A ratio is a comparison of the sizes of two or more quantities of the same kind by division. If a and b are two quantities of the same kind (in same units), then the fraction a/b is called the ratio of a to b. It is written as a : b. Thus, the ratio of a to b = a/b or a : b. The quantities a and b are called the terms of the ratio, a is called the first term or antecedent and b is called the second term or consequent.
For example, in the ratio 5 : 6, 5 & 6 are called terms of the ratio. 5 is called the first term and 6 is called second term.
Illustration I: 12 : 16 = 12/16 = (3 × 4)/(4 × 4) = 3/4 = 3 : 4
Illustration II:
3 : 4 is not same as 4 : 3.
Illustration III:
(i) There is no ratio between number of students in a class and the salary of a teacher.
(ii) There is no ratio between the weight of one child and the age of another child.
Illustration IV:
(i) Ratio between 150 gm and 2 kg
= Ratio between 150 gm and 2000 gm
= 150/2000 = 3/40 = 3 : 40
(ii) Ratio between 25 minutes and 45 seconds
= Ratio between (25 × 60) sec. and 45 sec.
= 1500/45 = 100/3 = 100 : 3
Illustration V:
(i) Ratio between 3 kg & 5 kg = 3/5
To compare two ratios, convert them into equivalent like fractions.
Illustration VI: To find which ratio is greater _____________
Sol:
3.6 : 4.8 = 3.6/4.8 = 36/48 = 3/4
L.C.M of 10 and 4 is 20.
So, 7/10 = (7 × 2)/(10 × 2) = 14/20
And 3/4 = (3 × 5)/(4 × 5) = 15/20
As 15 > 14 so, 15/20 > 14/20 i. e. 3/4 > 7/10
Hence, 3.6 : 4.8 is greater ratio.
The fraction by which the original quantity is multiplied to get a new quantity is called the factor multiplying ratio.
Illustration VII: Rounaq weighs 56.7 kg. If he reduces his weight in the ratio 7 : 6, find his new weight.
Sol: Original weight of Rounaq = 56.7 kg
He reduces his weight in the ratio 7 : 6
His new weight
Applications:
Example 1: Simplify the ratio 1/3 : 1/8 : 1/6
Sol: L.C.M. of 3, 8 and 6 is 24. 1/3 : 1/8 : 1/6 = 1 × 24/3 : 1 × 24/8 : 1 × 24/6 = 8 : 3 : 4
Example 2: The ratio of the number of boys to the number of girls in a school of 720 students is 3 : 5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the number of boys to the number of girls may change to 2 : 3.
Sol: The ratio of the number of boys to the number of girls = 3 : 5
Sum of the ratios = 3 + 5 = 8
So, the number of boys in the school = (3 × 720)/8 = 270
And the number of girls in the school = (5 × 720)/8 = 450
Let the number of new boys admitted be x, then the number of boys become (270 + x).
After admitting 18 new girls, the number of girls become 450 + 18 = 468
According to given description of the problem, (270 + x)/468 = 2/3
or, 3 (270 + x) = 2 x 468
or, 810 + 3x = 936 or, 3x = 126 or, x = 42.
Hence the number of new boys admitted = 42.
One ratio is the inverse of another if their product is 1. Thus a : b is the inverse of b : a and viceversa.
Some Properties of Ratios:
Applications:
Illustration I: The continued ratio of ₹ 200, ₹ 400 and ₹ 600 is ₹ 200 : ₹ 400 : ₹ 600 = 1 : 2 : 3.
Example 1: The monthly incomes of two persons are in the ratio 4 : 5 and their monthly expenditures are in the ratio 7 : 9. If each saves ₹ 50 per month, find their monthly incomes.
Sol: Let the monthly incomes of two persons be ₹ 4x and ₹ 5x so that the ratio is ₹ 4x : ₹ 5x = 4 : 5. If each saves ₹ 50 per month, then the expenditures of two persons are ₹ (4x – 50) and ₹ (5x – 50).
or, 36x – 35x = 450 – 350, or, x = 100
Hence, the monthly incomes of the two persons are ₹ 4 × 100 and ₹ 5 × 100 i.e. ₹ 400 and ₹ 500.
Example 2: The ratio of the prices of two houses was 16 : 23. Two years later when the price of the first has increased by 10% and that of the second by ₹ 477, the ratio of the prices becomes 11 : 20. Find the original prices of the two houses.
Sol: Let the original prices of two houses be ₹ 16x and ₹ 23x respectively. Then by the given conditions,
or, 352x – 253x = 5247, or, 99x = 5247; x = 53 Hence, the original prices of two houses are ₹ 16 × 53 and ₹ 23 × 53 i.e. ₹ 848 and ₹ 1,219.
Example 3: Find in what ratio will the total wages of the workers of a factory be increased or decreased if there be a reduction in the number of workers in the ratio 15 : 11 and an increment in their wages in the ratio 22 : 25.
Sol: Let x be the original number of workers and ₹ y the (average) wages per workers. Then the total wages before changes = ₹ xy. After reduction, the number of workers = (11x)/15 After increment, the (average) wages per workers = ₹ (25y)/22
∴ The total wages after changes
Thus, the total wages of workers get decreased from ₹ xy to ₹ 5xy/6
Hence, the required ratio in which the total wages decrease is
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