Number Theory & word problem

[zen]

1) If 'u' and 'v' are positive real numbers, is u > v?

A) u³ / v <1
B) u^1/3/v <1



2) A set of practice questions is 'L' cm long, M centimeter wide and N centimeter thick. These sets are shipped in a box which is L centimeters wide and M centimeters deep. how long does the box have to be to enable the shipping of 30 sets per box?

A) N = 5
B) M = 20 and L = 27.5

[Brent Hanneson]

Sorry, but it's hard to have worthwhile discussions in a thread, when there are two or more topics of discussion (i.e., problems)
You'll get more responses by having 1 question per thread.

Cheers,
Brent

[sandeep_chads]

The best way to answer question 1 is by putting numbers and validating the equations:

Given - u, v are positive integers
What we have to find is whether - u> v

Option 1 -> u^3 < v

put values for u and v

Case 1 -
u = 1/2
v = 1/7
U^3 = 1/8
u^3 < v therefore option 1 holds good and u>v

case 2 -
u=1/3
v=1/2
u^3 = 1/27
Here u^3 < v however u<v

Therefore option1 is not sufficient

Option 2 -> u^3 . v > 1
Here we can have u = 5, v = 1 or v = 5 and u = 1 and the equation will hold true. This one is also not sufficient


Combing option 1 and option 2 two we get:

1/v < u^3 < v

Put values:
V = 5, U =1
The equation above holds true and U<V

Put values:
V = 10, U =2
The equation above holds true and U<V

Put values:
V = 11, U =12
The equation above does not hold true

I observed that whenever I put U>V, the equation above fails therefore U<V. We need both a and b to answer the question therefore the right answer is (C)

[sandeep_chads]

For question 2,

Volume of the Set of questions = L *M * N
VOlume of the case = M*L*(x) where x = unknown length

30 sets need to fitted in the box, i.e.

LMN(30) = ML(x)
30N=x

We just need the value of N to answer the question. Therefore A is the right answer

[Red Bozo]

For Q1, this is what I did...seemed easier than taking values:

Since u and v are +ve real nos., we have:

A. v>u³
B. v>u^1/3

Case 1: u<1... so on the number line: 0___u³___u___u^1/3
Case 2: u>1... so on the number line: 0___u^1/3___u___u³
Case 3: u=1.

Now, in case 1: v can be either side of u and still satisfy statement A, so obviously A alone isn't sufficient.

Likewise, in case 2: v can be either side of u and still satisfy statement B, so obviously B alone isn't sufficient.

Considering both statements together, we see that v has to be greater than both u^1/3 and u³, but in both case 1 and 2, u is between u^1/3 and u³, so v is BOUND TO BE greater than u. So both together are sufficient.

Case 3 need not be considered since A and B individually get eliminated by Case 1 and 2.