Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 11 Ratio

Important Points

Ratio: The relation between two quantities (both of the same kind and in the same unit) obtained on dividing one quantity by the other, is called the ratio.
Keep in Mind :
(i) The ratio between two numbers or quantities is denoted by the colon “ : ”.
Thus, the ratio between two quantities p and q = p : q
(ii) The ratio between two quantities of same kind and in the same units is obtained on dividing one quantity by the other. Thus, the ratio between 20 kg and 80 kg = \( \frac{20}{80} = \frac{1}{4} = 1 : 4 \)
(iii) The first term of a ratio is called antecedent and its second term is called consequent. In the ratio 1:4; antecendent = 1 and consequent = 4.
(iv) A ratio must always by expressed in its lowest terms.
Proportion: A proportion is an expression which states that the two ratios are equal.
Keep in Mind :
In a proportion, its first and the fourth terms are called extremes whereas its second and the third terms are called means.
Thus, in 8 : 12 = 18 : 27; the terms 12 and 18 are means whereas 8 and 27 are extremes.
Also, Product of extremes = Product of Means

Question. Express each of the following ratios in its simplest form :
(a) (i) 4 : 6
(ii) 48 : 54
(iii) 200 : 250
(b) (i) 5 kg : 800 gm
(ii) 30 cm : 2 m
(iii) 3 m : 90 cm
(iv) 2 years : 9 months
(v) 1 hour : 45 minutes
(vi) 4 min : 45 sec
(c) (i) \( 1\frac{1}{2} : 2\frac{1}{2} \)
(ii) \( 3\frac{1}{2} : 7 \)
(iii) \( 2\frac{1}{3} : 3\frac{1}{2} : 1\frac{1}{4} \)
(iv) \( x^2 : 4x \)
(v) 2.5 : 1.5
(vi) 2.5 : 5

Answer:
(a) (i) 4 : 6 = \( \frac{4}{6} = \frac{2}{3} = 2 : 3 \)
(ii) 48 : 54 = \( \frac{48}{54} = \frac{8}{9} = 8 : 9 \)
(iii) 200 : 250 = \( \frac{200}{250} = \frac{4}{5} = 4 : 5 \)
(b) (i) 5 kg : 800 gm = 5000 : 800 (since 5 kg = 5000 gm)
\( \implies \frac{5000}{800} = \frac{50}{8} = \frac{25}{4} = 25 : 4 \)
(ii) 30 cm : 2 m = 30 cm : 200 cm (since 2 m = 200 cm)
\( \implies \frac{30}{200} = \frac{3}{20} = 3 : 20 \)
(iii) 3 m : 90 cm = 300 cm : 90 cm (since 3 m = 300 cm)
\( \implies \frac{300}{90} = \frac{10}{3} = 10 : 3 \)
(iv) 2 years : 9 months = 24 months : 9 months (since 2 years = 24 months)
\( \implies \frac{24}{9} = \frac{8}{3} = 8 : 3 \)
(v) 1 hour : 45 minutes = 60 minutes : 45 minutes (since 1 hour = 60 minutes)
\( \implies \frac{60}{45} = \frac{4}{3} = 4 : 3 \)
(vi) 4 min : 45 sec = 240 sec : 45 sec (since 4 min = 240 sec)
\( \implies \frac{240}{45} = \frac{16}{3} = 16 : 3 \)
(c) (i) \( 1\frac{1}{2} : 2\frac{1}{2} = \frac{3}{2} : \frac{5}{2} = \frac{3}{2} \div \frac{5}{2} = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5} = 3 : 5 \)
(ii) \( 3\frac{1}{2} : 7 = \frac{7}{2} : \frac{7}{1} = \frac{7}{2} \div \frac{7}{1} = \frac{7}{2} \times \frac{1}{7} = \frac{1}{2} = 1 : 2 \)
(iii) \( 2\frac{1}{3} : 3\frac{1}{2} : 1\frac{1}{4} = \frac{7}{3} : \frac{7}{2} : \frac{5}{4} \)
(L.C.M. of 3, 2 and 4 = 12)
\( \implies \frac{7}{3} \times 12 : \frac{7}{2} \times 12 : \frac{5}{4} \times 12 = 28 : 42 : 15 \)
(iv) \( x^2 : 4x = \frac{x^2}{4x} = \frac{x \times x}{4 \times x} = \frac{x}{4} = x : 4 \)
(v) 2.5 : 1.5 = \( \frac{25}{15} = \frac{5}{3} = 5 : 3 \)
(vi) 2.5 : 5 = \( \frac{2.5}{5} = \frac{2.5 \times 10}{5 \times 10} = \frac{25}{50} = \frac{1}{2} = 1 : 2 \)

In simple words: To simplify a ratio, first make sure both numbers are in the same unit. Then, write them as a fraction and divide both by their largest common factor until they cannot be divided anymore.

📝 Teacher's Note: When teaching ratios with different units, always use the ladder method to convert the larger unit to the smaller unit (e.g., kg to gm) to avoid working with messy decimals. Remind students that ratios themselves have no units.

🎯 Exam Tip: Always double-check if your final ratio can be simplified further. Leaving a ratio like 4:6 instead of 2:3 will often result in a loss of marks.

Question. A field is 80 m long and 60 m wide. Find the ratio of its width to its length.

Answer:
Width of field = 60 m
Length of field = 80 m
Ratio between width and length = \( \frac{60}{80} = \frac{3}{4} = 3 : 4 \)

In simple words: Write the width first and the length second, then simplify the fraction by dividing both numbers by 20.

📝 Teacher's Note: Pay close attention to the order requested in the question. If it asks for "width to length," width must be the first term (antecedent).

🎯 Exam Tip: Write down the given values clearly with their units before forming the ratio to ensure they are the same.

Question. State, true or false :
(i) A ratio equivalent to 7 : 9 is 27 : 21.
(ii) A ratio equivalent to 5 : 4 is 240 : 192.
(iii) A ratio of 250 gm and 3 kg is 1 : 12.

Answer:
(i) False. Correct: A ratio equivalent to 7 : 9 is 21 : 27 or 14 : 18.
(ii) True
(iii) True

In simple words: For (i), 27:21 simplifies to 9:7, which is the reverse of 7:9. For (iii), 3 kg is 3000 gm, and 250 goes into 3000 exactly 12 times.

📝 Teacher's Note: Use cross-multiplication to check equivalence. For 7:9 and 27:21, check if \( 7 \times 21 = 9 \times 27 \). If not, they are not equivalent.

🎯 Exam Tip: In True/False questions, if the statement is false, briefly write the correct reason to show the examiner you understood the concept.

Question. Is the ratio of 15 kg and 35 kg same as the ratio of 6 years and 14 years?

Answer:
Ratio of weight 15 kg and 35 kg = \( \frac{15 kg}{35 kg} = \frac{3}{7} = 3:7 \)
Ratio of 6 years and 14 years = \( \frac{6 years}{14 years} = \frac{3}{7} = 3:7 \)
Yes, the ratio in both cases are same.

In simple words: When we simplify both pairs of numbers, they both turn into 3:7, so they are the same.

📝 Teacher's Note: This illustrates that ratios compare pure numbers and can represent completely different physical quantities (weight vs. time) if their numerical proportions are identical.

🎯 Exam Tip: Show the simplification steps for both ratios separately before concluding with a "Yes" or "No" statement.

Question. Is the ratio of 6 g and 15 g same as the ratio of 36 cm and 90 cm ?

Answer:
Ratio of 6 g and 15 g = \( \frac{6}{15} = \frac{2}{5} = 2:5 \)
Ratio of 36 cm and 90 cm = \( \frac{36}{90} = \frac{18}{45} = \frac{6}{15} = 2:5 \)
Yes, the ratio in both cases are same.

In simple words: Both ratios simplify to 2:5 after dividing by their common factors.

📝 Teacher's Note: Encourage students to divide by the largest possible number (HCF) to reach the simplest form faster.

🎯 Exam Tip: Mention the simplified form of both ratios to prove they are equal.

Question. Find the ratio between 3.5 m, 475 cm and 2.8 m.

Answer:
The given values = 3.5 m, 475 cm and 2.8 m
= 3.5 x 100 cm : 475 cm : 2.8 x 100 cm
= 350 cm : 475 cm : 280 cm
= 70 cm : 95 cm : 56 cm (Dividing by 5)
The ratio is 70 : 95 : 56

In simple words: Convert everything to centimeters first, then divide all three numbers by 5 to get the simplest answer.

📝 Teacher's Note: When dealing with three terms, you must divide all three by the same number to simplify. If no number goes into all three, it is already in simplest form.

🎯 Exam Tip: Units must be eliminated in the final ratio. Write 70:95:56, not 70cm:95cm:56cm.

Question. Find the ratio between 5 dozen and 2 scores. [1 score = 20]

Answer:
Ratio between 5 dozen and 2 scores
Given = 1 score = 20
then, 2 scores = 2 x 20 = 40
and 1 dozen = 12,
5 dozen = 12 x 5 = 60
Then, ratio = 60 : 40 = 3 : 2

In simple words: Convert dozens to 60 items and scores to 40 items. Then simplify 60:40 by dividing by 20.

📝 Teacher's Note: This is a great real-world application. Ensure students know that 1 dozen = 12 before they attempt the calculation.

🎯 Exam Tip: Always show the conversion calculation (like 5 x 12) to get step-marks even if you make a division error later.

Exercise 11(B)

Question. The monthly salary of a person is Rs. 12,000 and his monthly expenditure is Rs. 8,500. Find the ratio of his:
(i) salary to expenditure
(ii) expenditure to savings
(iii) savings to salary

Answer:
Monthly salary of a person = Rs. 12,000
Monthly expenditure = 8,500
Saving of the person = (12,000 — 8500) = Rs. 3,500
(i) Ratio between salary and expenditure
12,000 : 8500 = \( \frac{12,000}{8,500} = \frac{120}{85} = \frac{24}{17} = 24 : 17 \)
(ii) Ratio between expenditure and savings
8500 : 3500 = \( \frac{8500}{3500} = \frac{85}{35} = \frac{17}{7} = 17 : 7 \)
(iii) Ratio between savings and salary
3,500 : 12,000 = \( \frac{3,500}{12,000} = \frac{35}{120} = \frac{7}{24} = 7 : 24 \)

In simple words: First, find the savings by subtracting spending from salary. Then, set up the ratios as requested and simplify them by canceling out the zeros and dividing by common factors.

📝 Teacher's Note: Use the "Salary = Expenditure + Savings" formula to help students find the missing value before they start the ratios.

🎯 Exam Tip: Clearly label part (i), (ii), and (iii) so the examiner can easily follow your different ratios.

Question. The strength of a class is 65, including 30 girls. Find the ratio of the number of:
(i) girls to boys
(ii) boys to the whole class
(iii) the whole class to girls.

Answer:
Total strength of class (including boys and girls) = 65
Number of girls = 30
Number of boys = (65 — 30) = 35
(i) Ratio between girls and boys = 30 : 35 = \( \frac{30}{35} = \frac{6}{7} = 6 : 7 \)
(ii) Ratio between boys and whole class = 35 : 65 = \( \frac{35}{65} = \frac{7}{13} = 7 : 13 \)
(iii) Whole class and girls = 65 : 30 = \( \frac{65}{30} = \frac{13}{6} = 13 : 6 \)

In simple words: Subtract 30 girls from the 65 students to find there are 35 boys. Then divide the numbers by 5 to simplify the ratios.

📝 Teacher's Note: Remind students that the "whole class" includes everyone. This is a common point of confusion.

🎯 Exam Tip: Re-read the question after finishing to ensure you didn't accidentally swap the terms in the ratio.

Question. The weekly expenses of a boy have increased from Rs. 1,500 to Rs. 2,250. Find the ratio of:
(i) increase in expenses to original expenses.
(ii) original expenses to increased expenses.
(iii) increased expenses to increase in expenses.

Answer:
Original expenses = Rs. 1500
Increased expenses = Rs. 2250
Increase in expenses = Rs. 2250 — Rs. 1500 = Rs. 750
Now,
(i) Ratio in increase in expenses to original expenses = Rs. 750 : Rs. 1500 = 1 : 2
(ii) Original expenses to increased expenses = Rs. 1500 : Rs. 2250
\( \implies \frac{1500}{750} = \frac{2250}{750} = 2 : 3 \)
(iii) Increased expenses to increase in expenses = Rs. 2250 : Rs. 750 = 3 : 1 (Dividing by 750)

In simple words: "Increase" is the difference between the new and old price (Rs. 750). Compare this 750 to the old 1500 and the new 2250 as asked.

📝 Teacher's Note: Distinguish clearly between "increased expenses" (the new total) and "increase in expenses" (just the extra amount added).

🎯 Exam Tip: Simplify large numbers by finding their HCF. Here, knowing that 750 goes into 1500 and 2250 makes the work very easy.

Question. Reduce each of the following ratios to their lowest terms :
(i) 1 hour 20 min : 2 hours
(ii) 4 weeks : 49 days
(iii) 3 years 4 months : 5 years 5 month.
(iv) 2 m 40 cm : 1 m 44 cm
(v) 5 kg 500 gm : 2 kg 750 gm

Answer:
(i) 1 hour 20 min : 2 hour
= (1 x 60 + 20) minutes : 2 x 60 minutes
= 80 minutes : 120 minutes
\( \implies \frac{80}{120} = \frac{2}{3} = 2 : 3 \)
(ii) 4 weeks : 49 days = (4 x 7) days : 49 days
= 28 : 49 = \( \frac{28}{49} = \frac{4}{7} = 4 : 7 \)
(iii) 3 years 4 months : 5 years 5 month
= (3 x 12 + 4) : (5 x 12 + 5)
= 40 : 65 = \( \frac{40}{65} = \frac{8}{13} = 8 : 13 \)
(iv) 2 m 40 cm : 1 m 44 cm
(2 x 100 + 40) : (1 x 100 + 44)
= 240 : 144 = \( \frac{240}{144} = \frac{10}{6} = \frac{5}{3} = 5 : 3 \)
(v) 5 kg 500 gm : 2 kg 750 gm
(5 x 1000 + 500) : (2 x 1000 + 750)
= 5500 : 2750 = \( \frac{5500}{2750} = \frac{2}{1} = 2 : 1 \)

In simple words: Convert the mixed units (like hours and minutes) into just one unit (like total minutes). Once both sides are in the same unit, simplify the numbers.

📝 Teacher's Note: Remind students of the basic conversions: 1 hour = 60 min, 1 week = 7 days, 1 year = 12 months, 1 m = 100 cm, 1 kg = 1000 gm.

🎯 Exam Tip: Always show the conversion steps. If you convert 3 years 4 months to 40 months correctly, you get marks even if you fail to simplify 40:65.

Question. Two numbers are in the ratio 9 : 2. If the smaller number is 320, find the larger number.

Answer:
Let the larger number = 9x
and smaller number = 2x
If smaller number is 320,
then, larger number will be = \( \frac{9x \times 320}{2x} = 1440 \)

In simple words: If 2 parts represent 320, then 1 part is 160. Multiply 160 by 9 to find that the larger number is 1440.

📝 Teacher's Note: The "x method" (2x = 320) is the most reliable way for students to solve these problems without confusion.

🎯 Exam Tip: Identify which part of the ratio represents which value. Since 2 is smaller than 9, 2 parts must equal the smaller number 320.

Question. A bus travels 180 km in 3 hours and a train travels 450 km in 5 hours. Find the ratio of speed of train to speed of bus.

Answer:
Distance travelled by bus = 180 km
Time taken = 3 hours
Speed = \( \frac{Distance}{Time} = \frac{180}{3} = 60 \) km/hr
Distance travelled by train = 450 km
Time taken = 5 hours
Speed = \( \frac{Distance}{Time} = \frac{450}{5} = 90 \) km/hr
Ratio of speed of train to speed of bus = 90 : 60 = 3 : 2

In simple words: First find how fast each one is going by dividing distance by time. Then compare the two speeds (90 and 60) and simplify.

📝 Teacher's Note: Use this to integrate Science/Physics concepts (Speed = Distance / Time) into Math class.

🎯 Exam Tip: Don't just find the speeds; remember to answer the final part of the question, which is the ratio.

Question. In winters, a school opens at 10 a.m. and closes at 3.30 p.m. If the lunch interval is of 30 minutes, find the ratio of lunch interval to total time of the class periods.

Answer:
Timing of a school (10 a.m. to 3.30 p.m) = 5 hours 30 minutes
Timing for lunch interval = 30 minutes
Total time of the class periods = 5 hours 30 minutes – 30 minutes
= 5 hours = 60 x 5 = 300 minutes
Ratio of lunch interval to total time of the class period = 30 minutes : 300 minutes = 1 : 10

In simple words: The school is open for 330 minutes total. Subtract the 30-minute lunch to get 300 minutes of class. The ratio of 30 to 300 is 1 to 10.

📝 Teacher's Note: This requires two steps: first finding the duration, then subtracting the lunch break to find the "class period" time.

🎯 Exam Tip: Be careful! "Total time of class periods" is NOT the total time the school is open; you must subtract the lunch break first.

Question. Rohit goes to school by car at 60 km per hour and Manoj goes to school by scooty at 40 km per hour. If they both live in the same locality, find the ratio between the time taken by Rohit and Manoj to reach school.

Answer:
Rohit travel by car, speed of the car = 60 km/hr
Manoj travel by scooty, speed of the scooty = 40 km/hr
Since, It is given that, they live in the same locality, let the distance be k.
Time taken by Rohit to reach school = \( \frac{Distance}{Speed} = \frac{k}{60} \)
Time taken by Manoj to reach school = \( \frac{k}{40} \)
The ratio between the time taken by Rohit and Manoj to reach school
= \( \frac{k}{60} : \frac{k}{40} \)
= \( \frac{1}{3} : \frac{1}{2} = 2 : 3 \)

In simple words: Since they travel the same distance, the person who goes faster (Rohit) takes less time. Their time ratio is the opposite of their speed ratio.

📝 Teacher's Note: Introduce the concept of inverse variation here. As speed increases, time decreases for the same distance.

🎯 Exam Tip: When distance is unknown, you can assume it is 'k' or '1' to set up the ratio calculation.

Question. In a club having 360 members, 40 play carrom, 96 play table tennis, 144 play badminton and remaining members play volley-ball. If no member plays two or more games, find the ratio of members who play :
(i) carrom to the number of those who play badminton.
(ii) badminton to the number of those who play table-tennis.
(iii) table-tennis to the number of those who play volley-ball.
(iv) volleyball to the number of those who play other games.

Answer:
Total members in a club = 360 members
Members who play carrom = 40
Members who play table tennis = 96
Members who play badminton = 144
Members who play volleyball = 360 – (40 + 96 + 144) = 360 – 280 = 80
(i) Ratio between the members who play carrom to the number of those who play badminton = 40 : 144 \( \implies \) 5 : 18
(ii) Ratio of members who play badminton to the number of those who play table-tennis = 144 : 96 \( \implies \) 6 : 4 = 3 : 2
(iii) Ratio of members who play table tennis to the number of those who play volley-ball = 96 : 80 = 6 : 5
(iv) Ratio of members who play volley-ball to the number of those who play other games = 80 : 280 \( \implies \) 4 : 14 = 2 : 7

In simple words: First find the volleyball players by subtracting everyone else from the total. Then compare the numbers as asked and simplify. "Other games" means the sum of carrom, table tennis, and badminton (280 people).

📝 Teacher's Note: This is a multi-step problem that tests both subtraction skills and the ability to interpret "other games" as a combined group.

🎯 Exam Tip: For part (iv), "other games" means everyone except volleyball players. Ensure you use 280, not the total 360.

Question. The length of a pencil is 18 cm and its radius is 4 cm. Find the ratio of its length to its diameter.

Answer:
Length of a pencil = 18 cm
Radius of a pencil = 4 cm
Diameter of a pencil = 2r = 2 x 4 = 8 cm
Ratio of length of a pencil to its diameter = 18 : 8 = 9 : 4

In simple words: The question gives you radius, but asks for diameter. Double the radius first (4 x 2 = 8), then find the ratio 18:8 and simplify it to 9:4.

📝 Teacher's Note: This is a classic "trap" question. Students often forget to convert radius to diameter before calculating the ratio.

🎯 Exam Tip: Always underline the keywords like "radius" or "diameter" in the question so you don't miss the required conversion.

Question. Ratio of distance of the school from A’s home to the distance of the school from B’s home is 2 : 1.
(i) Who lives nearer to the school ?
(ii) Complete the following table :

Answer:
Ratio of distance of school from A’s home to school from B’s home = 2 : 1
(i) B lives nearest to the school (because the distance for B is 1 part compared to 2 parts for A).
(ii) Let A’s home is 2x km from school and B’s home is x km from school.
Distance of school from A’s home = 2 x Distance of school from B’s home
\( \implies \) If A lives at a distance of 4 km, then B lives at a distance of \( \frac{1}{2} \times 4 = 2 \) km
\( \implies \) If B lives at a distance of 9 km then A lives at a distance of \( 2 \times 9 = 18 \) km
\( \implies \) If A lives at a distance of 8 km then B lives at a distance of \( \frac{1}{2} \times 8 = 4 \) km
\( \implies \) If B lives at a distance of 8 km, then A lives at a distance of \( 2 \times 8 = 16 \) km
\( \implies \) If A lives at a distance of 6 km, then B lives at a distance of \( \frac{1}{2} \times 6 = 3 \) km

In simple words: A lives twice as far as B. So if you know B's distance, multiply it by 2 to get A's. If you know A's distance, divide it by 2 to get B's.

📝 Teacher's Note: Use a simple drawing of a house, school, and another house to visualize the "twice as far" relationship.

🎯 Exam Tip: Clearly show the calculation for each row of the table to avoid simple calculation errors.

Question. The student-teacher ratio in a school is 45: 2. If there are 4050 students in the school, how many teachers must be there?

Answer:
Total number of students = 4050
Let total number of teachers = x
The student-teacher ratio in a school = 45 : 2
\( \therefore \) Required ratio = \( \frac{Total\ number\ of\ students}{Total\ number\ of\ teacher} \)
\( \implies \frac{45}{2} = \frac{4050}{x} \)
\( \implies x = \frac{4050 \times 2}{45} = 180 \) teachers
\( \therefore \) Number of teachers = 180

In simple words: Set up a proportion: \( \frac{45}{2} = \frac{4050}{x} \). Cross multiply and solve to find there are 180 teachers for the 4050 students.

📝 Teacher's Note: Explain that for every 45 students, there are 2 teachers. Dividing 4050 by 45 tells us there are 90 such "groups", and each group needs 2 teachers.

🎯 Exam Tip: Always specify what 'x' represents at the start of your answer to make your logic clear.

Exercise 11(C)

Question. Rs. 120 is to be divided between Hari and Gopi in the ratio 5: 3. How much does each get?

Answer:
Total amount = Rs. 120
Ratio in Hari and Gopi = 5 : 3
Sum of ratios = 5 + 3 = 8
Hari’s share = \( \frac{120 \times 5}{8} = Rs.\ 75 \)
and Gopi’s share = \( \frac{120 \times 3}{8} = Rs.\ 45 \)

In simple words: Total parts are 8 (5 + 3). One part is 120 divided by 8, which is 15. Hari gets 5 parts (Rs. 75) and Gopi gets 3 parts (Rs. 45).

📝 Teacher's Note: Teach the "Sum of Parts" method as a standard procedure for all "divide into ratio" problems.

🎯 Exam Tip: Check your work by adding the two shares together. They must equal the original total (75 + 45 = 120).

Question. Divide 72 in the ratio \( 2\frac{1}{2} : 1\frac{1}{2} \).

Answer:
Given ratio = \( 2\frac{1}{2} : 1\frac{1}{2} = \frac{5}{2} : \frac{3}{2} = 5 : 3 \)
Sum of ratio = 5 + 3 = 8
1st divide = \( \frac{5}{8} \times 72 = 45 \)
2nd divide = \( \frac{3}{8} \times 72 = 27 \)

In simple words: Simplify the mixed fraction ratio first. Since both have the same denominator, it becomes a simple 5:3 ratio. Then divide 72 by the total 8 parts.

📝 Teacher's Note: If the denominators were different, students would need to find a common denominator first to simplify the ratio.

🎯 Exam Tip: Convert mixed fractions to improper fractions correctly before doing anything else.

Question. Divide 81 into three parts in the ratio 2: 3: 4.

Answer:
Given ratio = 2 : 3 : 4
Sum of ratio = 2 + 3 + 4 = 9
1st part = \( \frac{2}{9} \times 81 = 18 \)
2nd part = \( \frac{3}{9} \times 81 = 27 \)
3rd part = \( \frac{4}{9} \times 81 = 36 \)

In simple words: The total number of parts is 9. Divide 81 by 9 to see that one part is 9. Then multiply 9 by 2, 3, and 4 to find the three pieces.

📝 Teacher's Note: The same logic for two-term ratios applies perfectly to three-term ratios as well.

🎯 Exam Tip: Writing the sum of the ratio terms (Sum = 9) is an important step that examiners look for.

Question. Divide Rs 10,400 among A, B and C in the ratio \( \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \).

Answer:
Given ratio = \( \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \)
= \( \frac{1}{2} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 \) (Since L. C. M. of 2, 3 and 4 = 12)
= 6 : 4 : 3
Sum of ratio = 6 + 4 + 3 = 13
A’s part = \( \frac{6}{13} \times 10400 = 6 \times 800 = 4800 \)
B’s part = \( \frac{4}{13} \times 10400 = 4 \times 800 = 3200 \)
C’s part = \( \frac{3}{13} \times 10400 = 3 \times 800 = 2400 \)

In simple words: First, turn the fractions into whole numbers by multiplying by the LCM (12). Then divide the total money using the new 6:4:3 ratio.

📝 Teacher's Note: Emphasize finding the LCM to convert fractional ratios into whole number ratios, which are much easier to work with.

🎯 Exam Tip: 13 is a prime number that often goes into large numbers in math problems. Notice that \( 13 \times 8 = 104 \), which simplifies the calculation significantly.

Question. A profit of Rs 2,500 is to be shared among three persons in the ratio 6 : 9 : 10. How much does each person get?

Answer:
Total profit = Rs 2,500
Given ratio = 6 : 9 : 10
Sum of ratio = 6 + 9 + 10 = 25
Share of 1st person = \( \frac{6}{25} \times 2500 = 6 \times 100 = Rs.\ 600 \)
Share of 2nd person = \( \frac{9}{25} \times 2500 = 9 \times 100 = Rs.\ 900 \)
Share of 3rd person = \( \frac{10}{25} \times 2500 = 10 \times 100 = Rs.\ 1000 \)

In simple words: Divide the profit into 25 equal parts. Each part is Rs. 100. Then give each person their number of parts.

📝 Teacher's Note: This is a standard business application of ratios. It helps students understand how partnerships work in real life.

🎯 Exam Tip: Always include the currency unit (Rs.) in your final answers for money problems.

Question. The angles of a triangle are in the ratio 3: 7: 8. Find the greatest and the smallest angles.

Answer:
Sum of angles of triangle = 180°
Given ratio 3 : 7 : 8
Sum of ratio = 3 + 7 + 8 = 18
Smallest angle = \( \frac{3}{18} \times 180^\circ = 30^\circ \)
Greatest angle = \( \frac{8}{18} \times 180^\circ = 80^\circ \)

In simple words: We know all three angles must add up to 180 degrees. Divide 180 by the total 18 parts, so each part is 10 degrees. The smallest is 3 parts (30°) and the biggest is 8 parts (80°).

📝 Teacher's Note: This combines geometry facts (angles of a triangle) with ratio calculation. It's a very common exam question type.

🎯 Exam Tip: Don't forget the degree (°) symbol! Angles must always be written with proper geometric units.

Question. The sides of a triangle are in the ratio 3: 2: 4. If the perimeter of the triangle is 27 cm, find the length of each side.

Answer:
Ratio in the sides of a triangle = 3 : 2 : 4
Sum of ratios = 3 + 2 + 4 = 9
Perimeter of triangle = 27 cm
Length of first side = \( \frac{27 \times 3}{9} = 9 \) cm
Length of second side = \( \frac{27 \times 2}{9} = 6 \) cm
Length of third side = \( \frac{27 \times 4}{9} = 12 \) cm

In simple words: Perimeter is the total length of all sides. Divide that 27 cm into 9 total parts, then multiply by the ratio numbers 3, 2, and 4.

📝 Teacher's Note: Ensure students understand that perimeter is just the sum of all sides, which matches the concept of "total amount" in ratio problems.

🎯 Exam Tip: Check that the sum of your calculated sides (9 + 6 + 12) equals the given perimeter (27).

Question. An alloy of zinc and copper weighs \( 12\frac{1}{2} \) kg. if in the alloy, the ratio of zinc and copper is 1 : 4, find the weight of copper in it.

Answer:
Weight of alloy = \( 12\frac{1}{2} \) kg = \( \frac{25}{2} \) kg.
Given ratio = 1: 4
Sum of ratio = 1 + 4 = 5
Weight of copper = \( \frac{4}{5} \times \frac{25}{2} \) kg = 2 x 5 = 10 kg

In simple words: The total alloy is 12.5 kg. Copper is 4 parts out of 5 total parts (which is 80% of the total). 80% of 12.5 kg is 10 kg.

📝 Teacher's Note: Using fractions (\( \frac{25}{2} \)) is often easier than using decimals (12.5) for mental calculation and simplification.

🎯 Exam Tip: The question only asks for the weight of copper. You don't need to calculate the weight of zinc, though you can do it to check your work (it should be 2.5 kg).

Question. How will Rs 31500 be shared between A, B and C ; if A gets the double of what B gets, and B gets the double of what C gets ?

Answer:
Let the share of C = 1
Share of B = double of C = 2 x 1 = 2
Share of A = double of B = 2 x 2 = 4
Given ratio (A : B : C) = 4 : 2 : 1
Sum of ratio = 4 + 2 + 1 = 7
A’s share = \( \frac{4}{7} \times Rs\ 31500 = 4 \times Rs\ 4500 = Rs\ 18000 \)
B’s share = \( \frac{2}{7} \times Rs\ 31500 = 2 \times Rs\ 4500 = Rs\ 9000 \)
C’s share = \( \frac{1}{7} \times Rs\ 31500 = 1 \times Rs\ 4500 = Rs\ 4500 \)

In simple words: Start with the smallest person (C). If C has 1 part, B must have 2, and A must have 4. Now divide the total money using this 4:2:1 ratio.

📝 Teacher's Note: This problem teaches how to "chain" ratios together based on descriptions. Always start by assigning a variable to the smallest quantity.

🎯 Exam Tip: Be careful with the "double" logic. If A is double of B, and B is double of C, A is actually four times C.

Question. Mr. Gupta divides Rs 81000 among his three children Ashok, Mohit and Geeta in such a way that Ashok gets four times what Mohit gets and Mohit gets 2.5 times what Geeta gets. Find the share of each of them.

Answer:
Let the share of Geeta = 1
Share of Mohit is (2.5 times of Geeta) = 2.5
Share of Ashok is (4 times of Mohit) = 4 x 2.5 = 10
Ratio = 1 : 2.5 : 10 = 1 x 2 : 2.5 x 2 : 10 x 2 = 2 : 5 : 20
Sum of ratio = 2 + 5 + 20 = 27
Share of Geeta = \( \frac{2}{27} \times Rs\ 81000 = 2 \times Rs\ 3000 = Rs\ 6000 \)
Share of Mohit = \( \frac{5}{27} \times Rs\ 81000 = 5 \times Rs\ 3000 = Rs\ 15000 \)
Share of Ashok = \( \frac{20}{27} \times Rs\ 81000 = 20 \times Rs\ 3000 = Rs\ 60000 \)

In simple words: If Geeta gets 1 part, Mohit gets 2.5, and Ashok gets 10. To make these whole numbers, multiply the whole ratio by 2 to get 2:5:20. Then divide the Rs. 81,000 using these parts.

📝 Teacher's Note: Converting the decimal ratio (1:2.5:10) to a whole number ratio (2:5:20) is a crucial step to make the calculation cleaner.

🎯 Exam Tip: Notice that 27 goes into 81 exactly 3 times. These "nice" numbers in exams usually mean you are on the right track.

Exercise 11(D)

Question. Which ratio is greater:
(i) \( \frac{8}{15} \) or \( \frac{5}{9} \)
(ii) \( \frac{3}{7} \) or \( \frac{6}{13} \)

Answer:
(i) \( \frac{8}{15} \) or \( \frac{5}{9} \)
\( \implies 8 \times 9 \) or \( 15 \times 5 \)
\( \implies 72 \) or \( 75 \)
Since, 75 > 72 \( \therefore \frac{5}{9} \) is greater.
(ii) \( \frac{3}{7} \) or \( \frac{6}{13} \)
\( \implies 3 \times 13 \) or \( 7 \times 6 \)
\( \implies 39 \) or \( 42 \)
Since, 42 > 39 \( \therefore \frac{6}{13} \) is greater.

In simple words: Use cross-multiplication. Multiply the top of the first ratio by the bottom of the second, and the top of the second by the bottom of the first. The bigger result tells you which ratio is larger.

📝 Teacher's Note: This is a faster alternative to finding a common denominator when comparing just two ratios.

🎯 Exam Tip: Write down the cross-multiplication results clearly (e.g., 72 vs 75) to justify your conclusion.

Question. Which ratio is smaller :
(i) \( \frac{9}{17} \) or \( \frac{8}{15} \)
(ii) \( \frac{7}{15} \) or \( \frac{15}{32} \)

Answer:
(i) \( \frac{9}{17} \) or \( \frac{8}{15} \)
\( \implies 9 \times 15 \) or \( 17 \times 8 \)
\( \implies 135 \) or \( 136 \)
Since, 135 < 136 \( \therefore \frac{9}{17} \) is smaller.
(ii) \( \frac{7}{15} \) or \( \frac{15}{32} \)
\( \implies 7 \times 32 \) or \( 15 \times 15 \)
\( \implies 224 \) or \( 225 \)
\( \implies 224 < 225 \therefore \frac{7}{15} \) is smaller.

In simple words: Just like the previous question, cross-multiply and see which result is lower.

📝 Teacher's Note: Remind students to multiply the numerator of a fraction with the denominator of the other to keep the results aligned with the correct fraction.

🎯 Exam Tip: Be careful! If the question asks for the "smaller" ratio, make sure your final sentence answers that, not the "greater" one.

Question. Increase 95 in the ratio 5 : 8.

Answer:
Given ratio = 5 : 8
The increased quantity = \( \frac{8}{5} \times \) given quantity
= \( \frac{8}{5} \times 95 = 152 \)

In simple words: Think of it as "5 becomes 8". To find the new number, multiply the original by the larger number and divide by the smaller one.

📝 Teacher's Note: When increasing a quantity, the larger term of the ratio must be in the numerator of the fraction you multiply by.

🎯 Exam Tip: Sanity check: if you are "increasing," your final answer must be larger than the starting number (152 > 95).

Question. Decrease 275 in the ratio 11 : 7.

Answer:
Given ratio = 11 : 7
The decrease quantity = \( \frac{7}{11} \times \) given quantity.
= \( \frac{7}{11} \times 275 = 175 \)

In simple words: Here, "11 becomes 7". Multiply the original by the smaller number and divide by the larger one to shrink the value.

📝 Teacher's Note: For decreasing, the smaller term goes on top. The original number always represents the first term in the ratio description (11 parts).

🎯 Exam Tip: Check your answer: if decreasing, the result must be smaller than the original (175 < 275).

Question. Decrease 850 in the ratio 17 : 6 and then increase the result in the ratio 5 : 9.

Answer:
The given quantity = 850 and decrease in the ratio = 17 : 6
The decreased quantity = \( \frac{6}{17} \times 850 = 300 \)
Now, the quantity is increased in the ratio 5 : 9
The resulting quantity = \( \frac{9}{5} \times 300 = 540 \)

In simple words: First shrink the 850 down to 300. Then take that 300 and grow it according to the second ratio to get 540.

📝 Teacher's Note: This is a two-step transformation. Make sure students finish the first calculation before starting the second.

🎯 Exam Tip: Don't try to combine both steps into one large fraction unless you are very confident; doing them one by one is safer.

Question. Decrease 850 in the ratio 17 : 6 and then decrease the resulting number again in 4 : 3.

Answer:
The given quantity = 850 and decrease in the ratio = 17 : 6
The decreased quantity = \( \frac{6}{17} \times 850 = 300 \)
Now, the quantity is decreased again in the ratio = 4 : 3
The resulting quantity = \( \frac{3}{4} \times 300 = 225 \)

In simple words: This is a double-decrease. Shrink 850 to 300, then shrink 300 again to 225.

📝 Teacher's Note: Remind students that the second ratio (4:3) applies to the *new* number (300), not the original 850.

🎯 Exam Tip: Both steps involve smaller numbers on top because both are "decreases."

Question. Increase 1200 in the ratio 2 : 3 and then decrease the resulting number in the ratio 10 : 3.

Answer:
The given quantity = 1200 and increase in the ratio is 2 : 3
The increased quantity = \( \frac{3}{2} \times 1200 = 1800 \)
Now, the quantity is decrease in the ratio 10 : 3
= \( \frac{3}{10} \times 1800 = 540 \)

In simple words: First grow the 1200 to 1800. Then shrink it down significantly to 540.

📝 Teacher's Note: Note how ratios can drastically change a value when applied sequentially.

🎯 Exam Tip: Check your steps: "Increase in ratio 2:3" means multiply by 3/2. "Decrease in ratio 10:3" means multiply by 3/10.

Question. Increase 1200 in the ratio 3 : 7 and then increase the resulting number again in the ratio 4 : 7.

Answer:
The given quantity = 1200 increase in the ratio is 3 : 7
The increased quantity = \( \frac{7}{3} \times 1200 = 2800 \)
Now, the quantity is increased again in the ratio 4 : 7
The resulting quantity = \( \frac{7}{4} \times 2800 = 4900 \)

In simple words: This is a double-increase. First 1200 becomes 2800, and then that becomes 4900.

📝 Teacher's Note: Point out that the final result is more than 4 times the original value. Ratios build on each other!

🎯 Exam Tip: For "Increase," the first number in the ratio is the 'old' part and the second is the 'new' part.

Question. The number 650 is decreased to 500 in the ratio a : b, find the ratio a : b.

Answer:
The given quantity = 650
and the decreased quantity = 500
The ratio (a : b) by which 650 is decreased to 500 = \( \frac{650}{500} = \frac{13}{10} \)
The resulting ratio = 13 : 10

In simple words: To find the ratio of change, put the original number on top and the new number on the bottom, then simplify.

📝 Teacher's Note: When a question says "the ratio a:b", it assumes the original value is 'a' parts and the new value is 'b' parts.

🎯 Exam Tip: Remember to simplify the ratio 65:50 by dividing both by 5 to get the final answer 13:10.

Question. The number 800 is increased to 960 in the ratio a : b, find the ratio a : b.

Answer:
The given quantity = 800
and the increased quantity = 960
The ratio (a : b) by which 800 is increased to 960 = \( \frac{800}{960} = \frac{5}{6} \)
The resulting ratio = 5 : 6

In simple words: Compare the original 800 to the new 960. Simplified, this is a 5 to 6 ratio increase.

📝 Teacher's Note: This is the reverse of previous problems. Here we have the before and after values and need to find the rule (ratio) that caused the change.

🎯 Exam Tip: For an increase, the first number 'a' should be smaller than the second number 'b'. For a decrease, 'a' should be larger than 'b'.