need help with ration, mixtures and alligation

[readsid]

Q1: the weights of 3 heaps of sand are in the ratio 5:6:7. the first two heaps were increased by a fractions of themselves such that the ration of the weights chaged to 7:6:5.

i) by what fraction the frist increased ?
1. 24/25
2. 2/5
3. 12/13
4. 19/20

ii) By what fraction was the second increased?
same options as above

no matter how many problems i work out i get stuck in some or the other problem related to this chaper.. i guess im missing on some basic funda ... can someone give me notes on this chapter or book that i could refer to???

[loi.fu.yogi]

The problem needs only the fraction. Hence we can substitute our values to get the fraction.

Any ratio will have greatest common factor removed.For example: if heaps weighed 50,60, and 70, we state the ratio as 5:6:7, removing the gcf 10.

In our problem we have two such ratios and a connecting link between them. we can substitute our own numbers for gcf and equate them to get the fraction.

The key is last heap's weight has remained constant but the ratio has changed. Assume weight of last heap is 7x. After changes to first two heap, the weight of last heap is 5y. Since last heap weight has not changed y= 7x/5.

Now we need to substitute a good number for x. Assume x=5. => y=7.

Now the initial weights of 3 heaps are as follows: 5x:6x:7x - Substitute x=5 - 25:30:35

Modified weight of 3 heaps : 7y:6y:5y - substitute y=7 - 49:42:35

Now cross check the last heap's weight, whether it has remained constant after substitution. Since it is 35 in both the case, we are good.

The difference in first heap = 49-25 = 24. Hence the fraction of change 24/25.
The difference in second heap = 42-30 =12. Hence the fraction of change 12/30 => 2/5.

[readsid]

thanks yogi....
hit and trial method... but will this work for all the problems??
can somebody break this down without using the hit and trial method ??