Blackbox
sashiim20
MathRevolution
If [x] is the least integer greater than or equal to x, and [x/2]=3, what is the scope of x?
A. 3<x≤4
B. 8<x≤9
C. 8<x≤10
D. 9<x≤12
E. 4<x≤6
[x/2]=3
maximum integer value for x could be 6. [6/2] = [3] = 3
substituting x = 5, we get = [5/2] = [2.5] = 3
substituting x = 4 , we get = [4/2] =[2] = 2
So x has to be greater than 4 and less than or equal to 6.
Therefore scope of x = 4<x≤6. Answer E...
Hello, Shashi
I am not sure I am following your approach. Could you please help me understand by giving a different explanation?
Hi Blackbox,
Question says [x] is the least integer greater than or equal to x. And [x/2]=3
So, maximum value of x when x/2 would be equal to 3 is 6.
When x = 6, [6/2] = 3
Now when we take x = 5 , 5/2 = 2.5
Now its given [x] is least integer greater than or equal to x.
Therefore for [2.5], least integer greater than or equal to 2.5 is = 3
Therefore x can have value of 6 and 5.
Now when we take x = 4 , 4/2 = 2
Therefore for [2], least integer greater than or equal to 2 is = 2. ------ Which is not equal to 3 as given in the equation. Hence x cannot be 4.
If we take x = 7, x/2 = 3.5.
For [3.5] least integer greater than or equal to 3.5 is = 4 ------ Which is not equal to 3 as given in the equation. Hence x cannot be 7.
Therefore x has to be greater than 4 and less than or equal to 6.
Given by x = 4<x≤6. Answer (E)...
Hope its clear now...