Percentages !!`

This topic has expert replies
Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

Percentages !!`

by vishubn » Sun Dec 14, 2008 1:19 am
If Jack’s and Kate’s annual salaried in 1985 were each 10 percent higher than their respective annual salaries in 1984, what was Jack’s annual salary in 1984?
(1) The sum of Jack’s and Kate’s annual salaries in 1984 was $50,000.
(2) The sum of Jack’s and Kate’s annual salaries in 1985 was $55,000.


My appp

A) j + k=50000

Insuff

B) 1.1j+1.1k=55000

together

Possible!! But Oa is Wrong it says E

any comments
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!
Source: — Data Sufficiency |

Legendary Member
Posts: 621
Joined: Wed Apr 09, 2008 7:13 pm
Thanked: 33 times
Followed by:4 members

by vittalgmat » Sun Dec 14, 2008 1:39 am
Vishubn
U almost got the correct solution... but fell into the classic
2 eqn 2 variable trap.

Stmt 1: j + k = 50k

insufficient.


stmt 2:
in 1985, each of their salaries were 10% more than their
corresponding prev yr salaries.

so
1.1j + 1.1k = 55k
=> 1.1(j+k) = 55k
=> j +k = 50k

insufficient.


combining them.
Both equations are identical.
hence no unique solutions.

therefore E

Here is the takeaway from this (from one of Ron/Stuart's comments)

The real "rule" in mathematics is this: If you have n *independent* *linear* equations and n unknowns, you can solve for all n of your unknowns. By "independent", we mean that you can't derive any one equation from the others.

Senior | Next Rank: 100 Posts
Posts: 45
Joined: Mon Dec 08, 2008 9:55 pm
Thanked: 4 times

by 720dreaming » Sun Dec 14, 2008 1:43 am
yup, gotta watch out for those!

Legendary Member
Posts: 574
Joined: Sun Jun 01, 2008 8:48 am
Location: Bangalore
Thanked: 28 times

by vishubn » Sun Dec 14, 2008 4:46 am
Damn :) !! laziness ! do cost crazy !!

thanks guys

vishu
KILL !! DIE !! or BEAT my FEAR !!! de@D END!!