ML Aggarwal Class 7 Maths Solutions Chapter 02 Fractions and Decimals

Question 1. What fraction of each of the following figure is shaded part?
Answer:
(i) Total parts = 8, Shaded parts = 2

Fraction = \( \frac{\text{Shaded parts}}{\text{Total parts}} = \frac{2}{8} = \frac{1}{4} \)

(ii) Total parts = 10, Shaded parts = 3

Fraction = \( \frac{\text{Shaded parts}}{\text{Total parts}} = \frac{3}{10} \)

(iii) Total parts = 12, Shaded parts = 5

Fraction = \( \frac{\text{Shaded parts}}{\text{Total parts}} = \frac{5}{12} \)

(iv) Total parts = 13, Shaded parts = 7

Fraction = \( \frac{\text{Shaded parts}}{\text{Total parts}} = \frac{7}{13} \)
In simple words: Count all the pieces in the shape. Count how many pieces are filled in with color. Write the colored pieces on top and all pieces on the bottom.

Exam Tip: Always identify the total number of equal parts and the shaded parts carefully - off-by-one errors are common mistakes.

Question 2. What fraction of an hour is 35 minutes?
Answer: We know 1 hour equals 60 minutes.

Fraction = \( \frac{35 \text{ minutes}}{60 \text{ minutes}} \)

\( \Rightarrow \frac{35}{60} \)

\( \Rightarrow \frac{35 \div 5}{60 \div 5} \)

\( \Rightarrow \frac{7}{12} \)

Hence, the required fraction = \( \frac{7}{12} \)
In simple words: Since an hour has 60 minutes, write 35 over 60. Then simplify by dividing both top and bottom by 5 to get 7/12.

Exam Tip: Always convert both quantities to the same unit before forming the fraction, then reduce it to lowest terms.

Question 3. Convert the following fractions into improper fractions:
(i) \( 2\frac{7}{9} \)
(ii) \( 5\frac{4}{11} \)
Answer: We apply the rule: improper fraction = \( \frac{(\text{natural number} \times \text{denominator}) + \text{numerator}}{\text{denominator}} \)

(i) \( 2\frac{7}{9} \)

\( \Rightarrow 2\frac{7}{9} = \frac{2 \times 9 + 7}{9} = \frac{18 + 7}{9} = \frac{25}{9} \)

Hence, the required fraction is \( \frac{25}{9} \)

(ii) \( 5\frac{4}{11} \)

\( \Rightarrow 5\frac{4}{11} = \frac{5 \times 11 + 4}{11} = \frac{55 + 4}{11} = \frac{59}{11} \)

Hence, the required fraction is \( \frac{59}{11} \)
In simple words: Multiply the whole number by the denominator, then add the numerator. Put this result on top and keep the denominator the same.

Exam Tip: Double-check your multiplication and addition before writing the final improper fraction to avoid calculation errors.

Question 4. Convert the following fractions into mixed fractions:
(i) \( \frac{73}{8} \)
(ii) \( \frac{94}{13} \)
Answer:
(i) \( \frac{73}{8} \)

Dividing 73 by 8, we get quotient = 9 and remainder = 1.

\( \Rightarrow \frac{73}{8} = 9\frac{1}{8} \)

Hence, \( \frac{73}{8} = 9\frac{1}{8} \)

(ii) \( \frac{94}{13} \)

Dividing 94 by 13, we get quotient = 7 and remainder = 3.

\( \Rightarrow \frac{94}{13} = 7\frac{3}{13} \)

Hence, \( \frac{94}{13} = 7\frac{3}{13} \)
In simple words: Divide the top number by the bottom number. The answer is the whole number part. The leftover is the new numerator, and the denominator stays the same.

Exam Tip: Always verify your answer by converting the mixed fraction back to improper form to confirm it matches the original.

Question 5. Fill in the missing numbers in the following equivalent fractions:
(i) \( \frac{3}{7} = \frac{\ldots}{35} \)
(ii) \( \frac{\ldots}{5} = \frac{30}{18} \)
(iii) \( \frac{\ldots}{9} = \frac{56}{72} \)
Answer:
(i) To convert 7 to 35, we multiply by 5. Therefore, multiply both numerator and denominator by 5.

\( \Rightarrow \frac{3}{7} = \frac{3 \times 5}{7 \times 5} = \frac{15}{35} \)

Hence, the missing number is 15.

(ii) To convert 30 to 5, we divide by 6. Therefore, divide both numerator and denominator by 6.

\( \Rightarrow \frac{30}{18} = \frac{30 \div 6}{18 \div 6} = \frac{5}{3} \)

Hence, the missing number is 3.

(iii) To convert 72 to 9, we divide by 8. Therefore, divide both numerator and denominator by 8.

\( \Rightarrow \frac{56}{72} = \frac{56 \div 8}{72 \div 8} = \frac{7}{9} \)

Hence, the missing number is 7.
In simple words: Find what number you multiply or divide by to change one denominator to the other. Do the same thing to both top and bottom.

Exam Tip: Check your work by cross-multiplying: 3 × 35 should equal 7 × 15 for equivalent fractions.

Question 6. Reduce the following fractions to their simplest form:
(i) \( \frac{48}{72} \)
(ii) \( \frac{276}{115} \)
(iii) \( \frac{72}{336} \)
Answer:
(i) By prime factorisation:

\( \Rightarrow \frac{48}{72} = \frac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 2 \times 3 \times 3} = \frac{2}{3} \)

Hence, \( \frac{48}{72} \) in simplest form is \( \frac{2}{3} \).

(ii) By prime factorisation:

\( \Rightarrow \frac{276}{115} = \frac{2 \times 2 \times 3 \times 23}{5 \times 23} = \frac{2 \times 2 \times 3}{5} = \frac{12}{5} \)

Hence, \( \frac{276}{115} \) in simplest form is \( \frac{12}{5} \).

(iii) By prime factorisation:

\( \Rightarrow \frac{72}{336} = \frac{2 \times 2 \times 2 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 3 \times 7} = \frac{3}{2 \times 7} = \frac{3}{14} \)

Hence, \( \frac{72}{336} \) in simplest form is \( \frac{3}{14} \).
In simple words: Find all the prime factors of the top and bottom. Cancel out the factors that appear in both. What's left is the simplest form.

Exam Tip: Use prime factorisation systematically rather than trial and error to avoid missing common factors.

Question 7. Convert the following fractions into equivalent like fractions:
(i) \( \frac{3}{4}, \frac{5}{6}, \frac{7}{8} \)
(ii) \( \frac{7}{25}, \frac{9}{10}, \frac{19}{40} \)
Answer:
(i) Find the LCM of 4, 6, and 8:

LCM of 4, 6, and 8 = 2 × 2 × 2 × 3 = 24.

\( \Rightarrow \frac{3}{4} = \frac{3 \times 6}{4 \times 6} = \frac{18}{24} \)

\( \Rightarrow \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24} \)

\( \Rightarrow \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24} \)

Hence, the equivalent like fractions are \( \frac{18}{24}, \frac{20}{24}, \frac{21}{24} \).

(ii) Find the LCM of 25, 10, and 40:

LCM of 25, 10, and 40 = 2 × 2 × 2 × 5 × 5 = 200.

\( \Rightarrow \frac{7}{25} = \frac{7 \times 8}{25 \times 8} = \frac{56}{200} \)

\( \Rightarrow \frac{9}{10} = \frac{9 \times 20}{10 \times 20} = \frac{180}{200} \)

\( \Rightarrow \frac{19}{40} = \frac{19 \times 5}{40 \times 5} = \frac{95}{200} \)

Hence, the equivalent like fractions are \( \frac{56}{200}, \frac{180}{200}, \frac{95}{200} \).
In simple words: Find the LCM of all the denominators. Change each fraction so it has this common denominator as the bottom number.

Exam Tip: Always use the LCM (not just any common multiple) to ensure you get fractions in the simplest equivalent form.

Question 8. Arrange the given fractions in descending order:
(i) \( \frac{2}{9}, \frac{2}{3}, \frac{8}{21} \)
(ii) \( \frac{1}{5}, \frac{3}{7}, \frac{7}{10} \)
Answer:
(i) Find the LCM of 9, 3, and 21: LCM = 3 × 3 × 7 = 63.

Convert to like fractions:

\( \Rightarrow \frac{2}{9} = \frac{2 \times 7}{9 \times 7} = \frac{14}{63} \)

\( \Rightarrow \frac{2}{3} = \frac{2 \times 21}{3 \times 21} = \frac{42}{63} \)

\( \Rightarrow \frac{8}{21} = \frac{8 \times 3}{21 \times 3} = \frac{24}{63} \)

Since 42 > 24 > 14: \( \frac{42}{63} > \frac{24}{63} > \frac{14}{63} \)

\( \Rightarrow \frac{2}{3} > \frac{8}{21} > \frac{2}{9} \)

Hence, the fractions in descending order are \( \frac{2}{3}, \frac{8}{21}, \frac{2}{9} \).

(ii) Find the LCM of 5, 7, and 10: LCM = 2 × 5 × 7 = 70.

Convert to like fractions:

\( \Rightarrow \frac{1}{5} = \frac{1 \times 14}{5 \times 14} = \frac{14}{70} \)

\( \Rightarrow \frac{3}{7} = \frac{3 \times 10}{7 \times 10} = \frac{30}{70} \)

\( \Rightarrow \frac{7}{10} = \frac{7 \times 7}{10 \times 7} = \frac{49}{70} \)

Since 49 > 30 > 14: \( \frac{49}{70} > \frac{30}{70} > \frac{14}{70} \)

\( \Rightarrow \frac{7}{10} > \frac{3}{7} > \frac{1}{5} \)

Hence, the fractions in descending order are \( \frac{7}{10}, \frac{3}{7}, \frac{1}{5} \).
In simple words: Convert all fractions to have the same denominator. Then compare the numerators - bigger numerator means bigger fraction. List them from largest to smallest.

Exam Tip: Always convert to like fractions first - comparing decimals or cross-multiplying is more error-prone than using a common denominator.

Question 9. Arrange the given fractions in ascending order:
(i) \( \frac{5}{7}, \frac{3}{8}, \frac{9}{14}, \frac{20}{21} \)
(ii) \( \frac{13}{18}, \frac{8}{15}, \frac{17}{24}, \frac{7}{12} \)
Answer:
(i) Find the LCM of 7, 8, 14, and 21: LCM = 2 × 2 × 2 × 3 × 7 = 168.

Convert to like fractions:

\( \Rightarrow \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168} \)

\( \Rightarrow \frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168} \)

\( \Rightarrow \frac{9}{14} = \frac{9 \times 12}{14 \times 12} = \frac{108}{168} \)

\( \Rightarrow \frac{20}{21} = \frac{20 \times 8}{21 \times 8} = \frac{160}{168} \)

Since 63 < 108 < 120 < 160: \( \frac{63}{168} < \frac{108}{168} < \frac{120}{168} < \frac{160}{168} \)

\( \Rightarrow \frac{3}{8} < \frac{9}{14} < \frac{5}{7} < \frac{20}{21} \)

Hence, the fractions in ascending order are \( \frac{3}{8}, \frac{9}{14}, \frac{5}{7}, \frac{20}{21} \).

(ii) Find the LCM of 18, 15, 24, and 12: LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360.

Convert to like fractions:

\( \Rightarrow \frac{13}{18} = \frac{13 \times 20}{18 \times 20} = \frac{260}{360} \)

\( \Rightarrow \frac{8}{15} = \frac{8 \times 24}{15 \times 24} = \frac{192}{360} \)

\( \Rightarrow \frac{17}{24} = \frac{17 \times 15}{24 \times 15} = \frac{255}{360} \)

\( \Rightarrow \frac{7}{12} = \frac{7 \times 30}{12 \times 30} = \frac{210}{360} \)

Since 192 < 210 < 255 < 260: \( \frac{192}{360} < \frac{210}{360} < \frac{255}{360} < \frac{260}{360} \)

\( \Rightarrow \frac{8}{15} < \frac{7}{12} < \frac{17}{24} < \frac{13}{18} \)

Hence, the fractions in ascending order are \( \frac{8}{15}, \frac{7}{12}, \frac{17}{24}, \frac{13}{18} \).
In simple words: Change all fractions to have the same denominator. Then look at the numerators from smallest to largest. Arrange them in that order.

Exam Tip: For ascending order, the fraction with the smallest numerator comes first. Double-check your LCM calculation to avoid mistakes in the final comparison.

Question 1. Evaluate the following:
(i) \( \frac{4}{3} + \frac{7}{8} \)
(ii) \( 8\frac{1}{2} - 3\frac{5}{8} \)
(iii) \( \frac{5}{12} + \frac{1}{18} - \frac{2}{9} \)
Answer:
(i) \( \frac{4}{3} + \frac{7}{8} \)

Find the LCM of 3 and 8: LCM = 2 × 2 × 2 × 3 = 24.

\( \Rightarrow \frac{4}{3} + \frac{7}{8} = \frac{4 \times 8}{3 \times 8} + \frac{7 \times 3}{8 \times 3} = \frac{32}{24} + \frac{21}{24} = \frac{32 + 21}{24} = \frac{53}{24} \)

Hence, \( \frac{4}{3} + \frac{7}{8} = \frac{53}{24} \)
In simple words: Find a common denominator. Change both fractions to use this denominator. Add the numerators and keep the denominator the same.

Exam Tip: Always use the LCM as the common denominator, and verify that the numerators are correctly converted before adding or subtracting.

Question 1. (ii) Add the following fractions:
\( \frac{4}{3} + \frac{7}{8} \)
Answer: To add these fractions, we must first find a common denominator. The LCM of 2 and 8 is 8.

\( \frac{4}{3} + \frac{7}{8} \)

\( \Rightarrow \frac{4 \times 8}{3 \times 8} + \frac{7 \times 3}{8 \times 3} \)

\( \Rightarrow \frac{32}{24} + \frac{21}{24} \)

\( \Rightarrow \frac{32 + 21}{24} \)

\( \Rightarrow \frac{53}{24} \)

\( \Rightarrow 2\frac{5}{24} \)

Therefore, \( \frac{4}{3} + \frac{7}{8} = 2\frac{5}{24} \)
In simple words: To add fractions with different denominators, change them to the same denominator first. Then add the numerators and keep the denominator the same.

Exam Tip: Always simplify your final answer and convert improper fractions to mixed numbers when needed - this shows complete understanding.

Question 1. (iii) Subtract the following fractions:
\( 8\frac{1}{2} - 3\frac{5}{8} \)
Answer: First, change the mixed numbers to improper fractions. The LCM of 2 and 8 is 8.

\( 8\frac{1}{2} - 3\frac{5}{8} \)

\( \Rightarrow \frac{17}{2} - \frac{29}{8} \)

\( \Rightarrow \frac{17 \times 4}{2 \times 4} - \frac{29}{8} \)

\( \Rightarrow \frac{68}{8} - \frac{29}{8} \)

\( \Rightarrow \frac{68 - 29}{8} \)

\( \Rightarrow \frac{39}{8} \)

\( \Rightarrow 4\frac{7}{8} \)

Therefore, \( 8\frac{1}{2} - 3\frac{5}{8} = 4\frac{7}{8} \)
In simple words: When subtracting mixed numbers, first convert them to improper fractions. Find a common denominator, then subtract the numerators.

Exam Tip: Check your work by adding your answer back to the smaller number - you should get the larger number if you subtracted correctly.

Question 1. (iv) Add and subtract the following fractions:
\( \frac{5}{12} + \frac{1}{18} - \frac{2}{9} \)
Answer: To work with these three fractions, find the LCM of 12, 18, and 9, which is 36. Convert each fraction to have 36 as the denominator.

\( \frac{5}{12} + \frac{1}{18} - \frac{2}{9} \)

\( \Rightarrow \frac{5 \times 3}{12 \times 3} + \frac{1 \times 2}{18 \times 2} - \frac{2 \times 4}{9 \times 4} \)

\( \Rightarrow \frac{15}{36} + \frac{2}{36} - \frac{8}{36} \)

\( \Rightarrow \frac{15 + 2 - 8}{36} \)

\( \Rightarrow \frac{9}{36} \)

\( \Rightarrow \frac{1}{4} \)

Therefore, \( \frac{5}{12} + \frac{1}{18} - \frac{2}{9} = \frac{1}{4} \)
In simple words: When you have more than two fractions, find the LCM of all the denominators first. Then convert each to this common denominator before adding and subtracting.

Exam Tip: Always simplify your final answer to its lowest terms by dividing both numerator and denominator by their greatest common factor.

Question 2. Simplify the following:
(i) \( 7\frac{3}{4} - 3\frac{5}{6} + \frac{7}{8} \)
(ii) \( 6\frac{1}{8} - 2\frac{1}{12} - 5\frac{1}{10} + 3\frac{7}{25} \)
Answer:
(i) Convert the mixed numbers to improper fractions. The LCM of 4, 6, and 8 is 24.

\( 7\frac{3}{4} - 3\frac{5}{6} + \frac{7}{8} \)

\( \Rightarrow \frac{31}{4} - \frac{23}{6} + \frac{7}{8} \)

\( \Rightarrow \frac{31 \times 6}{4 \times 6} - \frac{23 \times 4}{6 \times 4} + \frac{7 \times 3}{8 \times 3} \)

\( \Rightarrow \frac{186}{24} - \frac{92}{24} + \frac{21}{24} \)

\( \Rightarrow \frac{186 - 92 + 21}{24} \)

\( \Rightarrow \frac{115}{24} \)

\( \Rightarrow 4\frac{19}{24} \)

Therefore, \( 7\frac{3}{4} - 3\frac{5}{6} + \frac{7}{8} = 4\frac{19}{24} \)

(ii) Convert all mixed numbers to improper fractions. The LCM of 8, 12, 10, and 25 is 600.

\( 6\frac{1}{8} - 2\frac{1}{12} - 5\frac{1}{10} + 3\frac{7}{25} \)

\( \Rightarrow \frac{49}{8} - \frac{25}{12} - \frac{51}{10} + \frac{82}{25} \)

\( \Rightarrow \frac{49 \times 75}{8 \times 75} - \frac{25 \times 50}{12 \times 50} - \frac{51 \times 60}{10 \times 60} + \frac{82 \times 24}{25 \times 24} \)

\( \Rightarrow \frac{3675}{600} - \frac{1250}{600} - \frac{3060}{600} + \frac{1968}{600} \)

\( \Rightarrow \frac{3675 - 1250 - 3060 + 1968}{600} \)

\( \Rightarrow \frac{1333}{600} \)

\( \Rightarrow 2\frac{133}{600} \)

Therefore, \( 6\frac{1}{8} - 2\frac{1}{12} - 5\frac{1}{10} + 3\frac{7}{25} = 2\frac{133}{600} \)
In simple words: For problems with many fractions, convert them all to the same denominator. Then perform all the operations in order, left to right.

Exam Tip: Check each LCM calculation separately - a small error here causes the whole answer to be wrong. Write out the conversion clearly so you can verify each numerator.

Question 3. Aliyah studies for \( 5\frac{2}{3} \) hours daily. She devotes \( 2\frac{4}{5} \) hours of her time for science and mathematics. How much time does she devote for other subjects?
Answer: Total time Aliyah studies daily is \( 5\frac{2}{3} \) hours. Time spent on science and mathematics is \( 2\frac{4}{5} \) hours. To find time for other subjects, subtract the second amount from the first. The LCM of 3 and 5 is 15.

\( 5\frac{2}{3} - 2\frac{4}{5} \)

\( \Rightarrow \frac{17}{3} - \frac{14}{5} \)

\( \Rightarrow \frac{17 \times 5}{3 \times 5} - \frac{14 \times 3}{5 \times 3} \)

\( \Rightarrow \frac{85}{15} - \frac{42}{15} \)

\( \Rightarrow \frac{85 - 42}{15} \)

\( \Rightarrow \frac{43}{15} \)

\( \Rightarrow 2\frac{13}{15} \)

Aliyah devotes \( 2\frac{13}{15} \) hours for other subjects.
In simple words: To find how much time is left for other subjects, take away the science and maths time from the total study time.

Exam Tip: Always check: total time = science/maths time + other subjects time. This verification step takes 20 seconds and catches errors quickly.

Question 4. Ram solved \( \frac{2}{7} \) part of an exercise while Shwetha solved \( \frac{4}{5} \) of it. Who solved the lesser part? By how much?
Answer: Ram solved \( \frac{2}{7} \) and Shwetha solved \( \frac{4}{5} \). To compare these fractions, use cross multiplication: \( 2 \times 5 = 10 \) and \( 7 \times 4 = 28 \). Since 10 is less than 28, we have \( \frac{2}{7} < \frac{4}{5} \). This means Ram solved the lesser part.

To find how much less, calculate the difference. The LCM of 5 and 7 is 35.

\( \frac{4}{5} - \frac{2}{7} \)

\( \Rightarrow \frac{4 \times 7}{5 \times 7} - \frac{2 \times 5}{7 \times 5} \)

\( \Rightarrow \frac{28}{35} - \frac{10}{35} \)

\( \Rightarrow \frac{28 - 10}{35} \)

\( \Rightarrow \frac{18}{35} \)

Ram solved the lesser part by \( \frac{18}{35} \) of the exercise.
In simple words: To compare fractions with different bottoms, multiply the top of one by the bottom of the other. The fraction with the smaller product is the smaller fraction.

Exam Tip: Cross multiplication is a quick way to compare fractions without finding a common denominator - use it when the question asks you to compare two fractions.

Question 5. Sonali had Rs 35\(\frac{3}{5}\). She got Rs 16\(\frac{1}{15}\) from her mother and spent Rs 28\(\frac{2}{3}\) on food. How much money is left with her?
Answer: Sonali initially had Rs 35\(\frac{3}{5}\). She received Rs 16\(\frac{1}{15}\) from her mother, so she gained this amount. Then she spent Rs 28\(\frac{2}{3}\) on food. To find the remaining money, add what she had and received, then subtract what she spent. The LCM of 5, 15, and 3 is 15.

\( 35\frac{3}{5} + 16\frac{1}{15} - 28\frac{2}{3} \)

\( \Rightarrow \frac{178}{5} + \frac{241}{15} - \frac{86}{3} \)

\( \Rightarrow \frac{178 \times 3}{5 \times 3} + \frac{241}{15} - \frac{86 \times 5}{3 \times 5} \)

\( \Rightarrow \frac{534}{15} + \frac{241}{15} - \frac{430}{15} \)

\( \Rightarrow \frac{534 + 241 - 430}{15} \)

\( \Rightarrow \frac{345}{15} \)

\( \Rightarrow 23 \)

Rs 23 is left with Sonali.
In simple words: Add all the money coming in, then subtract all the money going out. What is left is the final amount she has.

Exam Tip: When a problem involves money, always write the currency symbol in your answer - examiners look for this as it shows careful work.

Exercise 2.3

Question 1. Evaluate the following:
(i) \( 7 \times \frac{3}{5} \)
(ii) \( 21 \times \frac{3}{14} \)
(iii) \( 3\frac{2}{5} \times 8 \)
(iv) \( 5 \times 6\frac{3}{4} \)
Answer:
(i) \( 7 \times \frac{3}{5} \)

\( \Rightarrow \frac{7 \times 3}{5} \)

\( \Rightarrow \frac{21}{5} \)

\( \Rightarrow 4\frac{1}{5} \)

Therefore, \( 7 \times \frac{3}{5} = 4\frac{1}{5} \)

(ii) \( 21 \times \frac{3}{14} \)

\( \Rightarrow \frac{21 \times 3}{14} \)

\( \Rightarrow \frac{21}{14} \times 3 \)

\( \Rightarrow \frac{3}{2} \times 3 \)

\( \Rightarrow \frac{9}{2} \)

\( \Rightarrow 4\frac{1}{2} \)

Therefore, \( 21 \times \frac{3}{14} = 4\frac{1}{2} \)

(iii) \( 3\frac{2}{5} \times 8 \)

\( \Rightarrow \frac{17}{5} \times 8 \)

\( \Rightarrow \frac{17 \times 8}{5} \)

\( \Rightarrow \frac{136}{5} \)

\( \Rightarrow 27\frac{1}{5} \)

Therefore, \( 3\frac{2}{5} \times 8 = 27\frac{1}{5} \)

(iv) \( 5 \times 6\frac{3}{4} \)

\( \Rightarrow 5 \times \frac{27}{4} \)

\( \Rightarrow \frac{5 \times 27}{4} \)

\( \Rightarrow \frac{135}{4} \)

\( \Rightarrow 33\frac{3}{4} \)

Therefore, \( 5 \times 6\frac{3}{4} = 33\frac{3}{4} \)
In simple words: To multiply a whole number by a fraction, multiply the whole number by the numerator and keep the denominator the same. Simplify by cancelling common factors before multiplying when possible.

Exam Tip: Always cancel (simplify) before you multiply - this makes the numbers smaller and easier to work with.

Question 2. Find:
(i) \( \frac{2}{3} \) of 18
(ii) \( \frac{1}{2} \) of \( 4\frac{2}{9} \)
(iii) \( \frac{5}{8} \) of \( 9\frac{2}{3} \)
Answer:
(i) \( \frac{2}{3} \) of 18

\( \Rightarrow \frac{2}{3} \times 18 \)

\( \Rightarrow \frac{2 \times 18}{3} \)

\( \Rightarrow 2 \times 6 \)

\( \Rightarrow 12 \)

Therefore, \( \frac{2}{3} \) of 18 = 12

(ii) \( \frac{1}{2} \) of \( 4\frac{2}{9} \)

\( \Rightarrow \frac{1}{2} \times 4\frac{2}{9} \)

\( \Rightarrow \frac{1}{2} \times \frac{38}{9} \)

\( \Rightarrow \frac{1 \times 38}{2 \times 9} \)

\( \Rightarrow \frac{38}{18} \)

\( \Rightarrow \frac{19}{9} \)

\( \Rightarrow 2\frac{1}{9} \)

Therefore, \( \frac{1}{2} \) of \( 4\frac{2}{9} = 2\frac{1}{9} \)

(iii) \( \frac{5}{8} \) of \( 9\frac{2}{3} \)

\( \Rightarrow \frac{5}{8} \times 9\frac{2}{3} \)

\( \Rightarrow \frac{5}{8} \times \frac{29}{3} \)

\( \Rightarrow \frac{5 \times 29}{8 \times 3} \)

\( \Rightarrow \frac{145}{24} \)

\( \Rightarrow 6\frac{1}{24} \)

Therefore, \( \frac{5}{8} \) of \( 9\frac{2}{3} = 6\frac{1}{24} \)
In simple words: "Of" means multiply. Change mixed numbers to improper fractions first, multiply the fractions, then simplify and convert back to a mixed number if needed.

Exam Tip: The word "of" in maths problems always means to multiply - remember this and the problem becomes straightforward.

Question 3. Evaluate the following:
(i) \( \frac{3}{7} \times \frac{5}{9} \)
(ii) \( \frac{2}{5} \times 5\frac{1}{4} \)
(iii) \( 2\frac{1}{3} \times 5\frac{4}{21} \)
(iv) \( 3\frac{1}{6} \times 7\frac{4}{23} \)
Answer:
(i) \( \frac{3}{7} \times \frac{5}{9} \)

\( \Rightarrow \frac{3 \times 5}{7 \times 9} \)

\( \Rightarrow \frac{3}{9} \times \frac{5}{7} \)

\( \Rightarrow \frac{1}{3} \times \frac{5}{7} \)

\( \Rightarrow \frac{5}{21} \)

Therefore, \( \frac{3}{7} \times \frac{5}{9} = \frac{5}{21} \)

(ii) \( \frac{2}{5} \times 5\frac{1}{4} \)

\( \Rightarrow \frac{2}{5} \times \frac{21}{4} \)

\( \Rightarrow \frac{2}{4} \times \frac{21}{5} \)

\( \Rightarrow \frac{1}{2} \times \frac{21}{5} \)

\( \Rightarrow \frac{21}{10} \)

\( \Rightarrow 2\frac{1}{10} \)

Therefore, \( \frac{2}{5} \times 5\frac{1}{4} = 2\frac{1}{10} \)

(iii) \( 2\frac{1}{3} \times 5\frac{4}{21} \)

\( \Rightarrow \frac{7}{3} \times \frac{109}{21} \)

\( \Rightarrow \frac{7}{21} \times \frac{109}{3} \)

\( \Rightarrow \frac{1}{3} \times \frac{109}{3} \)

\( \Rightarrow \frac{109}{9} \)

\( \Rightarrow 12\frac{1}{9} \)

Therefore, \( 2\frac{1}{3} \times 5\frac{4}{21} = 12\frac{1}{9} \)

(iv) \( 3\frac{1}{6} \times 7\frac{4}{23} \)

\( \Rightarrow \frac{19}{6} \times \frac{165}{23} \)

\( \Rightarrow \frac{19}{23} \times \frac{165}{6} \)

\( \Rightarrow \frac{19}{23} \times \frac{55}{2} \)

\( \Rightarrow \frac{19 \times 55}{23 \times 2} \)

\( \Rightarrow \frac{1045}{46} \)

\( \Rightarrow 22\frac{33}{46} \)

Therefore, \( 3\frac{1}{6} \times 7\frac{4}{23} = 22\frac{33}{46} \)
In simple words: When multiplying fractions, convert mixed numbers to improper fractions first. Cancel common factors between any numerator and any denominator before multiplying, then simplify.

Exam Tip: Cancelling before multiplying reduces the size of numbers dramatically and reduces calculation errors - always look for common factors between numerators and denominators.

Question 4. Find the value of:
(i) \( \frac{1}{3} \) of Rs 42
(ii) \( \frac{3}{7} \) of \( 4\frac{2}{3} \) kg
Answer:
(i) \( \frac{1}{3} \) of Rs 42

\( \Rightarrow \frac{1}{3} \times 42 \)

\( \Rightarrow \frac{42}{3} \)

\( \Rightarrow 14 \)

Therefore, \( \frac{1}{3} \) of Rs 42 = Rs 14

(ii) \( \frac{3}{7} \) of \( 4\frac{2}{3} \) kg

\( \Rightarrow \frac{3}{7} \times 4\frac{2}{3} \)

\( \Rightarrow \frac{3}{7} \times \frac{14}{3} \)

\( \Rightarrow \frac{3}{3} \times \frac{14}{7} \)

\( \Rightarrow 1 \times 2 \)

\( \Rightarrow 2 \)

Therefore, \( \frac{3}{7} \) of \( 4\frac{2}{3} \) kg = 2 kg
In simple words: Multiply the fraction by the quantity given. The "of" tells you to multiply, making the calculation straightforward.

Exam Tip: Always include the unit (Rs, kg, m, etc.) in your final answer - missing the unit costs marks even if your calculation is correct.

Question. \( 4\frac{1}{2} \) times of \( 5\frac{1}{2} \) metres
Answer: Start by converting each mixed number to an improper fraction: \( 4\frac{1}{2} = \frac{9}{2} \) and \( 5\frac{1}{2} = \frac{11}{2} \). Now multiply these fractions together:
\[ \frac{9}{2} \times \frac{11}{2} = \frac{9 \times 11}{2 \times 2} = \frac{99}{4} \]
Convert the result back to a mixed number: \( \frac{99}{4} = 24\frac{3}{4} \) metres.
In simple words: When you multiply \( 4\frac{1}{2} \) by \( 5\frac{1}{2} \), you get \( 24\frac{3}{4} \) metres.

Exam Tip: Always change mixed numbers to improper fractions before multiplying, then simplify the final result back to a mixed number if needed.

Question 5. Which is greater:
(i) \( \frac{2}{7} \) of \( \frac{3}{4} \) or \( \frac{3}{5} \) of \( \frac{5}{8} \)
(ii) \( \frac{1}{2} \) of \( \frac{6}{7} \) or \( \frac{2}{3} \) of \( \frac{3}{7} \)
Answer:
(i) First, find \( \frac{2}{7} \) of \( \frac{3}{4} \):
\[ \frac{2}{7} \times \frac{3}{4} = \frac{2 \times 3}{7 \times 4} = \frac{3}{14} \]
Next, find \( \frac{3}{5} \) of \( \frac{5}{8} \):
\[ \frac{3}{5} \times \frac{5}{8} = \frac{3 \times 5}{5 \times 8} = \frac{3}{8} \]
Now compare \( \frac{3}{14} \) and \( \frac{3}{8} \). Both have the same numerator 3. When fractions have the same numerator, the one with the smaller denominator is larger. Since 8 is less than 14, we have \( \frac{3}{8} > \frac{3}{14} \).
Therefore, \( \frac{3}{5} \) of \( \frac{5}{8} \) is greater.

(ii) First, find \( \frac{1}{2} \) of \( \frac{6}{7} \):
\[ \frac{1}{2} \times \frac{6}{7} = \frac{1 \times 6}{2 \times 7} = \frac{6}{14} = \frac{3}{7} \]
Next, find \( \frac{2}{3} \) of \( \frac{3}{7} \):
\[ \frac{2}{3} \times \frac{3}{7} = \frac{2 \times 3}{3 \times 7} = \frac{2}{7} \]
Now compare \( \frac{3}{7} \) and \( \frac{2}{7} \). These are like fractions (same denominator). When fractions have the same denominator, the one with the larger numerator is bigger. Since 3 is greater than 2, we have \( \frac{3}{7} > \frac{2}{7} \).
Therefore, \( \frac{1}{2} \) of \( \frac{6}{7} \) is greater.
In simple words: To compare fractions that are the result of multiplication, first work out each product. Then use the rule: if numerators are the same, smaller denominator wins; if denominators are the same, larger numerator wins.

Exam Tip: Reduce fractions to their simplest form before comparing to make the comparison clearer and avoid mistakes.

Question 6. If 1 metre cloth costs Rs \( 131\frac{3}{4} \), find the cost of \( 5\frac{1}{2} \) metres cloth.
Answer: Cost per metre = Rs \( 131\frac{3}{4} = \text{Rs } \frac{527}{4} \).

Cost of \( 5\frac{1}{2} \) metres = \( \frac{527}{4} \times 5\frac{1}{2} = \frac{527}{4} \times \frac{11}{2} \)
\[ = \frac{527 \times 11}{4 \times 2} = \frac{5797}{8} = 724\frac{5}{8} \]
The cost of \( 5\frac{1}{2} \) metres cloth = Rs \( 724\frac{5}{8} \).
In simple words: Multiply the cost per metre by the number of metres to get the total cost. Convert mixed numbers to fractions first, multiply, then convert back to a mixed number.

Exam Tip: Always write mixed numbers as improper fractions before doing multiplication with other fractions to avoid calculation errors.

Question 7. If the speed of a car is \( 105\frac{1}{5} \) km/h, find the distance covered by it in \( 3\frac{3}{5} \) hours.
Answer: Speed of the car = \( 105\frac{1}{5} = \frac{526}{5} \) km/h.

Time taken = \( 3\frac{3}{5} = \frac{18}{5} \) hours.

Distance covered = Speed × Time
\[ = \frac{526}{5} \times \frac{18}{5} = \frac{526 \times 18}{5 \times 5} = \frac{9468}{25} = 378\frac{18}{25} \]
The distance covered by the car = \( 378\frac{18}{25} \) km.
In simple words: To find distance, multiply speed by time. Convert each mixed number to an improper fraction, multiply them, and change the result back to a mixed number.

Exam Tip: Always check that the units match (speed in km/h, time in hours gives distance in km) before performing the calculation.

Question 8. A car runs 16 km using 1 litre of petrol. How much distance will it cover in \( 2\frac{3}{4} \) litres of petrol?
Answer: Distance covered in 1 litre of petrol = 16 km.

Distance covered in \( 2\frac{3}{4} \) litres = \( 16 \times 2\frac{3}{4} = 16 \times \frac{11}{4} \)
\[ = \frac{16 \times 11}{4} = \frac{176}{4} = 44 \text{ km} \]
The car will cover 44 km in \( 2\frac{3}{4} \) litres of petrol.
In simple words: To find how far the car travels with more petrol, multiply the distance it goes on 1 litre by the number of litres available.

Exam Tip: Reduce fractions or cancel common factors before multiplying to make arithmetic easier and cleaner.

Question 9. Sushant reads \( \frac{1}{3} \) part of a book in 1 hour. How much part of the book will he read in \( 2\frac{1}{5} \) hours?
Answer: Part of the book read in 1 hour = \( \frac{1}{3} \).

Part of the book read in \( 2\frac{1}{5} \) hours = \( \frac{1}{3} \times 2\frac{1}{5} = \frac{1}{3} \times \frac{11}{5} \)
\[ = \frac{1 \times 11}{3 \times 5} = \frac{11}{15} \]
Sushant will read \( \frac{11}{15} \) part of the book in \( 2\frac{1}{5} \) hours.
In simple words: Multiply the fraction of the book read per hour by the total number of hours to find how much he reads altogether.

Exam Tip: For word problems involving fractions and time, set up the multiplication with the rate and the duration clearly identified.

Question 10. An ornament is made of gold and copper and weighs 52 grams. If \( \frac{2}{13} \) of its part is copper, find the weight of pure gold in it.
Answer: Total weight of the ornament = 52 grams.

Fraction of copper = \( \frac{2}{13} \).

Weight of copper = \( \frac{2}{13} \times 52 = \frac{2 \times 52}{13} = \frac{104}{13} = 8 \) grams.

Weight of pure gold = Total weight - Weight of copper
\[ = 52 - 8 = 44 \text{ grams} \]
The weight of pure gold in the ornament = 44 grams.
In simple words: Find the copper weight by multiplying the copper fraction by the total weight. Subtract copper weight from the total to get the gold weight.

Exam Tip: Always ensure your final answer makes sense - the gold weight should be larger than the copper weight if copper is the smaller fraction.

Question 11. In a class of 40 students, \( \frac{1}{5} \) of the total number of students like to study English and \( \frac{2}{5} \) of the total number of students like to study Mathematics and the remaining like to study Science.
(i) How many students like to study English?
(ii) How many students like to study Mathematics?
(iii) What fraction of the total number of students like to study Science?
Answer: Total number of students = 40.

(i) Number of students who like to study English = \( \frac{1}{5} \times 40 = \frac{40}{5} = 8 \).
Hence, 8 students like to study English.

(ii) Number of students who like to study Mathematics = \( \frac{2}{5} \times 40 = \frac{2 \times 40}{5} = \frac{80}{5} = 16 \).
Hence, 16 students like to study Mathematics.

(iii) Fraction of students who like English and Mathematics = \( \frac{1}{5} + \frac{2}{5} = \frac{3}{5} \).

Fraction of students who like to study Science = \( 1 - \frac{3}{5} = \frac{5}{5} - \frac{3}{5} = \frac{2}{5} \).
Hence, \( \frac{2}{5} \) of the total number of students like to study Science.
In simple words: To find how many students like each subject, multiply the fraction by 40. For Science, add the English and Math fractions, then subtract from 1 (the whole class).

Exam Tip: Check that all three fractions add to 1 - this verifies your answer is correct.

Question 12. A rectangular sheet of paper is \( 12\frac{1}{2} \) cm long and \( 10\frac{2}{3} \) cm wide. Find its
(i) perimeter
(ii) area
Answer: Length of the sheet = \( 12\frac{1}{2} = \frac{25}{2} \) cm.

Width of the sheet = \( 10\frac{2}{3} = \frac{32}{3} \) cm.

(i) Perimeter = 2 × (Length + Width)
\[ = 2 \times \left( \frac{25}{2} + \frac{32}{3} \right) \]
To add these fractions, find a common denominator:
\[ = 2 \times \left( \frac{25 \times 3}{2 \times 3} + \frac{32 \times 2}{3 \times 2} \right) = 2 \times \left( \frac{75}{6} + \frac{64}{6} \right) = 2 \times \frac{139}{6} = \frac{139}{3} = 46\frac{1}{3} \text{ cm} \]
Hence, the perimeter of the sheet = \( 46\frac{1}{3} \) cm.

(ii) Area = Length × Width
\[ = \frac{25}{2} \times \frac{32}{3} = \frac{25 \times 32}{2 \times 3} = \frac{800}{6} = \frac{400}{3} = 133\frac{1}{3} \text{ sq. cm} \]
Hence, the area of the sheet = \( 133\frac{1}{3} \) sq. cm.
In simple words: Convert mixed numbers to improper fractions. For perimeter, add length and width, then multiply by 2. For area, just multiply length by width.

Exam Tip: Remember to give perimeter in linear units (cm) and area in square units (sq. cm) - this distinction is important for marks.

Question 13. In a school, \( \frac{25}{54} \) of the students are girls and the rest are boys. If the number of boys is 2030, find the number of girls.
Answer: Fraction of girls = \( \frac{25}{54} \).

Fraction of boys = \( 1 - \frac{25}{54} = \frac{54}{54} - \frac{25}{54} = \frac{29}{54} \).

Let the total number of students be x.

Given, number of boys = 2030.
\[ \frac{29}{54} \times x = 2030 \]
\[ x = 2030 \times \frac{54}{29} = 70 \times 54 = 3780 \]

Number of girls = \( \frac{25}{54} \times 3780 = 25 \times 70 = 1750 \).
Hence, the number of girls in the school = 1750.
In simple words: Find what fraction are boys (by subtracting the girls' fraction from 1). Use this to find the total by dividing the known number of boys by their fraction. Then multiply the total by the girls' fraction.

Exam Tip: Always verify your answer by checking that the number of girls plus the number of boys equals the total.

Question 14. In an orchard, \( \frac{1}{5} \) are orange trees, \( \frac{3}{13} \) are mango trees and the rest are banana trees. If the banana trees are 148 in number, find the total number of trees in the orchard.
Answer: Fraction of orange trees = \( \frac{1}{5} \).

Fraction of mango trees = \( \frac{3}{13} \).

Fraction of banana trees = \( 1 - \left( \frac{1}{5} + \frac{3}{13} \right) \).

LCM of 5 and 13 = 65.
\[ 1 - \left( \frac{1 \times 13}{5 \times 13} + \frac{3 \times 5}{13 \times 5} \right) = 1 - \left( \frac{13}{65} + \frac{15}{65} \right) = 1 - \frac{28}{65} = \frac{65}{65} - \frac{28}{65} = \frac{37}{65} \]

Let the total number of trees be x.

Given, number of banana trees = 148.
\[ \frac{37}{65} \times x = 148 \]
\[ x = 148 \times \frac{65}{37} = 4 \times 65 = 260 \]
Hence, the total number of trees in the orchard = 260.
In simple words: Add the fractions for oranges and mangoes. Subtract from 1 to get the banana fraction. Use this fraction to find the total from the known number of bananas.

Exam Tip: Check your work by computing how many orange, mango, and banana trees there are separately and verifying they sum to the total.

Exercise 2.4

Question 1. Find the reciprocal of each of the following:
(i) \( \frac{3}{7} \)
(ii) \( \frac{13}{9} \)
(iii) 8
Answer: The reciprocal of a fraction \( \frac{a}{b} \) is \( \frac{b}{a} \) - you flip the numerator and denominator upside down.

(i) The reciprocal of \( \frac{3}{7} \) is \( \frac{7}{3} \).

(ii) The reciprocal of \( \frac{13}{9} \) is \( \frac{9}{13} \).

(iii) Write 8 as \( \frac{8}{1} \). The reciprocal is \( \frac{1}{8} \).
In simple words: A reciprocal flips a fraction upside down - the top becomes the bottom and the bottom becomes the top. For whole numbers, put them over 1 first, then flip.

Exam Tip: The product of a number and its reciprocal always equals 1 - use this to check your answer is correct.