Method 3: Unitary method
This is most advisable type of approach towards almost all the different varieties of ratio problems.
Well, here we are not using any algebraic characters for the solution. Hence we can operate it by mind. No need of a little bit of rough work. However you have to improve in mental calculation and visualizing techniques.
Hence the answer is 25 students.
If p:q = 3 : 4, find;
- 2p + 3q
- (5p + 3q)/(5p - 3q)
- (3p2 + 4q) / (3p2 - 4q)
Given p : q = 3 : 4
Let p = 3x and q = 4x
- 2p + 3q = 2(3x) + 3(4x) = 18x
For each different value of 'x', we will get different values for the given expression. Hence a unique value for the given expression 2p + 3q can't determine.
- (5p + 3q) / (5p -3q) = (5(3x) + 3(4x)) / (5(3x) - 3(4x)) = 27x / 3x = 9
Hence this expression has a unique value for any different value of x.
- (3p2 + 4q) / (3p2 - 4q) = (3(3x)2 + 4(4x)) / (3(3x)2 - 4(4x))
= 9x2 + 16x / 9x2 - 16x = 9x + 16 / 9x - 16
For this expression also we will get different values for each different value of x. So it’s not possible to find a unique value for this expression.
Example 3: Ratio in Age Problems.
The ratio of the present ages of Ajay and Varun is 4:3. Five years ago it was 7:5. Find the present age of Varun.
- 40 years
- 35 years
- 30 years
- 25 years
And varun's present age = 3x
Five years ago;
Ajay's age = 4x - 5
Varun's age = 3x - 5
Given (4x - 5)/(3x - 5) = 7/5
X = 10
Varun's present age = 3x
= 3 * 10 = 30 years
Present age of Varun is represented by the ratio value of 3. Therefore normally Varun's age should be a multiple of three. Please note, in almost all the questions related to age problems, very rarely only ages expressed in the form of fractions, rest all in the form of integers. This favourable situation will help the student to reach a conclusion easily through the basic testing of divisibility. Here also our options consisting only one value as a multiple of three.i.e. 30. Hence this could be the possible answer.
If more than one of the options satisfying the basic testing condition (here it is the divisibility of 3), then student needs to make verification from the selected set of options and finally reach the conclusion.
Example 4: Ratio in Mixtures:
The ratio of the quantities of milk and water in a solution of 60 liters (consisting milk and water only) is 3:7. How much liters of additional milk should pour in to the mixture to get a resultant mixture which consisting the milk and water in the respective ratio of 2:3?
= 18 liters.
Present quantity of water in the solution = 60 - 18 = 42 liters
Let 'x' liters be the additional required quantity of milk.
Then; (18 + x) /42 = 2/3
Therefore x = 10 liters.
Example 5: Ratio in geometry
If the length and breadth of a rectangle are in the ratio of 4:3 and it has an area of 108 sq:c.m. find the length of diagonal in centimeters.
- Can't be determined.
Let length and breadth of the rectangles are 4x and 3x respectively.
Area = length * breadth
= 12 x2
12x2 = 108
X = ± 3
X = - 3 doesn't exist, because the dimensions of a rectangle is always a magnitude. Hence it should be non-negative.
So x = 3.
Therefore, length = 4x = 12 c.m.
Breadth = 3x = 9 c.m.
Diagonal of a rectangle = √(length2 + breadth2)
= √(122 + 92)
= 15 c.m.
Example 6: income - Expenditure - Savings problems.
Ratio of the respective incomes of A and B is 2:3 and the ratio of their respective expenditures is 5:8. Find the income of A, if each of them saves an equal amount of Rs.2000.
- Rs. 8,000.
- Rs. 10,000.
- Rs. 12,000.
- Rs. 8,500.
Income of A = 2i, Expenditure of A = 5e, Savings of A = 2i - 5e
Income of B = 3i, Expenditure of B = 8e, Savings of B = 3i - 8e
|Note: Income and expenditure are basically two different kinds of quantities, so while expressing them algebraically use different variables for income and expenditure.|
Given that each of them saves an equal amount.
Therefore 2i - 5e= 3i - 8e
i = 3e
Substitute this relation in the previous assumption.
Income of A = 2i= 2(3e) = 6e, Expenditure of A = 5e, Savings of A = 6e - 5e = e
Income of B = 3i= 3(3e) = 9e, Expenditure of B = 8e, Savings of B = 9e - 8e = e
Given that their individual savings is Rs.2,000.
I.e. e = Rs. 2,000
Income of A = 6e
= 6 * 2000 = Rs.12,000.
For this logical and numerical arrangement method require your numerical skills in the form of catching the mutual relations between values. Here it's given that each of them saves an equal amount. So we have to think that, which kind of a change in the ratio values will help us to make the difference of corresponding incomes and expenditures are equal.
In this initial step of arrangement we are not bothering about the original constant values which are available in the data. Finally with the help of available constant values and through a unitary method we can easily reach the solution. By a thorough practice will help the student make able to do these type of problems by mind, without doing rough works.
|Multiply the above ratio values by 3||6||9, These are the new ratio values
for their respective incomes.
|Savings = Income - Expenditure||1||1|
1 → Rs.2,000.
Therefore; 6 → 6 * 2000 = Rs. 12,000.
Method of algebraic constant:
It is a very effective method for the above type of problems.
|Two quantities are in the ratio of a:b, when a constant 'x' is add/subtract from both the quatities, the resultants are in the ratio of c:d.
Let us take the quantities are 'ak' and 'bk', where 'k' is the multiplicative constant. And, k = (c ∼ d) x / (ad ∼ bc)
Ratio of incomes = 2 : 3
Ratio of Expenditures = 5 : 8
Savings of each person = Rs. 2000
i.e. a:b = 2 : 3
c : d = 5 : 8
x = 2000
Hence, the multiplicative constant, k =
Therefore, Income of A = 2k = 2 * 6000 = Rs. 12000