• Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 822
  • Last Modified:

the age old problem

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?
0
shilpi84
Asked:
shilpi84
3 Solutions
 
ozoCommented:
27.5 and 16.5
rena was 3 times as old as seena when reena was 16.5 and seena was 5.5
0
 
shilpi84Author Commented:
how do we figure this out?
0
 
ozoCommented:
27.5 + 16.5 = 44
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
0
 
sunnycoderCommented:
Answers given by ozo are correct and here is a simple way to reach it:

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
0
 
Harisha M GCommented:
From the page http://72.14.203.104/search?q=cache:d3pyauuW1wEJ:www.puzzlesolver.com/board/viewtopic.php

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.
0

Featured Post

How to Use the Help Bell

Need to boost the visibility of your question for solutions? Use the Experts Exchange Help Bell to confirm priority levels and contact subject-matter experts for question attention.  Check out this how-to article for more information.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now