Two years ago, the age of A was three times the age of B. If B is currently 9 years old, then after how many years, the age of A will be double of the age of B1

examrobotsa's picture

Submitted by examrobotsa on

Q: 125 (CAPF/2020)
Two years ago, the age of A was three times the age of B. If B is currently 9 years old, then after how many years, the age of A will be double of the age of B1

question_subject: 

Polity

question_exam: 

CAPF

stats: 

0,9,4,3,1,0,9

keywords: 

{'age': [2, 1, 1, 2], 'b1': [0, 0, 0, 2], 'years': [1, 0, 0, 2], 'many years': [3, 0, 0, 4], 'times': [5, 2, 7, 3]}

To solve this problem, we need to figure out the age of A two years ago and the age of A at the present time.

We are given that two years ago, the age of A was three times the age of B. Since B is currently 9 years old, we can calculate the age of A two years ago:

Age of A two years ago = 3 * (Age of B two years ago)

Age of B two years ago = Age of B at present - 2 = 9 - 2 = 7

Age of A two years ago = 3 * 7 = 21

Next, we need to find the present age of A. We know that the age of A two years ago was 21, so we can calculate A`s present age:

Age of A at present = Age of A two years ago + 2 = 21 + 2 = 23

Now, we need to determine after how many years the age of A will be double the age of B. Let`s denote this number of years as "x".

After x years, the age of A will be A`s present age + x, and the age of B will be B`s present age + x.

According to the question,

Practice this on app