1. a let individual chairs cost $n and bunch of 6 chairs = $m

n= 0.9* 6m = 0.54m not sufficient.

b n = 5m+20 not sufficient.

a+b 5m+20 = 0.54m thus m = 0.04m = 20. Sufficient. thus C.

2

a can be (1,1) or (-1,-2,-3) not sufficient.

b can be (1,1) or (-1,1) not sufficient.

a+b

numbers cannot be (-1,1) or (-2,-1,1,2) meaning having both positive and negative integers.
numbers have to be either (-1,-2) or (1,2) and so on. meaning either completely positive or completely negative.

sufficient.

hence C.

First problem:
Set G: price of a set of 6 chairs, S: price of a single chair

Question stem asks, in other words, what is the value of G?

1) G = 90% * 6S.
We are not given S or G, so this equation cannot be solved. Insufficient.

2) G = 5S + 20.
We are not given S or G, so this equation cannot be solved. Insufficient.

1 & 2) We immediately see that we can substitute for G to create an equation with one unknown. Carrying out the operations just for illustration we see:
5S + 20 = 90% * 6S
20 = 0.4S
S = 20(5/2) = 50. Sufficient.

IMO C

Second problem:
The question stem asks whether the product of all integers in the list is positive. In other words, is the count of negative numbers in this list even?

1) This statement says that the smallest and greatest integers are the same sign, ie all integers in the list are positive or all are negative. But we could have {-5,-4,-3}, which yields a negative product, but {1,2,3,4,5}, which yields a positive one. Insufficient.

2) We could have {-2,-1,1,2}, which yields a positive number, or {-1,1,2,3} which yields negative. Insufficient.

1 & 2) Taking both statements, the list can be:
a) all negative numbers which can then be factored into the list (-1)^(k)*{a,b,c,d,...} where k is an even integer making the product positive.
b) all positive numbers, whose product will always equal a positive number. Sufficient.

IMO C

In the second question,
Can someone explain as to how we could deduce(from 1 and 2) that all elements in the list are either positive or negative?

Thanks.

For DS7 it shall be E

we need even number of equal signed numbers..

Hi treker,

The first statement in the second problem says "the product of the greatest and smallest integers in the list is positive." Algebraically, this is stated as xy > 0. By the way, I recommend memorizing this inequality as "x and y are the same sign" whenever you see it written this way, as this is quite popular in the GMAT I've noticed.

We can deduce that all numbers are either negative or positive because they all fall between the largest and the smallest number, which are both the same sign.

I hope this addresses your question.

DS8: Let Set = S, individual = I

[spoiler]A) S= 5.4I; S = .9 x 6I; neither I or S are given
B) S= 5I + 20; same as above (2 variables in eqn)
C) 5.4I = 5I + 20; equate the two together
D) invalid[/spoiler]


shouldn't DS7 be E?

1) product of low & high can be these combinations: neg/pos, neg/neg, pos/pos; insuff
2) even number of integers: neg neg pos pos; neg pos pos pos

combine: ex. (-2, 1, 2, 4) or something like that... or (pos x pos x pos x pos)