Maharashtra Board Class 5 Maths Chapter 5 Fractions Set 20 Solutions

Std 5 Maths Chapter 5 Fractions

Question 1. Add the following
(1) \( \frac{1}{5} + \frac{3}{5} \)
Answer: \( \frac{1}{5} + \frac{3}{5} = \frac{1+3}{5} = \frac{4}{5} \)
(2) \( \frac{2}{7} + \frac{4}{7} \)
Answer: \( \frac{2}{7} + \frac{4}{7} = \frac{2+4}{7} = \frac{6}{7} \)
(3) \( \frac{7}{12} + \frac{2}{12} \)
Answer: \( \frac{7}{12} + \frac{2}{12} = \frac{7+2}{12} = \frac{9}{12} \)
(4) \( \frac{2}{9} + \frac{7}{9} \)
Answer: \( \frac{2}{9} + \frac{7}{9} = \frac{2+7}{9} = \frac{9}{9} = 1 \)
(5) \( \frac{3}{15} + \frac{4}{15} \)
Answer: \( \frac{3}{15} + \frac{4}{15} = \frac{3+4}{15} = \frac{7}{15} \)
(6) \( \frac{2}{7} + \frac{1}{7} + \frac{3}{7} \)
Answer: \( \frac{2}{7} + \frac{1}{7} + \frac{3}{7} = \frac{2+1+3}{7} = \frac{6}{7} \)
(7) \( \frac{2}{10} + \frac{4}{10} + \frac{3}{10} \)
Answer: \( \frac{2}{10} + \frac{4}{10} + \frac{3}{10} = \frac{2+4+3}{10} = \frac{9}{10} \)
(8) \( \frac{4}{9} + \frac{1}{9} \)
Answer: \( \frac{4}{9} + \frac{1}{9} = \frac{4+1}{9} = \frac{5}{9} \)
(9) \( \frac{5}{8} + \frac{3}{8} \)
Answer: \( \frac{5}{8} + \frac{3}{8} = \frac{5+3}{8} = \frac{8}{8} = 1 \)
In simple words: To add fractions with the same bottom number (denominator), you simply add the top numbers (numerators) and keep the bottom number the same.

🎯 Exam Tip: Ensure correct addition of numerators while keeping the common denominator, and simplify the fraction if possible to its lowest terms or a whole number (like \( \frac{9}{9} = 1 \) or \( \frac{8}{8} = 1 \)).

Question 2. Mother gave \( \frac{3}{8} \) of one guava to Meena and \( \frac{2}{8} \) of the guava to Geeta. What part of the guava did she give them altogether?
Solution: Travelled by A + Travelled by B \( \frac{3}{8} + \frac{2}{8} = \frac{3+2}{8} = \frac{5}{8} \) given altogether
Answer: \( \frac{5}{8} \) part of guava given altogether
In simple words: To find the total part of guava given, we add the fractions given to Meena and Geeta, as they represent parts of the same whole guava.

🎯 Exam Tip: This is a basic addition of like fractions word problem. Clearly state the quantities given, the operation performed (addition), and the final answer as a simplified fraction.

Question 3. The girls of Std V cleaned \( \frac{3}{4} \) of a field while the boys cleaned \( \frac{1}{4} \). What part of the field was cleaned altogether?
Solution: Girls cleaned + Boys cleaned \( \frac{3}{4} + \frac{1}{4} = \frac{3+1}{4} = \frac{4}{4} = 1 \)
Answer: Full whole field cleaned altogether.
In simple words: To find the total part of the field cleaned, we add the fractions representing the parts cleaned by the girls and the boys.

🎯 Exam Tip: When a problem asks for "altogether" or "total," it usually implies addition. Ensure fractions are properly added and the result is simplified. A result of '1' means the whole field was cleaned.

Subtraction Of Like Fractions

A figure is divided into 5 equal parts and 4 of them are colored. That is, \( \frac{4}{5} \) part of the figure is coloured.

Now, we remove the colour from one of the coloured parts. That is, we subtract \( \frac{1}{5} \) from \( \frac{4}{5} \). The remaining coloured part is \( \frac{3}{5} \). Therefore, \( \frac{4}{5} - \frac{1}{5} = \frac{4-1}{5} = \frac{3}{5} \).

When subtracting a fraction from another like fraction, we write the difference between the numerators in the numerator and the common denominator in the denominator.

Example (1) Subtract : \( \frac{7}{13} - \frac{5}{13} \)

These two fractions have a common denominator. So, we shall subtract the second numerator from the first and write the denominator as it is.

\( \frac{7}{13} - \frac{5}{13} = \frac{7-5}{13} = \frac{2}{13} \)

Example (2) If Raju got \( \frac{5}{12} \) part of a sugarcane and Sanju got \( \frac{3}{12} \) part, how much was the extra part that Raju got?

To find out the difference, we must subtract.

\( \frac{5}{12} - \frac{3}{12} = \frac{5-3}{12} = \frac{2}{12} \). Thus, Raju got \( \frac{2}{12} \) extra.

Addition And Subtraction Problem Set 13 Additional Important Questions And Answers

(1) \( \frac{3}{6} + \frac{2}{6} + \frac{1}{6} \)
Answer: \( \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \)
In simple words: When adding several fractions with the same denominator, simply sum up all the numerators and place the total over the common denominator.

🎯 Exam Tip: For multiple fractions, add numerators carefully and ensure the common denominator remains unchanged. Simplify the final fraction if possible.

(2) \( \frac{4}{10} + \frac{1}{10} + \frac{3}{10} + \frac{2}{10} \)
Answer: \( \frac{4}{10} + \frac{1}{10} + \frac{3}{10} + \frac{2}{10} = \frac{4+1+3+2}{10} = \frac{10}{10} = 1 \)
In simple words: When adding several fractions with the same denominator, simply sum up all the numerators and place the total over the common denominator.

🎯 Exam Tip: For multiple fractions, add numerators carefully and ensure the common denominator remains unchanged. Simplify the final fraction if possible.

(3) \( \frac{1}{2} + \frac{1}{2} \)
Answer: \( \frac{1}{2} + \frac{1}{2} = \frac{1+1}{2} = \frac{2}{2} = 1 \)
In simple words: When adding fractions with the same denominator, simply sum up the numerators and place the total over the common denominator.

🎯 Exam Tip: Ensure correct addition of numerators and keep the common denominator. Simplify the fraction if possible to its lowest terms or a whole number.

Solve The Following Word Problems:

Question 1. \( \frac{3}{5} \) of journey travelled by A and \( \frac{2}{5} \) of journey travelled by B. What part of the journey travelled by both field was cleaned altogether?
Solution: Travelled by A + Travelled by B \( \frac{3}{5} + \frac{2}{5} = \frac{3+2}{5} = \frac{5}{5} = 1 \)
Answer: Full (whole) journey travelled by both.
In simple words: To find the total part of the journey travelled, we add the fractions representing the parts travelled by A and B.

🎯 Exam Tip: Similar to other word problems, identify the operation required (addition for "altogether"), perform it accurately, and state the simplified final answer. Note any unusual phrasing in the question, but answer based on the mathematical operation.