How to solve questions like these ?

[TkNeo]

As a bicycle salesperson, Norman earns a fixed salary of $20 per week plus $6 per bicycle for the first six bicycles he sells, $12 per bicycle for the next six bicycles he sells, and $18 per bicycle for every bicycle sold after the first 12. This week, Norman earned more than twice as much as he did last week. If he sold x bicycles last week and y bicycles this week, which of the following statements must be true?

I. y > 2x
II. y > x
III. y > 3


A. I only
B. II only
C. I and II
D. II and III
E. I, II, and III

OA:D

How do you proceed for questions like these. I have accross these types a few times now.

The way i go about is like this. To prove y > 2x , I will take a few scenarios and try to prove that y>2x is not essentially true. I I fail to do that then i would assume y >2x. But the problem is that since its not a standard problem i am never sure what data points would be representative enough. Lets say i take 5 set of data points and all of them prove y>2x but i miss a 6th type which proves y < 2x... So i don't feel confident...

What do you guys say ?

-Tk

[hopetobeat]

y>x is absolute, y>2 is qutie obvious because if y=3, salary of this week is 38, and last week must be <19>2x, if we find one senario y can be either >2x or not >2x, this answer should be opt out. Since 6 is a crutial sales number, I will start with x=6

x=6-->last week salary = 56 -->this week >112
If y=2x=12, this week salary=128, suffice; further, if y=11, 116>112 as well, so we can see when x=6, y can <or>2x. (3)is not correct.

IMO D

[camitava]

TkNeo,
I would like to take diff approach, I think -
Take some diff situation -
let x > 12 -> s = 18x - 128
x = 12 -> s = 128
6< x <12> s = 12x - 16
x = 6 -> s = 56
x <6> s = 20 + 6x
From the question, its obvious that we have to select II, III. But now our question comes whether I will be included or not. Let say x = 6 so sale = 56. But this week sale = 2 x 56 = 112 - which means y < 12. --- not holding I.
So IMO D. Hey got me, TKNeo?