
A and B are friends and A is older than B by 2 years. A’s father D is twice as old as A and B is twice as old as his sister C. The ages of C and D differ by 40 years. Find the age of A and B.
(a) A $=$ 18 years and B $=$ 24 years
(b) A $=$ 42 years and B $=$ 24 years
(c) A $=$ 56 years and B $=$ 24 years
(d) A $=$ 26 years and B $=$ 24 years
Answer
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Hint: At first take A’s age as x, B’s age as y then find relation between x, y; after find C, D is age in terms of x, y and hence find x, y to get the desired answer.
Complete step-by-step answer:
In the question we are told that A and B are friends and A is older than B by 2 years. A’s father D is twice as old as A and B is twice as old as sister C. The age difference of D and C are also given which is 40 years so we have to find ages of A and B.
So, now let’s suppose A is elder than B by 2 years so,
According to the question A is elder than B by 2 years so,
\[x\ =\ y+2\]
Hence, \[x-y\ =\ 2\] …………………..(i)
Also, it is given that A’s father D is twice as old as A.
So, we can write it as,
D’s age is equal to \[2\times \]A’s age
So, D’s age \[=\ 2\times x=2x\]
It is also given that B’s age is twice as old as his sister C.
So,
B’s age \[=2\times \] C’s age
We can also write it as,
\[\text{C }\!\!'\!\!\text{ s}\ \text{age = }\dfrac{\text{B }\!\!'\!\!\text{ s age}}{2}\]
So, \[\text{C }\!\!'\!\!\text{ s}\ \text{age = }\dfrac{y}{2}\]
As we know that age difference of D and C is 40 then,
\[2x-\dfrac{y}{2}\ =\ 40\]
On multiplying by 2 we get,
\[4x-y\ =\ 80\] …………………………………(ii)
So, the two equations we get is,
\[x-y\ =\ 2\] ………………………….(i)
\[4x-y\ =\ 80\] ……………………… (ii)
Now, we will subtract (i) from (ii) we get,
\[\left( 4x-y \right)\ -\ \left( x-y \right)=\ 80-2\]
On simplification we get,
\[3x=\ 78\]
Hence, \[x=\ 26\]
Now, substituting \[x\] in equation (i) we get,
\[26-y\ =\ 2\]
Hence, \[y\ =\ 24\]
Hence, the A’s age is 26 years old and B’s age is 24 years old.
Hence, the correct option is (d).
Note: Instead of taking y, once can also take B’s age as \[x+2\] and proceed to find the values. This will make the problem for linear equation make the problem for linear equation of 1 variable.
Complete step-by-step answer:
In the question we are told that A and B are friends and A is older than B by 2 years. A’s father D is twice as old as A and B is twice as old as sister C. The age difference of D and C are also given which is 40 years so we have to find ages of A and B.
So, now let’s suppose A is elder than B by 2 years so,
According to the question A is elder than B by 2 years so,
\[x\ =\ y+2\]
Hence, \[x-y\ =\ 2\] …………………..(i)
Also, it is given that A’s father D is twice as old as A.
So, we can write it as,
D’s age is equal to \[2\times \]A’s age
So, D’s age \[=\ 2\times x=2x\]
It is also given that B’s age is twice as old as his sister C.
So,
B’s age \[=2\times \] C’s age
We can also write it as,
\[\text{C }\!\!'\!\!\text{ s}\ \text{age = }\dfrac{\text{B }\!\!'\!\!\text{ s age}}{2}\]
So, \[\text{C }\!\!'\!\!\text{ s}\ \text{age = }\dfrac{y}{2}\]
As we know that age difference of D and C is 40 then,
\[2x-\dfrac{y}{2}\ =\ 40\]
On multiplying by 2 we get,
\[4x-y\ =\ 80\] …………………………………(ii)
So, the two equations we get is,
\[x-y\ =\ 2\] ………………………….(i)
\[4x-y\ =\ 80\] ……………………… (ii)
Now, we will subtract (i) from (ii) we get,
\[\left( 4x-y \right)\ -\ \left( x-y \right)=\ 80-2\]
On simplification we get,
\[3x=\ 78\]
Hence, \[x=\ 26\]
Now, substituting \[x\] in equation (i) we get,
\[26-y\ =\ 2\]
Hence, \[y\ =\ 24\]
Hence, the A’s age is 26 years old and B’s age is 24 years old.
Hence, the correct option is (d).
Note: Instead of taking y, once can also take B’s age as \[x+2\] and proceed to find the values. This will make the problem for linear equation make the problem for linear equation of 1 variable.
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