Question

Five years ago, the average age of A, B, and C was 41 years. If D joined them now,
the average of the ages of all four of them 46years. The present age of D is
​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

• Let the present age of A be 'x' years.

• Let the present age of B be 'y' years.

• Let the present age of C be 'z' years.

• Let the present age of D be 'w' years.

Now,

5 years ago,

• The age of A = x - 5 years

• The age of B = y - 5 years

• The age of C = z - 5 years.

Since,

$\large \underline{\tt \: \blue{ According \: to \: statement }}$

• The average of A, B & C was 41 years.

$\rm :\implies\:\dfrac{x - 5 + y - 5 + z - 5}{3} = 41$

$\rm :\implies\:x + y + z - 15 = 123$

$\rm :\longmapsto\: \boxed{ \pink{ \bf \:x + y + z = 138}} - - (1)$

$\large \underline{\tt \: \red{ According \: to \: statement }}$

When, D joined them, the average of the ages of all four of them is 46 years.

$\rm :\implies\:\dfrac{x + y + z + w}{4} = 46$

$\rm :\implies\:138 + w = 184$

$\rm :\longmapsto\: \boxed{ \pink{ \bf \: w = 46 \: years}}$

Answer 2

Solution:-

Here ,

Five years ago,

The average age of A, B and C was 41 years .

Let the age of the A, B and C be

( A - 5 ) , ( B - 5) and ( C - 5 )

Therefore,

According to the question,

(A - 5) + ( B - 5 ) + ( C - 5 ) / 3 = 41

A - 5 + B - 5 + C - 5 / 3 = 41

A + B + C - 15 = 41 * 3

A + B + C -15 = 123

A + B + C = 123 + 15

A + B + C = 138. ( 1 )

Now,

D joined them, So the average of all four of them is 46 years

Therefore,

A + B + C + D / 4 = 46

From ( 1 )

138 + D = 46 * 4

138 + D = 184

D = 184 - 138

D = 46 years

Hence, The present age of the D is 46 years $.$