The DAV Class 5 Maths Book Solutions Pdf and DAV Class 5 Maths Chapter 4 Worksheet 5 Solutions of Fractional Numbers offer comprehensive answers to textbook questions.
DAV Class 5 Maths Ch 4 Worksheet 5 Solutions
Question 1.
Solve the following word problems.
(a) Anjali spent \(\frac{1}{5}\) of her pocket money on comics and \(\frac{3}{4}\) on sweets. How much of her pocket money did she spend altogether?
Solution:
Anjali spent money on comics = \(\frac{1}{5}\)
Anjali spent money on sweets = \(\frac{3}{4}\)
Total money spent = \(\frac{1}{5}+\frac{3}{4}\)
= \(\frac{4}{20}+\frac{15}{20}\) (LCM = 20)
= \(\frac{4+15}{20}\)
= \(\frac{19}{20}\)
Anjali spend altogether \(\frac{19}{20}\) of her pocket money.
(b) In a high jump contest, Ramesh jumped \(3 \frac{8}{9} m\) and Rakesh jumped \(4 \frac{1}{3} m\). Who jumped more height and by how much more?
Solution:
Ramesh jumped = \(3 \frac{8}{9} m\) = \(\frac{35}{9}\) m
Rakesh jumped = \(4 \frac{1}{3} m\) = \(\frac{13}{3}\) m
\(\frac{35}{9}\), \(\frac{13}{3}\) (LCM = 9)
\(\frac{35}{9}<\frac{39}{9}\)
Ramesh jumped more height as 39 > 35
More height = \(\frac{39}{9}-\frac{35}{9}\) = \(\frac{4}{9}\) m
Rakesh jumped more height by \(\frac{4}{9}\) m.
(c) During examination, Sonal studied for 3\(\frac{1}{2}\) hours. She studied science for 1\(\frac{1}{4}\) hours and mathematics for the rest of the hours. How much time did she study mathematics?
Solution:
Total hours she studied = 3\(\frac{1}{2}\) hrs = \(\frac{7}{2}\) hrs
She studied science = 1\(\frac{1}{4}\) hr = \(\frac{5}{4}\) hr
She studied mathematics = \(\frac{7}{2}-\frac{5}{4}\) (LCM = 4)
= \(\frac{14}{4}-\frac{5}{4}\)
= \(\frac{14-5}{4}\)
= \(\frac{9}{4}\) hrs or 2\(\frac{1}{4}\) hrs
Sonal studied mathematics for 2\(\frac{1}{4}\) hrs.
(d) Mr. Gupta had 15\(\frac{2}{5}\) litres of petrol in his car. He went for a drive. By the time he reached home, he had only 2\(\frac{1}{3}\) litres of petrol left. How much petrol was used?
Solution:
Mr Gupta had petrol in his car = 15\(\frac{2}{5}\) = \(\frac{77}{5}\) litres
Petrol left in car = 2\(\frac{1}{3}\) l = \(\frac{7}{3}\) l
Petrol used = \(\frac{77}{5}-\frac{7}{3}\) (LCM = 15)
= \(\frac{231}{15}-\frac{35}{15}\)
= \(\frac{231-35}{15}\)
= \(\frac{196}{15}\) l
= 13\(\frac{1}{15}\) l
13\(\frac{1}{15}\) l petrol was used.
(e) Ms Kumar bought 2\(\frac{2}{5}\) kg potatoes, 2 kg onion and 1\(\frac{2}{5}\) kg tomatoes. Find the total weight of vegetables Ms Kumar bought.
Solution:
wt. of potatoes bought = 2\(\frac{2}{5}\) = \(\frac{12}{5}\) kg
wt. of onion bought = 2 kg
wt. of tomatoes bought = 1\(\frac{2}{5}\) kg = \(\frac{7}{5}\) kg
Total wt. of vegetables = \(\frac{12}{5}+\frac{7}{5}+\frac{2}{1}\) (LCM = 5)
= \(\frac{12+7+10}{5}\)
= \(\frac{29}{5}\) kg
= 5\(\frac{4}{5}\) kg
The total wt. of vegetables bought by Ms Kumar is 5\(\frac{4}{5}\) kg.
(f) Neha used 1\(\frac{1}{2}\) m red ribbon, \(\frac{3}{4}\) m yellow ribbon and 1 m white ribbon for her project. Find the total length of ribbon she used for her project.
Solution:
Length of red ribbon used = 1\(\frac{1}{2}\) m = \(\frac{3}{2}\) m
Length of yellow ribbon used = \(\frac{3}{4}\) m
Length of white ribbon used = 1 m
Total length of ribbon = \(\frac{3}{2}+\frac{3}{4}+\frac{1}{1}\) (LCM = 4)
= \(\frac{6}{4}+\frac{3}{4}+\frac{4}{4}\)
= \(\frac{6+3+4}{4}\)
= \(\frac{13}{4}\) m or 3\(\frac{1}{4}\) m
Neha used 3\(\frac{1}{4}\) m of ribbon for her project.
Word Problems
Example 1.
Rohit ate \(\frac{1}{3}\) of a cake on Tuesday 1 and \(\frac{1}{6}\) of the cake on Wednesday. What fraction of cake did he eat on these two days?
Solution:
He ate cake on Tuesday = \(\frac{1}{3}\)
He ate cake on Wednesday = \(\frac{1}{6}\)
Total cake ate by him = \(\frac{1}{3}+\frac{1}{6}\) (LCM = 6)
= \(\frac{2+1}{6}\)
= \(\frac{3}{6}\)
= \(\frac{1}{2}\)
Rohit ate \(\frac{1}{2}\) of whole cake on these two days.
Example 2.
Ritu has 12 metres long ribbon. She used 7\(\frac{1}{2}\) metres of it. How much ribbon is left with her?
Solution:
We subtract two lengths to find the length of the ribbon left.
Total length of ribbon = 12 m = \(\frac{12}{1}\) m
Ribbon used = 7\(\frac{1}{2}\) = \(\frac{15}{2}\) m
Ribbon left = \(\frac{12}{1}-\frac{15}{2}\) (LCM = 2)
= \(\frac{24}{2}-\frac{15}{2}\)
= \(\frac{24-15}{2}\)
= \(\frac{9}{2}\)
= 4\(\frac{1}{2}\) m
Ritu has 4\(\frac{1}{2}\) m ribbon left with her.