rena was 3 times as old as seena when reena was 16.5 and seena was 5.5

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Posted on 2006-04-20

the combined ages of reena and seens are 44 years and reena is twice as old as seena was when reena was half as old as seena will be when seena is 3 times as old as reena was when reena was 3 times as old as seena.how old r they?

5 Comments

reena was 3 times as old as seena when reena was 16.5 and seena was 5.5

3 time as old as that is 49.5

half as old as that is 24.75

seena was 13.75 when reena was 24.75

27.5 = 2*13.75

Start from end of question

>when reena was 3 times as old as seena

Let age of reena be r and age of seena be s ... Then

R S

--------------------------

3s s >>when reena was 3 times as old as seena

9s >>when seena is 3 times as old as reena

=>

11s 9s >>note that difference R-S is constant 2s, so at this point Reena is 11s yrs old

9/2s >>when reena was half as old as seena

=>

9/2s 9/2s - 2s >> constact difference

9s-4s >> reena is twice as old as seena was

=>

5s 3s

the combined ages of reena and seens are 44 years

=> 5s + 3s = 44

=> 8s = 44

=> s = 5.5

Hence their ages are

Reena 5 * 5.5 = 27.5

Seens 3 * 5.5 = 16.5

Cheers!

sunnycoder

First rearrange the question a bit and insert some variables (chosen to avoid awkward fractions):

1. The combined ages of Reena (call her age 2A) and Seena (call her age B) are 44 years (2A + B).

2. Reena (2A) is twice as old as:

3. Seena was (A) when Reena was (9C) half as old as:

4. Seena will be (18C) when Seena is (18C) three times as old as:

5. Reena was (6C) when Reena was (6C) three times as old as Seena (2C).

6. How old is Reena (now 2A)?

Ignore the variables and the text is exactly the same as the original question!

At each point of comparison, the difference in their ages (D) is the same, so:

D = 6C - 2C (line 5); D = 9C - A (line 3); and D = 2A - B (line 1)

So 4C = 9C - A, so 2A = 10C

Also 4C = 2A - B, so 4C = 10C - B, so B = 6C

So 2A/B = 10C/6C = 5/3

But 2A+B = 44

So 2A = 27.5

So Reena is 27 years 6 months.

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