Setting Up the Age Equations
This problem involves finding the sum of the ages of two people, A and B. We can represent their ages with variables. Let's denote the age of A as a years and the age of B as b years.
The problem gives us two pieces of information, which we can translate into mathematical equations:
- "The sum of the age of A and 5 times the age of B is 36 years." This translates to the equation: $a + 5b = 36 \quad (\text{Equation 1})$
- "When 3 times the age of A is added to 7 times the age of B, the result is 62 years." This translates to the equation: $3a + 7b = 62 \quad (\text{Equation 2})$
Our goal is to find the sum of their ages, which is a + b.
Solving the System of Equations
We have a system of two linear equations with two variables:
- \( a + 5b = 36 \)
- \( 3a + 7b = 62 \)
We can solve this system using the elimination method. The idea is to eliminate one of the variables (either a or b) by manipulating the equations.
Let's eliminate variable a. To do this, we can multiply Equation 1 by 3 so that the coefficient of a in both equations becomes the same.
Multiply Equation 1 by 3:
$3 \times (a + 5b) = 3 \times 36$ $3a + 15b = 108 \quad (\text{Equation 3})$Now we have:
| Equation 3: | \( 3a + 15b = 108 \) |
| Equation 2: | \( 3a + 7b = 62 \) |
Subtract Equation 2 from Equation 3 to eliminate a:
$(3a + 15b) - (3a + 7b) = 108 - 62$ $3a + 15b - 3a - 7b = 46$ $8b = 46$Now, solve for b:
$b = \frac{46}{8}$ $b = \frac{23}{4}$Now that we have the value of b, we can substitute it back into either Equation 1 or Equation 2 to find the value of a. Let's use Equation 1:
$a + 5b = 36$ $a + 5 \left( \frac{23}{4} \right) = 36$ $a + \frac{115}{4} = 36$To find a, subtract \(\frac{115}{4}\) from both sides:
$a = 36 - \frac{115}{4}$To subtract, find a common denominator, which is 4:
$a = \frac{36 \times 4}{4} - \frac{115}{4}$ $a = \frac{144}{4} - \frac{115}{4}$ $a = \frac{144 - 115}{4}$ $a = \frac{29}{4}$Calculating the Sum of Ages
We have found the values for a and b:
- Age of A, \( a = \frac{29}{4} \) years
- Age of B, \( b = \frac{23}{4} \) years
The question asks for the sum of the ages of A and B, which is \( a + b \).
$a + b = \frac{29}{4} + \frac{23}{4}$Since the denominators are the same, we can add the numerators:
$a + b = \frac{29 + 23}{4}$ $a + b = \frac{52}{4}$Simplify the fraction:
$a + b = 13$Therefore, the sum of the ages of A and B is 13 years.
Important Questions from Problem on Ages (Notes)
- At present, Sharad is three times as old as his son, and his daughter is 3 years younger than the son. If the sum of the ages of these three people 3 years ago was 63 years, then Sharad's present age (in years) is:
A father said to his son, "I was as old as you are when I became your father." If the current age of father is 52 years, the age of son after 10 years will be___________.
- The sum of the age of A and 5 times the age of B is 55 years. When 3 times the age of A is added to 7 times the age of B, the result is 99 years. The sum of the ages (in years) of A and B is:
- The sum of the age of A and 3 times the age of B is 43 years. When 5 times the age of A is added to 2 times the age of B, the result is 98 years. The sum of the ages (in years) of A and B is:
- Eight years ago, the age of the mother was four times the age of her son. Eight years hence, the mother's age will be two times the age of her son. Find the ratio of the present age of the mother to the age of the son.
