CBSE Class 6 Mathematics Revision Assignment 2

Practice Questions (Class VI) Mathematics

Question 1. Simplify the following :-
(i) \(18 - 5 + 16 - 9 + 2\)
(ii) \(75 \div 5 - 45 \times 2 of 5\)
(iii) \(28 \div 4 - 3 \times 16\)
Answer: (i) Using the BODMAS rule, perform operations from left to right for addition and subtraction:
\(= 18 - 5 + 16 - 9 + 2\)
\(= 13 + 16 - 9 + 2\)
\(= 29 - 9 + 2\)
\(= 20 + 2\)
\(= 22\)

(ii) Using the BODMAS rule, first solve 'of', then division, then multiplication:
\(= 75 \div 5 - 45 \times 2 \text{ of } 5\)
\(= 75 \div 5 - 45 \times 10\)
\(= 15 - 450\)
\(= -435\)

(iii) Using the BODMAS rule, perform division and multiplication first, then subtraction:
\(= 28 \div 4 - 3 \times 16\)
\(= 7 - 48\)
\(= -41\)
In simple words: Follow the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction) strictly in order to solve arithmetic expressions.

Exam Tip: Always evaluate 'of' before multiplication and division, and remember that subtraction of a larger number from a smaller number results in a negative integer.

Question 2. Arrange the following fractions in the ascending order: -
\(\frac{1}{6}, \frac{2}{5}, \frac{3}{4}, \frac{1}{3}\)
Answer: To arrange fractions in ascending order, first find the Least Common Multiple (LCM) of the denominators 6, 5, 4, and 3.
LCM of 6, 5, 4, 3 = 60
Convert each fraction to an equivalent fraction with a denominator of 60:
\(\frac{1}{6} = \frac{1 \times 10}{6 \times 10} = \frac{10}{60}\)
\(\frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60}\)
\(\frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60}\)
\(\frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60}\)
Comparing the numerators: \(10 < 20 < 24 < 45\)
Therefore, the ascending order is: \(\frac{10}{60}, \frac{20}{60}, \frac{24}{60}, \frac{45}{60}\), which simplifies to:
\(\frac{1}{6}, \frac{1}{3}, \frac{2}{5}, \frac{3}{4}\)
In simple words: Make the bottom numbers (denominators) the same by finding their LCM, then compare the top numbers (numerators) from smallest to largest.

Exam Tip: Always write the final answer in terms of the original fractions given in the question, not their converted forms.

Question 3. Compare the following fractions using \(>\), \(<\) or \(=\): -
(i) \(\frac{2}{5} \square \frac{1}{3}\)
(ii) \(\frac{1}{4} \square \frac{3}{5}\)
Answer: (i) To compare \(\frac{2}{5}\) and \(\frac{1}{3}\), use cross-multiplication:
\(2 \times 3 = 6\) and \(5 \times 1 = 5\)
Since \(6 > 5\), therefore \(\frac{2}{5} > \frac{1}{3}\).

(ii) To compare \(\frac{1}{4}\) and \(\frac{3}{5}\), use cross-multiplication:
\(1 \times 5 = 5\) and \(4 \times 3 = 12\)
Since \(5 < 12\), therefore \(\frac{1}{4} < \frac{3}{5}\).
In simple words: Cross-multiply the numerator of the first fraction with the denominator of the second, and vice versa, to easily see which fraction is larger.

Exam Tip: Cross-multiplication is a fast and foolproof method for comparing two fractions without needing to find a common denominator.

Question 4. A recipe requires \(\frac{3}{5}\) cup of milk and \(\frac{1}{6}\) cup of flour. How much more milk than flour is required?
Answer: Amount of milk = \(\frac{3}{5}\) cup
Amount of flour = \(\frac{1}{6}\) cup
To find how much more milk is required, subtract the amount of flour from the amount of milk:
\(\text{Difference} = \frac{3}{5} - \frac{1}{6}\)
LCM of denominators 5 and 6 = 30
\(\frac{3 \times 6}{5 \times 6} - \frac{1 \times 5}{6 \times 5} = \frac{18}{30} - \frac{5}{30}\)
\(= \frac{18 - 5}{30} = \frac{13}{30}\)
Therefore, \(\frac{13}{30}\) cup more milk than flour is required.
In simple words: Subtract the smaller fraction (flour) from the larger fraction (milk) after making their denominators equal.

Question 5. Simplify the following :-
(i) \(6\frac{1}{5} + 7\frac{1}{3} - 5\frac{1}{4}\)
(ii) \(20\frac{3}{4} - 5\frac{1}{3} - 11\frac{1}{4}\)
Answer: (i) Convert mixed fractions into improper fractions:
\(6\frac{1}{5} = \frac{31}{5}, \quad 7\frac{1}{3} = \frac{22}{3}, \quad 5\frac{1}{4} = \frac{21}{4}\)
Expression: \(\frac{31}{5} + \frac{22}{3} - \frac{21}{4}\)
LCM of 5, 3, 4 = 60
\(= \frac{31 \times 12}{60} + \frac{22 \times 20}{60} - \frac{21 \times 15}{60}\)
\(= \frac{372}{60} + \frac{440}{60} - \frac{315}{60}\)
\(= \frac{372 + 440 - 315}{60}\)
\(= \frac{812 - 315}{60} = \frac{497}{60} = 8\frac{17}{60}\)

(ii) Convert mixed fractions into improper fractions:
\(20\frac{3}{4} = \frac{83}{4}, \quad 5\frac{1}{3} = \frac{16}{3}, \quad 11\frac{1}{4} = \frac{45}{4}\)
Expression: \(\frac{83}{4} - \frac{16}{3} - \frac{45}{4}\)
Rearrange terms with the same denominator together:
\(= \left(\frac{83}{4} - \frac{45}{4}\right) - \frac{16}{3}\)
\(= \frac{83 - 45}{4} - \frac{16}{3}\)
\(= \frac{38}{4} - \frac{16}{3} = \frac{19}{2} - \frac{16}{3}\)
LCM of 2 and 3 = 6
\(= \frac{19 \times 3}{2 \times 3} - \frac{16 \times 2}{3 \times 2} = \frac{57}{6} - \frac{32}{6}\)
\(= \frac{57 - 32}{6} = \frac{25}{6} = 4\frac{1}{6}\)
In simple words: Always convert mixed numbers to improper fractions before performing addition and subtraction.

Exam Tip: Grouping fractions with common denominators together (as shown in part ii) makes subtraction much simpler and faster.

Question 6. A ribbon, which is \(6\frac{1}{2}\) m long, is cut into two pieces. The length of one piece is \(2\frac{2}{5}\) m. Find the length of the other piece.
Answer: Total length of the ribbon = \(6\frac{1}{2} \text{ m} = \frac{13}{2} \text{ m}\)
Length of one piece = \(2\frac{2}{5} \text{ m} = \frac{12}{5} \text{ m}\)
Length of the other piece = Total length - Length of one piece
\(= \frac{13}{2} - \frac{12}{5}\)
LCM of 2 and 5 = 10
\(= \frac{13 \times 5}{2 \times 5} - \frac{12 \times 2}{5 \times 2}\)
\(= \frac{65}{10} - \frac{24}{10}\)
\(= \frac{65 - 24}{10} = \frac{41}{10} \text{ m} = 4\frac{1}{10} \text{ m}\)
Therefore, the length of the other piece is \(4\frac{1}{10} \text{ m}\).
In simple words: Subtract the known piece from the total length to find the remaining piece.

Exam Tip: Convert your final improper fraction back into a mixed fraction if the original word problem used mixed fractions.

Question 7. Find the difference between the greatest and smallest fraction: -
\(\frac{1}{3}, \frac{2}{3}, \frac{5}{7}, \frac{9}{11}\)
Answer: To find the greatest and smallest fractions, convert each fraction into its decimal form or find a common denominator:
\(\frac{1}{3} \approx 0.333\)
\(\frac{2}{3} \approx 0.667\)
\(\frac{5}{7} \approx 0.714\)
\(\frac{9}{11} \approx 0.818\)
Comparing the values, the greatest fraction is \(\frac{9}{11}\) and the smallest fraction is \(\frac{1}{3}\).
Now, find the difference between them:
\(\text{Difference} = \frac{9}{11} - \frac{1}{3}\)
LCM of 11 and 3 = 33
\(= \frac{9 \times 3}{11 \times 3} - \frac{1 \times 11}{3 \times 11}\)
\(= \frac{27}{33} - \frac{11}{33}\)
\(= \frac{27 - 11}{33} = \frac{16}{33}\)
Therefore, the difference between the greatest and smallest fraction is \(\frac{16}{33}\).
In simple words: Convert fractions to decimals to easily identify which is largest and smallest, then subtract them.

Exam Tip: Decimal conversion is often faster than finding a common denominator when comparing four or more fractions with different denominators.

Question 8. Convert the following to fraction :-
(i) \(14.75\)
(ii) \(0.35\)
(iii) \(2.34\)
(iv) \(0.18\)
Answer: (i) \(14.75 = \frac{1475}{100}\)
Dividing numerator and denominator by 25:
\(= \frac{59}{4} \text{ or } 14\frac{3}{4}\)

(ii) \(0.35 = \frac{35}{100}\)
Dividing numerator and denominator by 5:
\(= \frac{7}{20}\)

(iii) \(2.34 = \frac{234}{100}\)
Dividing numerator and denominator by 2:
\(= \frac{117}{50} \text{ or } 2\frac{17}{50}\)

(iv) \(0.18 = \frac{18}{100}\)
Dividing numerator and denominator by 2:
\(= \frac{9}{50}\)
In simple words: Remove the decimal point, put the number over 1 followed by as many zeros as there are decimal places, and then reduce it to its lowest terms.

Exam Tip: Always reduce the resulting fraction to its simplest form by dividing the numerator and denominator by their highest common factor (HCF).

Question 9. Convert the following fractions to decimals :-
(i) \(5\frac{1}{2}\)
(ii) \(3\frac{1}{4}\)
(iii) \(\frac{11}{15}\)
(iv) \(\frac{5}{7}\)
Answer: (i) \(5\frac{1}{2} = \frac{11}{2}\)
Dividing 11 by 2: \(11 \div 2 = 5.5\)

(ii) \(3\frac{1}{4} = \frac{13}{4}\)
Dividing 13 by 4: \(13 \div 4 = 3.25\)

(iii) \(\frac{11}{15}\)
Dividing 11 by 15 using long division:
\(11 \div 15 \approx 0.7333...\) (or \(0.7\overline{3}\))

(iv) \(\frac{5}{7}\)
Dividing 5 by 7 using long division:
\(5 \div 7 \approx 0.714285...\)
In simple words: Divide the numerator (top number) by the denominator (bottom number) using long division.

Exam Tip: For non-terminating decimals, round off your answer to 3 decimal places unless stated otherwise.

Question 10. Simplify the following :-
(i) \(24 - 25.89 + 15.26\)
(ii) \(10 - 3.77 - 2.53\)
Answer: (i) Group positive decimal numbers together and subtract the negative decimal number:
\(= 24.00 + 15.26 - 25.89\)
\(= 39.26 - 25.89\)
\(= 13.37\)

(ii) Combine all negative decimal numbers:
\(= 10 - (3.77 + 2.53)\)
\(= 10 - 6.30\)
\(= 3.70\)
In simple words: Add numbers with the same signs first, and line up the decimal points carefully when adding or subtracting.

Exam Tip: Always pad whole numbers with decimal zeros (e.g., writing 24 as 24.00) to align decimal places correctly.

Question 11. The maximum temperature in Delhi on Thursday was \(44.5^{\circ}\text{C}\) and on Friday it was \(45.2^{\circ}\text{C}\). Find the temperature difference of these two days.
Answer: Maximum temperature on Friday = \(45.2^{\circ}\text{C}\)
Maximum temperature on Thursday = \(44.5^{\circ}\text{C}\)
Temperature difference = Friday temperature - Thursday temperature
\(= 45.2 - 44.5\)
\(= 0.7^{\circ}\text{C}\)
Therefore, the temperature difference between the two days is \(0.7^{\circ}\text{C}\).
In simple words: Subtract the lower temperature from the higher temperature to find the difference.

Exam Tip: Always include the correct unit (in this case, \(^{\circ}\text{C}\)) in your final answer.