Hi,

Can you please explain the solution?

Couldnt x be 1 and x-1 be 0

Isnt 0 also an integer?

No. Cannot be zero x.

Otherwise, the quantities were unuseful. if you consider x as zero (that is an integer) then in the quantities x/4 is impossible. x/0 is indefinite. And the question would fall apart.


Hope this helps

Multiples of 4 are 4 units apart each other. 4, 8, 12, 16 ...

Now x-1 and x are not multiples of 4. In a given set of 5 numbers where extremes are multiples of 4, like {4, 5, 6, 7, 8} x and x-1 are not multiples of 4. In the range, this can be 5,6 or 6,7. So x can be 5 or 6. Dividing them by 4, we get 1.50 and 1.75. In both the cases, tenth digit is greater.

Consider a second set for confirmation. {8, 9, 10, 11, 12}. x can be 10 or 11. 10/4 = 2.5 and 11/4 = 2.75. In this case too, tenth digit is greater. So the pattern repeats in every set and hence the tenth digit will ALWAYS be GREATER than the hundredth digit.

The answer is A.

I do not understand why you rule out the possibility of 1 as a remainder, Besides, Its seem all fine.

The question says "The integers x and (x - 1) are NOT divisible by 4." If 1 is a remainder, suppose for the number 9, then x= 9, and x-1 becomes (9-1)= 8, which IS DIVISIBLE by 4.
So we have to rule out the possibility of 1 as a remainder to comply with the question.

Hope this helps!

Sorry folks, but the correct answer is NOT A.
Since there's no information given that stipulates x must be positive, we must also consider negative values of x


The integers x and (x - 1) are not divisible by 4.
So, some possible values of x include: -6, -5, -2, -1, 2, 3, 6, 7, 10, 11, etc

Let's test a few values

If x = -6, we see that x/4 = -6/4 = -1.50. In this case, Quantity A = 5, and Quantity B = 0.
Here, Quantity A is greater than Quantity B

If x = -5, we see that x/4 = -5/4 = -1.25. In this case, Quantity A = 2, and Quantity B = 5
Here, Quantity B is greater than Quantity A

Answer: D

Exclude all multiples of 4 and their next successive numbers, i.e. include only 2,3 ... 6,7 ...10,11 ... 14,15 etc.
Let's focus on the required set of x containing 2,6,10,14, etc. We can see that every other number is added by 4, and we should only calculate a single case. This could be x=2 and 2/4=0.50

The tenths digit is always 5 and the hundredths digit is always 0

Answer is A

edited

Oh, I can see that negative values bring -1, -5, -9 as the x value possibilities and hence bifurcate answer choice A. Then it's D.

still answer is A, I think they forget to mention "positive integer."

When you divide an integer by 4, you know that your final value will have one of the following in decimal places: .00, 0.25, 0.50, or 0.75.

Just so you can trust me on this, let's check a few:
-4/4 = -1.00
-3/4 = -0.75
-2/4 = -0.50
-1/4 = -0.25
0/4 = 0.00
1/4 = 0.25
2/4 = 0.50
3/4 = 0.75
4/4 = 1.00
5/4 = 1.25
6/4 = 1.50
7/4 = 1.75
and so on...

Given that this pattern holds for all numbers, but then reverses when it goes positive to negative, let's just pick the simplest numbers we can from both positives and engatives. Looking at the start of our list above, what two numbers from 1-4 are next to each other and not divisible by 4? I see [1 & 2] or [2 & 3]. And then looking at the negatives, I see [-1 & -2] or [-2 & -3]

Option 1: x=2 and x-1 = 1
Option 2: x = 3 and x-1 = 2
Option 3: x = -1 and x-1 = -2
Option 4: x = -2 and x-1 = -3


So let's look at the remainders when these are divided by 4:
Option 1: 2/4 = 0.50 --> tenths place = 5, hundredths place = 0 --> the tenths place digit is greater (5 > 0)
Option 2: 3/4 = 0.75 --> tenths place = 7, hundredths place = 5 --> the tenths place digit is greater (7 > 5)
Option 3: -1/4 = -0.25 --> tenths place = 2, hundredths place = 5 --> ***the hundredths place digit is greater (5 > 2)***
Option 4: -2/4 = -0.50 --> tenths place = 5, hundredths place = 0 --> the tenths place digit is greater (5 > 0)

At first, when I only examined this problem using positive integers, I thought that quantity A was greater. However, when I realized integers can be negative, I realized that in option 3, the hundredths place is bigger, whereas in the other options, the tenths place is bigger. Thus, the answer is D - we cannot determine.

Carcass

I am curious about this question:
Consider the case of x=-1
You have -1*-2 -> not divisble by 4
but -1/4 is 0.25 which implies QB>QA

However if we take x=2 we have 2/4 -> 1/2 ->0.5 QA>QB

How can the OA be A if there are two conflicting cases?

But in this case the logic is the reverse of my explanation here https://gre.myprepclub.com/forum/the-in ... tml#p10456

If you take negative numbers such as -1 and -2 the number you divide is -2 NOT -1

-1 is our x-1 and -2 our x. It is the reverse.

-1/4 is not our case

-2/4 it is

Why would the number be -2? It says to divide x, and in our case x=-1, not x=-2. You can have x=-2 and x-1=-2-1=-3, which is also a valid case but so is x=-1.

Carcass
You are wrong if x=-1 x-1 is -2
so our case is -1/4

Thank you sir to point out this. appreciate

I said the other way around though