Ratio and Proportion
- Ratio of two quantities ‘a’ and ‘b’ having same units is simply a / b and is usually written as a:b
- The equivalence of two ratios is called proportion. If a : b = c : d, then a, b, c, d are said to be in proportion. Here, a x d = b x c
- Mean proportional is the geometric mean. For example, the mean proportional of ‘a’ and ‘b’ is square root of (a x b)
- If we have two ratios, say a : b and c : d, then (a x c) : (b x d) is called the compounded ratio
- If a : b = c : d, i.e., a/b = c/d, then (a + b) / (a – b) = (c + d) / (c – d)
This is called Componendo and Dividendo
- If we say that ‘a’ is directly proportional to ‘b’, it means that a = k x b, where ‘k’ is the constant of proportionality
- If we say that ‘a’ is inversely proportional to ‘b’, it means that a = k / b or a x b = k, where ‘k’ is the constant of proportionality
- If a ratio is multiplied or divided by a certain number, the properties of the ratio do not change. For example, if we multiply 1 : 2 by 5, we get 5 : 10, which is same as 1 : 2
- When more than one person is involved in a business, it is said to be running in partnership.
- The gains/losses from the business are divided in the ratio of their inputs, where input is calculated as the product of amount of investment and time period of investment.
If A and B invest Rs. V1 and Rs. V2 in a business for a time period of T1 and T2 respectively, then the profit/loss from the business is divided in the ratio (V1 x T1) : (V2 x T2)
The formula gets a summation if some amount is invested for a part of the total time period and some other amount is invested for the remaining time period.
- For same period of investment, the profit/loss from the business is divided in the ratio of value of investments, i.e., V1 : V2
- Ratio proportion and partnership | Set-2
- Mensuration 3D | Set-2
- Profit and loss | Set-2
- Permutation and Combination | Set-2
- Probability | Set-2
- Progressions (AP,GP, HP) | Set-2
- Compound Interest | Set-2
- Simple Interest | Set-2
- Algebra | Set-2
- Trigonometry & Height and Distances | Set-2
- Mixture and Alligation | Set 2
- Mensuration 2D | Set 2
- Problem on Numbers
- Capgemini Interview Experience 2018 (On-Campus for Full-Time)
Question 1 : If a : b = 5 : 9 and b : c = 7 : 4, then find a : b : c.
Solution : Here, we make the common term ‘b’ equal in both ratios.
Therefore, we multiply the first ratio by 7 and the second ratio by 9.
So, we have a : b = 35 : 63 and b : c = 63 : 36
Thus, a : b : c = 35 : 63 : 36
Question 2 : Find the mean proportional between 0.23 and 0.24 .
Solution : We know that the mean proportional between ‘a’ and ‘b’ is the square root of (a x b).
=> Required mean proportional = = 0.234946802
Question 3 : Divide Rs. 981 in the ratio 5 : 4
Solution : The given ratio is 5 : 4
Sum of numbers in the ratio = 5 + 4 = 9
We divide Rs. 981 in 9 parts.
981 / 9 = 109
Therefore, Rs. 981 in the ratio 5 : 4 = Rs. 981 in the ratio (5 / 9) : (4 / 9)
=> Rs. 981 in the ratio 5 : 4 = (5 x 109) : (4 x 109) = 545 : 436
Question 4 : A bag contains 50 p, 25 p and 10 p coins in the ratio 2 : 5 : 3, amounting to Rs. 510. Find the number of coins of each type.
Solution : Let the common ratio be 100 k.
Number of 50 p coins = 200 k
Number of 25 p coins = 500 k
Number of 10 p coins = 300 k
Value of 50 p coins = 0.5 x 200 k = 100 k
Value of 25 p coins = 0.25 x 500 k = 125 k
Value of 10 p coins = 0.1 x 300 k = 30 k
=> Total value of all coins = 100 k + 125 k + 30 k = 255 k = 510 (given)
=> k = 2
Therefore, Number of 50 p coins = 200 k = 400
Number of 25 p coins = 500 k = 1000
Number of 10 p coins = 300 k = 600
Question 5 : A mixture contains sugar solution and colored water in the ratio 4 : 3. If 10 liters of colored water is added to the mixture, the ratio becomes 4 : 5. Find the initial quantity of sugar solution in the given mixture.
Solution : The initial ratio is 4 : 3.
Let ‘k’ be the common ratio.
=> Initial quantity of sugar solution = 4 k
=> Initial quantity of colored water = 3 k
=> Final quantity of sugar solution = 4 k
=> Final quantity of colored water = 3 k + 10
Final ratio = 4 k : 3 k + 10 = 4 : 5
=> k = 5
Therefore, initial quantity of sugar solution in the given mixture = 4 k = 20 liters
Question 6 : Two friends A and B started a business with initial capital contribution of Rs. 1 lac and Rs. 2 lacs. At the end of the year, the business made a profit of Rs. 30,000. Find the share of each in the profit.
Solution : We know that if the time period of investment is same, profit/loss is divided in the ratio of value of investment.
=> Ratio of value of investment of A and B = 1,00,000 : 2,00,000 = 1 : 2
=> Ratio of share in profit = 1 : 2
=> Share of A in profit = (1/3) x 30,000 = Rs. 10,000
=> Share of B in profit = (2/3) x 30,000 = Rs. 20,000
Question 7 : Three friends A, B and C started a business, each investing Rs. 10,000. After 5 months A withdrew Rs. 3000, B withdrew Rs. 2000 and C invested Rs. 3000 more. At the end of the year, a total profit of Rs. 34,600 was recorded. Find the share of each.
Solution : We know that if the period of investment is not uniform, the gains/losses from the business are divided in the ratio of their inputs, where input is calculated as the product of amount of investment and time period of investment.
So, input = value of investment x period of investment, and here, period of investment would be broken into parts as the investment is not uniform throughout the time period.
A’s input = (10,000 x 5) + (7,000 x 7) = 99,000
B’s input = (10,000 x 5) + (8,000 x 7) = 1,06,000
C’s input = (10,000 x 5) + (13,000 x 7) = 1,41,000
=> A : B : C = 99000 : 106000 : 141000
=> A : B : C = 99 : 106 : 141
=> A : B : C = (99 / 346) : (106 / 346) : (141 / 346)
Thus, A’s share = (99 / 346) x 34600 = Rs. 9900
B’s share = (106 / 346) x 34600 = Rs. 10600
C’s share = (141 / 346) x 34600 = Rs. 14100
Question 8 : A invested Rs. 70,000 in a business. After few months, B joined him with Rs. 60,000. At the end of the year, the total profit was divided between them in ratio 2 : 1. After how many months did B join?
Solution : Let A work alone for ‘n’ months.
=> A’s input = 70,000 x 12
=> B’s input = 60,000 x (12 – n)
So, (70,000 x 12) / [60,000 x (12 – n)] = 2 / 1
=> (7 x 12) / [6 x (12 – n)] = 2 / 1
=> 12 – n = 7
=> n = 5
Therefore, B joined after 5 months.
This article has been contributed by Nishant Arora
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