Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 3 Fractions Including Problems

Points to Remember

• Fraction: A rational number written in the form \( \frac{a}{b} \), where \( a \) and \( b \) are integers. The top number \( a \) is the numerator, and the bottom number \( b \) is the denominator.
• Classification of Fractions:
  • Decimal fraction: A fraction with a denominator of 10 or any power of 10.
  • Vulgar fraction: A fraction with a denominator that is not 10 or a power of 10.
  • Proper fraction: A fraction where the denominator is larger than the numerator.
  • Improper fraction: A fraction where the denominator is smaller than the numerator.
  • Mixed fraction: A fraction made of a whole number combined with a proper fraction.
  • Note: When the numerator and denominator are equal, the fraction has a value of 1.
• Equivalent Fractions: Fractions that represent the same value are known as equivalent fractions.
• Simple and Complex Fractions:
  • Simple fraction: A fraction in which both the numerator and the denominator are whole numbers.
  • Complex fraction: A fraction where either the numerator, the denominator, or both are not whole numbers.
• Like and Unlike Fractions:
  • Like fractions: Fractions that share the same denominator.
  • Unlike fractions: Fractions that have different denominators.
• Converting Unlike Fractions into Like Fractions:
  • Find the lowest common multiple (LCM) of all the denominators.
  • Multiply the top and bottom of each fraction by the same number so that the bottom becomes equal to the LCM.
• Inserting a Fraction Between Two Fractions: Add the numerators together and the denominators together, then simplify the resulting fraction if needed.
• Addition and Subtraction of Fractions:
  • For like fractions, simply add or subtract the numerators while keeping the denominator unchanged.
  • For unlike fractions, first convert them into like fractions, then add or subtract their numerators.
• Multiplication: To multiply fractions, multiply all the numerators together and all the denominators together.
• Division: To divide by a fraction, multiply the first value by the reciprocal of the second value.
• Using 'BODMAS': The term 'BODMAS' stands for Bracket, Of, Division, Multiplication, Addition, and Subtraction. Calculations must be solved following this exact order of operations.
• Brackets and their removal: There are four types of brackets: bar bracket, circular brackets ( ), curly brackets { }, and square brackets [ ]. They should be simplified in this order: first the bar, followed by parentheses, then curly brackets, and finally square brackets, taking care of any negative signs outside them.

Exercise 3(A)

Question 1. Classify, each fraction given below, as decimal or vulgar fraction, proper or improper fraction and mixed fraction :
(i) \( \frac{3}{5} \)
(ii) \( \frac{11}{10} \)
(iii) \( \frac{13}{20} \)
(iv) \( \frac{18}{7} \)
(v) \( 3\frac{2}{9} \)
(vi) \( \frac{19}{10^3} \)
(vii) \( 2\frac{7}{10} \)
(viii) \( \frac{23}{500} \)
Answer:
(i) This is a vulgar and proper fraction.
(ii) This is a decimal and improper fraction.
(iii) This is a decimal and proper fraction.
(iv) This is a vulgar and improper fraction.
(v) This is a mixed fraction.
(vi) This is a decimal fraction.
(vii) This is a mixed and decimal fraction.
(viii) This is a vulgar and proper fraction.
In simple words: Decimal fractions have denominators like 10, 100, or 1000, while vulgar fractions do not. Proper fractions have a smaller number on top, improper fractions have a larger number on top, and mixed fractions combine a whole number with a fraction.

Exam Tip: Remember that any denominator containing powers of 10 makes the fraction a decimal fraction, even if it is written as an exponent like \( 10^3 \).

Question 2. Express the following improper fractions as mixed fractions :
(i) \( \frac{18}{5} \)
(ii) \( \frac{7}{4} \)
(iii) \( \frac{25}{6} \)
(iv) \( \frac{38}{5} \)
(v) \( \frac{22}{5} \)
Answer:
(i) Dividing 18 by 5 gives a quotient of 3 and a remainder of 3. So, \( \frac{18}{5} = 3\frac{3}{5} \).
(ii) Dividing 7 by 4 gives a quotient of 1 and a remainder of 3. So, \( \frac{7}{4} = 1\frac{3}{4} \).
(iii) Dividing 25 by 6 gives a quotient of 4 and a remainder of 1. So, \( \frac{25}{6} = 4\frac{1}{6} \).
(iv) Dividing 38 by 5 gives a quotient of 7 and a remainder of 3. So, \( \frac{38}{5} = 7\frac{3}{5} \).
(v) Dividing 22 by 5 gives a quotient of 4 and a remainder of 2. So, \( \frac{22}{5} = 4\frac{2}{5} \).
In simple words: To make a mixed fraction, divide the top number by the bottom number. The whole number you get is the main part, and the leftover remainder goes on top of the original bottom number.

Exam Tip: Always double check your answer by multiplying the whole number by the denominator and adding the numerator to see if you get back the original top number.

Question 3. Express the following mixed fractions as improper fractions :
(i) \( 2\frac{4}{9} \)
(ii) \( 7\frac{5}{13} \)
(iii) \( 3\frac{1}{4} \)
(iv) \( 2\frac{5}{48} \)
(v) \( 12\frac{7}{11} \)
Answer:
(i) \( 2\frac{4}{9} = \frac{2 \times 9 + 4}{9} = \frac{18 + 4}{9} = \frac{22}{9} \)
(ii) \( 7\frac{5}{13} = \frac{7 \times 13 + 5}{13} = \frac{91 + 5}{13} = \frac{96}{13} \)
(iii) \( 3\frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{12 + 1}{4} = \frac{13}{4} \)
(iv) \( 2\frac{5}{48} = \frac{2 \times 48 + 5}{48} = \frac{96 + 5}{48} = \frac{101}{48} \)
(v) \( 12\frac{7}{11} = \frac{12 \times 11 + 7}{11} = \frac{132 + 7}{11} = \frac{139}{11} \)
In simple words: To turn a mixed number into a regular fraction, multiply the whole number by the bottom number, then add the top number. Put this final value over the original bottom number.

Exam Tip: Keep the same denominator throughout your calculations. The denominator does not change when converting from a mixed fraction to an improper fraction.

Question 4. Reduce the given fractions to lowest terms
(i) \( \frac{8}{18} \)
(ii) \( \frac{27}{36} \)
(iii) \( \frac{18}{42} \)
(iv) \( \frac{35}{75} \)
(v) \( \frac{18}{45} \)
Answer:
(i) The highest common factor (HCF) of 8 and 18 is 2. Dividing both terms by 2, we get:
\( \frac{8}{18} = \frac{8 \div 2}{18 \div 2} = \frac{4}{9} \)
(ii) The HCF of 27 and 36 is 9. Dividing both terms by 9, we get:
\( \frac{27}{36} = \frac{27 \div 9}{36 \div 9} = \frac{3}{4} \)
(iii) The HCF of 18 and 42 is 6. Dividing both terms by 6, we get:
\( \frac{18}{42} = \frac{18 \div 6}{42 \div 6} = \frac{3}{7} \)
(iv) The HCF of 35 and 75 is 5. Dividing both terms by 5, we get:
\( \frac{35}{75} = \frac{35 \div 5}{75 \div 5} = \frac{7}{15} \)
(v) The HCF of 18 and 45 is 9. Dividing both terms by 9, we get:
\( \frac{18}{45} = \frac{18 \div 9}{45 \div 9} = \frac{2}{5} \)
In simple words: To simplify a fraction, find the biggest number that divides both the top and bottom numbers evenly. Then divide both by that number.

Exam Tip: If finding the highest common factor is difficult, you can simplify in smaller steps by dividing by smaller prime numbers (like 2, 3, or 5) until you cannot divide any further.

Question 5. State : true or false
(i) \( \frac{30}{40} \) and \( \frac{12}{16} \) are equivalent fractions.
(ii) \( \frac{10}{25} \) and \( \frac{25}{10} \) are equivalent fractions.
(iii) \( \frac{35}{49} \), \( \frac{20}{28} \), \( \frac{45}{63} \) and \( \frac{100}{140} \) are equivalent fractions.
Answer:
(i) True. Reducing both fractions to their simplest terms shows that \( \frac{30}{40} = \frac{3}{4} \) and \( \frac{12}{16} = \frac{3}{4} \). Since their values are equal, they are equivalent.
(ii) False. Simplifying each fraction gives \( \frac{10}{25} = \frac{2}{5} \) and \( \frac{25}{10} = \frac{5}{2} \). Since these simplest forms are not the same, they are not equivalent.
(iii) True. Simplifying each of these fractions yields \( \frac{35}{49} = \frac{5}{7} \), \( \frac{20}{28} = \frac{5}{7} \), \( \frac{45}{63} = \frac{5}{7} \), and \( \frac{100}{140} = \frac{5}{7} \). Since every fraction reduces to the same value, they are all equivalent.
In simple words: Fractions are equivalent if they simplify to the exact same value. If they turn into different numbers, they are not equivalent.

Exam Tip: To easily check if two fractions are equivalent, you can also use cross-multiplication: if the product of the first numerator and second denominator equals the product of the first denominator and second numerator, they are equivalent.

Question 6. Distinguish each of the following fractions, given below, as a simple fraction or a complex fraction :
(i) \( \frac{0}{8} \)
(ii) \( \frac{-3}{-8} \)
(iii) \( \frac{5}{-7} \)
(iv) \( \frac{3\frac{3}{5}}{18} \)
(v) \( \frac{-6}{2\frac{2}{5}} \)
(vi) \( \frac{3\frac{1}{3}}{7\frac{2}{7}} \)
(vii) \( \frac{-5\frac{2}{9}}{5} \)
(viii) \( \frac{-8}{0} \)
Answer:
(i) It is a simple fraction because both the numerator and the denominator are integers.
(ii) It is a simple fraction because the top and bottom values are integers.
(iii) It is a simple fraction because both numbers are integers.
(iv) It is a complex fraction because the numerator contains a mixed fraction, which is not an integer.
(v) It is a complex fraction because the denominator is a mixed fraction, not an integer.
(vi) It is a complex fraction since both the top and bottom values contain mixed fractions instead of integers.
(vii) It is a complex fraction because the numerator is a mixed fraction rather than an integer.
(viii) This is neither a simple nor a complex fraction because the bottom number is zero, making the fraction undefined.
In simple words: A simple fraction has only whole numbers (integers) on the top and bottom. If there is a fraction inside the top or bottom, it is called a complex fraction. A fraction cannot have zero on the bottom.

Exam Tip: Remember that fractions with a denominator of zero are undefined in mathematics. They cannot be classified as simple or complex.

Exercise 3(B)

Question 1. For each pair, given below, state whether it forms like fractions or unlike fractions :
(i) \( \frac{5}{8} \) and \( \frac{7}{8} \)
(ii) \( \frac{8}{15} \) and \( \frac{8}{21} \)
(iii) \( \frac{4}{9} \) and \( \frac{9}{4} \)
Answer:
(i) These are like fractions because they share the exact same denominator of 8.
(ii) These are unlike fractions because their denominators, 15 and 21, are different.
(iii) These are unlike fractions because their denominators, 9 and 4, are different.
In simple words: Like fractions have the same bottom number. If the bottom numbers are different, they are called unlike fractions.

Exam Tip: Only focus on the denominators (the bottom numbers) when deciding if fractions are like or unlike. The numerators do not affect this classification.

Question 2. Convert given fractions into fractions with equal denominators :
(i) \( \frac{5}{6} \) and \( \frac{7}{9} \)
(ii) \( \frac{2}{3} \), \( \frac{5}{6} \) and \( \frac{7}{12} \)
(iii) \( \frac{4}{5} \), \( \frac{17}{20} \), \( \frac{23}{40} \) and \( \frac{11}{16} \)
Answer:
(i) To find a common denominator for \( \frac{5}{6} \) and \( \frac{7}{9} \), calculate the LCM of 6 and 9, which is 18. Now convert each fraction:
\( \frac{5}{6} = \frac{5 \times 3}{6 \times 3} = \frac{15}{18} \)
\( \frac{7}{9} = \frac{7 \times 2}{9 \times 2} = \frac{14}{18} \)
The converted fractions are \( \frac{15}{18} \) and \( \frac{14}{18} \).
(ii) The LCM of the denominators 3, 6, and 12 is 12. Converting each fraction gives:
\( \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} \)
\( \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12} \)
\( \frac{7}{12} = \frac{7}{12} \)
The converted fractions are \( \frac{8}{12} \), \( \frac{10}{12} \), and \( \frac{7}{12} \).
(iii) The LCM of the denominators 5, 20, 40, and 16 is 80. Converting each fraction gives:
\( \frac{4}{5} = \frac{4 \times 16}{5 \times 16} = \frac{64}{80} \)
\( \frac{17}{20} = \frac{17 \times 4}{20 \times 4} = \frac{68}{80} \)
\( \frac{23}{40} = \frac{23 \times 2}{40 \times 2} = \frac{46}{80} \)
\( \frac{11}{16} = \frac{11 \times 5}{16 \times 5} = \frac{55}{80} \)
The converted fractions are \( \frac{64}{80} \), \( \frac{68}{80} \), \( \frac{46}{80} \), and \( \frac{55}{80} \).
In simple words: To make the bottom numbers the same, find the LCM of all the bottom numbers. Then multiply the top and bottom of each fraction by whatever number is needed to reach that LCM on the bottom.

Exam Tip: Be sure to multiply both the numerator and the denominator by the exact same value so you do not change the actual value of the fraction.

Question 3. Convert given fractions into fractions with equal numerators :
(i) \( \frac{8}{9} \) and \( \frac{12}{17} \)
(ii) \( \frac{6}{13} \), \( \frac{15}{23} \) and \( \frac{12}{17} \)
(iii) \( \frac{15}{19} \), \( \frac{25}{28} \), \( \frac{9}{11} \) and \( \frac{45}{47} \)
Answer:
(i) To make the numerators equal for \( \frac{8}{9} \) and \( \frac{12}{17} \), find the LCM of the numerators 8 and 12, which is 24. Converting each fraction:
\( \frac{8}{9} = \frac{8 \times 3}{9 \times 3} = \frac{24}{27} \)
\( \frac{12}{17} = \frac{12 \times 2}{17 \times 2} = \frac{24}{34} \)
The required fractions are \( \frac{24}{27} \) and \( \frac{24}{34} \).
(ii) The LCM of the numerators 6, 15, and 12 is 60. Converting each fraction:
\( \frac{6}{13} = \frac{6 \times 10}{13 \times 10} = \frac{60}{130} \)
\( \frac{15}{23} = \frac{15 \times 4}{23 \times 4} = \frac{60}{92} \)
\( \frac{12}{17} = \frac{12 \times 5}{17 \times 5} = \frac{60}{85} \)
The required fractions are \( \frac{60}{130} \), \( \frac{60}{92} \), and \( \frac{60}{85} \).
(iii) The LCM of the numerators 15, 25, 9, and 45 is 225. Converting each fraction:
\( \frac{15}{19} = \frac{15 \times 15}{19 \times 15} = \frac{225}{285} \)
\( \frac{25}{28} = \frac{25 \times 9}{28 \times 9} = \frac{225}{252} \)
\( \frac{9}{11} = \frac{9 \times 25}{11 \times 25} = \frac{225}{275} \)
\( \frac{45}{47} = \frac{45 \times 5}{47 \times 5} = \frac{225}{235} \)
The required fractions are \( \frac{225}{285} \), \( \frac{225}{252} \), \( \frac{225}{275} \), and \( \frac{225}{235} \).
In simple words: This time, we want the top numbers to be the same. Find the LCM of all the top numbers, and multiply the top and bottom of each fraction by the number that makes the top equal to that LCM.

Exam Tip: Remember that equalizing numerators is very similar to equalizing denominators. Just find the LCM of the top numbers instead of the bottom numbers, and multiply carefully.

Question 4. Put the given fractions in ascending order by making denominators equal :
(i) \( \frac{1}{3} \), \( \frac{2}{5} \), \( \frac{3}{4} \) and \( \frac{1}{6} \)
(ii) \( \frac{5}{6} \), \( \frac{7}{8} \), \( \frac{11}{12} \) and \( \frac{3}{10} \)
(iii) \( \frac{5}{7} \), \( \frac{3}{8} \), \( \frac{9}{14} \) and \( \frac{20}{21} \)
Answer:
(i) The LCM of the denominators 3, 5, 4, and 6 is 60. Converting each fraction:
\( \frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \)
\( \frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60} \)
\( \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \)
\( \frac{1}{6} = \frac{1 \times 10}{6 \times 10} = \frac{10}{60} \)
Comparing the numerators, we have:
\( \frac{10}{60} < \frac{20}{60} < \frac{24}{60} < \frac{45}{60} \)
So, the original fractions in ascending order are:
\( \frac{1}{6} < \frac{1}{3} < \frac{2}{5} < \frac{3}{4} \)
Thus, the correct order is \( \frac{1}{6} \), \( \frac{1}{3} \), \( \frac{2}{5} \), and \( \frac{3}{4} \).
(ii) The LCM of the denominators 6, 8, 12, and 10 is 240. Converting each fraction:
\( \frac{5}{6} = \frac{5 \times 40}{6 \times 40} = \frac{200}{240} \)
\( \frac{7}{8} = \frac{7 \times 30}{8 \times 30} = \frac{210}{240} \)
\( \frac{11}{12} = \frac{11 \times 20}{12 \times 20} = \frac{220}{240} \)
\( \frac{3}{10} = \frac{3 \times 24}{10 \times 24} = \frac{72}{240} \)
Comparing the numerators, we have:
\( \frac{72}{240} < \frac{200}{240} < \frac{210}{240} < \frac{220}{240} \)
So, the original fractions in ascending order are:
\( \frac{3}{10} < \frac{5}{6} < \frac{7}{8} < \frac{11}{12} \)
Thus, the correct order is \( \frac{3}{10} \), \( \frac{5}{6} \), \( \frac{7}{8} \), and \( \frac{11}{12} \).
(iii) The LCM of the denominators 7, 8, 14, and 21 is 168. Converting each fraction:
\( \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168} \)
\( \frac{3}{8} = \frac{3 \times 21}{8 \times 21} = \frac{63}{168} \)
\( \frac{9}{14} = \frac{9 \times 12}{14 \times 12} = \frac{108}{168} \)
\( \frac{20}{21} = \frac{20 \times 8}{21 \times 8} = \frac{160}{168} \)
Comparing the numerators, we have:
\( \frac{63}{168} < \frac{108}{168} < \frac{120}{168} < \frac{160}{168} \)
So, the original fractions in ascending order are:
\( \frac{3}{8} < \frac{9}{14} < \frac{5}{7} < \frac{20}{21} \)
Thus, the correct order is \( \frac{3}{8} \), \( \frac{9}{14} \), \( \frac{5}{7} \), and \( \frac{20}{21} \).
In simple words: Ascending order means sorting from smallest to largest. First, make the bottom numbers equal. Then, arrange the fractions based on their top numbers, from the smallest top number to the biggest.

Exam Tip: Be very careful when calculating large LCMs like 240 or 168. Write down your prime factorization steps clearly to avoid minor calculation errors.

Question 5. Arrange the given fractions in descending order by making numerators equal :
(i) \( \frac{5}{6} \), \( \frac{4}{15} \), \( \frac{8}{9} \) and \( \frac{1}{3} \)
(ii) \( \frac{3}{7} \), \( \frac{4}{9} \), \( \frac{5}{7} \) and \( \frac{8}{11} \)
(iii) \( \frac{1}{10} \), \( \frac{6}{11} \), \( \frac{8}{11} \) and \( \frac{3}{5} \)
Answer:
(i) The LCM of the numerators 5, 4, 8, and 1 is 40. Converting each fraction:
\( \frac{5}{6} = \frac{5 \times 8}{6 \times 8} = \frac{40}{48} \)
\( \frac{4}{15} = \frac{4 \times 10}{15 \times 10} = \frac{40}{150} \)
\( \frac{8}{9} = \frac{8 \times 5}{9 \times 5} = \frac{40}{45} \)
\( \frac{1}{3} = \frac{1 \times 40}{3 \times 40} = \frac{40}{120} \)
When numerators are equal, the fraction with the smaller denominator is larger. Comparing the denominators:
\( \frac{40}{45} > \frac{40}{48} > \frac{40}{120} > \frac{40}{150} \)
Thus, the original fractions in descending order are:
\( \frac{8}{9} > \frac{5}{6} > \frac{1}{3} > \frac{4}{15} \)
The sorted order is \( \frac{8}{9} \), \( \frac{5}{6} \), \( \frac{1}{3} \), and \( \frac{4}{15} \).
(ii) The LCM of the numerators 3, 4, 5, and 8 is 120. Converting each fraction:
\( \frac{3}{7} = \frac{3 \times 40}{7 \times 40} = \frac{120}{280} \)
\( \frac{4}{9} = \frac{4 \times 30}{9 \times 30} = \frac{120}{270} \)
\( \frac{5}{7} = \frac{5 \times 24}{7 \times 24} = \frac{120}{168} \)
\( \frac{8}{11} = \frac{8 \times 15}{11 \times 15} = \frac{120}{165} \)
With equal numerators, smaller denominators mean larger fractions. Sorting them:
\( \frac{120}{165} > \frac{120}{168} > \frac{120}{270} > \frac{120}{280} \)
Thus, the original fractions in descending order are:
\( \frac{8}{11} > \frac{5}{7} > \frac{4}{9} > \frac{3}{7} \)
The sorted order is \( \frac{8}{11} \), \( \frac{5}{7} \), \( \frac{4}{9} \), and \( \frac{3}{7} \).
(iii) The LCM of the numerators 1, 6, 8, and 3 is 24. Converting each fraction:
\( \frac{1}{10} = \frac{1 \times 24}{10 \times 24} = \frac{24}{240} \)
\( \frac{6}{11} = \frac{6 \times 4}{11 \times 4} = \frac{24}{44} \)
\( \frac{8}{11} = \frac{8 \times 3}{11 \times 3} = \frac{24}{33} \)
\( \frac{3}{5} = \frac{3 \times 8}{5 \times 8} = \frac{24}{40} \)
Comparing fractions with equal numerators, we order by smaller denominators first:
\( \frac{24}{33} > \frac{24}{40} > \frac{24}{44} > \frac{24}{240} \)
Thus, the original fractions in descending order are:
\( \frac{8}{11} > \frac{3}{5} > \frac{6}{11} > \frac{1}{10} \)
The sorted order is \( \frac{8}{11} \), \( \frac{3}{5} \), \( \frac{6}{11} \), and \( \frac{1}{10} \).
In simple words: Descending order means arranging from biggest to smallest. When you make the top numbers equal, the fraction with the smallest bottom number is actually the biggest.

Exam Tip: Be careful! With equal numerators, a smaller denominator represents a larger value. It is easy to accidentally reverse the order if you forget this rule.

Question 6. Find the greater fraction :
(i) \( \frac{3}{5} \) and \( \frac{11}{15} \)
(ii) \( \frac{4}{5} \) and \( \frac{3}{10} \)
(iii) \( \frac{6}{7} \) and \( \frac{5}{9} \)
(iv) \( \frac{3}{8} \) and \( \frac{4}{9} \)
(v) \( \frac{-2}{7} \) and \( \frac{-3}{10} \)
Answer:
(i) For \( \frac{3}{5} \) and \( \frac{11}{15} \), the LCM of 5 and 15 is 15.
\( \therefore \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \)
\( \frac{11}{15} = \frac{11}{15} \point_text \)
Looking at these, we can easily see that \( \frac{11}{15} > \frac{9}{15} \).
So, \( \frac{11}{15} \) is the bigger fraction.

(ii) For \( \frac{4}{5} \) and \( \frac{3}{10} \point_text \), the LCM of 5 and 10 is 10.
\( \therefore \frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10} \point_text \)
and \( \frac{3}{10} = \frac{3}{10} \point_text \)
From this, it is obvious that \( \frac{8}{10} > \frac{3}{10} \).
Thus, \( \frac{4}{5} \) is the greater fraction.

(iii) For \( \frac{6}{7} \) and \( \frac{5}{9} \), the LCM of 7 and 9 is 63.
\( \therefore \frac{6}{7} = \frac{6 \times 9}{7 \times 9} = \frac{54}{63} \)
and \( \frac{5}{9} = \frac{5 \times 7}{9 \times 7} = \frac{35}{63} \)
This shows us that \( \frac{54}{63} > \frac{35}{63} \).
Therefore, \( \frac{6}{7} \) is the larger fraction.

(iv) For \( \frac{3}{8} \) and \( \frac{4}{9} \), the LCM of 8 and 9 is 72.
\( \therefore \frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72} \)
\( \frac{4}{9} = \frac{4 \times 8}{9 \times 8} = \frac{32}{72} \)
We can clearly see that \( \frac{32}{72} > \frac{27}{72} \).
Consequently, \( \frac{4}{9} \) is the larger fraction.

(v) For \( \frac{-2}{7} \) and \( \frac{-3}{10} \), the LCM of 7 and 10 is 70.
\( \therefore \frac{-2}{7} = \frac{-2 \times 10}{7 \times 10} = \frac{-20}{70} \)
\( \frac{-3}{10} = \frac{-3 \times 7}{10 \times 7} = \frac{-21}{70} \)
This tells us that \( \frac{-20}{70} > \frac{-21}{70} \) (since negative numbers closer to zero are larger).
As a result, \( \frac{-2}{7} \) is the greater fraction.
In simple words: To find which fraction is bigger, change them so they both have the same bottom number. Then, just compare the top numbers. For negative fractions, the one closer to zero is always larger.

Exam Tip: Remember that for negative numbers, a smaller absolute value means a larger number (e.g., -20 is greater than -21). Always show the step where you find the Least Common Multiple (LCM) to gain full marks.

Question 7. Insert one fraction between :
(i) \( \frac{3}{7} \) and \( \frac{4}{9} \)
(ii) \( 2 \) and \( \frac{8}{3} \)
(iii) \( \frac{9}{17} \) and \( \frac{6}{13} \)
Answer:
(i) A fraction that lies between \( \frac{3}{7} \) and \( \frac{4}{9} \) is given by:
\( = \frac{3 + 4}{7 + 9} = \frac{7}{16} \)

(ii) To find a fraction between \( 2 \) (written as \( \frac{2}{1} \)) and \( \frac{8}{3} \):
\( = \frac{2 + 8}{1 + 3} = \frac{10}{4} = \frac{5}{2} = 2\frac{1}{2} \)

(iii) The fraction lying between \( \frac{9}{17} \) and \( \frac{6}{13} \) is calculated as follows:
\( = \frac{9 + 6}{17 + 13} = \frac{15}{30} = \frac{1}{2} \)
In simple words: To quickly find a fraction between two others, add their top numbers together to get the new top number, and add their bottom numbers together to get the new bottom number. Simplify the final fraction if possible.

Exam Tip: If one of the numbers is a whole number like 2, always write it as a fraction (like 2/1) before adding the numerators and denominators. This prevents common errors.

Question 8. Insert three fractions between
(i) \( \frac{2}{5} \) and \( \frac{4}{9} \)
(ii) \( \frac{1}{2} \) and \( \frac{5}{7} \)
(iii) \( \frac{3}{8} \) and \( \frac{6}{11} \)
(iv) \( \frac{11}{12} \) and \( \frac{2}{3} \)
(v) \( \frac{4}{7} \) and \( \frac{3}{4} \)
Answer:
(i) We first calculate a fraction that lies between \( \frac{2}{5} \) and \( \frac{4}{9} \):
\( = \frac{2 + 4}{5 + 9} = \frac{6}{14} = \frac{3}{7} \point_text \)

Next, let's find another fraction between \( \frac{2}{5} \) and our new fraction \( \frac{3}{7} \):
\( = \frac{2 + 3}{5 + 7} = \frac{5}{12} \point_text \)

Then, we find a third fraction situated between \( \frac{3}{7} \) and \( \frac{4}{9} \):
\( = \frac{3 + 4}{7 + 9} = \frac{7}{16} \point_text \)

Therefore, three fractions lying in between \( \frac{2}{5} \) and \( \frac{4}{9} \) are \( \frac{5}{12}, \frac{3}{7} \), and \( \frac{7}{16} \).

(ii) First, we find a fraction situated between \( \frac{1}{2} \) and \( \frac{5}{7} \):
\( = \frac{1 + 5}{2 + 7} = \frac{6}{9} = \frac{2}{3} \point_text \)

Next, we calculate a fraction between \( \frac{1}{2} \) and \( \frac{2}{3} \):
\( = \frac{1 + 2}{2 + 3} = \frac{3}{5} \point_text \)

Finally, we find one more fraction between \( \frac{2}{3} \) and \( \frac{5}{7} \):
\( = \frac{2 + 5}{3 + 7} = \frac{7}{10} \point_text \)

So, three fractions that fall between \( \frac{1}{2} \) and \( \frac{5}{7} \) are \( \frac{3}{5}, \frac{2}{3} \point_text \), and \( \frac{7}{10} \).

(iii) To start, let us determine a fraction between \( \frac{3}{8} \) and \( \frac{6}{11} \):
\( = \frac{3 + 6}{8 + 11} = \frac{9}{19} \point_text \)

Now, we find a fraction that lies between \( \frac{3}{8} \) and \( \frac{9}{19} \point_text \):
\( = \frac{3 + 9}{8 + 19} = \frac{12}{27} = \frac{4}{9} \point_text \)

After that, we locate a fraction between \( \frac{9}{19} \) and \( \frac{6}{11} \point_text \):
\( = \frac{9 + 6}{19 + 11} = \frac{15}{30} = \frac{1}{2} \point_text \)

Thus, three fractions that lie between \( \frac{3}{8} \) and \( \frac{6}{11} \) are \( \frac{4}{9}, \frac{9}{19} \point_text \), and \( \frac{1}{2} \).

(iv) First, we find a fraction in between \( \frac{11}{12} \) and \( \frac{2}{3} \point_text \):
\( = \frac{11 + 2}{12 + 3} = \frac{13}{15} \point_text \)

Following this, we find a fraction between \( \frac{11}{12} \) and \( \frac{13}{15} \point_text \):
\( = \frac{11 + 13}{12 + 15} = \frac{24}{27} = \frac{8}{9} \point_text \)

Lastly, we get a fraction between \( \frac{13}{15} \) and \( \frac{2}{3} \point_text \):
\( = \frac{13 + 2}{15 + 3} = \frac{15}{18} = \frac{5}{6} \point_text \)

Consequently, three fractions between \( \frac{11}{12} \) and \( \frac{2}{3} \) are \( \frac{8}{9}, \frac{13}{15} \point_text \), and \( \frac{5}{6} \point_text \).

(v) First, let us find a fraction lying between \( \frac{4}{7} \) and \( \frac{3}{4} \point_text \):
\( = \frac{4 + 3}{7 + 4} = \frac{7}{11} \point_text \)

Next, we find a fraction between \( \frac{4}{7} \) and \( \frac{7}{11} \point_text \):
\( = \frac{4 + 7}{7 + 11} = \frac{11}{18} \point_text \)

Then, we find a fraction between \( \frac{7}{11} \) and \( \frac{3}{4} \point_text \):
\( = \frac{7 + 3}{11 + 4} = \frac{10}{15} = \frac{2}{3} \point_text \)

In conclusion, three fractions that fall between \( \frac{4}{7} \) and \( \frac{3}{4} \) are \( \frac{11}{18}, \frac{7}{11} \point_text \), and \( \frac{2}{3} \).
In simple words: To find three fractions between two given fractions, first find one in the middle by adding their numerators and denominators. Then, find two more by repeating this process between the first fraction and the middle one, and the middle one and the second fraction.

Exam Tip: Always make sure to simplify your intermediate and final fractions to their simplest terms (e.g., reduce 6/14 to 3/7) before using them in subsequent steps, as this keeps calculations much easier.

Question 9. Insert two fractions between
(i) \( 1 \) and \( \frac{3}{11} \)
(ii) \( \frac{5}{9} \) and \( \frac{1}{4} \)
(iii) \( \frac{5}{6} \) and \( 1\frac{1}{5} \)
Answer:
(i) Write 1 as \( \frac{1}{1} \). Let's calculate a fraction that lies between \( \frac{1}{1} \) and \( \frac{3}{11} \):
\( = \frac{1 + 3}{1 + 11} = \frac{4}{12} = \frac{1}{3} \point_text \)

Now, we find another fraction between \( \frac{1}{3} \) and \( \frac{3}{11} \):
\( = \frac{1 + 3}{3 + 11} = \frac{4}{14} = \frac{2}{7} \point_text \)

Therefore, two fractions between 1 and \( \frac{3}{11} \) are \( \frac{1}{3} \) and \( \frac{2}{7} \).

(ii) First, we find a fraction in between \( \frac{5}{9} \) and \( \frac{1}{4} \):
\( = \frac{5 + 1}{9 + 4} = \frac{6}{13} \point_text \)

Next, we find a second fraction lying between \( \frac{6}{13} \) and \( \frac{1}{4} \point_text \):
\( = \frac{6 + 1}{13 + 4} = \frac{7}{17} \point_text \)

Thus, two fractions that fall between \( \frac{5}{9} \) and \( \frac{1}{4} \) are \( \frac{6}{13} \) and \( \frac{7}{17} \).

(iii) First, convert the mixed number \( 1\frac{1}{5} \) to an improper fraction, which is \( \frac{6}{5} \point_text \). Let's locate a fraction between \( \frac{5}{6} \) and \( \frac{6}{5} \point_text \):
\( = \frac{5 + 6}{6 + 5} = \frac{11}{11} = 1 \point_text \)

Now, we find a fraction between 1 (or \( \frac{1}{1} \)) and \( \frac{6}{5} \point_text \):
\( = \frac{1 + 6}{1 + 5} = \frac{7}{6} = 1\frac{1}{6} \point_text \)

As a result, two fractions between \( \frac{5}{6} \) and \( 1\frac{1}{5} \) are 1 and \( 1\frac{1}{6} \).
In simple words: To find two fractions between any two numbers, convert any whole or mixed numbers into simple fractions first. Then, add the tops and bottoms to find a middle fraction, and repeat this step to find a second one.

Exam Tip: Don't forget to rewrite mixed numbers like \(1\frac{1}{5}\) as improper fractions (\(\frac{6}{5}\)) before doing any operations. This is a very common place where students lose easy marks.

Exercise 3(C)

Question 1. Reduce to a single fraction :
(i) \( \frac{1}{2} + \frac{2}{3} \)
(ii) \( \frac{3}{5} - \frac{1}{10} \)
(iii) \( \frac{2}{3} - \frac{1}{6} \)
(iv) \( 1\frac{1}{3} + 2\frac{1}{4} \)
(v) \( \frac{1}{4} + \frac{5}{6} - \frac{1}{12} \)
(vi) \( \frac{2}{3} - \frac{3}{5} + 3 - \frac{1}{5} \)
(vii) \( \frac{2}{3} - \frac{1}{5} + \frac{1}{10} \)
(viii) \( 2\frac{1}{2} + 2\frac{1}{3} - 1\frac{1}{4} \)
(ix) \( 2\frac{5}{8} - 2\frac{1}{6} + 4\frac{3}{4} \)
Answer:
(i) We find the LCM of the denominators 2 and 3, which is 6.
\( = \frac{1 \times 3}{2 \times 3} + \frac{2 \times 2}{3 \times 2} \)
\( = \frac{3}{6} + \frac{4}{6} = \frac{3 + 4}{6} = \frac{7}{6} = 1\frac{1}{6} \)

(ii) Calculating the LCM of 5 and 10 gives 10.
\( = \frac{3 \times 2}{5 \times 2} - \frac{1}{10} = \frac{6}{10} - \frac{1}{10} \)
\( = \frac{6 - 1}{10} = \frac{5}{10} = \frac{1}{2} \point_text \)

(iii) The LCM for 3 and 6 is 6.
\( = \frac{2 \times 2}{3 \times 2} - \frac{1}{6} = \frac{4}{6} - \frac{1}{6} \)
\( = \frac{4 - 1}{6} = \frac{3}{6} = \frac{1}{2} \point_text \)

(iv) First, convert the mixed fractions to improper ones:
\( = \frac{4}{3} + \frac{9}{4} \)
Finding the LCM of 3 and 4 gives 12.
\( = \frac{4 \times 4}{3 \times 4} + \frac{9 \times 3}{4 \times 3} = \frac{16}{12} + \frac{27}{12} \)
\( = \frac{16 + 27}{12} = \frac{43}{12} = 3\frac{7}{12} \point_text \)

(v) Here, the LCM of 4, 6, and 12 is 12.
\( = \frac{1 \times 3}{4 \times 3} + \frac{5 \times 2}{6 \times 2} - \frac{1}{12} \)
\( = \frac{3}{12} + \frac{10}{12} - \frac{1}{12} \)
\( = \frac{3 + 10 - 1}{12} = \frac{13 - 1}{12} = \frac{12}{12} = 1 \point_text \)

(vi) The LCM of 3 and 5 is 15. We can write 3 as \( \frac{3}{1} \):
\( = \frac{2 \times 5}{3 \times 5} - \frac{3 \times 3}{5 \times 3} + \frac{3 \times 15}{15} - \frac{1 \times 3}{5 \times 3} \)
\( = \frac{10}{15} - \frac{9}{15} + \frac{45}{15} - \frac{3}{15} \)
\( = \frac{10 - 9 + 45 - 3}{15} = \frac{55 - 12}{15} = \frac{43}{15} = 2\frac{13}{15} \point_text \)

(vii) We find the LCM of 3, 5, and 10 to be 30:
\( = \frac{2 \times 10}{3 \times 10} - \frac{1 \times 6}{5 \times 6} + \frac{1 \times 3}{10 \times 3} \)
\( = \frac{20}{30} - \frac{6}{30} + \frac{3}{30} = \frac{20 - 6 + 3}{30} \)
\( = \frac{23 - 6}{30} = \frac{17}{30} \point_text \)

(viii) Convert the mixed fractions to improper fractions first:
\( = \frac{5}{2} + \frac{7}{3} - \frac{5}{4} \)
The LCM of 2, 3, and 4 is 12:
\( = \frac{5 \times 6}{2 \times 6} + \frac{7 \times 4}{3 \times 4} - \frac{5 \times 3}{4 \times 3} \)
\( = \frac{30}{12} + \frac{28}{12} - \frac{15}{12} \)
\( = \frac{30 + 28 - 15}{12} = \frac{58 - 15}{12} = \frac{43}{12} = 3\frac{7}{12} \point_text \)

(ix) Converting mixed numbers to improper fractions:
\( = \frac{21}{8} - \frac{13}{6} + \frac{19}{4} \)
The LCM of 8, 6, and 4 is 24:
\( = \frac{21 \times 3}{8 \times 3} - \frac{13 \times 4}{6 \times 4} + \frac{19 \times 6}{4 \times 6} \)
\( = \frac{63}{24} - \frac{52}{24} + \frac{114}{24} \)
\( = \frac{63 - 52 + 114}{24} = \frac{177 - 52}{24} = \frac{125}{24} = 5\frac{5}{24} \point_text \)
In simple words: To add or subtract fractions, make their bottom numbers the same by using their LCM. Then, combine the top numbers and write the result over the common bottom number. Turn any mixed numbers into simple fractions before you begin.

Exam Tip: When subtracting multiple fractions, be careful with the signs. Perform addition first, then subtraction, or group positive and negative terms separately to avoid arithmetic mistakes.

Question 2. Simplify :
(i) \( \frac{3}{4} \times 6 \)
(ii) \( \frac{2}{3} \times 15 \)
(iii) \( \frac{3}{4} \times \frac{1}{2} \)
(iv) \( \frac{9}{12} \times \frac{4}{7} \)
(v) \( 45 \times 2\frac{1}{3} \)
(vi) \( 36 \times 3\frac{1}{4} \)
(vii) \( 2 \div \frac{1}{3} \)
(viii) \( 3 \div \frac{2}{5} \)
(ix) \( 1 \div \frac{3}{5} \)
(x) \( \frac{1}{3} \div \frac{1}{4} \)
(xi) \( \frac{5}{8} \div \frac{3}{4} \)
(xii) \( 3\frac{3}{7} \div 1\frac{1}{14} \)
(xiii) \( 3\frac{3}{4} \times 1\frac{1}{5} \times \frac{20}{21} \)
Answer:
(i) \( \frac{3}{4} \times 6 = \frac{3 \times 6}{4 \times 1} = \frac{18}{4} = \frac{18 \div 2}{4 \div 2} = \frac{9}{2} = 4\frac{1}{2} \)

(ii) \( \frac{2}{3} \times 15 = \frac{2}{3} \times \frac{15}{1} = \frac{2 \times 15}{3 \times 1} = \frac{30}{3} = 10 \)

(iii) \( \frac{3}{4} \times \frac{1}{2} = \frac{3 \times 1}{4 \times 2} = \frac{3}{8} \point_text \)

(iv) \( \frac{9}{12} \times \frac{4}{7} = \frac{9 \times 4}{12 \times 7} = \frac{36}{84} \)
We find the highest common factor (HCF) of 36 and 84 to be 12:
\( \frac{36 \div 12}{84 \div 12} = \frac{3}{7} \point_text \)

(v) \( 45 \times 2\frac{1}{3} = \frac{45}{1} \times \frac{7}{3} = \frac{45 \times 7}{1 \times 3} = \frac{315}{3} = 105 \)

(vi) \( 36 \times 3\frac{1}{4} = \frac{36}{1} \times \frac{13}{4} = \frac{36 \times 13}{1 \times 4} = \frac{468}{4} = 117 \point_text \)

(vii) \( 2 \div \frac{1}{3} = \frac{2}{1} \times \frac{3}{1} = \frac{2 \times 3}{1 \times 1} = 6 \)

(viii) \( 3 \div \frac{2}{5} = \frac{3}{1} \times \frac{5}{2} = \frac{15}{2} = 7\frac{1}{2} \)

(ix) \( 1 \div \frac{3}{5} = 1 \times \frac{5}{3} = \frac{5}{3} = 1\frac{2}{3} \)

(x) \( \frac{1}{3} \div \frac{1}{4} = \frac{1}{3} \times \frac{4}{1} = \frac{1 \times 4}{3 \times 1} = \frac{4}{3} = 1\frac{1}{3} \)

(xi) \( \frac{5}{8} \div \frac{3}{4} = \frac{5}{8} \times \frac{4}{3} = \frac{5 \times 4}{8 \times 3} = \frac{20}{24} = \frac{20 \div 4}{24 \div 4} = \frac{5}{6} \)

(xii) \( 3\frac{3}{7} \div 1\frac{1}{14} = \frac{24}{7} \div \frac{15}{14} = \frac{24}{7} \times \frac{14}{15} = \frac{336}{105} \point_text \)
The highest common factor (HCF) of 336 and 105 is 21:
\( \frac{336 \div 21}{105 \div 21} = \frac{16}{5} = 3\frac{1}{5} \point_text \)

(xiii) \( 3\frac{3}{4} \times 1\frac{1}{5} \times \frac{20}{21} = \frac{15}{4} \times \frac{6}{5} \times \frac{20}{21} = \frac{15 \times 6 \times 20}{4 \times 5 \times 21} = \frac{1800}{420} \point_text \)
The highest common factor (HCF) of 1800 and 420 is 60:
\( \frac{1800 \div 60}{420 \div 60} = \frac{30}{7} = 4\frac{2}{7} \point_text \)
In simple words: To multiply fractions, just multiply the tops together and the bottoms together. To divide fractions, flip the second fraction upside down and multiply instead. Always simplify your final fraction by dividing by the highest common factor.

Exam Tip: Simplify the numbers during multiplication rather than at the end whenever possible. For example, in \( \frac{24}{7} \times \frac{14}{15} \), you can cancel 14 with 7 to get 2, making calculations much faster and reducing mistakes.

Question 3. Subtract :
(i) \( 2 \) from \( \frac{2}{3} \)
(ii) \( \frac{1}{8} \) from \( \frac{5}{8} \)
(iii) \( -\frac{2}{5} \) from \( \frac{2}{5} \)
(iv) \( -\frac{3}{7} \) from \( \frac{3}{7} \)
(v) \( 0 \) from \( -\frac{4}{5} \)
(vi) \( \frac{2}{9} \) from \( \frac{4}{5} \)
(vii) \( -\frac{4}{7} \) from \( -\frac{6}{11} \)
Answer:
(i) \( 2 \) from \( \frac{2}{3} \):
\( = \frac{2}{3} - \frac{2}{1} = \frac{2}{3} - \frac{2 \times 3}{1 \times 3} = \frac{2}{3} - \frac{6}{3} \)
\( = \frac{2 - 6}{3} = -\frac{4}{3} = -1\frac{1}{3} \point_text \)

(ii) \( \frac{1}{8} \) from \( \frac{5}{8} \):
\( = \frac{5}{8} - \frac{1}{8} = \frac{5 - 1}{8} = \frac{4}{8} = \frac{1}{2} \point_text \)

(iii) \( -\frac{2}{5} \) from \( \frac{2}{5} \):
\( = \frac{2}{5} - \left( -\frac{2}{5} \right) = \frac{2}{5} + \frac{2}{5} \point_text \)
\( = \frac{2 + 2}{5} = \frac{4}{5} \point_text \)

(iv) \( -\frac{3}{7} \) from \( \frac{3}{7} \point_text \):
\( = \frac{3}{7} - \left( -\frac{3}{7} \right) = \frac{3}{7} + \frac{3}{7} \point_text \)
\( = \frac{3 + 3}{7} = \frac{6}{7} \point_text \)

(v) \( 0 \) from \( -\frac{4}{5} \point_text \):
\( = -\frac{4}{5} - 0 = -\frac{4}{5} \point_text \)

(vi) \( \frac{2}{9} \) from \( \frac{4}{5} \point_text \):
\( = \frac{4}{5} - \frac{2}{9} \)
Finding the LCM of 5 and 9 gives 45:
\( = \frac{4 \times 9}{5 \times 9} - \frac{2 \times 5}{9 \times 5} = \frac{36}{45} - \frac{10}{45} \point_text \)
\( = \frac{36 - 10}{45} = \frac{26}{45} \point_text \)

(vii) \( -\frac{4}{7} \) from \( -\frac{6}{11} \point_text \):
\( = -\frac{6}{11} - \left( -\frac{4}{7} \right) = -\frac{6}{11} + \frac{4}{7} \point_text \)
The LCM of 7 and 11 is 77:
\( = \frac{-6 \times 7}{11 \times 7} + \frac{4 \times 11}{7 \times 11} \)
\( = \frac{-42}{77} + \frac{44}{77} = \frac{-42 + 44}{77} = \frac{2}{77} \point_text \)
In simple words: To subtract one fraction from another, always write the second fraction first, then a minus sign, and then the first fraction. Remember that subtracting a negative number is the same as adding it.

Exam Tip: "Subtract A from B" means writing B - A, not A - B. Reversing this order is the most common mistake made by students in subtraction problems.

Question 4. Find the value of
(i) \( \frac{1}{2} \) of 10 kg
(ii) \( \frac{3}{5} \) of 1 hour
(iii) \( \frac{4}{7} \) of \( 2\frac{1}{3} \) kg
(iv) \( 3\frac{1}{2} \) times of 2 metres
(v) \( \frac{1}{2} \) of \( 2\frac{2}{3} \)
(vi) \( \frac{5}{11} \) of \( \frac{4}{5} \) of 22 kg
Answer:
(i) \( \frac{1}{2} \) of 10 kg:
\( = \left( \frac{1}{2} \times 10 \right) \text{ kg} = 5 \text{ kg} \point_text \)

(ii) \( \frac{3}{5} \) of 1 hour:
Because 1 hour has 60 minutes, we can calculate:
\( = \left( \frac{3}{5} \times 60 \right) \text{ minutes} = 3 \times 12 \text{ minutes} = 36 \text{ minutes} \point_text \)

(iii) \( \frac{4}{7} \) of \( 2\frac{1}{3} \) kg:
\( = \left( \frac{4}{7} \times \frac{7}{3} \right) \text{ kg} = \frac{4}{3} \text{ kg} = 1\frac{1}{3} \text{ kg} \point_text \)

(iv) \( 3\frac{1}{2} \) times of 2 metres:
\( = \left( \frac{7}{2} \times 2 \right) \text{ metres} = 7 \text{ metres} \point_text \)

(v) \( \frac{1}{2} \) of \( 2\frac{2}{3} \):
\( = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3} = 1\frac{1}{3} \point_text \)

(vi) \( \frac{5}{11} \) of \( \frac{4}{5} \) of 22 kg:
\( = \left( \frac{5}{11} \times \frac{4}{5} \times 22 \right) \text{ kg} = (4 \times 2) \text{ kg} = 8 \text{ kg} \point_text \)
In simple words: The word 'of' or 'times' in math means multiplication. Convert units (like hours to minutes) if it makes calculating easier, and change mixed numbers to improper fractions before multiplying.

Exam Tip: Don't forget to include the correct unit (like kg, minutes, or metres) in your final answer. Leaving out units often results in losing a half mark.

Question 5. Simplify and reduce to a simple fraction :
(i) \( \frac{3}{3\frac{3}{4}} \)
(ii) \( \frac{\frac{3}{5}}{7} \)
(iii) \( \frac{3}{\frac{5}{7}} \)
(iv) \( \frac{2\frac{1}{5}}{1\frac{1}{10}} \)
(v) \( \frac{2}{5} \text{ of } \frac{6}{11} \times 1\frac{1}{4} \)
(vi) \( 2\frac{1}{4} \div \frac{1}{7} \times \frac{1}{3} \)
(vii) \( \frac{1}{3} \times 4\frac{2}{3} \div 3\frac{1}{2} \times \frac{1}{2} \)
(viii) \( \frac{2}{3} \times 1\frac{1}{4} \div \frac{3}{7} \text{ of } 2\frac{5}{8} \)
(ix) \( 0 \div \frac{8}{11} \)
(x) \( \frac{4}{5} \div \frac{7}{15} \text{ of } \frac{8}{9} \)
(xi) \( \frac{4}{5} \div \frac{7}{15} \times \frac{8}{9} \)
(xii) \( \frac{4}{5} \text{ of } \frac{7}{15} \div \frac{8}{9} \)
(xiii) \( \frac{1}{2} \text{ of } \frac{3}{4} \times \frac{1}{2} \div \frac{2}{3} \)
Answer:
(i) Change the mixed number in the denominator to an improper fraction first:
\( 3\frac{3}{4} = \frac{3 \times 4 + 3}{4} = \frac{15}{4} \)
Now, divide 3 by \( \frac{15}{4} \) by multiplying with the reciprocal:
\( \frac{3}{3\frac{3}{4}} = \frac{3}{\frac{15}{4}} = \frac{3 \times 4}{15} = \frac{12}{15} = \frac{4}{5} \)

(ii) To divide the fraction by 7, multiply it by the reciprocal of 7:
\( \frac{\frac{3}{5}}{7} = \frac{3}{5} \times \frac{1}{7} = \frac{3}{35} \)

(iii) To divide 3 by the fraction \( \frac{5}{7} \), multiply 3 by its reciprocal:
\( \frac{3}{\frac{5}{7}} = 3 \times \frac{7}{5} = \frac{21}{5} = 4\frac{1}{5} \)

(iv) First change both mixed numbers to improper fractions:
\( 2\frac{1}{5} = \frac{11}{5} \)
\( 1\frac{1}{10} = \frac{11}{10} \)
Now, divide the first fraction by the second fraction by multiplying by the reciprocal:
\( \frac{2\frac{1}{5}}{1\frac{1}{10}} = \frac{\frac{11}{5}}{\frac{11}{10}} = \frac{11}{5} \times \frac{10}{11} = \frac{10}{5} = 2 \)

(v) Convert the mixed fraction first:
\( 1\frac{1}{4} = \frac{5}{4} \)
According to the order of operations, solve 'of' first:
\( \frac{2}{5} \text{ of } \frac{6}{11} \times \frac{5}{4} = \left(\frac{2}{5} \times \frac{6}{11}\right) \times \frac{5}{4} = \frac{12}{55} \times \frac{5}{4} \)
Multiply the two fractions and simplify:
\( \frac{12 \times 5}{55 \times 4} = \frac{3}{11} \)

(vi) Convert mixed fractions to improper ones:
\( 2\frac{1}{4} = \frac{9}{4} \)
First, do the division by multiplying by the reciprocal of the divisor:
\( \frac{9}{4} \div \frac{1}{7} \times \frac{1}{3} = \frac{9}{4} \times \frac{7}{1} \times \frac{1}{3} = \frac{9 \times 7 \times 1}{4 \times 1 \times 3} = \frac{21}{4} = 5\frac{1}{4} \)

(vii) Write mixed fractions as improper ones:
\( 4\frac{2}{3} = \frac{14}{3} \)
\( 3\frac{1}{2} = \frac{7}{2} \)
Solve the division first by multiplying by the reciprocal:
\( \frac{1}{3} \times \frac{14}{3} \div \frac{7}{2} \times \frac{1}{2} = \frac{1}{3} \times \frac{14}{3} \times \frac{2}{7} \times \frac{1}{2} \)
Multiply the fractions and simplify:
\( = \frac{1 \times 14 \times 2 \times 1}{3 \times 3 \times 7 \times 2} = \frac{2}{9} \)

(viii) Change mixed numbers into improper fractions:
\( 1\frac{1}{4} = \frac{5}{4} \)
\( 2\frac{5}{8} = \frac{21}{8} \)
The expression is:
\( \frac{2}{3} \times \frac{5}{4} \div \frac{3}{7} \text{ of } \frac{21}{8} \)
Work out 'of' first:
\( \frac{3}{7} \text{ of } \frac{21}{8} = \frac{3}{7} \times \frac{21}{8} = \frac{9}{8} \)
Now replace this back into the expression:
\( \frac{2}{3} \times \frac{5}{4} \div \frac{9}{8} \)
Next, solve division by multiplying by the reciprocal:
\( = \frac{2}{3} \times \frac{5}{4} \times \frac{8}{9} = \frac{2 \times 5 \times 8}{3 \times 4 \times 9} = \frac{80}{108} = \frac{20}{27} \)

(ix) Any number dividing zero (except zero itself) is equal to zero:
\( 0 \div \frac{8}{11} = 0 \times \frac{11}{8} = 0 \)

(x) Apply the BODMAS rule to solve. We calculate 'of' first:
\( \frac{7}{15} \text{ of } \frac{8}{9} = \frac{7 \times 8}{15 \times 9} = \frac{56}{135} \)
Now, divide \( \frac{4}{5} \) by this result:
\( \frac{4}{5} \div \frac{56}{135} = \frac{4}{5} \times \frac{135}{56} = \frac{4 \times 135}{5 \times 56} = \frac{27}{14} = 1\frac{13}{14} \)

(xi) There is no 'of' here, so we do division first, then multiplication:
\( \frac{4}{5} \div \frac{7}{15} \times \frac{8}{9} = \frac{4}{5} \times \frac{15}{7} \times \frac{8}{9} \)
Simplify and multiply:
\( = \frac{4 \times 3 \times 8}{1 \times 7 \times 9} = \frac{32}{21} = 1\frac{11}{21} \)

(xii) Solve 'of' first using the BODMAS order:
\( \frac{4}{5} \text{ of } \frac{7}{15} = \frac{4 \times 7}{5 \times 15} = \frac{28}{75} \)
Next, do division by multiplying by the reciprocal:
\( \frac{28}{75} \div \frac{8}{9} = \frac{28}{75} \times \frac{9}{8} = \frac{7 \times 3}{25 \times 2} = \frac{21}{50} \)

(xiii) Follow BODMAS. Solve 'of' first:
\( \frac{1}{2} \text{ of } \frac{3}{4} = \frac{3}{8} \)
Now rewrite the expression:
\( \frac{3}{8} \times \frac{1}{2} \div \frac{2}{3} \)
Next, do the division:
\( \frac{1}{2} \div \frac{2}{3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \)
Now multiply the remaining parts:
\( \frac{3}{8} \times \frac{3}{4} = \frac{9}{32} \)
In simple words: When simplifying fractions with multiple operations, always follow the BODMAS rules. Solve 'of' first, then division, then multiplication, and change mixed numbers to improper ones before calculating.

Exam Tip: Be very careful with the order of operations (BODMAS). 'Of' must always be calculated before division or multiplication, as seen in the different results of sub-questions (x), (xi), and (xii).

Question 6. A bought \( 3\frac{3}{4} \) kg of wheat and \( 2\frac{1}{2} \) kg of rice. Find the total weight of wheat and rice bought.
Answer:
The weight of wheat bought is \( 3\frac{3}{4} \) kg, which is \( \frac{15}{4} \) kg.
The weight of rice bought is \( 2\frac{1}{2} \) kg, which is \( \frac{5}{2} \) kg.
To find the total, we add these two weights:
\( \text{Total Weight} = \frac{15}{4} + \frac{5}{2} \)
The LCM of 4 and 2 is 4. So we change \( \frac{5}{2} \) to \( \frac{10}{4} \):
\( \text{Total Weight} = \frac{15 \times 1}{4 \times 1} + \frac{5 \times 2}{2 \times 2} \)
\( = \frac{15 + 10}{4} = \frac{25}{4} = 6\frac{1}{4} \) kg.
So, the total weight of wheat and rice is \( 6\frac{1}{4} \) kg.
In simple words: To find the total weight, turn the mixed fractions into simple fractions. Then find a common bottom number to add them together.

Exam Tip: Always mention the unit of measurement, like kg, in your final answer to get full marks.

Question 7. Which is greater, \( \frac{3}{5} \) or \( \frac{7}{10} \) and by how much?
Answer:
We can compare the two fractions by cross multiplying.
Multiplying 3 by 10 gives 30, and multiplying 7 by 5 gives 35.
Since 30 is less than 35, the fraction \( \frac{3}{5} \) is smaller than \( \frac{7}{10} \).
To find the difference, we subtract \( \frac{3}{5} \) from \( \frac{7}{10} \). The LCM of 10 and 5 is 10.
\( \frac{7}{10} - \frac{3}{5} = \frac{7 \times 1}{10 \times 1} - \frac{3 \times 2}{5 \times 2} = \frac{7 - 6}{10} = \frac{1}{10} \).
Thus, \( \frac{7}{10} \) is larger than \( \frac{3}{5} \) by \( \frac{1}{10} \).
In simple words: Cross multiply to see which fraction is bigger. Then subtract the smaller fraction from the bigger one to find the difference.

Exam Tip: Using cross-multiplication is a quick way to compare two fractions, but always show the LCM subtraction steps to find the difference.

Question 8. What number should be added to \( 8\frac{2}{3} \) to \( 12\frac{5}{6} \)
Answer:
To find the missing number, we subtract the smaller mixed fraction from the larger one. We calculate \( 12\frac{5}{6} - 8\frac{2}{3} \).
First, convert both mixed fractions to improper ones:
\( 12\frac{5}{6} = \frac{77}{6} \)
\( 8\frac{2}{3} = \frac{26}{3} \)
The LCM of 3 and 6 is 6.
\( \frac{77 \times 1}{6 \times 1} - \frac{26 \times 2}{3 \times 2} = \frac{77 - 52}{6} = \frac{25}{6} = 4\frac{1}{6} \).
So, \( 4\frac{1}{6} \) must be added.
In simple words: To find what to add to a number to reach a goal, subtract the number you have from your goal.

Exam Tip: Make sure to convert mixed numbers to improper fractions before finding the LCM, to avoid calculation errors.

Question 9. What should be subtracted from \( 8\frac{3}{4} \) to get \( 2\frac{2}{3} \)
Answer:
To find the number we need to subtract, we take \( 2\frac{2}{3} \) away from \( 8\frac{3}{4} \).
Change these mixed fractions into improper fractions:
\( 8\frac{3}{4} = \frac{35}{4} \)
\( 2\frac{2}{3} = \frac{8}{3} \)
The LCM of 4 and 3 is 12.
\( \frac{35 \times 3}{4 \times 3} - \frac{8 \times 4}{3 \times 4} = \frac{105 - 32}{12} = \frac{73}{12} = 6\frac{1}{12} \).
So, the number is \( 6\frac{1}{12} \).
In simple words: To find out what to subtract from a big number to get a small number, just subtract the small number from the big one.

Exam Tip: Always double-check your subtraction by adding your answer back to the subtracted value to see if it equals the original number.

Question 10. A field is \( 16\frac{1}{2} \) m long and \( 12\frac{2}{5} \) m wide. Find the perimeter of the field.
Answer:
The length of the field is \( 16\frac{1}{2} \) m, and the width is \( 12\frac{2}{5} \) m.
We know that the perimeter of a rectangle is calculated as \( 2(l + b) \), where \( l \) is length and \( b \) is width.
Change both measurements to improper fractions:
\( 16\frac{1}{2} = \frac{33}{2} \) m
\( 12\frac{2}{5} = \frac{62}{5} \) m
Find the perimeter:
\( \text{Perimeter} = 2 \times \left(\frac{33}{2} + \frac{62}{5}\right) \)
Since the LCM of 2 and 5 is 10, we get:
\( \text{Perimeter} = 2 \times \left(\frac{33 \times 5}{2 \times 5} + \frac{62 \times 2}{5 \times 2}\right) \)
\( = 2 \times \left(\frac{165 + 124}{10}\right) \)
\( = 2 \times \frac{289}{10} \)
\( = \frac{289}{5} = 57\frac{4}{5} \) m.
Thus, the perimeter is \( 57\frac{4}{5} \) m.
In simple words: To find the boundary distance around a field, add the length and width together, then multiply by two.

Exam Tip: Make sure to write down the formula \( 2(l + b) \) clearly. Showing formulas helps you earn partial marks even if you make a calculation mistake later.

Question 11. Sugar costs Rs. \( 37\frac{1}{2} \) per kg. Find the cost of \( 8\frac{3}{4} \) kg sugar.
Answer:
The price of 1 kg of sugar is Rs. \( 37\frac{1}{2} \).
To find the cost of \( 8\frac{3}{4} \) kg of sugar, we multiply the quantity by the price per kg:
\( \text{Cost} = 37\frac{1}{2} \times 8\frac{3}{4} \)
Change both numbers into improper fractions:
\( 37\frac{1}{2} = \frac{75}{2} \)
\( 8\frac{3}{4} = \frac{35}{4} \)
Multiply the fractions:
\( \text{Total Cost} = \frac{75}{2} \times \frac{35}{4} = \frac{2625}{8} = \text{Rs. } 328\frac{1}{8} \).
So, the total cost is Rs. \( 328\frac{1}{8} \).
In simple words: To find the total cost, multiply the weight of the sugar by the price of one kilogram.

Exam Tip: In multiplication of fractions, you do not need to find a common denominator. Simply multiply the top numbers together and the bottom numbers together.

Question 12. A motor cycle runs \( 31\frac{1}{4} \) km consuming 1 litre of petrol. How much distance will it run consuming \( 1\frac{3}{5} \) liter of petrol?
Answer:
The motorcycle can travel \( 31\frac{1}{4} \) km on 1 litre of petrol.
To find how far it can go on \( 1\frac{3}{5} \) litres, we multiply the distance per litre by the total petrol consumed:
Convert the mixed numbers into improper fractions:
\( 31\frac{1}{4} = \frac{125}{4} \) km
\( 1\frac{3}{5} = \frac{8}{5} \) litres
Multiply the two values:
\( \text{Distance} = \frac{125}{4} \times \frac{8}{5} \)
Simplify by dividing 125 by 5 and 8 by 4:
\( \text{Distance} = \frac{1000}{20} = 50 \) km.
Therefore, the motorcycle will travel 50 km.
In simple words: If you know how far the bike goes on one litre, multiply that distance by the number of litres to get the total distance.

Exam Tip: Before multiplying, simplify the fractions by canceling common factors in the numerators and denominators to make the math easier.

Question 13. A rectangular park has length = \( 23\frac{2}{5} \) m and breadth = \( 16\frac{2}{3} \) m. Find the area of the park.
Answer:
The length of the park is \( 23\frac{2}{5} \) m, and the width is \( 16\frac{2}{3} \) m.
We calculate the area of a rectangle using the formula \( \text{Area} = \text{length} \times \text{breadth} \).
Change both mixed fractions to improper fractions first:
\( 23\frac{2}{5} = \frac{117}{5} \) m
\( 16\frac{2}{3} = \frac{50}{3} \) m
Now multiply these values to find the area:
\( \text{Area} = \frac{117}{5} \times \frac{50}{3} \)
We can simplify: 117 divided by 3 is 39, and 50 divided by 5 is 10.
\( \text{Area} = 39 \times 10 = 390 \text{ m}^2 \).
The park's total area is 390 square metres.
In simple words: To find the space inside a rectangular park, multiply the length by the width.

Exam Tip: Remember that area is measured in square units, so always write your final answer with \( \text{m}^2 \) or "square metres".

Question 14. Each of 40 identical boxes weighs \( 4\frac{4}{5} \) kg Find the total weight of all the boxes.
Answer:
The weight of a single box is \( 4\frac{4}{5} \) kg.
Convert this mixed number to an improper fraction:
\( 4\frac{4}{5} = \frac{24}{5} \) kg.
To find the total weight of 40 boxes, multiply the weight of one box by 40:
\( \text{Total Weight} = 40 \times \frac{24}{5} \)
Divide 40 by 5 to get 8, then multiply:
\( \text{Total Weight} = 8 \times 24 = 192 \) kg.
The total weight of all the boxes is 192 kg.
In simple words: Multiply the weight of one box by the total number of boxes to find the total weight.

Exam Tip: When multiplying a whole number by a fraction, divide the whole number by the denominator first if possible to keep the numbers small.

Question 15. Out of 24 kg of wheat, \( \frac{5}{6} \)th of wheat is consumed. Find, how much wheat is still left?
Answer:
The total amount of wheat available is 24 kg.
The fraction of wheat eaten is \( \frac{5}{6} \).
We find the weight of the wheat consumed:
\( \text{Consumed Wheat} = \frac{5}{6} \times 24 = 5 \times 4 = 20 \) kg.
To find out how much wheat is left, subtract the consumed wheat from the total:
\( \text{Leftover Wheat} = 24 - 20 = 4 \) kg.
So, 4 kg of wheat is still left.
In simple words: Find out how much wheat was eaten by multiplying the fraction by the total. Then, subtract that amount from the starting weight.

Exam Tip: You can also solve this by finding the remaining fraction first: \( 1 - \frac{5}{6} = \frac{1}{6} \), and then calculating \( \frac{1}{6} \times 24 = 4 \) kg. Both methods give the same correct answer!

Question 16. A rod of length \( 2\frac{2}{5} \) metre is divided into five equal parts. Find the length of each part so obtained.
Answer:
The total length of the rod is \( 2\frac{2}{5} \) metres.
Convert this mixed fraction into an improper fraction:
\( 2\frac{2}{5} = \frac{12}{5} \) m.
Since the rod is cut into 5 equal pieces, we divide the total length by 5:
\( \text{Length of each part} = \frac{12}{5} \div 5 \)
Multiply by the reciprocal of 5:
\( \text{Length of each part} = \frac{12}{5} \times \frac{1}{5} = \frac{12}{25} \) m.
So, each part is \( \frac{12}{25} \) metre long.
In simple words: To find the length of each piece, change the mixed number to an improper fraction, then divide it by the number of parts.

Exam Tip: Remember that dividing by a whole number like 5 is the same as multiplying by its reciprocal, \( \frac{1}{5} \).

Question 17. If A = \( 3\frac{3}{8} \) and B = \( 6\frac{5}{8} \) find :
(i) A \(\div\) B
(ii) B \(\div\) A
Answer:
We convert the mixed fractions for A and B into improper fractions first:
\( A = 3\frac{3}{8} = \frac{27}{8} \)
\( B = 6\frac{5}{8} = \frac{53}{8} \)

(i) Find \( A \div B \):
\( A \div B = \frac{27}{8} \div \frac{53}{8} \)
Multiply by the reciprocal:
\( = \frac{27}{8} \times \frac{8}{53} = \frac{27}{53} \)

(ii) Find \( B \div A \):
\( B \div A = \frac{53}{8} \div \frac{27}{8} \)
Multiply by the reciprocal:
\( = \frac{53}{8} \times \frac{8}{27} = \frac{53}{27} = 1\frac{26}{27} \)
In simple words: Turn both mixed numbers into simple fractions first. To divide, swap the top and bottom of the second fraction and multiply.

Exam Tip: Notice that \( B \div A \) is simply the reciprocal of \( A \div B \). You can save time by just flipping your answer from part (i) upside down!

Question 18. Cost of \( 3\frac{5}{7} \) litres of oil is Rs. \( 83\frac{1}{2} \). Find the cost of one litre oil.
Answer:
The price of \( 3\frac{5}{7} \) litres of oil is Rs. \( 83\frac{1}{2} \).
To find the price of 1 litre of oil, we divide the total cost by the total volume of oil:
\( \text{Cost per litre} = 83\frac{1}{2} \div 3\frac{5}{7} \)
Change both mixed fractions to improper fractions:
\( 83\frac{1}{2} = \frac{167}{2} \)
\( 3\frac{5}{7} = \frac{26}{7} \)
Now perform the division:
\( \text{Cost per litre} = \frac{167}{2} \div \frac{26}{7} \)
Multiply by the reciprocal of the second fraction:
\( = \frac{167}{2} \times \frac{7}{26} = \frac{1169}{52} = \text{Rs. } 22\frac{25}{52} \).
So, 1 litre of oil costs Rs. \( 22\frac{25}{52} \).
In simple words: To find the cost of just one litre, divide the total money spent by the total number of litres.

Exam Tip: For calculations with large numerators, multiply carefully. Double-check your final long division when converting back to a mixed fraction.

Question 19. The product of two numbers is \( 20\frac{5}{7} \). If one of these numbers is \( 6\frac{2}{3} \), find the other.
Answer:
The product of the two numbers is \( 20\frac{5}{7} \), and one of them is \( 6\frac{2}{3} \).
To find the other number, divide the product by the given number:
\( \text{Second number} = 20\frac{5}{7} \div 6\frac{2}{3} \)
Change both numbers to improper fractions:
\( 20\frac{5}{7} = \frac{145}{7} \)
\( 6\frac{2}{3} = \frac{20}{3} \)
Divide the fractions:
\( \text{Second number} = \frac{145}{7} \div \frac{20}{3} = \frac{145}{7} \times \frac{3}{20} \)
Simplify by dividing 145 and 20 by 5:
\( = \frac{29 \times 3}{7 \times 4} = \frac{87}{28} = 3\frac{3}{28} \).
Thus, the other number is \( 3\frac{3}{28} \).
In simple words: If you have the total product of two numbers and know one of them, divide the product by that number to find the second one.

Exam Tip: Always check if you can simplify fractions before multiplying the numerators and denominators to prevent having to deal with very large numbers.

Question 20. By what number should \( 5\frac{5}{6} \) be multiplied to get \( 3\frac{1}{3} \)?
Answer:
Let the required number be \( x \).
This means \( 5\frac{5}{6} \times x = 3\frac{1}{3} \).
To find \( x \), divide \( 3\frac{1}{3} \) by \( 5\frac{5}{6} \):
\( \text{Required number} = 3\frac{1}{3} \div 5\frac{5}{6} \)
Convert both to improper fractions:
\( 3\frac{1}{3} = \frac{10}{3} \)
\( 5\frac{5}{6} = \frac{35}{6} \)
Perform the division:
\( \text{Required number} = \frac{10}{3} \div \frac{35}{6} = \frac{10}{3} \times \frac{6}{35} \)
Simplify by dividing 10 and 35 by 5, and 6 by 3:
\( = \frac{2 \times 2}{1 \times 7} = \frac{4}{7} \).
So, the number is \( \frac{4}{7} \).
In simple words: To find what number to multiply by, divide the target answer by the starting number.

Exam Tip: When the target product is smaller than the starting number, your multiplier will be a proper fraction (less than 1).

Exercise 3(D)

Question 1. Simplify
Answer:
The expression to simplify is:
\( 6 + \left\{ \frac{4}{3} + \left( \frac{3}{4} - \frac{1}{3} \right) \right\} \)
According to the rules of brackets, we remove them. Since there are only additions and subtractions, we can open all brackets:
\( = \frac{6}{1} + \frac{4}{3} + \frac{3}{4} - \frac{1}{3} \)
The LCM of 1, 3, and 4 is 12.
Now convert each fraction to have a denominator of 12:
\( = \frac{6 \times 12}{12} + \frac{4 \times 4}{12} + \frac{3 \times 3}{12} - \frac{1 \times 4}{12} \)
\( = \frac{72 + 16 + 9 - 4}{12} \)
Combine the terms in the numerator:
\( = \frac{97 - 4}{12} = \frac{93}{12} \)
Simplify this fraction by dividing both numerator and denominator by 3:
\( = \frac{31}{4} = 7\frac{3}{4} \).
In simple words: Open the brackets first. Then, find the common denominator for all terms to add and subtract them together.

Exam Tip: When removing brackets, if there is a plus sign outside, the signs of the terms inside do not change.

Question 2. \( 8 - \left\{ \frac{3}{2} + \left( \frac{3}{5} - \frac{1}{2} \right) \right\} \)
Answer:
We simplify the expression step-by-step:
\( 8 - \left\{ \frac{3}{2} + \left( \frac{3}{5} - \frac{1}{2} \right) \right\} \)
First, open the parentheses inside the curly brackets:
\( = 8 - \left\{ \frac{3}{2} + \frac{3}{5} - \frac{1}{2} \right\} \)
Now, remove the curly brackets. Because there is a minus sign before the curly bracket, all signs inside the bracket will reverse:
\( = \frac{8}{1} - \frac{3}{2} - \frac{3}{5} + \frac{1}{2} \)
The LCM of 1, 2, and 5 is 10. Change each term to have 10 as the denominator:
\( = \frac{80}{10} - \frac{15}{10} - \frac{6}{10} + \frac{5}{10} \)
\( = \frac{80 - 15 - 6 + 5}{10} \)
Combine the terms in the numerator:
\( = \frac{85 - 21}{10} = \frac{64}{10} \)
Simplify by dividing the numerator and denominator by 2:
\( = \frac{32}{5} = 6\frac{2}{5} \).
In simple words: When you open a bracket with a minus sign in front of it, remember to swap every plus to minus and every minus to plus inside.

Exam Tip: A common mistake is forgetting to change the signs of all terms inside a bracket when expanding it with a negative sign outside. Pay close attention to this rule!

Question 3. \( \frac{1}{4}\left(\frac{1}{4} + \frac{1}{3}\right) - \frac{2}{5} \)
Answer:
First, evaluate the sum inside the inner brackets:
\[ \frac{1}{4} + \frac{1}{3} = \frac{3 + 4}{12} = \frac{7}{12} \]
Now, perform the multiplication outside the bracket:
\[ \frac{1}{4} \times \frac{7}{12} = \frac{7}{48} \]
Next, subtract \( \frac{2}{5} \) from this product:
\[ \frac{7}{48} - \frac{2}{5} \]
Find the common denominator of 48 and 5, which is 240:
\[ \frac{35 - 96}{240} = -\frac{61}{240} \]
In simple words: First solve the addition inside the bracket by finding a common denominator. Then multiply by the fraction outside, and finally subtract the remaining fraction to get the final result.

Exam Tip: Always solve terms inside parentheses first before doing multiplication and subtraction, which follows the BODMAS sequence of operations.

Question 4. \( 2\frac{3}{4} - \left[3\frac{1}{8} \div \left\{5 - \left(4\frac{2}{3} - \frac{11}{12}\right)\right\}\right] \)
Answer:
To begin, convert all mixed numbers to improper fractions:
\[ 2\frac{3}{4} = \frac{11}{4},\quad 3\frac{1}{8} = \frac{25}{8},\quad 4\frac{2}{3} = \frac{14}{3} \]
Substitute these back into the expression:
\[ \frac{11}{4} - \left[\frac{25}{8} \div \left\{5 - \left(\frac{14}{3} - \frac{11}{12}\right)\right\}\right] \]
Next, simplify the small round brackets first:
\[ \frac{14}{3} - \frac{11}{12} = \frac{56 - 11}{12} = \frac{45}{12} \]
Now, evaluate the curly brackets:
\[ 5 - \frac{45}{12} = \frac{60 - 45}{12} = \frac{15}{12} \]
Next, calculate the division inside the square brackets:
\[ \frac{25}{8} \div \frac{15}{12} = \frac{25}{8} \times \frac{12}{15} = \frac{5}{2} \]
Lastly, subtract this result from the initial fraction:
\[ \frac{11}{4} - \frac{5}{2} = \frac{11 - 10}{4} = \frac{1}{4} \]
In simple words: Change mixed numbers to improper fractions. Then work from the inside out: first simplify the round brackets, then the curly brackets, and then divide inside the square brackets. Finally, subtract to get the answer.

Exam Tip: When dividing fractions, remember to invert the second fraction and multiply. Always simplify the brackets in the correct order: round, curly, then square.

Question 5. \( 12\frac{1}{2} - \left[8\frac{1}{2} + \left\{9 - (5 - \overline{3-2})\right\}\right] \)
Answer:
Convert the mixed numbers into improper fractions:
\[ 12\frac{1}{2} = \frac{25}{2},\quad 8\frac{1}{2} = \frac{17}{2} \]
Start by evaluating the vinculum (the bar over the numbers):
\[ \overline{3-2} = 1 \]
This changes the innermost parenthesis as follows:
\[ 5 - 1 = 4 \]
Next, solve the terms inside the curly brackets:
\[ 9 - 4 = 5 \]
Now, work out the sum inside the square brackets:
\[ \frac{17}{2} + 5 = \frac{17 + 10}{2} = \frac{27}{2} \]
Finally, subtract this from the first term:
\[ \frac{25}{2} - \frac{27}{2} = \frac{-2}{2} = -1 \]
In simple words: Solve the bar over 3-2 first, which is 1. Then solve the subtraction in the round brackets, then the curly brackets, then add inside the square brackets, and finally subtract from the outer fraction.

Exam Tip: A bar over numbers acts as a vinculum, which is a bracket that must be solved before any other brackets in the expression.

Question 6. \( 1\frac{1}{5} \div \left\{2\frac{1}{3} - (5 + \overline{2-3})\right\} - 3\frac{1}{2} \)
Answer:
Convert the mixed numbers into improper fractions:
\[ 1\frac{1}{5} = \frac{6}{5},\quad 2\frac{1}{3} = \frac{7}{3},\quad 3\frac{1}{2} = \frac{7}{2} \]
First, calculate the term under the vinculum:
\[ \overline{2-3} = -1 \]
Substitute this to simplify the innermost parentheses:
\[ 5 + (-1) = 5 - 1 = 4 \]
Next, solve the subtraction inside the curly brackets:
\[ \frac{7}{3} - 4 = \frac{7 - 12}{3} = \frac{-5}{3} \]
Now, divide the first fraction by this result:
\[ \frac{6}{5} \div \left(\frac{-5}{3}\right) = \frac{6}{5} \times \frac{3}{-5} = -\frac{18}{25} \]
Finally, subtract the last fraction:
\[ -\frac{18}{25} - \frac{7}{2} = \frac{-36 - 175}{50} = \frac{-211}{50} = -4\frac{11}{50} \]
In simple words: First solve the subtraction under the bar, then calculate the terms inside the parentheses. After that, work out the curly bracket, divide, and finally subtract the remaining fraction to get the mixed number.

Exam Tip: Be very careful with signs when dealing with a vinculum and division of negative fractions. Keep track of negative numbers to avoid calculation errors.

Question 7. \( \left(\frac{1}{2} + \frac{2}{3}\right) \div \left(\frac{3}{4} - \frac{2}{9}\right) \)
Answer:
First, evaluate the terms inside each pair of parentheses:
For the first parenthesis, the common denominator for 2 and 3 is 6:
\[ \frac{1}{2} + \frac{2}{3} = \frac{3 + 4}{6} = \frac{7}{6} \]
For the second parenthesis, the common denominator for 4 and 9 is 36:
\[ \frac{3}{4} - \frac{2}{9} = \frac{27 - 8}{36} = \frac{19}{36} \]
Now, divide the first simplified fraction by the second one:
\[ \frac{7}{6} \div \frac{19}{36} = \frac{7}{6} \times \frac{36}{19} \]
Simplifying by dividing 36 by 6 gives:
\[ \frac{7 \times 6}{19} = \frac{42}{19} = 2\frac{4}{19} \]
In simple words: Work out the addition in the first set of brackets and the subtraction in the second set of brackets. Then divide the first answer by the second answer and simplify.

Exam Tip: Always find the least common multiple (LCM) for denominators when adding or subtracting fractions to ensure calculations are as simple as possible.

Question 8. \( \frac{6}{5}\text{ of }\left(3\frac{1}{3} - 2\frac{1}{2}\right) \div \left(2\frac{5}{21} - 2\right) \)
Answer:
First, convert the mixed numbers into improper fractions:
\[ 3\frac{1}{3} = \frac{10}{3},\quad 2\frac{1}{2} = \frac{5}{2},\quad 2\frac{5}{21} = \frac{47}{21} \]
Substitute these values back into the expression:
\[ \frac{6}{5}\text{ of }\left(\frac{10}{3} - \frac{5}{2}\right) \div \left(\frac{47}{21} - 2\right) \]
Now simplify inside the brackets:
For the first bracket (LCM is 6):
\[ \frac{10}{3} - \frac{5}{2} = \frac{20 - 15}{6} = \frac{5}{6} \]
For the second bracket:
\[ \frac{47}{21} - 2 = \frac{47 - 42}{21} = \frac{5}{21} \]
Substitute these back to simplify using the BODMAS rule (where "of" comes before division):
\[ \frac{6}{5}\text{ of }\frac{5}{6} \div \frac{5}{21} \]
Evaluate the "of" operation first:
\[ \left(\frac{6}{5} \times \frac{5}{6}\right) \div \frac{5}{21} = 1 \div \frac{5}{21} \]
Finally, compute the division by multiplying by the reciprocal:
\[ 1 \times \frac{21}{5} = \frac{21}{5} = 4\frac{1}{5} \]
In simple words: Change mixed numbers to improper fractions, then solve what is inside the two sets of brackets. Next, perform the "of" operation, and finally divide to get the final mixed fraction.

Exam Tip: Remember that in BODMAS, the word "of" stands for multiplication and must be calculated before any division or multiplication operations.

Question 9. \( 10\frac{1}{8}\text{ of }\frac{4}{5} \div \frac{35}{36}\text{ of }\frac{20}{49} \)
Answer:
Convert the mixed fraction to an improper fraction first:
\[ 10\frac{1}{8} = \frac{81}{8} \]
This gives us the expression:
\[ \frac{81}{8}\text{ of }\frac{4}{5} \div \frac{35}{36}\text{ of }\frac{20}{49} \]
Under BODMAS, evaluate the "of" parts before doing the division:
First "of" calculation:
\[ \frac{81}{8} \times \frac{4}{5} = \frac{81}{10} \]
Second "of" calculation:
\[ \frac{35}{36} \times \frac{20}{49} = \frac{5 \times 5}{9 \times 7} = \frac{25}{63} \]
Now, divide the first result by the second result:
\[ \frac{81}{10} \div \frac{25}{63} = \frac{81}{10} \times \frac{63}{25} = \frac{5103}{250} = 20\frac{103}{250} \]
In simple words: Convert the mixed fraction to an improper one. Solve both "of" parts first, and then divide the first resulting fraction by the second one to find the final mixed number.

Exam Tip: Never perform division before the "of" operation, as "of" has higher precedence in the order of operations.

Question 10. \( 5\frac{3}{4} - \frac{3}{7} \times 15\frac{3}{4} + 2\frac{2}{35} \div 1\frac{11}{25} \)
Answer:
Convert each mixed number to an improper fraction:
\[ 5\frac{3}{4} = \frac{23}{4},\quad 15\frac{3}{4} = \frac{63}{4},\quad 2\frac{2}{35} = \frac{72}{35},\quad 1\frac{11}{25} = \frac{36}{25} \]
Substitute these to get:
\[ \frac{23}{4} - \frac{3}{7} \times \frac{63}{4} + \frac{72}{35} \div \frac{36}{25} \]
Following BODMAS, perform division and multiplication first:
Evaluate the division term:
\[ \frac{72}{35} \div \frac{36}{25} = \frac{72}{35} \times \frac{25}{36} = \frac{10}{7} \]
Evaluate the multiplication term:
\[ \frac{3}{7} \times \frac{63}{4} = \frac{27}{4} \]
Now, substitute these back into the expression:
\[ \frac{23}{4} - \frac{27}{4} + \frac{10}{7} \]
Find the common denominator of 4 and 7, which is 28:
\[ \frac{161 - 189 + 40}{28} = \frac{201 - 189}{28} = \frac{12}{28} = \frac{3}{7} \]
In simple words: Turn the mixed fractions into improper ones. Solve the division and multiplication parts first. After that, find a common denominator of 28 to add and subtract the terms.

Exam Tip: Be sure to divide and multiply first before attempting any addition or subtraction, as per standard order of operations.

Question 11. \( \frac{3}{4}\text{ of }7\frac{3}{7} - 5\frac{3}{5} \div 3\frac{4}{15} \)
Answer:
Convert the mixed numbers into improper fractions:
\[ 7\frac{3}{7} = \frac{52}{7},\quad 5\frac{3}{5} = \frac{28}{5},\quad 3\frac{4}{15} = \frac{49}{15} \]
This simplifies the expression to:
\[ \frac{3}{4}\text{ of }\frac{52}{7} - \frac{28}{5} \div \frac{49}{15} \]
According to BODMAS rules, evaluate the "of" part first:
\[ \frac{3}{4} \times \frac{52}{7} = \frac{3 \times 13}{7} = \frac{39}{7} \]
Next, solve the division:
\[ \frac{28}{5} \div \frac{49}{15} = \frac{28}{5} \times \frac{15}{49} = \frac{4 \times 3}{7} = \frac{12}{7} \]
Finally, subtract the two results:
\[ \frac{39}{7} - \frac{12}{7} = \frac{27}{7} = 3\frac{6}{7} \]
In simple words: Turn the mixed fractions into improper ones, evaluate the "of" term, then the division term, and finally subtract the two to find the mixed number.

Exam Tip: Convert all mixed numbers to improper fractions before applying the BODMAS order of operations to prevent simple mathematical errors.

Exercise 3(E)

Question 1. A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is :
(i) The length of AB to that of CD ?
(ii) \( \frac{1}{2} \) the length of AB to that of \( \frac{1}{3} \) of CD ?
(iii) \( \frac{1}{5} \) of CD to that of AB ?
Answer:
Given:
Length of line AB = 6 cm
Length of line CD = 15 cm
(i) The ratio of AB's length to CD's length is:
\[ \frac{\text{Length of AB}}{\text{Length of CD}} = \frac{6}{15} = \frac{2}{5} \]
(ii) First, calculate half the length of AB:
\[ \frac{1}{2} \times 6 = 3\text{ cm} \]
Next, calculate one-third of CD's length:
\[ \frac{1}{3} \times 15 = 5\text{ cm} \]
Therefore, the required fraction is:
\[ \frac{3}{5} \]
(iii) First, calculate one-fifth of CD's length:
\[ \frac{1}{5} \times 15 = 3\text{ cm} \]
Therefore, the ratio of this length to AB's length is:
\[ \frac{3}{6} = \frac{1}{2} \]
In simple words: To find each fraction, work out the lengths as described for each part. Then write them as a ratio and simplify by dividing the top and bottom by their common factor.

Exam Tip: Be careful to put the correct quantity in the numerator and the denominator as specified by the wording in each part of the question.

Question 2. Subtract \( \frac{2}{7} - \frac{5}{21} \) from the sum of \( \frac{3}{4} \), \( \frac{5}{7} \) and \( \frac{7}{12} \)
Answer:
First, find the sum of the three given fractions:
\[ \frac{3}{4} + \frac{5}{7} + \frac{7}{12} \]
The LCM of 4, 7, and 12 is 84. Thus:
\[ \frac{63 + 60 + 49}{84} = \frac{172}{84} \]
Next, find the value of the subtraction term:
\[ \frac{2}{7} - \frac{5}{21} \]
The LCM of 7 and 21 is 21. Thus:
\[ \frac{6 - 5}{21} = \frac{1}{21} \]
Finally, subtract the second result from the first sum:
\[ \frac{172}{84} - \frac{1}{21} \]
Since the LCM of 84 and 21 is 84:
\[ \frac{172 - 4}{84} = \frac{168}{84} = 2 \]
In simple words: Add the three fractions together first by finding their common denominator. After that, subtract the second pair of fractions, and then subtract that result from your first sum to get 2.

Exam Tip: Keep your calculations neat and organized by splitting the problem into two distinct steps (summing and subtracting) before combining them.

Question 3. From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5\(\frac{1}{4}\) kg, 9\(\frac{1}{2}\) kg and 9\(\frac{3}{4}\) kg respectively.
(i) How many kg did he sell ?
(ii) How many kg are still left in the sack ?
Answer:
The total quantity of potatoes is 120 kg.
(i) To find the total weight sold, sum the individual weights:
\[ 6 + 5\frac{1}{4} + 9\frac{1}{2} + 9\frac{3}{4} \]
Convert mixed numbers to improper fractions:
\[ = 6 + \frac{21}{4} + \frac{19}{2} + \frac{39}{4} \]
Express with a common denominator of 4:
\[ = \frac{24 + 21 + 38 + 39}{4} = \frac{122}{4} = \frac{61}{2} = 30\frac{1}{2}\text{ kg} \]
(ii) To find the remaining potatoes, subtract the sold quantity from the total weight:
\[ 120 - 30\frac{1}{2} = \frac{120}{1} - \frac{61}{2} = \frac{240 - 61}{2} = \frac{179}{2} = 89\frac{1}{2}\text{ kg} \]
In simple words: Add all the sold quantities together to find that he sold 30 and a half kilograms. Then, subtract this from the starting 120 kilograms to find that 89 and a half kilograms remain.

Exam Tip: Be sure to write final answers with their proper units (kg) to earn full marks on word problems.

Question 4. If a boy works for six consecutive days for 8 hours, 7\(\frac{1}{2}\) hours, 8\(\frac{1}{4}\) hours, 6\(\frac{1}{4}\) 3hours, 6\(\frac{3}{4}\) hours and 7 hours respectively. How much money will he earn at the rate of Rs. 36 per hour ?
Answer:
Calculate the total hours worked by the boy across the six days:
\[ 8 + 7\frac{1}{2} + 8\frac{1}{4} + 6\frac{1}{4} + 6\frac{3}{4} + 7\text{ hours} \]
Convert these mixed numbers into improper fractions:
\[ = 8 + \frac{15}{2} + \frac{33}{4} + \frac{25}{4} + \frac{27}{4} + 7 \]
Express with a common denominator of 4:
\[ = \frac{32 + 30 + 33 + 25 + 27 + 28}{4} = \frac{175}{4}\text{ hours} = 43\frac{3}{4}\text{ hours} \]
Given that the rate of earning is Rs. 36 per hour:
\[ \text{Total earnings} = \text{Rs. } \frac{175}{4} \times 36 = \text{Rs. } 175 \times 9 = \text{Rs. } 1575 \]
In simple words: First add up all the daily hours worked to find that he worked 175/4 hours in total. Then multiply this total time by his hourly wage of Rs. 36 to get Rs. 1575.

Exam Tip: Simplify calculations by cancelling common factors between the numerator and denominator before performing the final multiplication.

Question 5. A student bought 4\(\frac{1}{3}\) m of yellow ribbon, 6\(\frac{1}{6}\) m of red ribbon and 3\(\frac{2}{9}\) m of blue ribbon for decorating a room. How many metres of ribbon did he buy ?
Answer:
Find the length of each color of ribbon bought:
Length of yellow ribbon = \( 4\frac{1}{3}\text{ m} = \frac{13}{3}\text{ m} \)
Length of red ribbon = \( 6\frac{1}{6}\text{ m} = \frac{37}{6}\text{ m} \)
Length of blue ribbon = \( 3\frac{2}{9}\text{ m} = \frac{29}{9}\text{ m} \)
Calculate the total length by adding them:
\[ \text{Total length} = \frac{13}{3} + \frac{37}{6} + \frac{29}{9} \]
The LCM of 3, 6, and 9 is 18:
\[ = \frac{78 + 111 + 58}{18} = \frac{247}{18} = 13\frac{13}{18}\text{ m} \]
In simple words: Convert the mixed meters of yellow, red, and blue ribbon into improper fractions, find their common denominator of 18, and add them up to find the total length of ribbon bought.

Exam Tip: When finding the LCM of multiple denominators, double-check that your numerator scaling is done correctly for each term.

Question 6. In a business, Ram and Deepak invest \(\frac{3}{5}\) and \(\frac{2}{5}\) of the total investment. If Rs. 40,000 is the total investment, calculate the amount invested by each ?
Answer:
Total amount invested in the business = Rs. 40,000
Calculate Ram's share of the investment:
\[ \text{Ram's investment} = \frac{3}{5}\text{ of Rs. } 40,000 = \text{Rs. } \frac{3}{5} \times 40,000 = \text{Rs. } 24,000 \]
Calculate Deepak's share of the investment:
\[ \text{Deepak's investment} = \frac{2}{5}\text{ of Rs. } 40,000 = \text{Rs. } \frac{2}{5} \times 40,000 = \text{Rs. } 16,000 \]
In simple words: Out of the total Rs. 40,000, Ram invests three-fifths, which is Rs. 24,000. Deepak invests the remaining two-fifths, which comes out to Rs. 16,000.

Exam Tip: Verify your final answers by adding the individual investments together to make sure they sum up to the total given investment (Rs. 24,000 + Rs. 16,000 = Rs. 40,000).

Question 7. Geeta had 30 problems for home work. She worked out \(\frac{2}{3}\) of them. How many problems were still left to be worked out by her ?
Answer:
Total number of problems assigned to Geeta = 30
Calculate the number of problems she completed:
\[ \text{Completed problems} = \frac{2}{3}\text{ of } 30 = \frac{2}{3} \times 30 = 20 \]
Find the number of incomplete problems:
\[ \text{Remaining problems} = 30 - 20 = 10 \]
In simple words: Geeta completed 20 of her 30 problems. Subtracting the completed tasks from the total shows that she has 10 problems left to finish.

Exam Tip: Read word problems carefully to identify whether they ask for the amount completed or the amount remaining, as this is a common area for silly mistakes.

Question 8. A picture was marked at Rs. 90. It was sold at \(\frac{3}{4}\) of its marked price. What was the sale price ?
Answer:
The marked price of the picture is Rs. 90.
Find the actual sale price:
\[ \text{Sale price} = \frac{3}{4}\text{ of Rs. } 90 = \text{Rs. } \frac{3}{4} \times 90 \]
Multiply and simplify:
\[ = \text{Rs. } \frac{270}{4} = \text{Rs. } 67\frac{1}{2} = \text{Rs. } 67.50 \]
In simple words: The item's marked price is Rs. 90, and it sold for three-quarters of that value, which works out to Rs. 67.50.

Exam Tip: When expressing financial values as decimals, always include two decimal places (such as Rs. 67.50 instead of Rs. 67.5) to keep standard currency formatting.

Question 9. Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10\(\frac{1}{2}\) kg ?
Answer:
Total number of parcels shipped = 15
Weight of each individual parcel = \( 10\frac{1}{2}\text{ kg} = \frac{21}{2}\text{ kg} \)
Calculate the total weight of all parcels combined:
\[ \text{Total weight} = 15 \times \frac{21}{2}\text{ kg} \]
\[ = \frac{315}{2}\text{ kg} = 157\frac{1}{2}\text{ kg} = 157.5\text{ kg} \]
In simple words: Multiplied the 15 parcels by the weight of a single parcel (10.5 kg) to get the total combined weight of 157.5 kg.

Exam Tip: Word problems with mixed numbers are often easiest to solve by first converting the weight into an improper fraction before multiplying.

Question 10. A rope is \( 25\frac{1}{2} \) m long. How many pieces , \( 1\frac{1}{2} \) each of length can be cut out from it?
Answer:
The total length of the rope is \( 25\frac{1}{2}\text{ m} \), which is equal to \( \frac{51}{2}\text{ m} \).
The length of each piece to be cut is \( 1\frac{1}{2}\text{ m} \), which is equal to \( \frac{3}{2}\text{ m} \).
To find the number of pieces we can get, we divide the total length by the length of each piece:
\( \text{Number of pieces} = \frac{51}{2} \div \frac{3}{2} \)
\( \implies \text{Number of pieces} = \frac{51}{2} \times \frac{2}{3} \)
\( \implies \text{Number of pieces} = 17 \)
Therefore, 17 pieces of rope can be cut.
In simple words: We divide the total length of the rope by the length of one piece to find that 17 pieces can be cut.

Exam Tip: When dividing fractions, always remember to multiply by the reciprocal of the second fraction and simplify before calculating the final value.

Question 11. The heights of two vertical poles, above the earth’s surface, are \( 14\frac{1}{4} \) m and \( 22\frac{1}{3} \) respectively. How much higher is the second pole as compared with the height of the first pole ?
Answer:
The height of the first vertical pole is \( 14\frac{1}{4}\text{ m} \), which can be written as \( \frac{57}{4}\text{ m} \).
The height of the second vertical pole is \( 22\frac{1}{3}\text{ m} \), which can be written as \( \frac{67}{3}\text{ m} \).
To find how much higher the second pole is, we subtract the first pole's height from the second pole's height:
\( \text{Difference} = 22\frac{1}{3} - 14\frac{1}{4} \)
\( \implies \text{Difference} = \frac{67}{3} - \frac{57}{4} \)
The LCM of 3 and 4 is 12. Converting both to like fractions, we get:
\( \implies \text{Difference} = \frac{268 - 171}{12} \)
\( \implies \text{Difference} = \frac{97}{12}\text{ m} \)
\( \implies \text{Difference} = 8\frac{1}{12}\text{ m} \)
So, the second pole is \( 8\frac{1}{12}\text{ m} \) higher than the first pole.
In simple words: We subtract the height of the shorter pole from the taller pole to see how much taller it is.

Exam Tip: When subtracting mixed fractions, convert them to improper fractions first and find the common denominator to avoid simple subtraction errors.

Question 12. Vijay weighed \( 65\frac{1}{2} \) kg. He gained \( 1\frac{2}{5} \) kg during the first week, \( 1\frac{1}{4} \) kg during the second week, but lost \( \frac{5}{16} \) kg during the third week. What was his weight after the third week ?
Answer:
Vijay's initial weight was \( 65\frac{1}{2}\text{ kg} \), which is \( \frac{131}{2}\text{ kg} \).
The weight gained in the first week was \( 1\frac{2}{5}\text{ kg} \), which is \( \frac{7}{5}\text{ kg} \).
The weight gained in the second week was \( 1\frac{1}{4}\text{ kg} \), which is \( \frac{5}{4}\text{ kg} \).
The weight lost in the third week was \( \frac{5}{16}\text{ kg} \).
To calculate his weight after the third week, we add the gained weights and subtract the lost weight:
\( \text{Final weight} = 65\frac{1}{2} + 1\frac{2}{5} + 1\frac{1}{4} - \frac{5}{16} \)
\( \implies \text{Final weight} = \frac{131}{2} + \frac{7}{5} + \frac{5}{4} - \frac{5}{16} \)
Finding the LCM of 2, 5, 4, and 16, which is 80, we get:
\( \implies \text{Final weight} = \frac{131 \times 40 + 7 \times 16 + 5 \times 20 - 5 \times 5}{80} \)
\( \implies \text{Final weight} = \frac{5240 + 112 + 100 - 25}{80} \)
\( \implies \text{Final weight} = \frac{5452 - 25}{80} \)
\( \implies \text{Final weight} = \frac{5427}{80}\text{ kg} \)
\( \implies \text{Final weight} = 67\frac{67}{80}\text{ kg} \)
So, Vijay's weight after the third week was \( 67\frac{67}{80}\text{ kg} \).
In simple words: We find his final weight by starting with his original weight, adding what he gained, and subtracting what he lost.

Exam Tip: For expressions involving both addition and subtraction of multiple fractions, find a single common denominator (LCM) for all terms to solve the problem in fewer steps.

Question 13. A man spends \( \frac{2}{5} \) of his salary on food and \( \frac{3}{10} \) on house rent, electricity, etc. What fraction of his salary is still left with him ?
Answer:
Let the man's total salary be represented as 1.
The fraction of salary spent on food is \( \frac{2}{5} \).
The fraction of salary spent on rent, electricity, and other things is \( \frac{3}{10} \).
First, we find the total fraction spent by adding these two parts:
\( \text{Total spent} = \frac{2}{5} + \frac{3}{10} \)
The LCM of 5 and 10 is 10.
\( \implies \text{Total spent} = \frac{4 + 3}{10} = \frac{7}{10} \)
Now, we subtract the total spent from the whole salary (1) to find the remaining fraction:
\( \text{Fraction left} = 1 - \frac{7}{10} \)
\( \implies \text{Fraction left} = \frac{10 - 7}{10} = \frac{3}{10} \)
Thus, the fraction of his salary left is \( \frac{3}{10} \).
In simple words: We add up the fractions spent on food and rent, then subtract that total from 1 to find the leftover portion.

Exam Tip: When a problem mentions spending fractions of the total salary (and not the "remaining" salary), you can add the fractions directly before subtracting from 1.

Question 14. A man spends \( \frac{2}{5} \) of his salary on food and \( \frac{3}{10} \) of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him ?
Answer:
Let the man's total salary be represented as Rs. 1.
The fraction of salary spent on food is \( \frac{2}{5} \).
The fraction of salary left after this spending is:
\( \text{Remaining salary} = 1 - \frac{2}{5} = \frac{3}{5} \).
He spends \( \frac{3}{10} \) of this remaining salary on house rent and electricity:
\( \text{Spent on rent} = \frac{3}{10} \text{ of } \frac{3}{5} \)
\( \implies \text{Spent on rent} = \frac{3}{10} \times \frac{3}{5} = \frac{9}{50} \)
To find the final fraction left, we subtract this second expenditure from the remaining salary:
\( \text{Fraction left} = \frac{3}{5} - \frac{9}{50} \)
The LCM of 5 and 50 is 50.
\( \implies \text{Fraction left} = \frac{30 - 9}{50} = \frac{21}{50} \)
Therefore, the fraction of salary still left with him is \( \frac{21}{50} \).
In simple words: We first find the money left after spending on food. Then, we find \( \frac{3}{10} \) of that amount and subtract it to get the final leftover portion.

Exam Tip: Be very careful with the word "remaining" in word problems. It means you must calculate the intermediate leftover amount first before multiplying by the next fraction.

Question 15. Shyam bought a refrigerator for Rs. 5000. He paid \( \frac{1}{10} \) of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month ?
Answer:
The total price of the refrigerator is Rs. 5000.
The cash payment made by Shyam is \( \frac{1}{10} \) of the price:
\( \text{Cash payment} = \frac{1}{10} \times 5000 = \text{Rs. } 500 \)
The remaining amount to be paid is:
\( \text{Balance amount} = 5000 - 500 = \text{Rs. } 4500 \)
This remaining balance is paid in 12 equal monthly instalments:
\( \text{Monthly instalment} = 4500 \div 12 \)
\( \implies \text{Monthly instalment} = \text{Rs. } 375 \)
Thus, Shyam has to pay Rs. 375 each month.
In simple words: Shyam pays Rs. 500 first. The leftover Rs. 4500 is split into 12 equal monthly parts of Rs. 375 each.

Exam Tip: Always subtract the initial cash payment from the total cost before dividing the remaining balance into equal monthly payments.

Question 16. A lamp post has half of its length in mud, and \( \frac{1}{3} \) of its length in water.
(i) What fraction of its length is above the water ?
(ii) If \( 3\frac{1}{3} \) m of the lamp post is above the water, find the whole length of the lamp post.
Answer:
(i) Let the total length of the lamp post be 1.
The fraction of the post in mud is \( \frac{1}{2} \).
The fraction of the post in water is \( \frac{1}{3} \).
The sum of the parts in mud and water is:
\( \text{Covered fraction} = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6} \)
The fraction of the lamp post above the water is:
\( \text{Fraction above water} = 1 - \frac{5}{6} = \frac{1}{6} \)
So, \( \frac{1}{6} \) of the length is above the water.

(ii) We are given that \( 3\frac{1}{3}\text{ m} \), which is \( \frac{10}{3}\text{ m} \), is above the water.
Let the total length of the post be \( L \). Since \( \frac{1}{6} \) of the total length is above water, we write:
\( \frac{1}{6} \text{ of } L = \frac{10}{3}\text{ m} \)
\( \implies L = \frac{10}{3} \times 6 \)
\( \implies L = 20\text{ m} \)
Therefore, the whole length of the lamp post is 20 m.
In simple words: (i) We find that \( \frac{5}{6} \) of the post is buried in mud or water, leaving \( \frac{1}{6} \) above water. (ii) If this \( \frac{1}{6} \) is \( 3\frac{1}{3}\text{ m} \), the total length is 20 m.

Exam Tip: Treat the total length as 1 to easily determine the fractional part that is exposed. Use the reciprocal of this fraction to solve for the actual total length.

Question 17. I spent \( \frac{3}{5} \) of my savings and still have Rs. 2,000 left. What were my savings ?
Answer:
Let the total savings be represented as 1.
The fraction of savings spent is \( \frac{3}{5} \).
The fraction of savings remaining is:
\( \text{Fraction left} = 1 - \frac{3}{5} = \frac{2}{5} \).
We are given that the actual amount left is Rs. 2000, which means:
\( \frac{2}{5} \text{ of total savings} = \text{Rs. } 2000 \)
\( \implies \text{Total savings} = 2000 \times \frac{5}{2} \)
\( \implies \text{Total savings} = 1000 \times 5 = \text{Rs. } 5000 \)
Thus, the total savings were Rs. 5000.
In simple words: If \( \frac{3}{5} \) of the savings are spent, then \( \frac{2}{5} \) is left. Since \( \frac{2}{5} \) is Rs. 2000, the total savings must be Rs. 5000.

Exam Tip: When given a remaining value, find the remaining fraction first. Set this fraction equal to the given value to find the starting total.

Question 18. In a school, \( \frac{4}{5} \) of the children are boys. If the number of girls is 200, find the number of boys.
Answer:
Let the total number of children in the school be represented as 1.
The fraction of boys is \( \frac{4}{5} \).
The fraction of girls is:
\( \text{Fraction of girls} = 1 - \frac{4}{5} = \frac{1}{5} \).
The actual number of girls is 200, so:
\( \frac{1}{5} \text{ of total children} = 200 \)
\( \implies \text{Total children} = 200 \times 5 = 1000 \)
Now, we calculate the number of boys:
\( \text{Number of boys} = \frac{4}{5} \text{ of } 1000 \)
\( \implies \text{Number of boys} = \frac{4}{5} \times 1000 = 800 \)
So, there are 800 boys in the school.
In simple words: Since \( \frac{4}{5} \) are boys, the remaining \( \frac{1}{5} \) are girls. If \( \frac{1}{5} \) is 200, there are 1000 kids in total, meaning 800 of them are boys.

Exam Tip: You can also find the boys directly by noting that the number of boys is 4 times the number of girls since the ratio of boys to girls is 4 to 1.

Question 19. If \( \frac{4}{5} \) of an estate is worth Rs. 42,000, find the worth of whole estate. Also, find the value of \( \frac{3}{7} \) of it.
Answer:
We are given that \( \frac{4}{5} \) of the estate has a value of Rs. 42000.
To find the total value of the estate, we divide the given value by the fraction:
\( \text{Total value} = 42000 \div \frac{4}{5} \)
\( \implies \text{Total value} = 42000 \times \frac{5}{4} \)
\( \implies \text{Total value} = 10500 \times 5 = \text{Rs. } 52500 \)
Now, we find the value of \( \frac{3}{7} \) of this total estate value:
\( \text{Value of } \frac{3}{7} \text{ of estate} = \frac{3}{7} \times 52500 \)
\( \implies \text{Value of } \frac{3}{7} \text{ of estate} = 3 \times 7500 = \text{Rs. } 22500 \)
Thus, the complete estate is worth Rs. 52,500, and \( \frac{3}{7} \) of it is worth Rs. 22,500.
In simple words: We find the full price of the estate by multiplying Rs. 42000 by \( \frac{5}{4} \). Then, we multiply that full price by \( \frac{3}{7} \) to get the final answer.

Exam Tip: Be sure to write down both parts of the final answer clearly, labeling the total worth of the estate and the value of the specific fraction.

Question 20. After going \( \frac{3}{4} \) of my journey, I find that I have covered 16 km. How much Journey is still left ?
Answer:
Let the total journey length be represented as \( J \).
The fraction of the journey covered is \( \frac{3}{4} \), which is equal to 16 km:
\( \frac{3}{4} \text{ of } J = 16\text{ km} \)
\( \implies J = 16 \times \frac{4}{3} = \frac{64}{3}\text{ km} \)
The journey distance remaining is the total distance minus the distance covered:
\( \text{Remaining journey} = \frac{64}{3} - 16 \)
\( \implies \text{Remaining journey} = \frac{64 - 48}{3} = \frac{16}{3}\text{ km} \)
\( \implies \text{Remaining journey} = 5\frac{1}{3}\text{ km} \)
So, there is \( 5\frac{1}{3}\text{ km} \) of the journey still left.
In simple words: Since \( \frac{3}{4} \) of the journey is 16 km, the total journey is \( \frac{64}{3}\text{ km} \). Subtracting the 16 km already covered leaves \( 5\frac{1}{3}\text{ km} \) left to travel.

Exam Tip: Alternatively, since \( \frac{3}{4} \) of the journey is completed, \( \frac{1}{4} \) is left. You can find the remaining journey directly by calculating \( \frac{1}{4} \) of the total journey using the proportion of the covered 16 km.

Question 21. When Krishna travelled 25 km, he found that \( \frac{3}{5} \) of his journey was still left. What was the length of the whole journey.
Answer:
Let the total journey length be represented as 1.
The fraction of the journey still left is \( \frac{3}{5} \).
Therefore, the fraction of the journey already travelled is:
\( \text{Fraction travelled} = 1 - \frac{3}{5} = \frac{2}{5} \).
Since Krishna has actually travelled 25 km, we set this fraction equal to the distance:
\( \frac{2}{5} \text{ of total journey} = 25\text{ km} \)
\( \implies \text{Total journey} = 25 \times \frac{5}{2} \)
\( \implies \text{Total journey} = \frac{125}{2}\text{ km} = 62\frac{1}{2}\text{ km} \).
Thus, the length of the whole journey was \( 62\frac{1}{2}\text{ km} \).
In simple words: Since \( \frac{3}{5} \) of the trip is left, \( \frac{2}{5} \) of it has been completed. If \( \frac{2}{5} \) is 25 km, then the total trip is \( 62\frac{1}{2}\text{ km} \).

Exam Tip: Read carefully to distinguish whether the given distance is the part completed or the part left. Here, the distance is the completed part, which corresponds to the fraction \( 1 - \frac{3}{5} = \frac{2}{5} \).

Question 22. From a piece of land, one-third is bought by Rajesh and one-third of remaining is bought by Manoj. If 600 m² land is still left unsold, find the total area of the piece of land.
Answer:
Let the total area of the land be represented as 1.
The portion bought by Rajesh is \( \frac{1}{3} \).
The remaining portion of the land is:
\( \text{Remaining land} = 1 - \frac{1}{3} = \frac{2}{3} \).
Manoj buys one-third of this remaining land:
\( \text{Portion bought by Manoj} = \frac{1}{3} \text{ of } \frac{2}{3} = \frac{2}{9} \).
The unsold portion of the land is:
\( \text{Unsold land} = \frac{2}{3} - \frac{2}{9} \)
The LCM of 3 and 9 is 9.
\( \implies \text{Unsold land} = \frac{6 - 2}{9} = \frac{4}{9} \).
Since the area of the unsold land is \( 600\text{ m}^2 \), we set this equal to the unsold fraction:
\( \frac{4}{9} \text{ of total land} = 600\text{ m}^2 \)
\( \implies \text{Total area} = 600 \times \frac{9}{4} \)
\( \implies \text{Total area} = 150 \times 9 = 1350\text{ m}^2 \).
Thus, the total area of the land is \( 1350\text{ m}^2 \).
In simple words: Rajesh buys \( \frac{1}{3} \), leaving \( \frac{2}{3} \). Manoj buys \( \frac{1}{3} \) of that, which is \( \frac{2}{9} \). This leaves \( \frac{4}{9} \) of the land unsold, which equals \( 600\text{ m}^2 \), giving a total of \( 1350\text{ m}^2 \).

Exam Tip: Be sure to keep intermediate calculations as fractions rather than converting them to decimals, as fractions preserve exact numbers for the final division step.

Question 23. A boy spent \( \frac{3}{5} \) of his money on buying 1 cloth and \( \frac{1}{4} \) of the remaining on buying shoes. If initially he has Rs. 2,400; how much did he spend on shoes?
Answer:
The initial amount of money the boy has is Rs. 2400.
The amount spent on buying the cloth is \( \frac{3}{5} \) of his total money:
\( \text{Spent on cloth} = \frac{3}{5} \times 2400 = 3 \times 480 = \text{Rs. } 1440 \).
The remaining money after this purchase is:
\( \text{Remaining money} = 2400 - 1440 = \text{Rs. } 960 \).
The amount spent on shoes is \( \frac{1}{4} \) of this remaining money:
\( \text{Spent on shoes} = \frac{1}{4} \times 960 = \text{Rs. } 240 \).
Thus, he spent Rs. 240 on shoes.
In simple words: The boy spends Rs. 1440 on cloth. From the remaining Rs. 960, he spends \( \frac{1}{4} \), which is Rs. 240, on shoes.

Exam Tip: Remember to calculate the remaining balance after the first purchase before taking the next fraction to find the amount spent on shoes.

Question 24. A boy spent \( \frac{3}{5} \) of his money on buying cloth and \( \frac{1}{4} \) of his money on buying shoes. If initially he has Rs. 2,400; how much did he spend on shoes?
Answer:
The initial amount of money the boy has is Rs. 2400.
The fraction of money spent on buying cloth is \( \frac{3}{5} \).
The fraction of money spent on buying shoes is \( \frac{1}{4} \) of his total money.
To find the amount spent on shoes, we calculate \( \frac{1}{4} \) of the initial Rs. 2400:
\( \text{Spent on shoes} = \frac{1}{4} \times 2400 = \text{Rs. } 600 \).
Thus, the boy spent Rs. 600 on shoes.
In simple words: Since the boy spent \( \frac{1}{4} \) of his total initial money on shoes, we calculate \( \frac{1}{4} \) of Rs. 2400, which gives Rs. 600.

Exam Tip: Pay close attention to whether the second fraction applies to the "remaining" money or the "total" money. In this question, it is \( \frac{1}{4} \) of the total money, so we calculate it directly from Rs. 2400.