Study Notes || Ratio, Proportion and Partnership

Ratio , Proportion and Partnership are important topics for the MBA entrance exams.

Concepts of Ratio & Proportion

Example 1: If A: B = 2:5 and B:C = 7:3 then find A:B:C
Solution:
A : B = 2 : 5
B : C = 7 : 3
In this, the value of B has to be same so to equate the value of B, we can take LCM of both the values of B i.e. (5×7) = 35
So multiply (A : B) by 7 and (B : C) by 5
Hence, A : B : C = 14 : 35 : 15
Example 2: If the ratio of A : B : C is given in reciprocals, then convert it.
Solution: Suppose, if A : B : C = (1/2): (1/3): (1/5)
take any number which is multiple of the product (2×3×5) = 30
Now multiply it in the numerator part
A:B:C = (30/2) : (30/3) : (30/5) = 15:10:6
Example 3: The sum of the ages of Akhil and Binay is 48 years. Akhil is 4 years older than Chetan. The ratio of the ages of Binay and Chetan is 4: 7. What was the age of Akhil 5 years back?
Solution:
Given Age of Akhil + Binay = 48 and Akhil is 4 yrs  older than Chetan
So, Binay + Chetan = 44
and Binay : Chetan = 4 : 7
Comparing B + C = 11
So, 11 = 44
1 = 4
Binay = 4×4 = 16 yrs
Chetan = 7×4 = 28 yrs
Akhil = 32 yrs
Age of Akhil 5yrs back = 32-5 = 27 yrs
Example 4: The ratio of the ages of A and B is 6 : 5. The difference between the ages of C and A is more than 3 years. The age of D is a prime number between the ages of A and B. The ratio of the ages of B and C is 2 : 3. If the ages of all four are integers, what is the difference between the ages of C and D?
Solution: Given ratio of A : B = 6 : 5 and B : C = 2 : 3
So, A : B : C = 12 : 10 : 15
Now let’s use conditions given i.e.
(i) All ages of A, B, C and D should be integers.
(ii) difference of age between A and C should be more than 3 yrs but from the above ratio difference is only 3 yrs.
So, we have to take A : B : C = 24 : 20 : 30
(iii) age of D lies between A and B and it should be prime number
Numbers between 24 and 20 are = 23, 22, 21 and 23 is the prime number. So, the age of D is 23
Difference of C and D = 30 – 23 = 7yrs
Example 5: Ten years ago, the ages of Adam and Parker were in the ratio 6 : 7. After six years, Adam's age would be 9.09% less than Parker's age. What would be age of Parker after 9 years?
Solution: In percentage topic we have learnt that 9.09% =  1/11, if age of Parker after 6 yrs is X then age of Adam will be X – (1/11)X = (10/11)X hence ratio of Adam and Parker = (10/11)X : X = 10 : 11
                         adam : parker
10yrs ago           6     :     7
after 6yrs          10    :    11
Difference between both these are 16 yrs
So,  4 = 16
1 = 4
hence, age of parker after 6 yrs = 11×4 = 44
age of parker after 9 yrs = 44 + 3 = 47 yrs
Example 6: Five years ago, the ages of a father and son were in the ratio 7 : 2. After three years, their ages would be in the ratio 5 : 2. What was father's age 7 years back?
Solution: Father : Son
5yrs ago       7  :   2
3yrs after     5   :  2
The difference between the age of the father at 5yrs ago and 3 yrs after should be equal to the difference between age of son at 5yrs ago and 3 yrs. To achieve this condition, we manipulate the ratio.
               Father : Son
5yrs ago    21     :    6
3yrs after  25     :   10
now, we can see that 25-21 = 4 and 10-6 = 4
difference between 3yrs after and 5 yrs ago = 8yrs
So, 4 = 8 yrs
      1 = 2 yrs
hence the age of father’s 5 yrs ago is 42 yrs and 7yrs ago = 40 yrs.
Example 7: A stick is broken up into two parts. The ratio of the lengths of the bigger part and the smaller part is equal to the ratio of the lengths of the full stick and the bigger part. What is this ratio?
Solution: Let the length of bigger part of the rod  is 1 m and the length of a smaller part of the rod is x m, the total length of the rod is (1+x) m
Given, the ratio of bigger to smaller is equal to the ratio of the length of the full stick to a bigger part
So, 1/x = (1+x) /1
x² + x = 1
x² + x – 1 = 0
using quadratics x = (-1 ± √ 5)/2
but x = (-1+√ 5)/2 is the answer because another will give the negative answer.
Example 8: The contents of two vessels containing wine and water in the ratio 2 : 3 and 5: 6 are mixed in the ratio 10: 7. What would be the ratio of wine and water in the final mixture?
Solution:         Wine        Water
vessel 1              2       :        3
vessel 2              5       :        6
to make the total quantity same in both vessel we have to multiply the ratio of vessel 1 by (5+6= 11) and vessel 2 by (2+3 = 5)
Now, Vessel 1    22: 33
          Vessel 2    25: 30
But, vessel 1 and vessel 2 are mixed in the ratio of 10: 7.
So, wine = 22×10+25×7 = 395
  water    = 33×10+30×7 = 540  
It is given that two vessels containing wine and water in the ratio 2 : 3 and 5: 6 but both vessels have the same quantity of total mixture.
So, ratio of wine and water in final mixture = 395 : 540 = 79 : 108
How to express data in a smarter way.
               Wine: water
vessel 1     (2 : 3) × 11 × 10
vessel2      (5 : 6) × 5 × 7
vessel 1     (2 : 3) × 110 = (2 : 3)× 22 = 44 : 66
vessel2      (5 : 6) × 35   = (5 : 6)× 7 = 35 : 42
Hence, ratio will be (44+35) : (66+42) = 79 : 108
Example 9: A mixture contains wine and water in the ratio 3: 2 and another contains them in the ratio 4: 5. How many litres of the former must be mixed with 15 litres of the latter so that the resultant mixture contains equal quantities of wine and water?
Solution: In this, after mixing of both different mixture, the quantity of wine and water becomes equal.
Let X litre of mixture 1 is mixed with 15 litres of mixture 2, then
(3/5)X + (4/9)15 = (2/5)X + (5/9)15
X/5 = 15/9
X = 75/9 = 25/3
How to do it by using the ratio concept and above-discussed methodology?
Ultimately we have to equal the ratio of wine and water in the final mixture.
               Wine:   Water
Mix 1        3:     2
Mix 2        4:     5
We have to make the sum of the quantity of wine in mix 1 and mix 2 equal to the sum of the quantity of water in mix 1 and mix2.
But in the above question, it is already given so, we will mix it in the same quantity.
If we mix 5 litres of mixture1 and 9 litres of mixture2, then the ratio of quantity will be the same.
So, if it is 15 litres of mixture2, then mixture1 = (15×5)/9 = 25/3 litres.
Example 10: A mixture contains wine and water in the ratio 3: 2 and another contains them in the ratio 7 : 3. In what ratio should the two be mixed to get a resultant mixture with wine and water in the ratio 17: 8?
Solution: Let these are mixed in a ratio of 1:X, we will not assume a: b because there will be two variables and will make the calculation harder.
Now,        wine: water
mixture1       3 : 2
mixture2       7 : 3
First, we will solve it as done in 8.
So,(3:2)×(10)×1 = (3:2)×2
      (7:3)×(5)× X  = (7:3)×X
the ratio of wine and water in the final mixture is (3×5+7X):(2×5+3X)
(6+7X) / (4+3X) = 17/8
solving it, X = 4
So, it will be mixed in the ratio of 1:4
Wrong Approach
Some of you will mark answer (1:2), why?
  3 : 2 × 1 = 3 : 2
  7 : 3 × 2 = 14: 6
So, (3+14): (2+6)= 17: 8. this will not be applicable here because the quantity of mixtures 1 & 2 is not equal.
Allegation Approach: we will discuss it in detail in the next article.
In this approach, the allegation will be applied only on one object throughout the solution.
So, let’s apply it on wine
in mixture1, % of wine is 60% in solution
In mixture2, % of wine is 70% in solution
In the final mixture, % of the wine is 68%
So,  60--------68-------70
        60    8    68     2   70, you can see the difference of 68 and 60 is 8 and 70 and 68 is 2.
For calculating the ratio of mixing, the answer will be the reciprocal of the ratio of these differences.
ratio in which it is mixed = 2 : 8 = 1 : 4

Partnerships:

This is a very prominent part of most of the exams. These questions are really scoring as well because these questions are really scoring. Check below what this topic is about and how are typical questions solved.

What is Partnership?

When two or more people join hands to create a start-up or some business it’s called a ‘partnership’ business. Typically, they invest some capital and earn some profit. And, this profit is distributed among partners either in some prefixed ratio or the ratio of their investment. In exams, mostly profit distribution is based on later type. The calculation for profit based on investment is done as shown below:
P₁: P₂ = C₁×T₁: C₂×T₂
Here, P₁ = Profit for Partner 1.
C₁ = Capital by Partner 1.
T₁ = Time period for which Partner 1 invested his capital.
P₂ = Profit for Partner 2.
C₂ = Capital by Partner 2.
T₂ = Time period for which Partner 2 invested his capital.

Typical Question Methodology:

1) We are given the amount of investment by two partners for particular time like A and B invested Rs.1500 and Rs.2000 respectively. After 4 months, they admit C with contribution of Rs.2250. And, B withdraws his contribution after 9 months ⇒ A invested his capital for 12 months while C invested his capital for (12 – 4=) 8 months and B invested for 9 months.
So, according to ratio formula, A: B: C = 1500×12: 2000×9: 2250×8 = 180: 180: 180 (We’ve cancelled out the equal number of zeroes from all the numbers)
⇒ A: B: C = 1: 1: 1
2) Second, we are given the profit at the end of year was Rs 900. What’d B get?
Now, A: B: C =  1: 1: 1
⇒ B’s share = {1/(1+1+1)} =1/3rd of profits = (1/3)*900 =  Rs. 300
We can see that calculating profit isn’t that tricky so we’ll be solving sample questions in a way that would help you in calculating the ratio of profits among partners.

Sample Questions:

Q1. A contributes one-forth capital for one fourth of time. B contributes one fifth of capital for half of the time. C contributes the remaining capital for the whole time. Find their profit-sharing ratio.
Sol. Part of Capital by:  A = 1/4 ; B = 1/5 ; C gave remaining ⇒ C’s share in capital = 1 – (1/4 + 1/5) = 11/20
Time investment by: A = 1/4; B = 1/2; C = 1 (whole time means full one 1 year)
Using, P₁: P₂ = C₁×T₁: C₂×T₂

⇒ A: B : C =

Now we’ll take LCM of 16, 10 and 20 which is 80 and divided it by each denominator and multiply number thus obtained by each of the respective denominator. 
⇒ A: B: C = 5: 8: 44.

Q2. A begun a business with Rs.450, after some months B joined him with capital of Rs.300. In the end, the profits were distributed in the ratio of 2:1. After how long did B join?
Sol. Capital by: A = 450, B = 300
Time capital stay invested by: A = 12, B = n(say)
Using, P₁: P₂ = C₁×T₁: C₂×T₂

_{(2/1) = (450 ×12)/(300n)}
⇒ n = 9 months
So, B’s capital was invested for 9 months and he joined after (12 – 9 =) 3 months.

Q3. A and B rent a pasture for 10 months. A puts in 100 cows for 8 months. How many can B put in for remaining 2 months, if he pays half as much again as A?
Sol. Here, B has to pay half as much again as A ⇒ If A pays 1 then B pays (1+1/2=) 3/2
So, required ratio = 1: 3/2 ⇒ A: B = 2: 3
Also, number of cows is considered as capital. Let cows brought in by B be ‘n’
So, using, P₁: P₂ = C₁×T₁: C₂×T₂
2: 3 = 100×8: n×2

⇒ n = 600 cows

Q4. A, B and C invested capitals in the ratio of 5: 6: 8.At the end of business term, they received the profits in the ratio of 5: 3: 12. Find the ratio of time for which they contributed their capital.
Sol. To solve such questions use formula below:
T_A: T_B: T_C = P_A÷C_A: P_B÷C_B: P_C÷C_C
Here, T_A = Time by A, P_A = Profit for A, C_A=  Capital by A.
T_B= Time by B, P_B = Profit for B, C_B=  Capital by B.
T_C = Time by C, P_C = Profit for C, C_C=  Capital by C.
⇒ T_A: T_B: T_C = 5÷5: 3÷6: 12÷8 = 1: 1/2: 3/2 = 2: 1: 3

Q5. Three partners A,B and C had business. A, whose money had been used for 4 months claimed 1/8 profit. B whose money was used for 6 month claimed 1/3 of profit. C had invested Rs. 1560 for 8 months. How much did A and B contribute?
Sol.⇒ n = Rs. 720
So, A invested Rs 720. Similarly, we can find the amount of investment by B.

Q6. A and B invested in the ratio of 3: 2. If 5% of total profit goes to charity and A got Rs 855 as his share then find the total profit?
Sol. In such questions, use this formula,

So, total profit is Rs. 1500.

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