A and B are friends and their ages differ by 2 years. A's father D is twice as old as A and B is twice as old as his
sister C. The age of D and C differ by 40 years. Find the ages of A and B.
Answers
Let the Age of A be : P
Let the Age of B be : Q
● A and B ages differ by 2 years
P - Q = 2 ---------- (1)
● A's father D is twice as old as A
Age of A's Father (D) = 2P
● B is twice as old as his Sister C
B's Sister C is half of the Age of B
● The Age of D and C differ by 40
D - C = 40
Subtract Equation (1) from Equation (2)
[4P - Q] - [P - Q] = 80 - 2
4P - Q - P + Q = 78
3P = 78
P = 26
Substitute value of P in Equation (1)
26 - Q = 2
Q = 26 - 2
Q = 24
Ages of A and B are 26 and 24 respectively
Let the Age of A be : P
Let the Age of B be : Q
● A and B ages differ by 2 years
\longrightarrow⟶ P - Q = 2 ---------- (1)
● A's father D is twice as old as A
\longrightarrow⟶ Age of A's Father (D) = 2P
● B is twice as old as his Sister C
\longrightarrow⟶ B's Sister C is half of the Age of B
\mathrm{\longrightarrow C = \dfrac{Q}{2}}⟶C=
2
Q
● The Age of D and C differ by 40
\longrightarrow⟶ D - C = 40
\mathrm{\longrightarrow 2P - \dfrac{Q}{2} = 40}⟶2P−
2
Q
=40
\mathrm{\longrightarrow \dfrac{2(2P) - Q}{2} = 40}⟶
2
2(2P)−Q
=40
\mathrm{\longrightarrow \dfrac{4P - Q}{2} = 40}⟶
2
4P−Q
=40
\mathrm{\longrightarrow {4P - Q} = 80\;-----\;(2)}⟶4P−Q=80−−−−−(2)
Subtract Equation (1) from Equation (2)
\longrightarrow⟶ [4P - Q] - [P - Q] = 80 - 2
\longrightarrow⟶ 4P - Q - P + Q = 78
\longrightarrow⟶ 3P = 78
\longrightarrow⟶ P = 26
Substitute value of P in Equation (1)
\longrightarrow⟶ 26 - Q = 2
\longrightarrow⟶ Q = 26 - 2
\longrightarrow⟶ Q = 24
\longrightarrow⟶ Ages of A and B are 26 and 24 respectively