Selina Concise Solutions for ICSE Class 7 Mathematics Chapter 6 Ratio and Proportion Including Sharing in a Ratio

Points to Remember

1. Ratio
A ratio is a way to compare two amounts of the same kind and with the same unit by dividing the first amount by the second. We use the colon symbol (:) to show the ratio between two quantities, such as \( a : b \).
Note:
(i) A ratio is just a number and does not have any unit.
(ii) We must always write a ratio in its simplest or lowest form.
(iii) If we multiply or divide both parts of a ratio by the same number, the ratio does not change.

2. Proportion
A proportion is when two ratios are equal, written as \( a : b = c : d \). This means the ratio of the first two terms is equal to the ratio of the last two terms.
Note:
(i) The terms \( a \) and \( d \) are called extreme terms, while \( b \) and \( c \) are called mean terms. Their relation is always \( a \times d = b \times c \).
(ii) The fourth term is called the fourth proportional.

3. Continued Proportion
Three quantities are in continued proportion if the ratio of the first to the second equals the ratio of the second to the third, which is written as \( a : b = b : c \). Here, \( b \) is called the mean proportional between \( a \) and \( c \), and \( c \) is called the third proportional to \( a \) and \( b \).

Exercise 6(A)

Question 1. Express each of the given ratio in its simplest form :
(i) 22 : 66
(ii) 1.5 : 2.5
(iii) \( 6\frac{1}{4} : 12\frac{1}{2} \)
(iv) 40 kg : 1 quintal
(v) 10 paise : Rs. 1
(vi) 200 m : 5 km
(vii) 3 hours : 1 day
(viii) 6 months : \( 1\frac{1}{3} \) years
(ix) \( 1\frac{1}{3} : 2\frac{1}{4} : 2\frac{1}{2} \)
Answer:
(i) \( 22 : 66 = \frac{22}{66} = \frac{22 \div 22}{66 \div 22} = \frac{1}{3} = 1 : 3 \) (since the biggest number that divides both is 22).
(ii) \( 1.5 : 2.5 = \frac{15}{25} = \frac{15 \div 5}{25 \div 5} = \frac{3}{5} = 3 : 5 \) (by dividing by 5).
(iii) \( 6\frac{1}{4} : 12\frac{1}{2} = \frac{25}{4} : \frac{25}{2} = \frac{25}{4} \times \frac{2}{25} = \frac{2}{4} = \frac{1}{2} = 1 : 2 \).
(iv) \( 40 \text{ kg} : 1 \text{ quintal} \). Since \( 1 \text{ quintal} = 100 \text{ kg} \), the ratio is \( 40 \text{ kg} : 100 \text{ kg} = \frac{40 \div 20}{100 \div 20} = \frac{2}{5} = 2 : 5 \).
(v) \( 10 \text{ paise} : \text{Rs. } 1 \). Since \( \text{Rs. } 1 = 100 \text{ paise} \), we write \( 10 : 100 = 1 : 10 \).
(vi) \( 200 \text{ m} : 5 \text{ km} \). Since \( 1 \text{ km} = 1000 \text{ m} \), we have \( 200 \text{ m} : 5000 \text{ m} = \frac{200 \div 200}{5000 \div 200} = 1 : 25 \).
(vii) \( 3 \text{ hours} : 1 \text{ day} \). Since \( 1 \text{ day} = 24 \text{ hours} \), the ratio is \( 3 : 24 = 1 : 8 \).
(viii) \( 6 \text{ months} : 1\frac{1}{3} \text{ years} \). Since \( 1\frac{1}{3} \text{ years} = \frac{4}{3} \times 12 = 16 \text{ months} \text{, the ratio is } 6 : 16 = 3 : 8 \).
(ix) \( 1\frac{1}{3} : 2\frac{1}{4} : 2\frac{1}{2} = \frac{4}{3} : \frac{9}{4} : \frac{5}{2} \). The LCM of 3, 4, and 2 is 12. Multiplying each part by 12 gives \( 16 : 27 : 30 \).
In simple words: To write ratios in their simplest form, first make sure both sides have the same unit. Next, divide both parts by their biggest common factor until they cannot be divided any further.

Exam Tip: Never include units like kg, meters, or hours in your final ratio answer, as ratios are pure numbers.

Question 2. Divide 64 cm long string into two parts in the ratio 5 : 3.
Answer:
The sum of the ratio terms is \( 5 + 3 = 8 \).
The length of the first portion is \( \frac{5}{8} \times 64 \text{ cm} = 40 \text{ cm} \).
The length of the second portion is \( \frac{3}{8} \times 64 \text{ cm} = 24 \text{ cm} \).
In simple words: Add the ratio numbers together to find the total shares. Divide the main value by this total, and then multiply to find the size of each part.

Exam Tip: You can quickly check your answer by adding both parts together to see if they equal the original length (40 + 24 = 64).

Question 3. Rs. 720 is divided between x and y in the ratio 4:5. How many rupees will each get?
Answer:
The total sum of money is Rs. 720, shared in the ratio \( 4 : 5 \).
Adding the ratio terms gives \( 4 + 5 = 9 \).
The share of x is \( \frac{4}{9} \times 720 = \text{Rs. } 320 \).
The share of y is \( \frac{5}{9} \times 720 = \text{Rs. } 400 \).
In simple words: Add the parts of the ratio together to get 9 total shares. Find each person's share by dividing the total money into 9 parts and multiplying by their share number.

Exam Tip: Always write the correct currency units (Rs.) before your final values to avoid losing presentation marks.

Question 4. The angles of a triangle are in the ratio 3 :2 : 7. Find each angle.
Answer:
The ratio of the three angles is \( 3 : 2 : 7 \).
The sum of these ratio terms is \( 3 + 2 + 7 = 12 \).
Since the sum of all angles in any triangle is always \( 180^\circ \):
The first angle is \( \frac{3}{12} \times 180^\circ = 45^\circ \).
The second angle is \( \frac{2}{12} \times 180^\circ = 30^\circ \).
The third angle is \( \frac{7}{12} \times 180^\circ = 105^\circ \).
In simple words: All three angles in a triangle must add up to 180 degrees. Use the ratio to divide 180 degrees into three matching parts.

Exam Tip: Remember to write the degree symbol (\( ^\circ \)) next to each angle value in your final answer.

Question 5. A rectangular field is 100 m by 80 m. Find the ratio of (i) length to its breadth (ii) breadth to its perimeter.
Answer:
The length \( (l) = 100 \text{ m} \) and the breadth \( (b) = 80 \text{ m} \).
The perimeter is calculated as \( 2(l + b) = 2(100 + 80) \text{ m} = 2 \times 180 = 360 \text{ m} \).
(i) The ratio of length to breadth is \( 100 : 80 \). Dividing both by their greatest common divisor (20) gives \( 5 : 4 \).
(ii) The ratio of breadth to the perimeter is \( 80 : 360 \). Dividing both by 40 gives \( 2 : 9 \).
In simple words: First find the perimeter of the rectangle. Then write down the fraction for each comparison and reduce it to its simplest form.

Exam Tip: Pay close attention to the order of the words in the question. "Breadth to perimeter" means breadth is on top and perimeter is on the bottom.

Question 6. The sum of three numbers, whose ratios are \( 3\frac{1}{3} : 4\frac{1}{5} : 6\frac{1}{8} \) is 4917.Find the numbers.
Answer:
The ratio of the numbers is \( 3\frac{1}{3} : 4\frac{1}{5} : 6\frac{1}{8} = \frac{10}{3} : \frac{21}{5} : \frac{49}{8} \).
The least common multiple (LCM) of the denominators 3, 5, and 8 is 120.
Multiplying each term by 120 simplifies the ratio to:
\( 400 : 504 : 735 \).
The sum of these simplified ratio terms is \( 400 + 504 + 735 = 1639 \).
The first number is \( \frac{400}{1639} \times 4917 = 400 \times 3 = 1200 \).
The second number is \( \frac{504}{1639} \times 4917 = 504 \times 3 = 1512 \).
The third number is \( \frac{735}{1639} \times 4917 = 735 \times 3 = 2205 \).
In simple words: Convert the mixed fractions to regular fractions first. Find a common denominator to clear the fractions, and then divide the main number using the new ratio.

Exam Tip: To avoid large calculations, observe that \( 4917 \) is exactly \( 1639 \times 3 \). Simplifying this step first makes the multiplication much easier.

Question 7. The ratio between two quantities is 3 : 4. If the first is Rs. 810, find the second.
Answer:
The ratio of the two quantities is \( 3 : 4 \).
Let the unknown second quantity be \( x \).
This gives us the proportion \( \frac{3}{4} = \frac{810}{x} \).
\( \implies x = \frac{810 \times 4}{3} \)
\( \implies x = 270 \times 4 = 1080 \).
Thus, the second quantity is Rs. 1080.
In simple words: The first number in the ratio represents Rs. 810. Since 3 parts equal 810, each single part is worth 270. Multiply 270 by 4 to get the second quantity.

Exam Tip: You can use cross-multiplication for ratio fractions to quickly find the value of any unknown variable.

Question 8. Two numbers are in the ratio 5 : 7. Their difference is 10. Find the numbers.
Answer:
Let the two numbers be \( 5x \) and \( 7x \).
The difference between them is \( 10 \), so:
\( 7x - 5x = 10 \)
\( \implies 2x = 10 \)
\( \implies x = 5 \).
Therefore, the first number is \( 5 \times 5 = 25 \), and the second number is \( 7 \times 5 = 35 \).
In simple words: The difference between the ratio parts is 2. Since the actual difference is 10, each share is worth 5. Multiply both 5 and 7 by 5 to find the numbers.

Exam Tip: Using a common variable like \( x \) to write your ratios makes setting up word equations very easy.

Question 9. Two numbers are in the ratio 10 : 11. Their sum is 168. Find the numbers.
Answer:
Let the two numbers be \( 10x \) and \( 11x \).
Their total sum is \( 168 \), so:
\( 10x + 11x = 168 \)
\( \implies 21x = 168 \)
\( \implies x = \frac{168}{21} = 8 \).
Thus, the first number is \( 10 \times 8 = 80 \), and the second number is \( 11 \times 8 = 88 \).
In simple words: Add the ratio terms together to get 21. Since the total sum is 168, divide 168 by 21 to find that one part is 8. Multiply each ratio term by 8 to get the numbers.

Exam Tip: When dividing numbers, make sure to write down the final values clearly so that the examiner can easily see both answers.

Question 10. A line is divided in two parts in the ratio 2.5 : 1.3. If the smaller one is 35.1 cm, find the length of the line.
Answer:
We can simplify the ratio \( 2.5 : 1.3 \) by multiplying both sides by 10, which gives \( 25 : 13 \).
The sum of these simplified ratio terms is \( 25 + 13 = 38 \).
The smaller part is represented by the ratio term 13, and its length is \( 35.1 \text{ cm} \).
Let the total length of the line be \( L \).
We can set up the relation:
\( \frac{13}{38} \times L = 35.1 \)
\( \implies L = \frac{35.1 \times 38}{13} \)
\( \implies L = 2.7 \times 38 = 102.6 \text{ cm} \).
So, the total length of the line is \( 102.6 \text{ cm} \).
In simple words: Turn the decimal ratio into whole numbers. Since the smaller part of 13 shares is 35.1 cm, each share is 2.7 cm. Multiply 2.7 cm by the total of 38 shares to find the whole length.

Exam Tip: Removing decimals from a ratio right at the start helps prevent calculation mistakes later in the problem.

Question 11. In a class, the ratio of boys to the girls is 7:8. What part of the whole class are girls.
Answer:
The ratio of boys to girls is \( 7 : 8 \).
Adding the ratio parts gives the total class shares as \( 7 + 8 = 15 \).
Since girls have 8 shares out of the total 15, the fraction of the class made up of girls is \( \frac{8}{15} \).
In simple words: Together, the boys and girls make up 15 parts of the class. Because there are 8 parts for girls, they represent 8 out of 15 parts of the total class.

Exam Tip: A fractional part answer should always be simplified to its lowest terms, though \( \frac{8}{15} \) is already in its simplest form.

Question 12. The population of a town is 1,80,000, out of which males are 1/3 of the whole population. Find the number of females. Also, find the ratio of the number of females to the whole population.
Answer:
The total population of the town is \( 180,000 \).
The number of males is \( \frac{1}{3} \times 180,000 = 60,000 \).
The number of females is \( 180,000 - 60,000 = 120,000 \).
The ratio of females to the entire population is \( 120,000 : 180,000 = \frac{120,000}{180,000} = \frac{2}{3} = 2 : 3 \).
In simple words: Find the male population first, and subtract it from the total to find the number of females. Then, write the number of females as a ratio to the total population and simplify.

Exam Tip: Read carefully whether the final ratio asks for females compared to males or females compared to the entire population.

Question 13. Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. Find the ratio between :
(i) the weights of metals A and B in the alloy.
(ii) the weight of metal B and the weight of the alloy.
Answer:
The total weight of the alloy is \( 10 \text{ gm} \), and metal A weighs \( 7.5 \text{ gm} \).
The weight of metal B is \( 10 - 7.5 = 2.5 \text{ gm} \).
(i) The ratio of the weight of metal A to metal B is \( 7.5 : 2.5 = 75 : 25 = 3 : 1 \).
(ii) The ratio of the weight of metal B to the entire alloy is \( 2.5 : 10 = 25 : 100 = 1 : 4 \).
In simple words: Find how much of metal B is in the alloy. Then, use this weight to find both ratios and simplify them into whole numbers.

Exam Tip: When simplifying decimal weights, write them as whole fractions before reducing them to avoid easy mistakes.

Question 14. The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide Rs. 3,150 in the ratio of their ages.
Answer:
First, let us convert both ages into months:
Age of boy A = \( (6 \times 12) + 8 = 72 + 8 = 80 \text{ months} \).
Age of boy B = \( (7 \times 12) + 4 = 84 + 4 = 88 \text{ months} \).
The ratio of their ages is \( 80 : 88 = 10 : 11 \) (by dividing both parts by 8).
The sum of these ratio terms is \( 10 + 11 = 21 \).
The share of boy A is \( \frac{10}{21} \times 3150 = \text{Rs. } 1500 \).
The share of boy B is \( \frac{11}{21} \times 3150 = \text{Rs. } 1650 \).
In simple words: Convert both ages to months so they are in the same units. Simplify this ratio to 10:11, and then use it to divide the Rs. 3,150.

Exam Tip: Never calculate ratios using mixed units like years and months. Always convert everything to the smaller unit (months) first.

Question 15. Three persons start a business and spend Rs. 25,000; Rs. 15,000 and Rs. 40,000 respectively. Find the share of each out of a profit of Rs. 14,400 in a year.
Answer:
The ratio of their investments is:
\( 25000 : 15000 : 40000 = 25 : 15 : 40 = 5 : 3 : 8 \).
The sum of these ratio parts is \( 5 + 3 + 8 = 16 \).
The share of partner A is \( \frac{5}{16} \times 14400 = 5 \times 900 = \text{Rs. } 4500 \).
The share of partner B is \( \frac{3}{16} \times 14400 = 3 \times 900 = \text{Rs. } 2700 \).
The share of partner C is \( \frac{8}{16} \times 14400 = 8 \times 900 = \text{Rs. } 7200 \).
In simple words: Simplify the ratio of the starting money they each put in. Then divide the yearly profit of Rs. 14,400 using this simplified ratio.

Exam Tip: Profits in partnership problems are always distributed in the exact ratio of the initial capitals invested by each person.

Question 16. A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. Find the share of each.
Answer:
Let the share of the second person be \( x \).
This means the share of the first person is \( \frac{3}{5}x \).
The ratio of their shares is \( \frac{3}{5}x : x = 3 : 5 \).
The sum of these ratio terms is \( 3 + 5 = 8 \).
The first person's share of land is \( \frac{3}{8} \times 600 = 225 \text{ sq m} \).
The second person's share of land is \( \frac{5}{8} \times 600 = 375 \text{ sq m} \).
In simple words: If one person gets 3 parts, the other gets 5 parts. Together they have 8 parts. Divide the total area of 600 sq m using this 3:5 ratio.

Exam Tip: If the problem gives fractional relations, write them as whole numbers (like 3 and 5) to keep the arithmetic simple.

Question 17. Two poles of different heights are standing vertically on a horizontal field. At a particular time, the ratio between the lengths of their shadows is 2 :3. If the height of the smaller pole is 7.5 m, find the height of the other pole.
Answer:
The heights of vertical objects and their shadows are directly proportional.
The ratio of the shadow lengths is \( 2 : 3 \).
Let the height of the taller pole be \( h \).
We are given the smaller pole height is \( 7.5 \text{ m} \), so we can write:
\( \frac{2}{3} = \frac{7.5}{h} \)
\( \implies h = \frac{7.5 \times 3}{2} \)
\( \implies h = \frac{22.5}{2} = 11.25 \text{ m} \).
So, the height of the taller pole is \( 11.25 \text{ m} \).
In simple words: The ratio of the actual heights matches the ratio of their shadows. Use this proportion to calculate the height of the taller pole.

Exam Tip: State clearly in your answer that object height and shadow length are directly proportional, as this is the main geometric rule used here.

Question 18. Two numbers are in the ratio 4 : 7. If their L.C.M. is 168, find the numbers.
Answer:
Let the two numbers be \( 4x \) and \( 7x \).
The least common multiple (LCM) of \( 4x \) and \( 7x \) is \( 4 \times 7 \times x = 28x \).
Since their LCM is 168:
\( 28x = 168 \)
\( \implies x = \frac{168}{28} = 6 \).
The numbers are \( 4 \times 6 = 24 \) and \( 7 \times 6 = 42 \).
In simple words: Write the numbers as 4x and 7x. Their LCM is 28x, which equals 168. Find x, and use it to get the final numbers.

Exam Tip: For any two numbers written as \( ax \) and \( bx \) where \( a \) and \( b \) have no common factors, their LCM is always \( abx \).

Question 19. Rs. 300 is divided between A and B in such a way that A gets half of B. Find :
(i) the ratio between the shares of A and B.
(ii) the share of A and the share of B.
Answer:
Let B's share be \( x \).
Then A's share is \( \frac{1}{2}x \).
(i) The ratio of their shares is \( \frac{1}{2}x : x = 1 : 2 \).
(ii) The sum of the ratio parts is \( 1 + 2 = 3 \).
A's share is \( \frac{1}{3} \times 300 = \text{Rs. } 100 \).
B's share is \( \frac{2}{3} \times 300 = \text{Rs. } 200 \).
In simple words: Since A gets half of what B gets, their sharing ratio is 1:2. Use this simple ratio to divide the Rs. 300.

Exam Tip: Writing the ratio directly as 1 : 2 avoids dealing with fractions and keeps the calculations quick and clean.

Question 20. The ratio between two numbers is 5 : 9. Find the numbers, if their H.C.F. is 16.
Answer:
Let the numbers be \( 5x \) and \( 9x \).
The highest common factor (HCF) of \( 5x \) and \( 9x \) is \( x \).
Since the HCF is given as 16, we have \( x = 16 \).
The first number is \( 5 \times 16 = 80 \).
The second number is \( 9 \times 16 = 144 \).
In simple words: The HCF is the common multiplier. Just multiply both parts of the ratio by the HCF (16) to find the two numbers.

Exam Tip: If you are given the HCF of two numbers in a ratio, you can find the actual numbers directly by multiplying each ratio term by that HCF.

Question 21. A bag contains Rs. 1,600 in the form of Rs. 10 and Rs. 20 notes. If the ratio between the numbers of Rs. 10 and Rs. 20 notes is 2 : 3; find the total number of notes in all.
Answer:
Let the number of Rs. 10 notes be \( 2x \) and the number of Rs. 20 notes be \( 3x \).
The value from Rs. 10 notes is \( 10 \times 2x = 20x \).
The value from Rs. 20 notes is \( 20 \times 3x = 60x \).
The total value is Rs. 1600, so:
\( 20x + 60x = 1600 \)
\( \implies 80x = 1600 \)
\( \implies x = 20 \).
The number of Rs. 10 notes is \( 2 \times 20 = 40 \).
The number of Rs. 20 notes is \( 3 \times 20 = 60 \).
The total number of notes is \( 40 + 60 = 100 \).
In simple words: Set the number of notes using the ratio. Find the total value by multiplying the notes by their values, solve for the multiplier, and then add the notes together.

Exam Tip: Remember to multiply the number of notes by their face value (10 or 20) to set up your value equation correctly.

Question 22. The ratio between the prices of a scooter and a refrigerator is 4 : 1. If the scooter costs Rs. 45,000 more than the refrigerator, find the price of the refrigerator.
Answer:
Let the price of the scooter be \( 4x \) and the price of the refrigerator be \( 1x \).
According to the given problem, the scooter costs Rs. 45,000 more than the refrigerator.
So, we can write:
\( 4x - 1x = 45000 \)
\( \implies 3x = 45000 \)
\( \implies x = \frac{45000}{3} \)
\( \implies x = 15000 \)
So, the price of the refrigerator is Rs. 15,000.
In simple words: Since the scooter is four times the price of the fridge, the difference is three times the fridge's price. This difference is Rs. 45,000, which means one part (the fridge) is Rs. 15,000.
Exam Tip: Always define your variables using the given ratio to easily set up and solve the equation.

Exercise 6(B)

Question 1. Check whether the following quantities form a proportion or not?
(i) \( 3x, 7x, 24 \) and \( 56 \)
(ii) \( 0.8, 3, 2.4 \) and \( 9 \)
(iii) \( 1\frac{1}{2}, 3\frac{1}{4}, 4\frac{1}{2} \) and \( 9\frac{3}{4} \)
(iv) \( 0.4, 0.5, 2.9 \) and \( 3.5 \)
(v) \( 2\frac{1}{2}, 5\frac{1}{2}, 3.0 \) and \( 6.0 \)
Answer:
(i) Let us test if the product of the first and last terms matches the product of the middle two terms.
Product of extremes = \( 3x \times 56 = 168x \)
Product of means = \( 7x \times 24 = 168x \)
Since the product of extremes equals the product of means, the given numbers are in proportion.
(ii) For the numbers \( 0.8, 3, 2.4, \) and \( 9 \):
Product of outer terms = \( 0.8 \times 9 = 7.2 \)
Product of inner terms = \( 3 \times 2.4 = 7.2 \)
Because both products are equal, these quantities form a proportion.
(iii) First, let's convert the mixed fractions into improper fractions:
\( 1\frac{1}{2} = \frac{3}{2} \)
\( 3\frac{1}{4} = \frac{13}{4} \)
\( 4\frac{1}{2} = \frac{9}{2} \)
\( 9\frac{3}{4} = \frac{39}{4} \)
Now, let's find the product of the extremes and means:
Product of extremes = \( \frac{3}{2} \times \frac{39}{4} = \frac{117}{8} \)
Product of means = \( \frac{13}{4} \times \frac{9}{2} = \frac{117}{8} \)
As the two values are equal, the terms are in proportion.
(iv) Let's find the products for \( 0.4, 0.5, 2.9, \) and \( 3.5 \):
Product of extremes = \( 0.4 \times 3.5 = 1.4 \)
Product of means = \( 0.5 \times 2.9 = 1.45 \)
Since the product of extremes is not equal to the product of means (\( 1.4 \neq 1.45 \)), these numbers do not form a proportion.
(v) Converting the mixed fractions gives \( \frac{5}{2} \) and \( \frac{11}{2} \).
Now, let's check the products:
Product of extremes = \( \frac{5}{2} \times 6 = 15 \)
Product of means = \( \frac{11}{2} \times 3 = \frac{33}{2} = 16.5 \)
Since the products are different (\( 15 \neq 16.5 \)), these numbers are not in proportion.
In simple words: To check if four numbers are in proportion, multiply the first and last numbers, and then multiply the middle two numbers. If the two answers are the same, they are in proportion.

Exam Tip: Remember that for four terms \( a, b, c, \) and \( d \) to be in proportion, the relation \( a \times d = b \times c \) must always hold true.

Question 2. Find the fourth proportional of
(i) 3, 12 and 4
(ii) 5, 9 and 45
(iii) 2.1, 1.5 and 8.4
(iv) \( \frac{1}{3}, \frac{2}{5} \) and 8.4
(v) 4 hours 40 minutes, 1 hour 10 minutes and 16 hours
Answer:
(i) Let the fourth proportional be \( x \).
Then, \( 3 : 12 = 4 : x \)
\( \implies 3 \times x = 12 \times 4 \)
\( \implies x = \frac{12 \times 4}{3} = 16 \)
So, the fourth proportional is 16.
(ii) Let \( x \) represent the fourth proportional.
Then, \( 5 : 9 = 45 : x \)
\( \implies 5 \times x = 9 \times 45 \)
\( \implies x = \frac{9 \times 45}{5} = 81 \)
So, the fourth proportional is 81.
(iii) Let \( x \) be the fourth proportional.
Then, \( 2.1 : 1.5 = 8.4 : x \)
\( \implies 2.1 \times x = 1.5 \times 8.4 \)
\( \implies x = \frac{1.5 \times 8.4}{2.1} \)
Since \( \frac{8.4}{2.1} = 4 \), we have:
\( \implies x = 1.5 \times 4 = 6 \)
So, the fourth proportional is 6.
(iv) Let \( x \) be the fourth proportional.
Then, \( \frac{1}{3} : \frac{2}{5} = 8.4 : x \)
\( \implies \frac{1}{3} \times x = \frac{2}{5} \times 8.4 \)
\( \implies x = \frac{\frac{2}{5} \times 8.4}{\frac{1}{3}} \)
\( \implies x = \frac{2}{5} \times 8.4 \times 3 \)
\( \implies x = \frac{2 \times 8.4 \times 3}{5} \)
\( \implies x = \frac{50.4}{5} = 10.08 \)
So, the fourth proportional is 10.08.
(v) First, convert each time duration into minutes:
4 hours 40 minutes = \( 4 \times 60 + 40 = 280 \) minutes
1 hour 10 minutes = \( 1 \times 60 + 10 = 70 \) minutes
16 hours = \( 16 \times 60 = 960 \) minutes
Let \( x \) (in minutes) be the fourth proportional.
Then, \( 280 : 70 = 960 : x \)
\( \implies 280 \times x = 70 \times 960 \)
\( \implies x = \frac{70 \times 960}{280} \)
\( \implies x = \frac{960}{4} = 240 \) minutes
Now, convert 240 minutes back into hours:
\( \frac{240}{60} = 4 \) hours
So, the fourth proportional is 4 hours.
In simple words: To find the fourth proportional, let it be \( x \). Set up the ratio as the first number over the second equals the third over \( x \), and solve for \( x \).

Exam Tip: When finding proportions with mixed units like hours and minutes, always convert all terms to the smallest unit (minutes) before calculating.

Question 3. Find the third proportional of
(i) 27 and 9
(ii) 2 m 40cm and 40cm
(iii) 1.8 and 0.6
(iv) \( \frac{1}{7} \) and \( \frac{3}{14} \)
(v) 1.6 and 0.8
Answer:
(i) Let the third proportional be \( x \).
Then, \( 27 : 9 = 9 : x \)
\( \implies 27 \times x = 9 \times 9 \)
\( \implies x = \frac{9 \times 9}{27} = \frac{81}{27} = 3 \)
So, the third proportional is 3.
(ii) First, convert 2 m 40 cm into centimeters:
2 m 40 cm = \( 2 \times 100 + 40 = 240 \) cm
Let the third proportional be \( x \) cm.
Then, \( 240 : 40 = 40 : x \)
\( \implies 240 \times x = 40 \times 40 \)
\( \implies x = \frac{40 \times 40}{240} \)
\( \implies x = \frac{1600}{240} = \frac{20}{3} = 6\frac{2}{3} \) cm
So, the third proportional is \( 6\frac{2}{3} \) cm.
(iii) Let \( x \) be the third proportional.
Then, \( 1.8 : 0.6 = 0.6 : x \)
\( \implies 1.8 \times x = 0.6 \times 0.6 \)
\( \implies x = \frac{0.6 \times 0.6}{1.8} \)
\( \implies x = \frac{0.36}{1.8} = 0.2 \)
So, the third proportional is 0.2.
(iv) Let \( x \) be the third proportional.
Then, \( \frac{1}{7} : \frac{3}{14} = \frac{3}{14} : x \)
\( \implies \frac{1}{7} \times x = \frac{3}{14} \times \frac{3}{14} \)
\( \implies \frac{1}{7} \times x = \frac{9}{196} \)
\( \implies x = \frac{9}{196} \times 7 = \frac{9}{28} \)
So, the third proportional is \( \frac{9}{28} \).
(v) Let the third proportional be \( x \).
Then, \( 1.6 : 0.8 = 0.8 : x \)
\( \implies 1.6 \times x = 0.8 \times 0.8 \)
\( \implies x = \frac{0.8 \times 0.8}{1.6} \)
\( \implies x = \frac{0.64}{1.6} = 0.4 \)
So, the third proportional is 0.4.
In simple words: The third proportional of two numbers means you repeat the second number in the middle. So, first number over second equals second number over the unknown number, then solve.

Exam Tip: For any two numbers \( a \) and \( b \), the third proportional is always calculated as \( \frac{b^2}{a} \). Keeping this formula in mind helps avoid errors.

Question 4. Find the mean proportional between
(i) 16 and 4
(ii) 3 and 27
(iii) 0.9 and 2.5
(iv) 0.6 and 9.6
(v) \( \frac{1}{4} \) and \( \frac{1}{16} \)
Answer:
(i) The mean proportional is the square root of the product of the two numbers.
Mean proportional = \( \sqrt{16 \times 4} = \sqrt{64} = 8 \)
(ii) Mean proportional = \( \sqrt{3 \times 27} = \sqrt{81} = 9 \)
(iii) Mean proportional = \( \sqrt{0.9 \times 2.5} \)
We can write this as:
\( \sqrt{\frac{9}{10} \times \frac{25}{10}} = \sqrt{\frac{225}{100}} = \frac{15}{10} = 1.5 \)
(iv) Mean proportional = \( \sqrt{0.6 \times 9.6} \)
This can be written as:
\( \sqrt{\frac{6}{10} \times \frac{96}{10}} = \sqrt{\frac{576}{100}} = \frac{24}{10} = 2.4 \)
(v) Mean proportional = \( \sqrt{\frac{1}{4} \times \frac{1}{16}} = \sqrt{\frac{1}{64}} = \frac{1}{8} \)
In simple words: To find the mean proportional of two numbers, multiply them together and then find the square root of the result.

Exam Tip: If you are working with decimals, convert them to fractions first. This makes it much easier to find the square root without making a decimal mistake.

Question 5.
(i) If A : B = 3 : 5 and B : C = 4 : 7, find A : B : C
(ii) If x : y = 2 : 3 and y : z = 5 : 7, find x : y : z
(iii) If m : n = 4 : 9 and n : s = 3 : 7, find m : s
(iv) If P : Q = \( \frac{1}{2} : \frac{1}{3} \) and Q : R = \( 1\frac{1}{2} : 1\frac{1}{3} \), find P : R.
(v) If a : b = 1.5 : 3.5 and b : c = 5 : 6, find a : c.
(vi) If \( 1\frac{1}{4} : 2\frac{1}{3} = p : q \) and \( q : r = 4\frac{1}{2} : 5\frac{1}{4} \); find p : r
Answer:
(i) To combine the ratios, we make the term \( B \) equal in both.
We are given:
\( A : B = 3 : 5 \)
\( B : C = 4 : 7 \)
The common term is \( B \). The LCM of 5 and 4 is 20.
Multiply the first ratio by 4:
\( A : B = (3 \times 4) : (5 \times 4) = 12 : 20 \)
Multiply the second ratio by 5:
\( B : C = (4 \times 5) : (7 \times 5) = 20 : 35 \)
Now, combining them, we get:
\( A : B : C = 12 : 20 : 35 \)
(ii) Here, the common term is \( y \). The LCM of 3 and 5 is 15.
Multiply the terms in \( x : y \) by 5:
\( x : y = (2 \times 5) : (3 \times 5) = 10 : 15 \)
Multiply the terms in \( y : z \) by 3:
\( y : z = (5 \times 3) : (7 \times 3) = 15 : 21 \)
Therefore, the combined ratio is:
\( x : y : z = 10 : 15 : 21 \)
(iii) To find the ratio of \( m \) to \( s \), we multiply the two given ratios:
\( \frac{m}{s} = \frac{m}{n} \times \frac{n}{s} \)
\( \implies \frac{m}{s} = \frac{4}{9} \times \frac{3}{7} \)
\( \implies \frac{m}{s} = \frac{4 \times 1}{3 \times 7} = \frac{4}{21} \)
Thus, the ratio \( m : s = 4 : 21 \).
(iv) Let's simplify the ratios first:
\( \frac{P}{Q} = \frac{\frac{1}{2}}{\frac{1}{3}} = \frac{3}{2} \)
Next, convert the mixed fractions for \( Q : R \):
\( Q : R = 1\frac{1}{2} : 1\frac{1}{3} = \frac{3}{2} : \frac{4}{3} \)
\( \implies \frac{Q}{R} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} \)
Now, find \( P : R \) by multiplying these two fractions:
\( \frac{P}{R} = \frac{P}{Q} \times \frac{Q}{R} = \frac{3}{2} \times \frac{9}{8} = \frac{27}{16} \)
So, \( P : R = 27 : 16 \).
(v) First, simplify the ratio \( a : b \):
\( \frac{a}{b} = \frac{1.5}{3.5} = \frac{15}{35} = \frac{3}{7} \)
We also have:
\( \frac{b}{c} = \frac{5}{6} \)
To find \( a : c \), multiply the fractions:
\( \frac{a}{c} = \frac{a}{b} \times \frac{b}{c} = \frac{3}{7} \times \frac{5}{6} \)
\( \implies \frac{a}{c} = \frac{1 \times 5}{7 \times 2} = \frac{5}{14} \)
Therefore, \( a : c = 5 : 14 \).
(vi) First, change the mixed fractions to improper fractions:
\( p : q = 1\frac{1}{4} : 2\frac{1}{3} = \frac{5}{4} : \frac{7}{3} \)
\( \implies \frac{p}{q} = \frac{5}{4} \times \frac{3}{7} = \frac{15}{28} \)
Now, do the same for \( q : r \):
\( q : r = 4\frac{1}{2} : 5\frac{1}{4} = \frac{9}{2} : \frac{21}{4} \)
\( \implies \frac{q}{r} = \frac{9}{2} \times \frac{4}{21} = \frac{3 \times 2}{1 \times 7} = \frac{6}{7} \)
To find \( p : r \), multiply these two results:
\( \frac{p}{r} = \frac{p}{q} \times \frac{q}{r} = \frac{15}{28} \times \frac{6}{7} \)
\( \implies \frac{p}{r} = \frac{15 \times 3}{14 \times 7} = \frac{45}{98} \)
Thus, \( p : r = 45 : 98 \).
In simple words: To find the ratio between the first and last letters (like \( p \) and \( r \)), you just multiply the two separate fraction ratios together.

Exam Tip: For three-term ratios like \( A : B : C \), always make sure the middle term has the same value by using their lowest common multiple.

Question 6. If x : y = 5 : 4 and 2 : x = 3 : 8, find the value of y.
Answer:
From the second given ratio, we can write:
\( \frac{2}{x} = \frac{3}{8} \)
By cross-multiplying, we solve for \( x \):
\( 3 \times x = 2 \times 8 \)
\( \implies 3x = 16 \)
\( \implies x = \frac{16}{3} \)
Now, using the first ratio:
\( \frac{x}{y} = \frac{5}{4} \)
\( \implies y = x \times \frac{4}{5} \)
Substitute the value of \( x = \frac{16}{3} \) into this equation:
\( y = \frac{16}{3} \times \frac{4}{5} = \frac{64}{15} \)
Therefore, the value of \( y \) is \( \frac{64}{15} \) (or \( 4\frac{4}{15} \)).
In simple words: First find the value of \( x \) from the second ratio. Then, put that number into the first ratio to find \( y \).

Exam Tip: When given multiple ratios, identify which equation contains only one unknown variable first. Solve that one first before substituting its value into the other equations.

Question 7. Find the value of x, when 2.5 : 4 = x : 7.5.
Answer:
Since the ratios form a proportion, we have:
\( 4 \times x = 2.5 \times 7.5 \)
\( \implies x = \frac{2.5 \times 7.5}{4} \)
Let's convert the decimal numbers into fractions to make calculations simpler:
\( x = \frac{25 \times 75}{4 \times 100} \)
\( \implies x = \frac{25 \times 3}{4 \times 4} = \frac{75}{16} \)
Converting \( \frac{75}{16} \) into a mixed fraction gives:
\( x = 4\frac{11}{16} \)
Thus, the value of \( x \) is \( 4\frac{11}{16} \) (or \( 4.6875 \)).
In simple words: Multiply the outer numbers and make them equal to the middle numbers multiplied together. Then divide to get the value of \( x \).

Exam Tip: If you are working with decimals, converting them to fractions like \( \frac{25}{10} \) and \( \frac{75}{10} \) simplifies the division process and reduces errors.

Question 8. Show that 2, 12 and 72 are in continued proportion.
Answer:
For three numbers \( a, b, \) and \( c \) to be in a continued proportion, the ratio of the first to the second must be equal to the ratio of the second to the third:
\( a : b = b : c \)
Here, the numbers are \( 2, 12, \) and \( 72 \). Let's check their ratios:
First ratio: \( \frac{2}{12} = \frac{1}{6} \)
Second ratio: \( \frac{12}{72} = \frac{1}{6} \)
Since the ratio \( \frac{2}{12} \) is equal to \( \frac{12}{72} \), the numbers 2, 12, and 72 form a continued proportion.
In simple words: To show three numbers are in continued proportion, divide the first by the second, and the second by the third. If the two fractions are equal, they are in continued proportion.

Exam Tip: Alternatively, you can prove continued proportion by checking if the square of the middle term is equal to the product of the first and third terms (\( b^2 = a \times c \)). Here, \( 12^2 = 144 \) and \( 2 \times 72 = 144 \).