2 years ago, the ratio of X and Y's age was 7 ∶ 5 and after 2 years their age will be in the ratio 9 ∶ 7. What is the present age of Y?
12 years
Solving Age Ratio Word Problems
This question is a classic age word problem involving ratios. We are given the ratio of the ages of two individuals, X and Y, at two different points in time relative to the present. We need to find the present age of Y.
Let's define variables for the present ages:
- Let the present age of X be $x$ years.
- Let the present age of Y be $y$ years.
Now, let's translate the given information into mathematical equations based on the ratios provided.
Ages 2 Years Ago
According to the question, 2 years ago:
- X's age was $x - 2$ years.
- Y's age was $y - 2$ years.
The ratio of their ages 2 years ago was 7 ∶ 5. This can be written as an equation:
\(\frac{x-2}{y-2} = \frac{7}{5}\)
Cross-multiplying gives us:
\(5(x-2) = 7(y-2)\)
\(5x - 10 = 7y - 14\)
Rearranging the terms to form a linear equation:
\(5x - 7y = -14 + 10\)
\(5x - 7y = -4\) (Equation 1)
Ages 2 Years From Now
According to the question, after 2 years from now:
- X's age will be $x + 2$ years.
- Y's age will be $y + 2$ years.
The ratio of their ages after 2 years will be 9 ∶ 7. This can be written as an equation:
\(\frac{x+2}{y+2} = \frac{9}{7}\)
Cross-multiplying gives us:
\(7(x+2) = 9(y+2)\)
\(7x + 14 = 9y + 18\)
Rearranging the terms to form another linear equation:
\(7x - 9y = 18 - 14\)
\(7x - 9y = 4\) (Equation 2)
Solving the System of Linear Equations
Now we have a system of two linear equations with two variables, $x$ and $y$:
1) \(5x - 7y = -4\)
2) \(7x - 9y = 4\)
We can solve this system using various methods, such as substitution or elimination. Let's use the elimination method. To eliminate \(x\), we can multiply Equation 1 by 7 and Equation 2 by 5:
Multiply Equation 1 by 7:
\(7 \times (5x - 7y) = 7 \times (-4)\)
\(35x - 49y = -28\) (Equation 3)
Multiply Equation 2 by 5:
\(5 \times (7x - 9y) = 5 \times (4)\)
\(35x - 45y = 20\) (Equation 4)
Now, subtract Equation 3 from Equation 4:
\((35x - 45y) - (35x - 49y) = 20 - (-28)\)
\(35x - 45y - 35x + 49y = 20 + 28\)
\(4y = 48\)
Now, solve for \(y\):
\(y = \frac{48}{4}\)
\(y = 12\)
So, the present age of Y is 12 years.
We can also find the present age of X by substituting the value of \(y\) into either Equation 1 or Equation 2. Using Equation 1:
\(5x - 7(12) = -4\)
\(5x - 84 = -4\)
\(5x = -4 + 84\)
\(5x = 80\)
\(x = \frac{80}{5}\)
\(x = 16\)
The present age of X is 16 years.
Verification
Let's check if these present ages satisfy the conditions given in the problem:
- Present ages: X = 16, Y = 12
- 2 years ago: X's age = $16 - 2 = 14$, Y's age = $12 - 2 = 10$. Ratio $14/10 = 7/5$. This matches the first condition.
- 2 years from now: X's age = $16 + 2 = 18$, Y's age = $12 + 2 = 14$. Ratio $18/14 = 9/7$. This matches the second condition.
Both conditions are satisfied, confirming that our calculated ages are correct.
The question asks for the present age of Y, which is 12 years.
| Time Period | X's Age | Y's Age | Ratio (X:Y) | Equation |
|---|---|---|---|---|
| 2 years ago | \(x-2\) | \(y-2\) | 7:5 | \(\frac{x-2}{y-2} = \frac{7}{5} \Rightarrow 5x - 7y = -4\) |
| Present | \(x\) | \(y\) | - | - |
| 2 years from now | \(x+2\) | \(y+2\) | 9:7 | \(\frac{x+2}{y+2} = \frac{9}{7} \Rightarrow 7x - 9y = 4\) |
Revision Table: Key Concepts in Age Problems
| Concept | Explanation | Example |
|---|---|---|
| Defining Variables | Represent unknown ages (usually present age) with variables (e.g., x, y). | Let present age be $P$. Age after 5 years is $P+5$. Age 3 years ago is $P-3$. |
| Setting up Equations from Ratios | If the ratio of ages A and B is \(a:b\), then \(\frac{\text{Age of A}}{\text{Age of B}} = \frac{a}{b}\). | Ratio of X and Y is 3:4 $\Rightarrow \frac{X}{Y} = \frac{3}{4}$. |
| Handling Time Shifts | Ages change linearly with time. Add/subtract the number of years to the present age. | If present age is $P$, age after $t$ years is $P+t$. Age $t$ years ago was $P-t$. |
| Solving Simultaneous Equations | Age ratio problems often result in a system of linear equations that need to be solved. | Methods include substitution and elimination. |
Additional Information: Types of Age Problems
Age problems in quantitative aptitude often involve different scenarios:
- Simple Age Changes: Problems where only current age and future/past ages are related.
- Age Ratios: Problems like this one, where ratios of ages are given at different points in time.
- Sum/Difference of Ages: Problems providing sums or differences of ages at different times.
- Product of Ages: Less common, but can involve setting up equations based on the product of ages.
- Average Age: Problems involving the average age of a group of people.
Solving these problems requires careful reading, defining variables correctly, and setting up accurate algebraic equations based on the given information. Practice with various types helps build proficiency.
Important Questions from Ratio and Proportion
In a mixture of 156 litres, the ratio of milk and water is 7 : 6. How much water should be added to make the ratio 14 : 13?
If the ratio of the first to second number is 3 : 4 and that of the second to the third number is 8 : 5, and sum of three numbers is 190 then the third number is:
The third proportional to 9 and 15 is:
The average age of 3 persons is 30 years.If their ages are in the ratio of 3 : 5 : 7 respectively, then the age of the eldest person is:
The income of A and B are in the ratio 5 : 3. The expenses of A, B and C are in the ratio of 8 : 5 : 2. If C spends 2000 and B saves ₹ 700, then A saves:
