6} \\times \\frac{1}{5}{\/tex}
          {tex}\\frac{1}{6}{\/tex} (multiplier) {tex}\\times \\frac{1}{5}{\/tex} (multiplicand)
          Number of rows {tex}={\/tex} Denominator of the multiplicand {tex}=5{\/tex}
          Number of columns = Denominator of the multiplier = 6
          Thus, the whole is divided into {tex}(5 \\times 6)=30{\/tex} equal parts
          So, {tex}\\frac{1}{6} \\times \\frac{1}{5}=\\frac{1}{6 \\times 5}=\\frac{1}{30}{\/tex}.<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          Q.7: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
          {tex}\\frac{2}{3} \\times \\frac{4}{5}{\/tex}<\/p>\n\n\n\n

          Solution:<\/p>\n\n\n\n

          {tex}\\frac{2}{3} \\times \\frac{4}{5}{\/tex}
          First, the whole is divided into 5 rows and 3 columns creating {tex}15(5 \\times 3){\/tex} equal parts.
          The value we get by dividing {tex}\\frac{4}{5}{\/tex} into 3 equal parts is {tex}\\frac{4}{3 \\times 5}{\/tex}.
          Thus, we multiply this result by 2 to get the product. This is {tex}\\frac{2 \\times 4}{3 \\times 5}{\/tex}.
          So, {tex}\\frac{2}{3} \\times \\frac{4}{5}=\\frac{2 \\times 4}{3 \\times 5}=\\frac{8}{15}{\/tex}.
          <\/p>\n\n\n\n

          \n\n\n\n

          Q.8: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
          {tex}\\frac{1}{4} \\times \\frac{2}{3}{\/tex}<\/p>\n\n\n\n

          Solution:<\/p>\n\n\n\n

          {tex}\\frac{1}{4} \\times \\frac{2}{3}{\/tex}
          First, the whole is divided into 3 rows and 4 columns creating {tex}12(3 \\times 4){\/tex} equal parts.
          The value we get by dividing {tex}\\frac{2}{3}{\/tex} into 4 equal parts is {tex}\\frac{2}{4 \\times 3}{\/tex}.
          So, {tex}\\frac{1}{4} \\times \\frac{2}{3}=\\frac{1 \\times 2}{4 \\times 3}=\\frac{2}{12}{\/tex}.
          <\/p>\n\n\n\n

          \n\n\n\n

          Q.9: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
          {tex}\\frac{3}{5} \\times \\frac{1}{2}{\/tex}<\/p>\n\n\n\n

          Solution:<\/p>\n\n\n\n

          {tex}\\frac{3}{5} \\times \\frac{1}{2}{\/tex}
          First, the whole is divided into 2 rows and 5 columns creating {tex}10(2 \\times 5){\/tex} equal parts.
          The value we get by dividing {tex}\\frac{1}{2}{\/tex} into 5 equal parts is {tex}\\frac{1}{5 \\times 2}{\/tex}.
          Thus, we multiply this result by 3 to get the product. This is {tex}\\frac{3 \\times 1}{5 \\times 2}{\/tex}.
          So, {tex}\\frac{3}{5} \\times \\frac{1}{2}=\\frac{3 \\times 1}{5 \\times 2}=\\frac{3}{10}{\/tex}.
          <\/p>\n\n\n\n

          \n\n\n\n

          Q.10: Find the product. Use a unit square as a whole for representing the fractions and carrying out the operations.
          {tex}\\frac{4}{6} \\times \\frac{3}{5}{\/tex}<\/p>\n\n\n\n

          Solution:<\/p>\n\n\n\n

          {tex}\\frac{4}{6} \\times \\frac{3}{5}{\/tex}
          First, the whole is divided into 5 rows and 6 columns creating {tex}30(5 \\times 6){\/tex} equal parts.
          The value we get by dividing {tex}\\frac{3}{5}{\/tex} into 6 equal parts is {tex}\\frac{3}{6 \\times 5}{\/tex}.
          Thus, we multiply this result by 4 to get the product. This is {tex}\\frac{4 \\times 3}{6 \\times 5}{\/tex}.
          So, {tex}\\frac{4}{6} \\times \\frac{3}{5}=\\frac{4 \\times 3}{6 \\times 5}=\\frac{12}{30}{\/tex}.<\/p>\n\n\n\n

          \n\n\n\n

          Q.11: A water tank is filled from a tap. If the tap is open for 1 hour, {tex}\\frac{7}{10}{\/tex} of the tank gets filled. How much of the tank is filled if the tap is open for<\/p>\n\n\n\n

            \n
          1. {tex}\\frac{1}{3}{\/tex} hour ________<\/li>\n\n\n\n
          2. {tex}\\frac{2}{3}{\/tex} hour\u00a0________<\/li>\n\n\n\n
          3. {tex}\\frac{3}{4}{\/tex} hour\u00a0________<\/li>\n\n\n\n
          4. {tex}\\frac{7}{10}{\/tex} hour\u00a0________<\/li>\n\n\n\n
          5. For the tank to be full, how long should the tap be running?<\/li>\n<\/ol>\n\n\n\n

            Solution:<\/p>\n\n\n\n

            Part of the tank filled in 1 hour {tex}=\\frac{7}{10}{\/tex}<\/p>\n\n\n\n

              \n
            1. Part of the tank filled in {tex}\\frac{1}{3}{\/tex} hour {tex}=\\frac{1}{3} \\times \\frac{7}{10}=\\frac{7}{30}{\/tex}.<\/li>\n\n\n\n
            2. Part of the tank filled in {tex}\\frac{2}{3}{\/tex} hour {tex}=\\frac{2}{3} \\times \\frac{7}{10}=\\frac{14}{30}=\\frac{7}{15}{\/tex}.<\/li>\n\n\n\n
            3. Part of the tank filled in {tex}\\frac{3}{4}{\/tex} hour {tex}=\\frac{3}{4} \\times \\frac{7}{10}=\\frac{21}{40}{\/tex}.<\/li>\n\n\n\n
            4. Part of the tank filled in {tex}\\frac{7}{10}{\/tex} hour {tex}=\\frac{7}{10} \\times \\frac{7}{10}=\\frac{49}{100}{\/tex}.<\/li>\n\n\n\n
            5. Part of the tank filled in 1 hour {tex}=\\frac{7}{10}{\/tex}.
              or Time required to fill {tex}\\frac{7}{10}{\/tex} of a tank {tex}=1{\/tex} hour.<\/li>\n<\/ol>\n\n\n\n

              Time required to fill 1 tank {tex}=1 \\div \\frac{7}{10}=1 \\times \\frac{10}{7}=\\frac{10}{7}{\/tex} hours {tex}=1 \\frac{3}{7}{\/tex} hours.<\/p>\n\n\n\n

              \n\n\n\n

              Q.12: The government has taken {tex}\\frac{1}{6}{\/tex} of Somu’s land to build a road. What part of the land remains with Somu now? She gives half of the remaining part of the land to her daughter, Krishna, and {tex}\\frac{1}{3}{\/tex} it to her son Bora. After giving them their shares, she kept the remaining land for herself.<\/p>\n\n\n\n

                \n
              1. What part of the original land did Krishna get?<\/li>\n\n\n\n
              2. What part of the original land did Bora get?<\/li>\n\n\n\n
              3. What part of the original land did Somu keep for herself?<\/li>\n<\/ol>\n\n\n\n

                Solution:<\/p>\n\n\n\n

                Part of Somu’s land acquired by the government {tex}=\\frac{1}{6}{\/tex}
                Somu’s original land {tex}=1-\\frac{1}{6}=\\frac{6-1}{6}=\\frac{5}{6}{\/tex}<\/p>\n\n\n\n

                  \n
                1. Part of the land given to Krishna {tex}=\\frac{1}{2}{\/tex} of original land {tex}=\\frac{1}{2} \\times \\frac{5}{6}=\\frac{5}{12}{\/tex}.<\/li>\n\n\n\n
                2. Part of the land given to Bora {tex}=\\frac{1}{3}{\/tex} of original land {tex}=\\frac{1}{3} \\times \\frac{5}{6}=\\frac{5}{18}{\/tex}.<\/li>\n\n\n\n
                3. Part of the land Somu kept for herself {tex}=\\frac{5}{6}-\\left(\\frac{5}{12}+\\frac{5}{18}\\right){\/tex}
                  {tex} =\\frac{5}{6}-\\left(\\frac{3 \\times 5+5 \\times 2}{36}\\right) {\/tex}
                  {tex} =\\frac{5}{6}-\\left(\\frac{15+10}{36}\\right) {\/tex}
                  {tex} =\\frac{5}{6}-\\left(\\frac{25}{36}\\right) {\/tex}
                  {tex} =\\frac{6 \\times 5-25}{36} {\/tex}
                  {tex} =\\frac{30-25}{36}=\\frac{5}{36} . {\/tex}<\/li>\n<\/ol>\n\n\n\n
                  \n\n\n\n

                  Q.13: Find the area of a rectangle of sides {tex}3 \\frac{3}{4} {ft}{\/tex} and {tex}9 \\frac{3}{5} {ft}{\/tex}.<\/p>\n\n\n\n

                  Solution:<\/p>\n\n\n\n

                  {tex} \\text { Area of rectangle }=3 \\frac{3}{4} {ft} \\times 9 \\frac{3}{5} {ft} =\\frac{15}{4} {ft} \\times \\frac{48}{5} {ft} {\/tex}
                  {tex}= \\frac {15}{4} \\times \\frac {48}{5}{\/tex} = 3 {tex}=3\\times 12{\/tex} = 36 sq. ft<\/p>\n\n\n\n

                  \n\n\n\n

                  Q.14: Tsewang plants four saplings in a row in his garden. The distance between two saplings is {tex}\\frac{3}{4}{\/tex}m. Find the distance between the first and last sapling.<\/p>\n\n\n\n

                  Solution:<\/p>\n\n\n\n

                  No. of saplings in a row in garden {tex}=4{\/tex}
                  Distance between two saplings {tex}=\\frac{3}{4}{\/tex}
                  Distance between first and last sapling {tex}=\\frac{3}{4}+\\frac{3}{4}+\\frac{3}{4}=\\frac{3+3+3}{4}=\\frac{9}{4} {~m}=2 \\frac{1}{4} {~m}{\/tex}.
                  <\/p>\n\n\n\n

                  \n\n\n\n

                  Q.15: Which is heavier: {tex}\\frac{12}{15}{\/tex} of 500 grams or {tex}\\frac{3}{20}{\/tex} of {tex}4 {~kg} ?{\/tex}<\/p>\n\n\n\n

                  Solution:<\/p>\n\n\n\n

                  {tex}\\frac{12}{15}{\/tex} of 500 grams {tex}=\\frac{12}{15} \\times 500 {~g}{\/tex}
                  {tex}=\\frac {12}{15}\\times 500 = 4 \\times 100 = 400 g{\/tex}
                  {tex}\\frac{3}{20}{\/tex} of {tex}4 {~kg}=\\frac{3}{20} \\times 4000 {~g}{\/tex}
                  {tex}= \\frac {3}{20} \\times 4000 = 3 \\times 200 = 600 g{\/tex}
                  Hence, {tex}\\frac{3}{20}{\/tex} of 4 kg is heavier than {tex}\\frac{12}{15}{\/tex} of 500 grams.<\/p>\n\n\n\n

                  \n\n\n\n

                  Q.16: Is the Product Always Greater than the Numbers Multiplied?
                  What can you conclude about the relationship between the numbers multiplied and the product? Fill in the blanks:<\/p>\n\n\n\n