Five years ago, A was three times as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B (in years).
50,20
A = Present age of A
B = Present age of B
Five years ago person A was A - 5 yrs old.
Person B was B - 5 yrs old.
Five years ago, A was three times as old as B mean:
A - 5 = 3 ( B - 5 )
Ten years later person A shal be:
A - 5 + 10 = A + 5 yrs old
Person B shal be:
B - 5 + 10 = B + 5 yrs old
Ten years later, A shall be twice as old as B mean:
A + 5 = 2 ( B + 5 )
Now you must solve system:
A - 5 = 3 ( B - 5 )
A + 5 = 2 ( B + 5 )
A - 5 = 3 * B - 3 * 5
A - 5 = 3 B - 15
A + 5 = 2 ( B + 5 )
A + 5 = 2 * B + 2 * 5
A + 5 = 2 B + 10
Your system become:
A - 5 = 3 B - 15
A + 5 = 2 B + 10
Try to solve this system.
The solutions are: A = 35 , B = 15
Proof:
Five years ago person A was 35 - 5 = 30 yrs old.
Five years ago person B was 15 - 5 = 10 yrs old.
30 / 10 = 3
Ten years later person A shal be:
30 + 10 = 40 yrs old
Person B shal be:
10 + 10 = 20 yrs old
40 / 20 = 2
To solve this problem, let's assign variables to the present ages of A and B. Let A represent A's present age and B represent B's present age.
The problem states that five years ago, A was three times as old as B. We can express this relationship using an equation:
A - 5 = 3(B - 5)
Next, the problem states that ten years later, A shall be twice as old as B. We can also express this relationship using an equation:
A + 10 = 2(B + 10)
Now we have a system of two equations. We can solve this system using algebraic methods to find the values of A and B. Let's solve it step by step:
Step 1: Solve the first equation for A.
A - 5 = 3B - 15
A = 3B - 15 + 5
A = 3B - 10
Step 2: Substitute the value of A in the second equation.
3B - 10 + 10 = 2(B + 10)
3B = 2B + 20
3B - 2B = 20
B = 20
Step 3: Substitute the value of B in the first equation to find A.
A = 3(20) - 10
A = 60 - 10
A = 50
Therefore, the present age of A is 50 years, and the present age of B is 20 years