Let the present age of A be $A$ years and the present age of B be $B$ years.
From the first statement, "A is 6 years older than B":
$ A = B + 6 \quad \quad (1) $
From the second statement, "10 years ago":
The condition is "B's age was three quarters of A's age":
$ B - 10 = \frac{3}{4}(A - 10) \quad \quad (2) $
Substitute $B$ from equation (1) into equation (2). From (1), we get $B = A - 6$.
Substitute $B = A - 6$ into equation (2):
$ (A - 6) - 10 = \frac{3}{4}(A - 10) $
Simplify the equation:
$ A - 16 = \frac{3}{4}(A - 10) $
Multiply both sides by 4 to eliminate the fraction:
$ 4(A - 16) = 3(A - 10) $
Expand both sides:
$ 4A - 64 = 3A - 30 $
Rearrange the terms to solve for $A$:
$ 4A - 3A = 64 - 30 $
$ A = 34 $
The present age of A is 34 years.
The age of Dr. Pandey is four times the age of his son. After 10 years, the age of Dr. Pandey will be twice the age of his son. The present age of Dr. Pandey's son is?
Three years ago, the average age of a family of six members was 19 years. Since then, a boy has been born, and the average age of the family is the same today as it was three years ago. What is the age of the boy?
Amita is 2 years older than her friend Amrita. Amita's father is twice as old as Amita, and Amrita is twice as old as her sister. The ages of Amita's father and Amrita's sister differ by 43 years. The sum of the ages (in years) of Amita and Amrita is: