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
- Placements | QA | Mixture and Alligation
- Placements | QA | Percentages
- Placements | QA | Profit and Loss
- Placements | QA | Algebra
- Placements | QA | Work and Wages
- Cognizant Placement Paper | Aptitude Set 4
- Cognizant Placement Paper | Aptitude Set 3
- English Reading Comprehension | Set 5
- English Reading Comprehension | Set 4
- English Reading Comprehension | Set 3
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
Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above