Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 14 Fractions

Fractions

Important Points

I. Fraction:

A fraction is a quantity which expresses a part of the whole.

Fraction = \( \frac{\text{Numerator}}{\text{Denominator}} \)

Type of Fractions:

1. Proper Fraction: A fraction, whose numerator is less than its denominator, is called a proper fraction. e.g., \( \frac{3}{5} \), \( \frac{4}{6} \) etc.

2. Improper Fraction: A fraction, whose numerator is greater than or equal to its denominator, is called an improper fraction. e.g., \( \frac{8}{6} \), \( \frac{24}{13} \), \( \frac{2}{2} \), \( \frac{3}{3} \), \( \frac{6}{6} \) etc.

3. Mixed Fraction: A mixed fraction consists of two parts:
(i) an integer and (ii) a proper fraction
e.g., \( 5 \frac{2}{3} \) is a mixed fraction, consisting of an integer (5) and a proper fraction \( (\frac{2}{3}) \).

4. Like and Unlike Fractions: Fraction having the same denominator but different numerators are called like fractions e.g., \( \frac{2}{5} \), \( \frac{1}{5} \), \( \frac{3}{5} \), \( \frac{7}{5} \) etc. are like fractions.
If denominator of the given fractions are not same, the fractions are called unlike fractions e.g., \( \frac{1}{4} \), \( \frac{3}{8} \), \( \frac{6}{9} \), \( \frac{7}{10} \) etc.

5. Equivalent Fractions: If two or more fractions have the same value, they are called equivalent or equal fractions.
e.g., \( \frac{1}{2} \), \( \frac{2}{4} \), \( \frac{6}{12} \), \( \frac{8}{16} \) etc. are equivalent fractions as \( \frac{1}{2} = \frac{2}{4} = \frac{6}{12} = \frac{8}{16} \).

Convertion Of Fractions:

(i) Mixed Fraction into an Improper Fraction:- Multiply the integral part by the denominator and to this product add the numerator e.g., \( 2 \frac{5}{15} \)
the required improper fraction = \( \frac{2 \times 15 + 5}{15} = \frac{35}{15} \)

(ii) Improper Fraction into Mixed Fraction:- Divide numerator by the denominator. The quotient of this division is the integral part and the remainder obtained is numerator of the required mixed fraction.
For example: \( \frac{23}{3} = \text{Quotient} \frac{\text{Remainder}}{\text{Numerator}} = 7 \frac{2}{3} \)

(iii) Unlike Fraction into Like Fractions:
(1) Find L.C.M. of the denominators of all given fractions.
(2) Divide L.C.M. by the denominator and multiply the quotient to numerator and denominator of fraction.
e.g., \( \frac{2}{7} \), \( \frac{3}{5} \) and \( \frac{1}{3} \)
L.C.M. of denominator 7, 5, 3 = 105
Now, in \( \frac{2}{7} \) dividing L.C.M. by 7 Quotient = 15
\( \implies \frac{2 \times 15}{7 \times 15} = \frac{30}{105} \)
\( \frac{3}{5} \) dividing L.C.M. by 5 Quotient = 21
\( \implies \frac{3 \times 21}{5 \times 21} = \frac{63}{105} \)
\( \frac{1}{3} \) dividing L.C.M. by 3 Quotient = 35
\( \implies \frac{1 \times 35}{3 \times 35} = \frac{35}{105} \)
\( \therefore \frac{2}{7} \), \( \frac{3}{5} \) and \( \frac{1}{3} = \frac{30}{105} \), \( \frac{63}{105} \) and \( \frac{35}{105} \)

Exercise 14(A)

Question 1. For each expression, given below, write a fraction :
(i) 2 out of 7 = ………..
(ii) 5 out of 17 = ………..
(iii) three-fifths = ………
Answer:
(i) 2 out of 7 = \( \frac{2}{7} \)
(ii) 5 out of 17 = \( \frac{5}{17} \)
(iii) three-fifths = \( \frac{3}{5} \)
In simple words: A fraction is a way to show how many parts of something you have. The top number (numerator) is the number of parts you're talking about, and the bottom number (denominator) is the total number of parts.

📝 Teacher's Note: Use a pizza or a chocolate bar analogy to explain "out of". If a pizza has 7 slices and you take 2, you have 2 out of 7, written as \( \frac{2}{7} \).

🎯 Exam Tip: Always place the "total" amount in the denominator (bottom) and the "part" in the numerator (top).

Question 2. Fill in the blanks :
(i) \( \frac{5}{8} \) is ………..fraction.
(ii) \( \frac{8}{5} \) is ……….. fraction.
(iii) \( \frac{-15}{-15} \) is ……….. fraction.
(iv) The value of \( \frac{5}{5} = \) ……….. .
(v) The value of \( \frac{5}{-5} = \) ……….. .
(vi) \( 3 \frac{3}{10} \) is ……….. fraction.
(vii) \( \frac{2}{15} \) and \( \frac{7}{15} \) are ………..fractions.
(viii) \( \frac{23}{12} \) and \( \frac{23}{15} \) are ……….. fractions.
(ix) \( \frac{6}{15} \) and \( \frac{28}{70} \) are ……….. fractions.
(x) \( \frac{8}{24} \) and \( \frac{8}{32} \) are not ……….. fractions.
(xi) \( 3 \frac{2}{13} = \frac{3 \times 13 + \dots}{13} = \dots \)
(xii) \( -4 \frac{3}{5} = \dots = \dots \)
Answer:
(i) Proper
(ii) Improper
(iii) Improper
(iv) 1
(v) -1
(vi) Mixed
(vii) Like
(viii) Unlike fraction
(ix) Equal fraction
(x) Like
(xi) \( 3 \frac{2}{13} = \frac{3 \times 13 + 2}{13} = \frac{41}{13} \)
(xii) \( -4 \frac{3}{5} = - \frac{(4 \times 5 + 3)}{5} = - \frac{23}{5} \)
In simple words: This exercise checks if you know the names of different types of fractions. Proper fractions are "normal" (top smaller), improper are "top-heavy" (top bigger), and mixed fractions have a whole number and a fraction together.

📝 Teacher's Note: Remind students that "Like fractions" must have the exact same denominator. Also, emphasize that any number divided by itself is 1.

🎯 Exam Tip: Pay attention to negative signs in mixed fractions. The negative applies to the whole value, so convert to improper first, then add the negative sign.

Question 3. From the following fractions, separate :
(i) Proper fractions
(ii) Improper fractions :
\( \frac{2}{9}, \frac{4}{3}, \frac{7}{15}, \frac{11}{20}, \frac{20}{11}, \frac{18}{23} \) and \( \frac{27}{35} \).
Answer:
We know that proper fraction is a fraction whose numerator is less than its denominator and improper fraction is the fraction whose numerator is greater than its denominator :
(i) Proper fractions: \( \frac{2}{9}, \frac{7}{15}, \frac{11}{20}, \frac{18}{23} \) and \( \frac{27}{35} \)
(ii) Improper fractions: \( \frac{4}{3}, \frac{20}{11} \)
In simple words: Look at the top and bottom numbers. If the top is smaller, it's proper. If the top is bigger or equal, it's improper.

📝 Teacher's Note: Use the "weight" analogy. A proper fraction is stable because the small number is on top. An improper fraction is "top-heavy" and can be turned into a mixed fraction.

🎯 Exam Tip: Just check the numerator vs the denominator. Don't let the size of the numbers confuse you; just compare them.

Question 4. Change the following mixed fractions to improper fractions :
(i) \( 2 \frac{1}{5} \)
(ii) \( 3 \frac{1}{4} \)
(iii) \( 7 \frac{1}{8} \)
(iv) \( 2 \frac{1}{11} \)
Answer:
(i) \( 2 \frac{1}{5} = \frac{2 \times 5 + 1}{5} = \frac{10 + 1}{5} = \frac{11}{5} \)
(ii) \( 3 \frac{1}{4} = \frac{3 \times 4 + 1}{4} = \frac{12 + 1}{4} = \frac{13}{4} \)
(iii) \( 7 \frac{1}{8} = \frac{7 \times 8 + 1}{8} = \frac{56 + 1}{8} = \frac{57}{8} \)
(iv) \( 2 \frac{1}{11} = \frac{2 \times 11 + 1}{11} = \frac{22 + 1}{11} = \frac{23}{11} \)
In simple words: To change a mixed fraction, multiply the big whole number by the bottom number, add the top number, and keep the bottom number the same.

📝 Teacher's Note: Practice the "Circle Rule": Start at the denominator, multiply by the whole number, add the numerator, and put it all over the same denominator.

🎯 Exam Tip: Double-check your multiplication (like 7 × 8) to avoid silly calculation errors.

Question 5. Change the following improper fractions to mixed fractions :
(i) \( \frac{100}{17} \)
(ii) \( \frac{81}{11} \)
(iii) \( \frac{209}{7} \)
(iv) \( \frac{113}{15} \)
Answer:
(i) \( \frac{100}{17} = 5 \frac{15}{17} \)
(ii) \( \frac{81}{11} = 7 \frac{4}{11} \)
(iii) \( \frac{209}{7} = 29 \frac{6}{7} \)
(iv) \( \frac{113}{15} = 7 \frac{8}{15} \)
In simple words: Divide the top number by the bottom number. The answer is the whole number, and the remainder is the new top number.

📝 Teacher's Note: Show long division. The quotient becomes the whole number, the remainder becomes the numerator, and the divisor stays the denominator.

🎯 Exam Tip: Verify by converting your mixed fraction back to improper. It should match the original question.

Question 6. Change the following groups of fractions to like fractions :
(i) \( \frac{1}{3}, \frac{2}{5}, \frac{3}{4}, \frac{1}{6} \)
(ii) \( \frac{5}{6}, \frac{7}{8}, \frac{11}{12}, \frac{3}{10} \)
(iii) \( \frac{2}{7}, \frac{7}{8}, \frac{5}{14}, \frac{9}{16} \)
Answer:
(i) \( \frac{1}{3}, \frac{2}{5}, \frac{3}{4}, \frac{1}{6} \)
L.C.M. of denominators 3, 5, 4, 6 = 60

= \( 2 \times 3 \times 1 \times 5 \times 2 \times 1 = 60 \)
Now, \( \frac{1}{3} = \frac{1 \times 20}{3 \times 20} = \frac{20}{60} \);
\( \frac{2}{5} = \frac{2 \times 12}{5 \times 12} = \frac{24}{60} \); \( \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \)
\( \frac{1}{6} = \frac{1 \times 10}{6 \times 10} = \frac{10}{60} \)
\( \therefore \frac{1}{3}, \frac{2}{5}, \frac{3}{4} \) and \( \frac{1}{6} = \frac{20}{60}, \frac{24}{60}, \frac{45}{60}, \frac{10}{60} \)

(ii) \( \frac{5}{6}, \frac{7}{8}, \frac{11}{12}, \frac{3}{10} \)
L.C.M. of denominators 6, 8, 12, 10 = 120

= \( 2 \times 2 \times 3 \times 2 \times 5 = 120 \)
Now, \( \frac{5}{6} = \frac{5 \times 20}{6 \times 20} = \frac{100}{120} \);
\( \frac{7}{8} = \frac{7 \times 15}{8 \times 15} = \frac{105}{120} \); \( \frac{11}{12} = \frac{11 \times 10}{12 \times 10} = \frac{110}{120} \)
\( \frac{3}{10} = \frac{3 \times 12}{10 \times 12} = \frac{36}{120} \)
\( \therefore \frac{5}{6}, \frac{7}{8}, \frac{11}{12}, \frac{3}{10} = \frac{100}{120}, \frac{105}{120}, \frac{110}{120}, \frac{36}{120} \)

(iii) \( \frac{2}{7}, \frac{7}{8}, \frac{5}{14}, \frac{9}{16} \)
L.C.M. of denominators 7, 8, 14, 16 = 112

= \( 2 \times 7 \times 4 \times 2 = 112 \)
Now, \( \frac{2}{7} = \frac{2 \times 16}{7 \times 16} = \frac{32}{112} \); \( \frac{7}{8} = \frac{7 \times 14}{8 \times 14} = \frac{98}{112} \);
\( \frac{5}{14} = \frac{5 \times 8}{14 \times 8} = \frac{40}{112} \); \( \frac{9}{16} = \frac{9 \times 7}{16 \times 7} = \frac{63}{112} \)
\( \therefore \frac{2}{7}, \frac{7}{8}, \frac{5}{14}, \frac{9}{16} = \frac{32}{112}, \frac{98}{112}, \frac{40}{112}, \frac{63}{112} \)
In simple words: To make unlike fractions look the same, find a common denominator using LCM. Then, multiply each fraction so their bottom numbers all match that LCM.

📝 Teacher's Note: The division method is the most efficient way to find LCM for multiple numbers. Ensure students align their work clearly to avoid missing any prime factors.

🎯 Exam Tip: After converting, check if the denominators are all the same. If even one is different, re-calculate your LCM or the multiplication step.

Exercise 14(B)

Question 1. Reduce the given fractions to their lowest terms :
(i) \( \frac{8}{10} \)
(ii) \( \frac{50}{75} \)
(iii) \( \frac{18}{81} \)
(iv) \( \frac{40}{120} \)
(v) \( \frac{105}{70} \)
Answer:
(i) \( \frac{8}{10} = \frac{8 \div 2}{10 \div 2} = \frac{4}{5} \)
(ii) \( \frac{50}{75} = \frac{50 \div 25}{75 \div 25} = \frac{2}{3} \)
(iii) \( \frac{18}{81} = \frac{18 \div 9}{81 \div 9} = \frac{2}{9} \)
(iv) \( \frac{40}{120} = \frac{40 \div 40}{120 \div 40} = \frac{1}{3} \)
(v) \( \frac{105}{70} = \frac{105 \div 35}{70 \div 35} = \frac{3}{2} \)
In simple words: To simplify a fraction, divide both the top and bottom numbers by the same biggest number that goes into both of them perfectly.

📝 Teacher's Note: Encourage students to find the HCF (Highest Common Factor) of the numerator and denominator to simplify in a single step.

🎯 Exam Tip: Keep simplifying until you can't find any more numbers that divide into both the numerator and denominator. A common trap is stopping too early.

Question 2. State, whether true or false ?
(i) \( \frac{2}{5} = \frac{10}{15} \)
(ii) \( \frac{35}{42} = \frac{5}{6} \)
(iii) \( \frac{5}{4} = \frac{4}{5} \)
(iv) \( \frac{7}{9} = 1 \frac{1}{7} \)
(v) \( \frac{9}{7} = 1 \frac{1}{7} \)
Answer:
(i) \( \frac{2}{5} = \frac{10}{15} \)
\( \frac{10}{15} = \frac{10 \div 5}{15 \div 5} = \frac{2}{3} \)
\( \therefore \frac{2}{5} \neq \frac{2}{3} \), False

(ii) \( \frac{35}{42} = \frac{5}{6} \)
\( \frac{35}{42} = \frac{35 \div 7}{42 \div 7} = \frac{5}{6} \)
\( \therefore \frac{5}{6} = \frac{5}{6} \), True

(iii) \( \frac{5}{4} = \frac{4}{5} \), False

(iv) \( \frac{7}{9} = 1 \frac{1}{7} \)
Now, \( 1 \frac{1}{7} = \frac{7 \times 1 + 1}{7} = \frac{8}{7} \)
\( \frac{7}{9} \neq \frac{8}{7} \), False

(v) \( \frac{9}{7} = 1 \frac{1}{7} \)
Now, \( 1 \frac{1}{7} = \frac{7 \times 1 + 1}{7} = \frac{8}{7} \)
\( \frac{9}{7} \neq \frac{8}{7} \), False
In simple words: Simplify both sides of the equals sign. If the final fractions are the same, it's True. If they are different, it's False.

📝 Teacher's Note: This is a great way to practice equivalent fractions. Remind students that \( \frac{5}{4} \) (improper) can never equal \( \frac{4}{5} \) (proper).

🎯 Exam Tip: Always convert mixed numbers to improper fractions before comparing them to other fractions.

Question 3. Which fraction is greater ?
(i) \( \frac{3}{5} \) or \( \frac{2}{3} \)
(ii) \( \frac{5}{9} \) or \( \frac{3}{4} \)
(iii) \( \frac{11}{14} \) or \( \frac{26}{35} \)
Answer:
(i) \( \frac{3}{5} \) or \( \frac{2}{3} \)
L.C.M. of 5, 3 = 15
Now, \( \frac{3}{5} = \frac{3 \times 3}{5 \times 3} = \frac{9}{15} \); \( \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \)
\( \frac{10}{15} > \frac{9}{15} \implies \frac{2}{3} > \frac{3}{5} \)

(ii) \( \frac{5}{9} \) or \( \frac{3}{4} \)
Converting in like fraction,
\( \frac{5 \times 4}{9 \times 4} = \frac{20}{36} \); \( \frac{3 \times 9}{4 \times 9} = \frac{27}{36} \)
\( \frac{27}{36} > \frac{20}{36} \implies \frac{3}{4} > \frac{5}{9} \)

(iii) \( \frac{11}{14} \) or \( \frac{26}{35} \)
Converting in like fraction,
\( \frac{11 \times 5}{14 \times 5} = \frac{55}{70} \); \( \frac{26 \times 2}{35 \times 2} = \frac{52}{70} \)
\( \frac{55}{70} > \frac{52}{70} \implies \frac{11}{14} > \frac{26}{35} \)
In simple words: To see which fraction is bigger, change them so they have the same bottom number. Then, just compare the top numbers.

📝 Teacher's Note: Cross-multiplication is another quick method for comparing two fractions. Multiply \( 3 \times 3 = 9 \) and \( 5 \times 2 = 10 \). Since 10 is bigger, \( \frac{2}{3} \) is bigger.

🎯 Exam Tip: Write down the final answer in terms of the original fractions given in the question, not just the common-denominator versions.

Question 4. Which fraction is smaller?
(i) \( \frac{3}{8} \) or \( \frac{4}{5} \)
(ii) \( \frac{8}{15} \) or \( \frac{4}{7} \)
(iii) \( \frac{7}{26} \) or \( \frac{10}{39} \)
Answer:
(i) \( \frac{3}{8} \) or \( \frac{4}{5} \)
Converting in like fraction
\( \frac{3 \times 5}{8 \times 5} = \frac{15}{40} \); \( \frac{4 \times 8}{5 \times 8} = \frac{32}{40} \)
\( \frac{15}{40} < \frac{32}{40} \implies \frac{3}{8} < \frac{4}{5} \)

(ii) \( \frac{8}{15} \) or \( \frac{4}{7} \)
Converting into like fraction
\( \frac{8 \times 7}{15 \times 7} = \frac{56}{105} \); \( \frac{4 \times 15}{7 \times 15} = \frac{60}{105} \)
\( \frac{56}{105} < \frac{60}{105} \implies \frac{8}{15} < \frac{4}{7} \)

(iii) \( \frac{7}{26} \) or \( \frac{10}{39} \)
Converting the like fraction
\( \frac{7 \times 3}{26 \times 3} = \frac{21}{78} \); \( \frac{10 \times 2}{39 \times 2} = \frac{20}{78} \)
\( \frac{20}{78} < \frac{21}{78} \implies \frac{10}{39} < \frac{7}{26} \)
In simple words: This is just like finding the "greater" fraction, but this time we look for the smaller top number.

📝 Teacher's Note: When finding common denominators for large numbers like 26 and 39, finding the LCM first saves a lot of multiplication work.

🎯 Exam Tip: Read the question carefully to see if it asks for "smaller" or "greater" - students often find the right values but pick the wrong answer.

Exercise 14(E)

Question 1. From a rope of \( 10 \frac{1}{2} \) m long, \( 4 \frac{5}{8} \) m is cut off. Find the length of the remaining rope.
Answer:
Length of rope = \( 10 \frac{1}{2} \) m
Length of cut off rope = \( 4 \frac{5}{8} \) m
Remaining rope = \( (10 \frac{1}{2} - 4 \frac{5}{8}) \) m
\( \implies \frac{21}{2} - \frac{37}{8} \)
\( \implies \frac{84 - 37}{8} = \frac{47}{8} = 5 \frac{7}{8} \) m.
In simple words: If you have a rope and cut a piece away, you just subtract the small piece from the total length.

📝 Teacher's Note: This is a real-life subtraction problem. Convert both mixed fractions to improper fractions first to make the subtraction easier.

🎯 Exam Tip: Don't forget the units (m for meters) in your final answer. Teachers often deduct marks for missing units.

Question 2. A piece of cloth is 5 metre long. After washing, it shrinks by \( \frac{1}{25} \) of its length. What is the length of the cloth after washing?
Answer:
Length of a piece of cloth = 5 m
After washing, it is shrinked = \( \frac{1}{25} \) of 5 m = \( \frac{1}{5} \) m
Length of cloth after washing
\( \implies (5 - \frac{1}{5}) \) m
\( \implies \frac{25 - 1}{5} = \frac{24}{5} = 4 \frac{4}{5} \) m
In simple words: First, find out how much the cloth shrunk (like 1 cm or 1 inch). Then take that away from the starting length.

📝 Teacher's Note: "Shrinks by \( \frac{1}{25} \)" means we calculate \( \frac{1}{25} \times \text{original length} \) first, then subtract it.

🎯 Exam Tip: Be careful with the phrase "of its length" - it always implies multiplication.

Exercise 14(E) (Continued)

Question 3. I bought wheat worth Rs. \( 12 \frac{1}{2} \), rice worth Rs. \( 25 \frac{3}{4} \) and vegetables worth Rs. \( 10 \frac{1}{4} \). If I gave a hundred-rupee note to the shopkeeper; how much did he return to me?
Answer:
Money given to Shopkeeper = Rs. 100
Total Amount of goods bought = Rs. \( (12 \frac{1}{2} + 25 \frac{3}{4} + 10 \frac{1}{4}) \)
\( \implies \frac{25}{2} + \frac{103}{4} + \frac{41}{4} \)
\( \implies \frac{50 + 103 + 41}{4} = \text{Rs. } \frac{194}{4} \)
Money returned by shopkeeper = Rs. \( (100 - \frac{194}{4}) \)
\( \implies \frac{400 - 194}{4} \)
\( \implies \frac{206}{4} = \frac{103}{2} = \text{Rs. } 51 \frac{1}{2} \).
In simple words: First, add up the cost of everything you bought. Then, subtract that total from the Rs. 100 you gave the shopkeeper to find your change.

📝 Teacher's Note: When adding fractions with different denominators like 2 and 4, encourage students to convert all to the common denominator (4) immediately to simplify the addition.

🎯 Exam Tip: In word problems involving money, ensure you show the subtraction step clearly after finding the total expenditure to get full marks.

Question 4. Out of 500 oranges in a box, \( \frac{3}{25} \) are rotten and \( \frac{1}{5} \) are kept for some guests. How many oranges are left in the box?
Answer:
Number of oranges = 500
Bad oranges = \( \frac{3}{25} \) of 500
\( \implies \frac{3}{25} \times 500 = 60 \)
Kept for guests = \( \frac{1}{5} \) of 500
\( \implies \frac{1}{5} \times 500 = 100 \)
No. of oranges which can be used = \( 500 - 60 - 100 \)
\( \implies 500 - 160 = 340 \).
In simple words: Calculate the actual number of rotten oranges and the number kept for guests. Subtract both these numbers from the total to find what remains.

📝 Teacher's Note: This problem teaches students how to find the value of a fraction of a whole number. Remind them that "of" means multiplication.

🎯 Exam Tip: Always state what each calculated number represents (e.g., "Bad oranges = 60") to keep your steps organized and clear for the examiner.

Question 5. An ornament piece is made of gold and copper. Its total weight is 96g. If \( \frac{1}{12} \) of the ornament is copper, find the weight of gold in it.
Answer:
Total weight = 96 g
Weight of copper = \( \frac{1}{12} \) of 96
\( \implies \frac{1}{12} \times 96 = 8 \text{ gm} \)
Weight of gold = Total weight - weight of copper
\( \implies 96\text{g} - 8\text{g} = 88\text{g} \).
In simple words: Since you know how much copper is in the ornament, calculate its weight in grams and subtract it from the total weight to find the gold's weight.

📝 Teacher's Note: This is a classic "parts of a whole" problem. You can also explain that if \( \frac{1}{12} \) is copper, then \( \frac{11}{12} \) must be gold.

🎯 Exam Tip: Units are essential. Make sure to write "g" or "gm" throughout the solution and in the final answer.

Question 6. A girl did half of some work on Monday and one-third of it on Tuesday. How much will she have to do on Wednesday in order to complete the work?
Answer:
Let total work done = 1
Work done on Monday = \( \frac{1}{2} \)
Work done on Tuesday = \( \frac{1}{3} \)
Work done on Wednesday = remaining work
\( \implies 1 - (\frac{1}{2} + \frac{1}{3}) \)
\( \implies 1 - (\frac{3 + 2}{6}) = 1 - \frac{5}{6} \)
\( \implies \frac{6 - 5}{6} = \frac{1}{6} \).
Work done on Wednesday = \( \frac{1}{6} \) of work.
In simple words: Treat the whole job as "1". Add the fractions of work done on the first two days, and subtract that total from 1 to see what's left for Wednesday.

📝 Teacher's Note: Use a circular "work clock" or pie chart to show how much is eaten up by \( \frac{1}{2} \) and \( \frac{1}{3} \), leaving a small slice for Wednesday.

🎯 Exam Tip: When the total is not given, always assume the "whole" is equal to 1.

Question 7. A man spends \( \frac{3}{8} \) of his money and still has Rs. 720 left with him. How much money did he have at first?
Answer:
Let a man has money = Re. 1
Amount spent = \( \frac{3}{8} \) of Re. 1 = Rs. \( \frac{3}{8} \)
Amount left = \( 1 - \frac{3}{8} = \frac{8 - 3}{8} = \text{Re. } \frac{5}{8} \)
\( \implies \frac{5}{8} \) of his total money = Rs. 720
Total money = Rs. \( \frac{720 \times 8}{5} \)
\( \implies \text{Rs. } 144 \times 8 = \text{Rs. } 1152 \).
In simple words: If he spent \( \frac{3}{8} \), he still has \( \frac{5}{8} \) left. We are told that this \( \frac{5}{8} \) is actually Rs. 720, so we use that to find the full original amount.

📝 Teacher's Note: This introduces simple algebraic reasoning. Help students understand that "Leftover Fraction = Leftover Value" is the key equation here.

🎯 Exam Tip: To find the "whole" from a "part," multiply the value by the reciprocal (flipped version) of the fraction.

Question 8. In a school, \( \frac{4}{5} \) of the students are boys, and the number of girls is 100. Find the number of boys.
Answer:
Let the total number of boys and girls = x
Total number of boys = \( \frac{4}{5} \) of x = \( \frac{4x}{5} \)
According to question, total strength of School,
\( \implies x - \frac{4x}{5} = 100 \)
\( \implies \frac{5x - 4x}{5} = 100 \)
\( \implies \frac{x}{5} = 100 \implies x = 500 \)
Number of boys = total strength - girls
\( \implies 500 - 100 = 400 \).
In simple words: If \( \frac{4}{5} \) are boys, then \( \frac{1}{5} \) must be girls. Since we know there are 100 girls, we find the total number of students first, then subtract the girls to get the boys.

📝 Teacher's Note: Alternatively, students can see that \( \frac{1}{5} \) of the total is 100, so the whole is 500. Then \( 500 \times \frac{4}{5} = 400 \).

🎯 Exam Tip: Make sure you answer the specific question asked-here it's the "number of boys," not just the "total strength."

Question 9. After finishing \( \frac{3}{4} \) of my journey, I find that 12 km of my journey is covered. How much distance is still left to be covered?
Answer:
Let the total journey = x,
distance covered = \( \frac{3}{4} \) = 12 km
Then, according to question \( \frac{3}{4} \) of x = 12 km
\( \implies x = 12 \times \frac{4}{3} \implies x = 16 \text{ km} \)
Distance left = total distance - distance covered
\( \implies 16 - 12 = 4 \text{ km} \).
In simple words: You know 12 km is three-quarters of the trip. Find the full trip length first, then subtract the 12 km you already walked to find what's left.

📝 Teacher's Note: This is a good opportunity to use mental math. If 3 parts equal 12 km, then 1 part equals 4 km. Since the whole is 4 parts, the total is 16 km.

🎯 Exam Tip: Double check if the question asks for the "total distance" or the "remaining distance."

Question 10. When Ajit travelled 15 km, he found that one-fourth of his journey was still left. What was the full length of the journey?
Answer:
Let the total length of journey = x
Journey travelled = 15 km
Journey still left = \( \frac{1}{4} \) of x
Now, according to question,
\( \implies x - 15 = \frac{1}{4} \text{ of } x \)
\( \implies x - 15 = \frac{x}{4} \)
\( \implies x - \frac{x}{4} = 15 \)
\( \implies \frac{4x - x}{4} = 15 \)
\( \implies 3x = 15 \times 4 \)
\( \implies x = \frac{15 \times 4}{3} = 20 \text{ km} \)
Total length of the journey = 20 km.
In simple words: If one-fourth is left, that means he has already finished three-fourths of the trip. So, 15 km is actually three-quarters of the whole journey.

📝 Teacher's Note: This question tests logic. If \( \frac{1}{4} \) is left, \( 1 - \frac{1}{4} = \frac{3}{4} \) is completed. Setting up \( \frac{3}{4}x = 15 \) is the fastest way to solve this.

🎯 Exam Tip: Drawing a simple bar representing the journey and shading three out of four parts helps visualize the solution correctly.

Question 11. In a particular month, a man earns Rs. 7,200. Out of this income, he spends \( \frac{3}{10} \) on food, \( \frac{1}{4} \) on house rent, \( \frac{1}{10} \) on insurance and \( \frac{2}{25} \) on holidays. How much did he save in that month?
Answer:
Earning of a man in a particular month = Rs. 7200
Amount spent on food = \( \frac{3}{10} \) of Rs. 7200 = Rs. 2160
Amount spent on house rent = \( \frac{1}{4} \) of Rs. 7200 = Rs. 1800
Amount spent on insurance = \( \frac{1}{10} \) of Rs. 7200 = Rs. 720
Amount spent on holidays = \( \frac{2}{25} \) of Rs. 7200 = Rs. \( 2 \times 288 \) = Rs. 576
Total amount spent = Rs. \( (2160 + 1800 + 720 + 576) \) = Rs. 5256
Amount saved = Rs. 7200 - Rs. 5256 = Rs. 1944.
In simple words: Calculate exactly how many rupees were spent on each category separately. Add them all up and subtract that total from the monthly salary to find the savings.

📝 Teacher's Note: You can also solve this by adding all the fractions first, then multiplying the total fraction of spending by 7200, but calculating individual amounts is less prone to error for many students.

🎯 Exam Tip: Be very careful with the division during calculation (like \( 7200 \div 25 \)). Scratch work on the side is helpful.

Revision Exercise

Question 1. Show that \( \frac{3}{7} \) lies between \( \frac{2}{5} \) and \( \frac{5}{7} \).
Answer:
\( \frac{3}{7} \) will lie between \( \frac{2}{5} \) and \( \frac{5}{7} \) if \( \frac{2}{5} < \frac{3}{7} < \frac{5}{7} \) or \( \frac{2}{5} > \frac{3}{7} > \frac{5}{7} \)
Now, comparing \( \frac{2}{5}, \frac{3}{7}, \frac{5}{7} \)
L.C.M. of 5 and 7 = 35
\( \implies \frac{2}{5} = \frac{2 \times 7}{5 \times 7} = \frac{14}{35} \)
\( \implies \frac{3}{7} = \frac{3 \times 5}{7 \times 5} = \frac{15}{35} \)
and \( \frac{5}{7} = \frac{5 \times 5}{7 \times 5} = \frac{25}{35} \)
\( \because \frac{14}{35} < \frac{15}{35} < \frac{25}{35} \)
\( \implies \frac{2}{5} < \frac{3}{7} < \frac{5}{7} \)
\( \frac{3}{7} \) lies between \( \frac{2}{5} \) and \( \frac{5}{7} \).
In simple words: To show a fraction is in the middle of two others, convert all three to have the same denominator. Then you can easily see if the middle one's numerator is between the other two.

📝 Teacher's Note: "Between" means the value must be strictly greater than one and strictly less than the other.

🎯 Exam Tip: Use the "less than" symbols (\( < \)) clearly to demonstrate the order of the values.

Question 2. Show that \( \frac{4}{5} \) lies between \( \frac{3}{4} \) and \( \frac{5}{6} \).
Answer:
\( \frac{3}{4} > \frac{4}{5} > \frac{5}{6} \) or \( \frac{3}{4} < \frac{4}{5} < \frac{5}{6} \)
Now L.C.M. of 4, 5, 6 = 60
\( \implies \frac{3}{4} = \frac{3 \times 15}{4 \times 15} = \frac{45}{60} \)
\( \implies \frac{4}{5} = \frac{4 \times 12}{5 \times 12} = \frac{48}{60} \)
\( \implies \frac{5}{6} = \frac{5 \times 10}{6 \times 10} = \frac{50}{60} \)
\( \because \frac{45}{60} < \frac{48}{60} < \frac{50}{60} \)
\( \implies \frac{3}{4} < \frac{4}{5} < \frac{5}{6} \)
Hence \( \frac{4}{5} \) lies between \( \frac{3}{4} \) and \( \frac{5}{6} \).
In simple words: Just like the previous question, use a common denominator (60) to check the order of the numerators. Since 48 is between 45 and 50, the fraction is in the middle.

📝 Teacher's Note: Finding the lowest common multiple (LCM) of 4, 5, and 6 is crucial. 60 is the smallest number all three can go into.

🎯 Exam Tip: Always state your conclusion clearly with "Hence" or "Therefore" to finish the proof.

Question 3. Evaluate :
(i) \( 3 \frac{5}{6} - 1 \frac{4}{15} - (3 \frac{2}{9} - 1 \frac{3}{5}) \)
(ii) \( \frac{3}{4} \text{ of } 1 \frac{1}{2} + 4 \frac{1}{2} \)
(iii) \( \frac{5}{6} \text{ of } \frac{3}{4} + \frac{7}{8} \times 1 \frac{1}{2} \)
(iv) \( \frac{1}{3} + \frac{7}{9} \div (\frac{7}{10} \times 1 \frac{1}{4}) \)
(v) \( 1 \frac{4}{13} \text{ of } 2 \frac{2}{7} \div \frac{68}{91} - (1 \frac{1}{2} - 1 \frac{1}{3}) \)
(vi) \( 8 - \{5 \frac{1}{3} - (3 - 2 \frac{1}{2})\} \)
Answer:
(i) \( 3 \frac{5}{6} - 1 \frac{4}{15} - (3 \frac{2}{9} - 1 \frac{3}{5}) \)
\( \implies \frac{23}{6} - \frac{19}{15} - (\frac{29}{9} - \frac{8}{5}) \)
\( \implies \frac{23}{6} - \frac{19}{15} - \frac{29}{9} + \frac{8}{5} \)
\( \implies \frac{345 - 114 - 290 + 144}{90} \) (L.C.M. of 6, 15, 9, 5 = 90)
\( \implies \frac{345 + 144 - 114 - 290}{90} = \frac{489 - 404}{90} \)
\( \implies \frac{85}{90} = \frac{85 \div 5}{90 \div 5} = \frac{17}{18} \)

(ii) \( \frac{3}{4} \text{ of } 1 \frac{1}{2} + 4 \frac{1}{2} \)
\( \implies \frac{3}{4} \text{ of } \frac{3}{2} + \frac{9}{2} \)
\( \implies \frac{9}{8} + \frac{9}{2} \) (first remove 'of')
\( \implies \frac{9}{8} + \frac{36}{8} = \frac{45}{8} = 5 \frac{5}{8} \)

(iii) \( \frac{5}{6} \text{ of } \frac{3}{4} + \frac{7}{8} \times 1 \frac{1}{2} \)
\( \implies \frac{5}{6} \text{ of } \frac{3}{4} + \frac{7}{8} \times \frac{3}{2} \)
\( \implies \frac{5}{8} + \frac{7}{8} \times \frac{3}{2} \) (remove 'of')
\( \implies \frac{5}{8} + \frac{21}{16} = \frac{10 + 21}{16} = \frac{31}{16} = 1 \frac{15}{16} \)

(iv) \( \frac{1}{3} + \frac{7}{9} \div (\frac{7}{10} \times 1 \frac{1}{4}) \)
\( \implies \frac{1}{3} + \frac{7}{9} \div (\frac{7}{10} \times \frac{5}{4}) \)
\( \implies \frac{1}{3} + \frac{7}{9} \div \frac{7}{8} \) (remove bracket)
\( \implies \frac{1}{3} + \frac{7}{9} \times \frac{8}{7} \) (remove \( \div \))
\( \implies \frac{1}{3} + \frac{8}{9} = \frac{3 + 8}{9} = \frac{11}{9} = 1 \frac{2}{9} \)

(v) \( 1 \frac{4}{13} \text{ of } 2 \frac{2}{7} \div \frac{68}{91} - (1 \frac{1}{2} - 1 \frac{1}{3}) \)
\( \implies \frac{17}{13} \text{ of } \frac{16}{7} \div \frac{68}{91} - (\frac{3}{2} - \frac{4}{3}) \)
\( \implies \frac{17}{13} \text{ of } \frac{16}{7} \div \frac{68}{91} - (\frac{9 - 8}{6}) \)
\( \implies \frac{17}{13} \text{ of } \frac{16}{7} \div \frac{68}{91} - \frac{1}{6} \) (remove bracket)
\( \implies \frac{272}{91} \div \frac{68}{91} - \frac{1}{6} \) (remove 'of')
\( \implies \frac{272}{91} \times \frac{91}{68} - \frac{1}{6} = 4 - \frac{1}{6} \)
\( \implies \frac{24 - 1}{6} = \frac{23}{6} = 3 \frac{5}{6} \)

(vi) \( 8 - \{5 \frac{1}{3} - (3 - 2 \frac{1}{2})\} \)
\( \implies 8 - \{ \frac{16}{3} - (3 - \frac{5}{2})\} \)
\( \implies 8 - \{ \frac{16}{3} - \frac{6 - 5}{2}\} \)
\( \implies \frac{8}{1} - \frac{16}{3} + \frac{1}{2} \)
\( \implies \frac{48 - 32 + 3}{6} = \frac{19}{6} = 3 \frac{1}{6} \).
In simple words: Always follow the BODMAS rule! Solve brackets first, then "of," then division/multiplication, and finally addition/subtraction. Convert all mixed numbers to improper fractions before starting.

📝 Teacher's Note: Emphasize that "of" should be treated like multiplication but must be solved before any standard multiplication or division in the sequence.

🎯 Exam Tip: Work step-by-step. Don't try to solve multiple operations at once, as this is where most arithmetic errors happen.

Question 4. Mr. Mehra gave one-third of his money to his son, one-fifth of his money to his daughter and the remaining amount to his wife. If his wife got Rs. 91,000, how much money did Mr. Mehra have originally?
Answer:
Let Mr. Mehra has money = 1
Money given to his son = \( \frac{1}{3} \)
and money given to his daughter = \( \frac{1}{5} \)
Remaining money given to his wife
\( \implies 1 - (\frac{1}{3} + \frac{1}{5}) \)
\( \implies 1 - \frac{5 + 3}{15} = 1 - \frac{8}{15} \)
\( \implies \frac{15 - 8}{15} = \frac{7}{15} \)
\( \frac{7}{15} \) of his money = Rs. 91,000
Total money = Rs. \( \frac{91000 \times 15}{7} \)
\( \implies \text{Rs. } 13,000 \times 15 = \text{Rs. } 1,95,000 \).
In simple words: Find the fraction of money left for the wife after the son and daughter get their shares. Since that fraction equals Rs. 91,000, use it to calculate the original full amount.

📝 Teacher's Note: This is a multi-step problem. Encourage students to solve the bracket first to find the total portion given away before finding the remaining portion.

🎯 Exam Tip: Check your final multiplication carefully. Multiplying \( 13,000 \times 15 \) requires focus to keep track of the zeros.

Question 5. A sum of Rs. 84,000 is divided among three persons A, B and C. If A gets one-fourth of it and B gets one-fifth of it; how much did C get?
Answer:
Total money = Rs. 84,000
A gets = \( \frac{1}{4} \) of 84,000 = Rs. 21,000
B gets = \( \frac{1}{5} \) of 84,000 = Rs. 16,800
C gets remaining money
C's share = Rs. 84,000 - (Rs. 21,000 + Rs. 16,800)
\( \implies \text{Rs. } 84,000 - 37,800 = \text{Rs. } 46,200 \).
In simple words: Calculate the actual rupee amounts for person A and person B. Add those amounts together and subtract them from the starting Rs. 84,000 to see what's left for person C.

📝 Teacher's Note: Show that you can also solve this by adding the fractions (\( \frac{1}{4} + \frac{1}{5} = \frac{9}{20} \)) and then finding \( 1 - \frac{9}{20} = \frac{11}{20} \) of the total.

🎯 Exam Tip: Label each person's share clearly to avoid confusing "who gets what."

Question 6. In one hour Rohit walks \( 3 \frac{2}{5} \) km. How much distance will he cover in \( 2 \frac{1}{2} \) hours?
Answer:
Distance covered in 1 hour = \( 3 \frac{2}{5} = \frac{17}{5} \) km
Distance covered in \( 2 \frac{1}{2} \) hours
\( \implies 3 \frac{2}{5} \times 2 \frac{1}{2} \text{ km} = \frac{17}{5} \times \frac{5}{2} \text{ km} \)
\( \implies \frac{17}{2} = 8 \frac{1}{2} \text{ km} \).
In simple words: If you know how far he walks in one hour, just multiply that distance by the number of hours he walks.

📝 Teacher's Note: Multiplication of mixed numbers must always be done by converting them to improper fractions first. Notice how the "5"s cancel out nicely in this problem.

🎯 Exam Tip: Look for opportunities to cancel numbers in the numerator and denominator before multiplying to keep the numbers small and easy.

Question 7. An 84 m long string is cut into pieces each of length \( 5 \frac{1}{4} \) m. How many pieces are obtained?
Answer:
Length of string = 84 m
Length of each piece = \( 5 \frac{1}{4} \text{ m} = \frac{21}{4} \text{ m} \)
Number of pieces = \( 84 \div \frac{21}{4} \)
\( \implies 84 \times \frac{4}{21} = 4 \times 4 = 16 \).
In simple words: To find how many small pieces fit into a big length, divide the total length by the length of one piece.

📝 Teacher's Note: Remind students that dividing by a fraction is the same as multiplying by its reciprocal. "Keep, Change, Flip."

🎯 Exam Tip: Make sure the final answer is a whole number of pieces. If you get a fraction, re-check your division steps.

Question 8. In buying a ready made shirt-two-fifths of my pocket money is spent If Rs. 540 is still left with me, find :
(i) The money I had before I bought the shirt.
(ii) The cost of the shirt
Answer:
Let total money in the pocket = 1
Amount spent on shirt = \( \frac{2}{5} \)
Balance amount = \( 1 - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5} \)
Now \( \frac{3}{5} \) of total money = Rs. 540
(i) Total money = Rs. \( 540 \times \frac{5}{3} \)

\( \implies \) Rs. \( 180 \times 5 = \text{Rs. } 900 \)
(ii) Cost of shirt = \( \frac{2}{5} \) of Rs. 900

\( \implies \) Rs. \( 2 \times 180 = \text{Rs. } 360 \).
In simple words: If you spend two-fifths of your money, you have three-fifths left. Since that leftover three-fifths is Rs. 540, we can work backwards to find the original amount (Rs. 900) and then find how much the shirt cost (Rs. 360).

📝 Teacher's Note: This is a two-part problem that links a remaining fraction to a real value. Help students see that finding the "whole" first makes part (ii) very straightforward.

🎯 Exam Tip: Always state what the fraction represents (e.g., "Balance amount") before setting up the equation with the rupee value.

Question 9. Mohan leaves Rs. 1,20,000 to his wife and three children such that two-fifths of this money is given to his wife and the remaining is distributed equally among the children. Find, how much each child gets ?
Answer:
Total amount = Rs. 1,20,000
Amount given to his wife = \( \frac{2}{5} \) of Rs. 1,20,000

\( \implies \) Rs. \( 2 \times 24,000 = \text{Rs. } 48,000 \)
Remaining amount = Rs. 1,20,000 - Rs. 48,000 = Rs. 72,000
This amount is distributed among three children equally.
Each's share = Rs. \( 72,000 \times \frac{1}{3} = \text{Rs. } 24,000 \).
In simple words: First, calculate the wife's share (Rs. 48,000) and take it away from the total. Then, take the remaining Rs. 72,000 and divide it by 3 to find out what each child receives.

📝 Teacher's Note: Remind students to perform the division by 5 first (\( 120,000 \div 5 \)) when calculating the wife's share to keep the numbers manageable.

🎯 Exam Tip: Read carefully - the remaining money is shared by three children, so don't forget to divide by 3 at the end.

Question 10. Simplify :
(i) \( 3 \frac{5}{8} \text{ of } 2 \frac{2}{3} \div 1 \frac{3}{8} \)
(ii) \( (1 \div 3 \frac{1}{3}) \times 3 \frac{1}{3} \text{ of } 7 \frac{2}{9} - 6 \)
(iii) \( \frac{3}{4} \times 1 \frac{1}{3} \div \frac{3}{7} \text{ of } 2 \frac{5}{8} \)
Answer:
(i) \( 3 \frac{5}{8} \text{ of } 2 \frac{2}{3} \div 1 \frac{3}{8} \)

\( \implies \frac{29}{8} \text{ of } \frac{8}{3} \div \frac{11}{8} \)

\( \implies \frac{29}{3} \div \frac{11}{8} \) (Removing 'of')

\( \implies \frac{29}{3} \times \frac{8}{11} = \frac{232}{33} = 7 \frac{1}{33} \)

(ii) \( (1 \div 3 \frac{1}{3}) \times 3 \frac{1}{3} \text{ of } 7 \frac{2}{9} - 6 \)

\( \implies (1 \div \frac{10}{3}) \times \frac{10}{3} \text{ of } \frac{65}{9} - 6 \)

\( \implies (1 \times \frac{3}{10}) \times \frac{10}{3} \text{ of } \frac{65}{9} - 6 \)

\( \implies \frac{3}{10} \times \frac{10}{3} \text{ of } \frac{65}{9} - 6 \) (Removing bracket)

\( \implies \frac{3}{10} \times \frac{650}{27} - \frac{6}{1} \) (Removing 'of')

\( \implies \frac{65}{9} - \frac{6}{1} \) (Removing '×')

\( \implies \frac{65 - 54}{9} = \frac{11}{9} = 1 \frac{2}{9} \)

(iii) \( \frac{3}{4} \times 1 \frac{1}{3} \div \frac{3}{7} \text{ of } 2 \frac{5}{8} \)

\( \implies \frac{3}{4} \times \frac{4}{3} \div \frac{3}{7} \text{ of } \frac{21}{8} \)

\( \implies \frac{3}{4} \times \frac{4}{3} \div \frac{9}{8} \) (Removing 'of')

\( \implies \frac{3}{4} \times \frac{4}{3} \times \frac{8}{9} \)

\( \implies 1 \times \frac{8}{9} = \frac{8}{9} \).
In simple words: These problems use BODMAS with fractions. Always handle mixed numbers and "of" statements before doing standard divisions and subtractions.

📝 Teacher's Note: These long expressions are great for testing if students truly understand the priority of operations. Remind them that division turns into multiplication by a flipped fraction (reciprocal).

🎯 Exam Tip: Cross-cancelling common factors between numerators and denominators (like the 3 and 4 here) will save a massive amount of time during exams.