In word problems on fraction we will solve different types of problems on multiplication of fractional numbers and division of fractional numbers.
1. Nairitee took \(\frac{7}{8}\) hour to paint a table and \(\frac{2}{3}\) hour to paint a chair. How much time did he take in painting both items?
Solution:
Total time taken in painting both items = \(\frac{7}{8}\) h + \(\frac{2}{3}\) h
= (\(\frac{7}{8}\) + \(\frac{2}{3}\)) h
= (\(\frac{21 + 16}{24}\)) h
= \(\frac{37}{24}\) h
= 1\(\frac{13}{24}\) h
Therefore, Nairitee took 1\(\frac{13}{24}\) hours in painting both items.
2. Nitheeya and Nairitee \(\frac{3}{10}\) and \(\frac{1}{6}\) of a cake respectively. What portion of the cake did they eat together?
Solution:
The portion of cake ate by Nitheeya = \(\frac{3}{10}\)
The portion of cake ate by Nitheeya = \(\frac{1}{6}\)
The portion they ate together = \(\frac{3}{10}\) + \(\frac{1}{6}\)
= \(\frac{9}{30}\) + \(\frac{5}{30}\); [Since, LCM of 10 and 6 = 30]
= \(\frac{9 + 5}{30}\)
= \(\frac{14}{30}\)
= \(\frac{7}{15}\)
Therefore, together Nitheeya and Nairitee ate \(\frac{7}{15}\) of the cake.
3. Rachel took \(\frac{1}{2}\) hour to paint a table and \(\frac{1}{3}\) hour to paint a chair. How much time did she take in all?
Solution:
Time taken to paint a table = \(\frac{1}{2}\) hour Time taken to paint a chair = \(\frac{1}{3}\) hour Total time taken = \(\frac{1}{2}\) hour + \(\frac{1}{3}\) hour = \(\frac{5}{6}\) hour |
\(\frac{1}{2}\) + \(\frac{1}{3}\) L.C.M. of 2, 3 is 6. = \(\frac{3}{6}\) + \(\frac{2}{6}\) \(\frac{1 × 3}{2 × 3}\) = \(\frac{3}{6}\) \(\frac{1 × 2}{3 × 2}\) = \(\frac{2}{6}\) |
1. Out of \(\frac{12}{17}\) m of cloth given to a tailor, \(\frac{1}{5}\) m were used. Find the length of cloth unused.
Solution:
Length of the cloth given to the tailors = \(\frac{12}{17}\) m
Length of cloth used = \(\frac{1}{5}\) m
Length of the unused cloth = \(\frac{12}{17}\) m - \(\frac{1}{5}\) m
= (\(\frac{12}{17}\) - \(\frac{1}{5}\)) m
= (\(\frac{12 × 5}{17 × 5}\) - \(\frac{1 × 17}{5 × 17}\)) m; [Since, LCM of 17 and 5 = 85]
= (\(\frac{60}{85}\) - \(\frac{17}{85}\)) m
= (\(\frac{60 - 17}{85}\) m
= (\(\frac{43}{85}\) m
2. Nairitee has $6\(\frac{4}{7}\). She gives $4\(\frac{2}{3}\) to her mother. How much money does she have now?
Solution:
Money with Nairitee = $6\(\frac{4}{7}\)
Money given to her mother = $4\(\frac{2}{3}\)
Money left with Nairitee = $6\(\frac{4}{7}\) - $4\(\frac{2}{3}\)
= $(6\(\frac{4}{7}\) - 4\(\frac{2}{3}\))
= $(\(\frac{46}{7}\) - \(\frac{14}{3}\))
= $(\(\frac{46 × 3}{7 × 3}\) - \(\frac{14 × 7}{3 × 7}\)); [Since, LCM of 7 and 3 = 21]
= $(\(\frac{138}{21}\) - \(\frac{98}{21}\))
= $\(\frac{40}{21}\)
= $1\(\frac{19}{21}\)
Therefore, Nairitee has $1\(\frac{19}{21}\).
3. If 3\(\frac{1}{2}\) m of wire is cut from a piece of 10 m long wire, how much of wire is left?
Total length of the wire = 10 m
Fraction of the wire cut out = 3\(\frac{1}{2}\) m = \(\frac{7}{2}\) m
Length of the wire left = 10 m – 3\(\frac{1}{2}\) m
= [\(\frac{10}{1}\) - \(\frac{7}{2}\)] m, [L.C.M. of 1, 2 is 2]
= [\(\frac{20}{2}\) - \(\frac{7}{2}\)] m, [\(\frac{10}{1}\) × \(\frac{2}{2}\)]
= [\(\frac{20 - 7}{2}\)] m
= \(\frac{13}{2}\) m
= 6\(\frac{1}{2}\) m
1. \(\frac{4}{7}\) of a number is 84. Find the number.
Solution:
According to the problem,
\(\frac{4}{7}\) of a number = 84
Number = 84 × \(\frac{7}{4}\)
[Here we need to multiply 84 by the reciprocal of \(\frac{4}{7}\)]
= 21 × 7
= 147
Therefore, the number is 147.
2. One half of the students in a school are girls, \(\frac{3}{5}\) of these girls are studying in lower classes. What fraction of girls are studying in lower classes?
Solution:
Fraction of girls studying in school = \(\frac{1}{2}\)
Fraction of girls studying in lower classes = \(\frac{3}{5}\) of \(\frac{1}{2}\)
= \(\frac{3}{5}\) × \(\frac{1}{2}\)
= \(\frac{3 × 1}{5 × 2}\)
= \(\frac{3}{10}\)
Therefore, \(\frac{3}{10}\) of girls studying in lower classes.
3. Maddy reads three-fifth of 75 pages of his lesson. How many more pages he need to complete the lesson?
Solution:
Maddy reads = \(\frac{3}{5}\) of 75
= \(\frac{3}{5}\) × 75
= 45 pages.
Maddy has to read = 75 – 45.
= 30 pages.
Therefore, Maddy has to read 30 more pages.
1. A herd of cows gives 4 litres of milk each day. But each cow gives one-third of total milk each day. They give 24 litres milk in six days. How many cows are there in the herd?
Solution:
A herd of cows gives 4 litres of milk each day.
Each cow gives one-third of total milk each day = \(\frac{1}{3}\) of 4
Therefore, each cow gives \(\frac{4}{3}\) of milk each day.
Total no. of cows = 4 ÷ \(\frac{4}{3}\)
= 4 × \(\frac{3}{4}\)
= 3
Therefore there are 3 cows in the herd.
Worksheet on Word problems on Fractions:
1. Shelly walked \(\frac{1}{3}\) km. Kelly walked \(\frac{4}{15}\) km. Who walked farther? How much farther did one walk than the other?
2. A frog took three jumps. The first jump was \(\frac{2}{3}\) m long, the second was \(\frac{5}{6}\) m long and the third was \(\frac{1}{3}\) m long. How far did the frog jump in all?
3. A vessel contains 1\(\frac{1}{2}\) l of milk. John drinks \(\frac{1}{4}\) l of milk; Joe drinks \(\frac{1}{2}\) l of milk. How much of milk is left in the vessel?
4. Between 4\(\frac{2}{3}\)and 3\(\frac{2}{3}\) which is greater and by how much?
5. What must be subtracted from 5\(\frac{1}{6}\) to get 2\(\frac{1}{8}\)?
● Multiplication is Repeated Addition.
● Multiplication of Fractional Number by a Whole Number.
● Multiplication of a Fraction by Fraction.
● Properties of Multiplication of Fractional Numbers.
● Worksheet on Multiplication on Fraction.
● Division of a Fraction by a Whole Number.
● Division of a Fractional Number.
● Division of a Whole Number by a Fraction.
● Properties of Fractional Division.
● Worksheet on Division of Fractions.
● Simplification of Fractions.
● Worksheet on Simplification of Fractions.
● Worksheet on Word Problems on Fractions.
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