CBSE Class 7 Mathematics Fractions and Decimals Assignment Set 04

VII - Mathematics Assignment No - 07 - Fractions and Decimals

Q1. Cost of 1 litre disel is Rs \(33 \frac{1}{4}\). What is the cost of 25 litres of disel ?
Answer: Cost of 1 litre diesel = Rs \(33 \frac{1}{4} = \frac{133}{4}\)
Cost of 25 litres of diesel = \(25 \times \frac{133}{4}\)
\(= \frac{3325}{4} = 831 \frac{1}{4}\) Rs (or Rs 831.25)
In simple words: Multiply the cost of 1 litre by the total number of litres.

Exam Tip: Convert mixed fractions into improper fractions before multiplying.

Q2. A man can walk \(6 \frac{3}{4}\text{ km}\) in an hour. How far does he walk in \(7 \frac{5}{6}\text{ hours}\)?
Answer: Distance walked in 1 hour = \(6 \frac{3}{4}\text{ km} = \frac{27}{4}\text{ km}\)
Total time = \(7 \frac{5}{6}\text{ hours} = \frac{47}{6}\text{ hours}\)
Total distance = \(\frac{27}{4} \times \frac{47}{6}\)
\(= \frac{9 \times 47}{4 \times 2} = \frac{423}{8} = 52 \frac{7}{8}\text{ km}\)
In simple words: Multiply the speed per hour by the total hours to find the total distance.

Exam Tip: Cancel common factors in numerator and denominator before final multiplication to simplify calculations.

Q3. The cost of 1 m of cloth is Rs \(48 \frac{1}{4}\) find the cost of \(2 \frac{1}{3}\text{ m}\) cloth.
Answer: Cost of 1 m cloth = Rs \(48 \frac{1}{4} = \frac{193}{4}\)
Length of cloth = \(2 \frac{1}{3}\text{ m} = \frac{7}{3}\text{ m}\)
Total cost = \(\frac{193}{4} \times \frac{7}{3}\)
\(= \frac{1351}{12} = 112 \frac{7}{12}\) Rs
In simple words: Multiply the cost per metre by the total length required.

Exam Tip: Ensure units are consistent before multiplying fractions.

Q4. Find the area of a rectangular plot of land whose length is \(18 \frac{1}{4}\text{ m}\) and width is \(10 \frac{3}{4}\text{ m}\).
Answer: Length of rectangular plot (\(l\)) = \(18 \frac{1}{4}\text{ m} = \frac{73}{4}\text{ m}\)
Width of rectangular plot (\(b\)) = \(10 \frac{3}{4}\text{ m} = \frac{43}{4}\text{ m}\)
Area = \(l \times b\)
Area = \(\frac{73}{4} \times \frac{43}{4}\)
\(= \frac{3139}{16} = 196 \frac{3}{16}\text{ m}^2\)
In simple words: Multiply length by width to get the area of a rectangle.

Exam Tip: Always write the correct square units (\(\text{m}^2\)) for area calculations.

Q5. Find the area of a square field if its one side is \(50 \frac{1}{3}\text{ m}\) long.
Answer: Side of square field (\(s\)) = \(50 \frac{1}{3}\text{ m} = \frac{151}{3}\text{ m}\)
Area of square = \(s^2\)
Area = \(\frac{151}{3} \times \frac{151}{3}\)
\(= \frac{22801}{9} = 2533 \frac{4}{9}\text{ m}^2\)
In simple words: Multiply the side length by itself to find the area of a square.

Exam Tip: Squaring a fraction means squaring both its numerator and denominator independently after converting to improper form.

Q6. A tin box contains \(15 \frac{1}{2}\text{ litres}\) of oil. How much oil is needed to fill \(\frac{7}{8}\) part of tin box?
Answer: Total capacity of tin box = \(15 \frac{1}{2}\text{ litres} = \frac{31}{2}\text{ litres}\)
Part of tin box to be filled = \(\frac{7}{8}\)
Oil needed = \(\frac{7}{8} \times \frac{31}{2}\)
\(= \frac{217}{16} = 13 \frac{9}{16}\text{ litres}\)
In simple words: Multiply the total capacity by the fractional part needed.