14 years
This problem involves finding the present age of the younger of two people, A and B, based on two pieces of information: the difference in their present ages and a relationship between their ages six years ago.
We need to set up mathematical equations to represent the given conditions and then solve them to find the required age.
Let's denote the present age of the younger person as \(Y\) years and the present age of the elder person as \(E\) years.
According to the first condition:
So, we can write the equation:
\(E - Y = 16\)
From this, we can express the elder's age in terms of the younger's age:
\(E = Y + 16\)
Now, let's consider the ages 6 years ago:
According to the second condition:
So, we can write the equation:
\(E - 6 = 3 \times (Y - 6)\)
We now have a system of two equations:
We can substitute the expression for \(E\) from equation (1) into equation (2):
\((Y + 16) - 6 = 3(Y - 6)\)
Now, simplify and solve for \(Y\):
\(Y + 10 = 3Y - 18\)
To isolate \(Y\) terms on one side and constants on the other, add 18 to both sides and subtract \(Y\) from both sides:
\(10 + 18 = 3Y - Y\)
\(28 = 2Y\)
Now, divide by 2 to find the value of \(Y\):
\(Y = \frac{28}{2}\)
\(Y = 14\)
The value of \(Y\) represents the present age of the younger person.
Let's check if our answer satisfies both conditions:
Both conditions are satisfied, so the present age of the younger person is indeed 14 years.
Based on our calculations, the present age of the younger person is 14 years.
| Present Age | Age 6 Years Ago | |
|---|---|---|
| Younger (Y) | 14 | 14 - 6 = 8 |
| Elder (E) | 30 | 30 - 6 = 24 |
Checking the conditions:
| Concept | Description | How it Applies Here |
|---|---|---|
| Representing Ages | Use variables (e.g., \(x\), \(y\)) for unknown ages. | Used \(Y\) for younger, \(E\) for elder. |
| Ages in the Past/Future | To find age \(n\) years ago, subtract \(n\). To find age \(n\) years in future, add \(n\). | Used \(Y-6\) and \(E-6\) for ages 6 years ago. |
| Translating Words to Equations | Convert sentences into mathematical expressions (e.g., "differ by" means subtraction, "times as old" means multiplication). | Converted "differ by 16" to \(E-Y=16\) and "elder was 3 times as old as younger" to \(E-6 = 3(Y-6)\). |
| Solving Simultaneous Equations | Use substitution or elimination to find the values of the variables. | Used substitution to solve for \(Y\). |
Solving word problems, especially age problems or problems involving linear equations, often follows a structured approach:
Age problems are common examples of linear equations in one or two variables and are fundamental in algebra.
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