Question

Three years ago, the average age of A, B and C was 27

years and that of B and C 5 years ago was20 years. A’s

present age is
​

Answer 1

Step-by-step explanation:

Let present age of A, B and C be x, y and z years

3

(x−3)+(y−3)+(z−3)

=27

⇒x+y+z=90

Also,

2

y−5+z−5

=20

⇒y+z=50

⇒x=90−50=40 years.

Answer 2

GIVEN :-

• Three years ago, average age of A , B and C was 27 .

• Five years ago, average age of B and C was 20 .

TO FIND :-

• Present age of A .

SOLUTION :-

Let ,

• Present age of A = x years

• Present age of B = y years

• Present age of C = z years

 

Three years ago ,

• Age of A = ( x - 3 ) years

• Age of B = ( y - 3 ) years

• Age of C = ( z - 3 ) years

Five years ago ,

• Age of A = ( x - 5 ) years

• Age of B = ( y - 5 ) years

• Age of C = ( z - 5 ) years

 According to the first condition ,

     

     $\longrightarrow \sf \dfrac{(x-3)+(y-3)+(z-3)}{3} =27 \\\\\longrightarrow \dfrac{x+y+x-3-3-3}{3} =27 \\\\\longrightarrow x+y+z-9 = 27 \times 3\\\\\longrightarrow x+y+z -9 = 81 \\\\\longrightarrow\bf x+y+z =90 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(1)$

According to the second condition ,

   

  $\longrightarrow \sf \dfrac{(y-5)+(z-5)}{2} =20 \\\\\longrightarrow \dfrac{y+z-5-5}{2}=20 \\\\\longrightarrow y+z-10 = 2 \times 20 \\\\\longrightarrow y+z-10=40\\\\\longrightarrow \bf y+z = 50 \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..................(2)$

Put  $\sf y+z=50$  in Eq (1) ,

  $\longrightarrow \sf x+50=90\\\\\longrightarrow x=90-50\\\\\longrightarrow \bf x = 40$

• Present age of A = 40 years