Question

In a group of 4 friends, ratio of present age of A and D
is 4:5 and that B to C is 3:4. Calculate the present
average age of A, B and C, if 4 years ago, A was 20
years younger than C and at present C is twice of age
of A.​

Answer 1

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Answer 2

Answer:

let the age of A is x.

at present C is twice of A.

so age of C is 2x.

4 years ago age of A was x-4 and age of C was 2x-4.

given that 4 years ago,A was 20 year younger than C

so mathematically,

x-4=2x-4-20

=>x-4=2x-24

=>2x-x=24-4

=>x=20

so present age of A is 20years and present age of C is 2x=2×20=40 years.

according to the question

ratio of present age of A and D is 4:5

mathematically

$\frac{present \: age \: of \: A}{present \: age \: of \: D} = \frac{4}{5} \\ = > \frac{20}{present \: age \: of \: D} = \frac{4}{5} \\ = > present \: age \: of \: D \times 4 = 20 \times 5 \\ = > present \: age \: of \: D = \frac{100}{4} = 25$

again ratio of present age of B and C is 3:4

$\frac{present \: age \: of \: B}{present \: age \: of \: C} = \frac{3}{4} \\ = > \frac{present \: age \: of \: B}{40} = \frac{3}{4} \\ = > present \: age \: of \: B \times 4 = 3 \times 40 \\ = > present \: age \: of \: B = \frac{120}{4} = 30$

so present age of A, B,C is 20,30,40 respectively

so average age of A B and C is

$\frac{20 + 30 + 40}{3} \\ = \frac{90}{3} = 30$