Word Prob

1997->k
1998-> k(1+ x/100)
1999- > k(1+ x/100) - y/100 k(1+x/100)
= k(1+x/100) (1-y/100)
= k (100+x/100) (100-y/100)


Q: Is k (100+x/100) (100-y/100)> k

Stmt I

Clearly insuff since we can pick x and y were left hand side of the question may oy may not be greater than k

x=1% y=99%
x=99% y=.0000000001%

INSUFF

Stmt II
xy/100 < x – y

Let try to manipulate what wee need to find and see if we can find an answer

Is k (100+x/100) (100-y/100)> k (k's cancel)

(100+x) (100-y) > (100)^2
100^2+100x-100y - xy > 100^2
100(x-y) > xy

Is x-y > xy/100??

From stm II YES

Hence SUFF

B)

beater wrote:The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y
(2) xy/100 < x – y

Let R = rent collected in 1997

In 1998 rent collected = R[(100+x)/100]

In 1999 rent collected = R[(100+x)/100] * [1-(y/100)]

1997 < 1998 > 1998

R < R[(100+x)/100] > R[(100+x)/100] * [1-(y/100)]

Divide by R:

1 < [(100+x)/100] > [(100+x)/100] * [1-(y/100)]


Statement (1) x>y


x = 1, y = 0

1 < (101/100)*(1) True

x=201, y = 200

1< negative False

Insufficient

Statement (2).... anyone else?

IMO B

IMO D

1 x> Y
2 On the same lines x- y - xy/100 which implies x>y

If z is the rent in 1997,

then rent in 1998 is z+(xz/100)
and the rent in 1999 is [z+(xz/100)] -(y/100)[z+(xz/100)]
=[z+(xz/100)] - (yz/100) - (xyz/10000) = z+(z/100)(x-y-(xy/100))

The question requires us to compare z+(z/100)(x-y-(xy/100)) (1999)w/ z (1997)

Stm1: x>y
This gives us the diff values of x-y-xy (if x=1 and y=0; or x=2, y=1)
--Insuff

Stm2: (xy/100)<x-y implies that z+(z/100)(x-y-(xy/100)) is greater than z--Suff
so ans is B