# Ratio Proportion and Partnership

### 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

### Partnership

- 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
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### Sample Problems

**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.

### Problems on Ratio proportion and partnership | Set-2

This article has been contributed by **Nishant Arora**

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