# DAV Class 5 Maths Chapter 4 Worksheet 5 Solutions

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.