A ratio is a relationship between two quantities of the same kind and with the same units, obtained by dividing the first quantity by the second.
The ratio of $a$ to $b$ is written as $a : b$ or $\frac{a}{b}$ (read as “$a$ is to $b$“).
Terms of a Ratio: In the ratio $a : b$:
$a$ is called the Antecedent (first term).
$b$ is called the Consequent (second term).
Examples:
Comparing Marks:
Student 1 gets 2 marks.
Student 2 gets 4 marks.
Ratio $= \frac{2}{4} = \frac{1}{2} = 1:2$.
Comparing Ages:
Ratio between 18 years and 24 years $= \frac{18}{24} = \frac{3}{4} = 3:4$.
Different Units (Must Convert First):
Ratio of 18 mm to 3 cm?
Convert 3 cm to mm: $3 \text{ cm} = 30 \text{ mm}$.
Ratio $= \frac{18}{30} = \frac{3}{5} = 3:5$.
A ratio $a:b$ is in its simplest form if the Highest Common Factor (H.C.F.) of $a$ and $b$ is 1.
Example: $12 : 16$ is not in simplest form.
Divide both by their H.C.F (4): $\frac{12}{16} = \frac{3}{4} = 3:4$.
Properties:
To divide a total quantity $N$ into parts based on a ratio $a:b$ (or $a:b:c$):
If a number $N$ is divided into two parts in the ratio $a : b$:
First Part $= \frac{a}{a+b} \times N$
Second Part $= \frac{b}{a+b} \times N$
If a number $N$ is divided into three parts in the ratio $a : b : c$:
First Part $= \frac{a}{a+b+c} \times N$
Second Part $= \frac{b}{a+b+c} \times N$
Third Part $= \frac{c}{a+b+c} \times N$
Example 1: Distribution of Money
Divide Rs 2400 into three shares in the ratio $1 : 2 : 3$.
Solution:
Total parts $= 1 + 2 + 3 = 6$.
1st Share $= \frac{1}{6} \times 2400 = 400$
2nd Share $= \frac{2}{6} \times 2400 = 800$
3rd Share $= \frac{3}{6} \times 2400 = 1200$
Answer: Rs 400, Rs 800, Rs 1200.
Example 2: Numbers and Squares
The ratio of three numbers is $3 : 4 : 5$ and the sum of their squares is 1250. Find the sum of the numbers.
Solution:
Let the numbers be $3k, 4k, 5k$.
$(3k)^2 + (4k)^2 + (5k)^2 = 1250$
$9k^2 + 16k^2 + 25k^2 = 1250$
$50k^2 = 1250 \implies k^2 = 25 \implies k = 5$.
Sum of numbers $= 3k + 4k + 5k = 12k$.
$12 \times 5 = 60$.
Answer: 60.
Example 3: Ratio Change
Salaries of Ravi and Sumit are in the ratio $2:3$. If the salary of each is increased by Rs 4000, the new ratio becomes $40:57$. What is Sumit’s present salary?
Solution:
Let salaries be $2k$ and $3k$.
$\frac{2k + 4000}{3k + 4000} = \frac{40}{57}$
Cross multiply: $57(2k + 4000) = 40(3k + 4000)$
$114k + 228000 = 120k + 160000$
$6k = 68000 \implies 3k = 34000$
Sumit’s salary ($3k$) is Rs 34,000.
New Salary $= 34000 + 4000 = 38000$. (Note: Question asks for present salary, which is usually the original before increase, but slide solves for the increased amount. Based on standard interpretation, original is Rs 34,000).
Example 4: Comparing Fractions
If $A:B:C = 1\frac{3}{4} : 2\frac{1}{2} : 1\frac{1}{6}$, find the relationship between A, B, and C.
Solution:
Convert to improper fractions: $\frac{7}{4} : \frac{5}{2} : \frac{7}{6}$.
L.C.M. of denominators (4, 2, 6) is 12. Multiply all terms by 12.
$\frac{7 \times 12}{4} : \frac{5 \times 12}{2} : \frac{7 \times 12}{6}$
$21 : 30 : 14$.
Answer: $B > A > C$.
A statement of equality of two ratios is called a Proportion.
If $a : b = c : d$, then $a, b, c, d$ are in proportion.
It is written as $a : b :: c : d$ (read as “$a$ is to $b$ as $c$ is to $d$“).
In the proportion $a : b :: c : d$:
Extremes: The first ($a$) and fourth ($d$) terms.
Means: The second ($b$) and third ($c$) terms.
Fourth Proportional: If $a:b :: c:d$, then $d$ is the fourth proportional.
Continued Proportion: If $a:b :: b:c$, we say $a, b, c$ are in continued proportion.
$c$ is the Third Proportional.
$b$ is the Mean Proportional between $a$ and $c$.
Formula: $b^2 = ac$ or $b = \sqrt{ac}$.
Addendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+c}{b+d} = \frac{a}{b} = \frac{c}{d}$.
Subtrahendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a-c}{b-d} = \frac{a}{b} = \frac{c}{d}$.
Componendo: $\frac{a+b}{b} = \frac{c+d}{d}$.
Dividendo: $\frac{a-b}{b} = \frac{c-d}{d}$.
Componendo & Dividendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.
Example 1: Finding an Unknown Term
If $5, 15, x, 90$ are in proportion, find $x$.
Solution:
$5 : 15 :: x : 90$
Product of Means = Product of Extremes
$15 \times x = 5 \times 90$
$15x = 450 \implies x = 30$.
Example 2: Mean Proportional
Find the mean proportional between 8 and 98.
Solution:
Mean Proportional $= \sqrt{8 \times 98}$
$= \sqrt{16 \times 49} = 4 \times 7 = 28$.
Example 3: Continued Proportion
If $81, x, x, 256$ are in proportion, find $x$.
Solution:
$\frac{81}{x} = \frac{x}{256}$
$x^2 = 81 \times 256$
$x = \sqrt{81} \times \sqrt{256} = 9 \times 16 = 144$.
A method in which the value of a single unit is found first to find the value of any required quantity.
Example: If 5 pens cost Rs 30, what is the cost of 8 pens?
Cost of 1 pen $= \frac{30}{5} = 6$.
Cost of 8 pens $= 6 \times 8 = 48$.
Two quantities $x$ and $y$ are in direct proportion if an increase in one causes a corresponding increase in the other.
Notation: $x \propto y$.
Equation: $x = ky$ or $\frac{x}{y} = k$ (constant).
Rule: $\frac{x_1}{y_1} = \frac{x_2}{y_2}$.
Example:
The cost of 5 meters of cloth is Rs 210. What is the cost of 2 meters?
Solution:
Let cost be $y$.
$5 : 210 :: 2 : y$ (or use Unitary Method: $1m = 42$).
$5y = 210 \times 2$
$y = \frac{420}{5} = 84$.
Answer: Rs 84.
Two quantities are in inverse proportion if an increase in one causes a corresponding decrease in the other.
Notation: $x \propto \frac{1}{y}$.
Equation: $x = \frac{k}{y}$ or $xy = k$ (constant).
Rule: $x_1 y_1 = x_2 y_2$.
Example 1: Men and Days
If 56 men can do a piece of work in 42 days, how many men will do the same work in 14 days?
Solution:
(Men $\times$ Days) is constant.
$56 \times 42 = x \times 14$.
$x = \frac{56 \times 42}{14} = 56 \times 3 = 168$.
Answer: 168 men.
Example 2: Mixed Variation
$A$ varies directly as $B$ and inversely as $C$. $A$ is 12 when $B$ is 6 and $C$ is 2. What is the value of $A$ when $B$ is 12 and $C$ is 3?
Solution:
Formula: $A = \frac{k \times B}{C}$
Find $k$: $12 = \frac{k \times 6}{2} \implies 12 = 3k \implies k = 4$.
Find new A: $A = \frac{4 \times 12}{3} = \frac{48}{3} = 16$.
Answer: 16.
The ratio of the area of a square field of side 14m to that of a rectangular field is 4:3. If the breadth of the rectangular field is 7m, find the ratio of their perimeters.
Solution:
Area of Square $= 14^2 = 196$.
$\frac{\text{Area Square}}{\text{Area Rectangle}} = \frac{4}{3} \implies \frac{196}{\text{Area Rect}} = \frac{4}{3}$.
Area Rect $= \frac{196 \times 3}{4} = 49 \times 3 = 147$.
Area Rect $= l \times b \implies 147 = l \times 7 \implies l = 21$.
Perimeter Square $= 4 \times 14 = 56$.
Perimeter Rect $= 2(l+b) = 2(21+7) = 56$.
Ratio of Perimeters $= 56 : 56 = 1 : 1$.