Maharashtra Board Class 6 Maths Chapter 4 Operations on Fractions Set 10 Solutions

Operations On Fractions Class 6 Maths Chapter 4 Practice Set 10 Solutions Maharashtra Board

Std 6 Maths Practice Set 10 Solutions Answers

Question 1. Add:
(i) \(6\frac{1}{3} + 2\frac{1}{3}\)
(ii) \(1\frac{1}{4} + 3\frac{1}{2}\)
(iii) \(5\frac{1}{5} + 2\frac{1}{7}\)
(iv) \(3\frac{1}{5} + 2\frac{1}{3}\)
Answer:
(i) \(6\frac{1}{3} + 2\frac{1}{3}\)
\(6\frac{1}{3} + 2\frac{1}{3} = \frac{6 \times 3 + 1}{3} + \frac{2 \times 3 + 1}{3}\)
\( = \frac{18 + 1}{3} + \frac{6 + 1}{3}\)
\( = \frac{19}{3} + \frac{7}{3}\)
\( = \frac{19 + 7}{3}\)
\( = \frac{26}{3}\)
\(\therefore 6\frac{1}{3} + 2\frac{1}{3} = 8\frac{2}{3}\)
(ii) \(1\frac{1}{4} + 3\frac{1}{2}\)
\(1\frac{1}{4} + 3\frac{1}{2} = \frac{1 \times 4 + 1}{4} + \frac{3 \times 2 + 1}{2}\)
\( = \frac{4 + 1}{4} + \frac{6 + 1}{2}\)
\( = \frac{5}{4} + \frac{7}{2}\)
\( = \frac{5}{4} + \frac{7 \times 2}{2 \times 2}\)
\( = \frac{5}{4} + \frac{14}{4}\)
\( = \frac{5 + 14}{4}\)
\( = \frac{19}{4}\)
\(\therefore 1\frac{1}{4} + 3\frac{1}{2} = 4\frac{3}{4}\)
(iii) \(5\frac{1}{5} + 2\frac{1}{7}\)
\(5\frac{1}{5} + 2\frac{1}{7} = \frac{5 \times 5 + 1}{5} + \frac{2 \times 7 + 1}{7}\)
\( = \frac{25 + 1}{5} + \frac{14 + 1}{7}\)
\( = \frac{26}{5} + \frac{15}{7}\)
\( = \frac{26 \times 7}{5 \times 7} + \frac{15 \times 5}{7 \times 5}\)
\( = \frac{182}{35} + \frac{75}{35}\)
\( = \frac{182 + 75}{35}\)
\( = \frac{257}{35}\)
\(\therefore 5\frac{1}{5} + 2\frac{1}{7} = 7\frac{12}{35}\)
(iv) \(3\frac{1}{5} + 2\frac{1}{3}\)
\(3\frac{1}{5} + 2\frac{1}{3} = \frac{3 \times 5 + 1}{5} + \frac{2 \times 3 + 1}{3}\)
\( = \frac{15 + 1}{5} + \frac{6 + 1}{3}\)
\( = \frac{16}{5} + \frac{7}{3}\)
\( = \frac{16 \times 3}{5 \times 3} + \frac{7 \times 5}{3 \times 5}\)
\( = \frac{48}{15} + \frac{35}{15}\)
\( = \frac{48 + 35}{15}\)
\( = \frac{83}{15}\)
\(\therefore 3\frac{1}{5} + 2\frac{1}{3} = 5\frac{8}{15}\)
In simple words: To add mixed fractions, first convert them to improper fractions, find a common denominator, add the numerators, and then convert the result back to a mixed fraction if needed.

🎯 Exam Tip: Always simplify the resulting fraction to its lowest terms or convert it to a mixed number for full marks.

Question 2. Subtract:
(i) \(3\frac{1}{3} - 1\frac{1}{4}\)
(ii) \(5\frac{1}{2} - 3\frac{1}{3}\)
(iii) \(7\frac{1}{8} - 6\frac{1}{10}\)
(iv) \(7\frac{1}{2} - 3\frac{1}{5}\)
Answer:
(i) \(3\frac{1}{3} - 1\frac{1}{4}\)
\(3\frac{1}{3} - 1\frac{1}{4} = \frac{3 \times 3 + 1}{3} - \frac{1 \times 4 + 1}{4}\)
\( = \frac{9 + 1}{3} - \frac{4 + 1}{4}\)
\( = \frac{10}{3} - \frac{5}{4}\)
\( = \frac{10 \times 4}{3 \times 4} - \frac{5 \times 3}{4 \times 3}\)
\( = \frac{40}{12} - \frac{15}{12}\)
\( = \frac{40 - 15}{12}\)
\( = \frac{25}{12}\)
\(\therefore 3\frac{1}{3} - 1\frac{1}{4} = 2\frac{1}{12}\)
(ii) \(5\frac{1}{2} - 3\frac{1}{3}\)
\(5\frac{1}{2} - 3\frac{1}{3} = \frac{5 \times 2 + 1}{2} - \frac{3 \times 3 + 1}{3}\)
\( = \frac{10 + 1}{2} - \frac{9 + 1}{3}\)
\( = \frac{11}{2} - \frac{10}{3}\)
\( = \frac{11 \times 3}{2 \times 3} - \frac{10 \times 2}{3 \times 2}\)
\( = \frac{33}{6} - \frac{20}{6}\)
\( = \frac{33 - 20}{6}\)
\( = \frac{13}{6}\)
\(\therefore 5\frac{1}{2} - 3\frac{1}{3} = 2\frac{1}{6}\)
(iii) \(7\frac{1}{8} - 6\frac{1}{10}\)
\(7\frac{1}{8} - 6\frac{1}{10} = \frac{7 \times 8 + 1}{8} - \frac{6 \times 10 + 1}{10}\)
\( = \frac{56 + 1}{8} - \frac{60 + 1}{10}\)
\( = \frac{57}{8} - \frac{61}{10}\)
\( = \frac{57 \times 5}{8 \times 5} - \frac{61 \times 4}{10 \times 4}\)
\( = \frac{285}{40} - \frac{244}{40}\)
\( = \frac{285 - 244}{40}\)
\( = \frac{41}{40}\)
\(\therefore 7\frac{1}{8} - 6\frac{1}{10} = 1\frac{1}{40}\)
(iv) \(7\frac{1}{2} - 3\frac{1}{5}\)
\(7\frac{1}{2} - 3\frac{1}{5} = \frac{7 \times 2 + 1}{2} - \frac{3 \times 5 + 1}{5}\)
\( = \frac{14 + 1}{2} - \frac{15 + 1}{5}\)
\( = \frac{15}{2} - \frac{16}{5}\)
\( = \frac{15 \times 5}{2 \times 5} - \frac{16 \times 2}{5 \times 2}\)
\( = \frac{75}{10} - \frac{32}{10}\)
\( = \frac{75 - 32}{10}\)
\( = \frac{43}{10}\)
\(\therefore 7\frac{1}{2} - 3\frac{1}{5} = 4\frac{3}{10}\)
In simple words: To subtract mixed fractions, convert them to improper fractions, find a common denominator, subtract the numerators, and then convert the result back to a mixed fraction.

🎯 Exam Tip: Pay close attention to finding the Least Common Multiple (LCM) for denominators to avoid complex calculations later.

Question 3. Solve:
(i) Suyash bought \(2\frac{1}{2}\) kg of sugar and Ashish bought \(3\frac{1}{2}\) kg. How much sugar did they buy altogether? If sugar costs Rs. 32 per kg, how much did they spend on the sugar they bought?
Answer:
i. Sugar bought by Suyash = \(2\frac{1}{2}\) kg
Sugar bought by Ashish = \(3\frac{1}{2}\) kg
\(\therefore\) Total sugar bought by both = \(2\frac{1}{2} + 3\frac{1}{2}\)
\( = \frac{2 \times 2 + 1}{2} + \frac{3 \times 2 + 1}{2}\)
\( = \frac{4 + 1}{2} + \frac{6 + 1}{2}\)
\( = \frac{5}{2} + \frac{7}{2}\)
\( = \frac{5 + 7}{2}\)
\( = \frac{12}{2}\)
\( = 6\) kg
Cost of 1 kg of sugar = Rs. 32
\(\therefore\) Cost of 6 kg of sugar = \(32 \times 6\)
\( = \) Rs. 192
\(\therefore\) They bought 6 kg sugar altogether and the total money spent on sugar is Rs. 192.
In simple words: We added the amounts of sugar bought by Suyash and Ashish to find the total, then multiplied the total quantity by the cost per kg to find the total expenditure.

🎯 Exam Tip: For word problems, clearly state the given values and what needs to be found, then show all steps of the calculation.

(ii) Aradhana grows potatoes in \(\frac{2}{5}\) part of her garden, greens in \(\frac{1}{3}\) part and brinjals in the remaining part. On how much of her plot did she plant brinjals?
Answer:
ii. Part of garden occupied by potatoes = \(\frac{2}{5}\)
Part of garden occupied by greens = \(\frac{1}{3}\)
Since brinjals are planted in the remaining part,
\(\therefore\) (Part occupied by potatoes) + (part occupied by greens) + (part occupied by brinjals) = 1 entire garden.
\(\therefore\) Part of garden occupied by brinjals = 1 - (part of garden occupied by potatoes + part of garden occupied by greens)
\( = 1 - (\frac{2}{5} + \frac{1}{3})\)
\( = 1 - (\frac{2 \times 3}{5 \times 3} + \frac{1 \times 5}{3 \times 5})\)
\( = 1 - (\frac{6}{15} + \frac{5}{15})\)
\( = 1 - (\frac{6 + 5}{15})\)
\( = 1 - \frac{11}{15}\)
\( = \frac{1 \times 15}{1 \times 15} - \frac{11}{15}\)
\( = \frac{15}{15} - \frac{11}{15}\)
\( = \frac{15 - 11}{15}\)
\( = \frac{4}{15}\)
\(\therefore\) Aradhana planted brinjals on \(\frac{4}{15}\) part of her plot.
In simple words: We summed the fractions of the garden used for potatoes and greens, then subtracted this sum from 1 (representing the whole garden) to find the fraction used for brinjals.

🎯 Exam Tip: When dealing with "remaining part" problems, assume the whole is 1 and subtract the known parts from it.

(iii) Sandeep filled water in \(\frac{4}{7}\) of an empty tank. After that, Ramakant filled \(\frac{1}{4}\) part more of the same tank. Then Umesh used \(\frac{3}{14}\) part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?
Answer:
iii. Part of tank filled by Sandeep = \(\frac{4}{7}\)
Part of tank filled by Ramakant = \(\frac{1}{4}\)
\(\therefore\) Part of tank filled by both of them together = \(\frac{4}{7} + \frac{1}{4}\)
\( = \frac{4 \times 4}{7 \times 4} + \frac{1 \times 7}{4 \times 7}\)
\( = \frac{16}{28} + \frac{7}{28}\)
\( = \frac{16 + 7}{28}\)
\( = \frac{23}{28}\)
Part of tank used by Umesh = \(\frac{3}{14}\)
\(\therefore\) Part of tank filled with water = \(\frac{23}{28} - \frac{3}{14}\)
\( = \frac{23}{28} - \frac{3 \times 2}{14 \times 2}\)
\( = \frac{23}{28} - \frac{6}{28}\)
\( = \frac{23 - 6}{28}\)
\( = \frac{17}{28}\)
Since maximum capacity of tank is 560 litres
\(\therefore\) Quantity of water left in tank = \(\frac{17}{28} \times 560 = 340\) litres
\(\therefore\) The quantity of water left in the tank is 340 litres.
In simple words: We first added the fractions of water filled by Sandeep and Ramakant, then subtracted the fraction used by Umesh to find the remaining fraction. Finally, we multiplied this fraction by the total tank capacity to get the remaining volume in litres.

🎯 Exam Tip: Ensure all fractions have a common denominator before performing addition or subtraction, and remember to convert the final fraction to a quantity if a total capacity is given.

Maharashtra Board Class 6 Maths Chapter 4 Operations On Fractions Practice Set 10 Intext Questions And Activities

Question 1. How to do this subtraction: \(4\frac{1}{4} - 2\frac{1}{2}\)? Is it same as \([4-2 + \frac{1}{4} - \frac{1}{2}]\)? (Textbook pg. no. 23)
Answer:
\(4\frac{1}{4} - 2\frac{1}{2}\)
\(4\frac{1}{4} - 2\frac{1}{2} = \frac{4 \times 4 + 1}{4} - \frac{2 \times 2 + 1}{2}\)
\( = \frac{17}{4} - \frac{5}{2}\)
\( = \frac{17}{4} - \frac{5 \times 2}{2 \times 2}\)
\( = \frac{17}{4} - \frac{10}{4}\)
\( = \frac{7}{4}\)
\( = 1\frac{3}{4}\)
Now for \([4-2 + \frac{1}{4} - \frac{1}{2}]\)
\(4-2 + \frac{1}{4} - \frac{1}{2} = 2 + \frac{1}{4} - \frac{1 \times 2}{2 \times 2}\)
\( = 2 + \frac{1}{4} - \frac{2}{4}\)
\( = 2 + \frac{1 - 2}{4}\)
\( = 2 + \frac{-1}{4}\)
\( = \frac{2 \times 4}{1 \times 4} - \frac{1}{4}\)
\( = \frac{8}{4} - \frac{1}{4}\)
\( = \frac{8 - 1}{4}\)
\( = \frac{7}{4}\)
\( = 1\frac{3}{4}\)
The subtraction \(4\frac{1}{4} - 2\frac{1}{2}\) is the same as \([4 - 2 + \frac{1}{4} - \frac{1}{2}]\).
In simple words: The two methods yield the same result. The first method converts mixed numbers to improper fractions and subtracts. The second method separates the whole numbers and fractional parts, performs subtraction on each, and then combines them, showing that both approaches are valid for subtracting mixed fractions.

🎯 Exam Tip: Understanding that mixed number subtraction can be done by separating whole and fractional parts can simplify calculations, especially when the fractional part of the first number is smaller than the second.

Std 6 Maths Digest