|Table of contents|
|Properties of Proportion|
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After reading this unit a student will learn:
In the case when ratio of boys and girls in a school is given and the total no. of student is also given, then if we know the no. of boys in the school, we can find out the no. of girls of that school by using ratios.
A ratio is a comparison of the sizes of two or more quantities of the same kind by division.
Example: Simplify the ratio 1/3 : 1/8 : 1/6
Solution: 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: The ratio of the no. of boys to the no. 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 no. of boys to the no. of girls may change to 2 : 3.
Solution: The ratio of the no. of boys to the no. of girls = 3 : 5
Sum of the ratios = 3 + 5 = 8
So, the no. of boys in the school = (3 × 720)/8 = 270
And the no. of girls in the school = (5 × 720)/8 = 450
Let the no. of new boys admitted be x, then the no. of boys become (270 + x).
After admitting 18 new girls, the no. 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 no. of new boys admitted = 42.
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 Rs. 50 per month, find their monthly incomes.
Solution: Let the monthly incomes of two persons be Rs. 4x and Rs. 5x so that the ratio is Rs. 4x : Rs. 5x = 4 : 5. If each saves Rs. 50 per month, then the expenditures of two persons are Rs. (4x – 50) and Rs. (5x – 50).
Hence, the monthly incomes of the two persons are Rs. 4 × 100 and Rs. 5 × 100 i.e. Rs. 400 and Rs. 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 Rs. 477, the ratio of the prices becomes 11 : 20. Find the original prices of the two houses.
Solution: Let the original prices of two houses be Rs. 16x and Rs. 23x respectively. Then by the given conditions,
or 320x + 32x = 253x + 5247
or 352x - 253x = 5247, or 99x = 5247 ; ∴ x = 53
Hence, the original prices of two houses are Rs. 16 × 53 and Rs. 23 × 53 i.e. Rs. 848 and Rs. 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.
Solution: Let x be the original number of workers and Rs. y the (average) wages per workers.
Then the total wages before changes = Rs. xy.
After reduction, the number of workers = (11 x)/15
After increment, the (average) wages per workers = Rs. (25 y)/22
∴ The total wages after changes
Thus, the total wages of workers get decreased from Rs. xy to Rs. 5xy/6
Hence, the required ratio in which the total wages decrease is (xy : (5xy/ 6) = 6:5)
The quantities a, b, c, d are called terms of the proportion; a, b, c and d are called its first, second, third and fourth terms respectively. If the income of a man is increased in the given ratio and if the increase in his income is given then to find out his new income, Proportion problem is used. Again if the ages of two men are in the given ratio and if the age of one man is given, we can find out the age of another man by Proportion.
Example 1: The nos. 2.4, 3.2, 1.5, 2 are in proportion because these nos. satisfy the property the product of extremes = product of means.
Here 2.4 × 2 = 4.8 and 3.2 × 1.5 = 4.8
Example 2: Find the value of x if 10/3 : x : : 5/2 : 5/4
Solution: 10/3 : x = 5/2 : 5/4
Using cross product rule, x × 5/2 = (10/3) × 5/4
Or, x = (10/3) × (5/4) × (2/5) = 5/3
Example 3: Find the fourth proportional to 2/3, 3/7, 4
Solution: Let the fourth proportional be x then 2/3, 3/7, 4, x are in proportion.
Using cross product rule, (2/3) × x = (3 × 4)/7
Or, x = (3 × 4 × 3)/(7 × 2) = 18/7.
Example 4: Find the third proportion to 2.4 kg, 9.6 kg
Solution: Let the third proportion to 2.4 kg, 9.6 kg be x kg.
Then 2.4 kg, 9.6 kg and x kg are in continued proportion since b2 = ac
So, 2.4/9.6 = 9.6/x or, x = (9.6 × 9.6)/2.4 = 38.4
Hence the third proportional is 38.4 kg.
Example 5: Find the mean proportion between 1.25 and 1.8
Solution: Mean proportion between 1.25 and 1.8 is
1. If a : b = c : d, then ad = bc
Proof. ; ∴ ad= bc (By cross -multiplication)
2. If a : b = c : d, then b : a = d : c (Invertendo)
Hence, b : a = d : c.
3. If a : b = c : d, then a : c = b : d (Alternendo)
Proof. or, ad= bc
Dividing both sides by cd, we get
4. If a : b = c : d, then a + b : b = c + d : d (Componendo)
5. If a : b = c : d, then a – b : b = c – d : d (Dividendo)
6. If a : b = c : d, then a + b : a – b = c + d : c – d (Componendo and Dividendo)
Dividing (1) by (2) we get
7. If a : b = c : d = e : f = ………………..….., then each of these ratios (Addendo) is equal (a + c + e + ……..) : (b + d + f + …….)
Example 1: If a : b = c : d = 2.5 : 1.5, what are the values of ad : bc and a+c : b+d?
Solution: we have
From (1) ad = bc, or, ad/bc=1 , i.e. ad : bc = 1:1
Hence, the values of ad : bc and a + c : b + d are 1 : 1 and 5 : 3 respectively.
Example 2: A dealer mixes tea costing Rs. 6.92 per kg. with tea costing Rs. 7.77 per kg. and sells the mixture at Rs. 8.80 per kg. and earns a profit of % on his sale price. In what proportion does he mix them?
Solution: Let us first find the cost price (C.P.) of the mixture. If S.P. is Rs. 100, profit is
∴ C.P. = Rs. (100 - = Rs. = Rs. 165/2
If S.P. is Rs. 8.80, C.P. is (165 × 8.80)/(2 × 100) = Rs. 7.26
∴ C.P. of the mixture per kg = Rs. 7.26
2nd difference = Profit by selling 1 kg. of 2nd kind @ Rs. 7.26
= Rs. 7.77 – Rs. 7.26 = 51 paise
1st difference = Rs. 7.26 – Rs. 6.92 = 34 paise
We have to mix the two kinds in such a ratio that the amount of profit in the first case must balance the amount of loss in the second case.
Hence, the required ratio = (2nd diff) : (1st diff.) = 51 : 34 = 3 : 2.