Sunil01 wrote:
Vyshak wrote:
Is x < y?
St1: sqrt(x) < y --> x < y^2
Since sqrt(x) is positive, y can take only positive values
If x = 1; y = 3; then is x < y? Yes
If x = 5; y = 3; then is x < y? No
Not sufficient
St2: x^2 < y
This statement seems to be fine if x is an integer. But what if x is a fraction?
If x is an integer and x^2 < y then x < y
If x and y are fraction, say, x = 1/2; y = 1/2 then is x < y? No
Not sufficient
Combining St1 and St2:
x^2 < y --> x < y when x is an integer
x < y^2 --> y is positive and x < y when x and y are fraction
Sufficient
Answer: C
Hi Vyshak,
Not getting how you combined statement 1 and statement 2.
Thanks & regards,
Sunil01
Hi Sunil,
Combining St1 and St2 we can derive the following conclusions:
1. y is positive
2. x^2 < y --> x is definitely less than y when x is an integer. If we check St1 we had a case in which x < y^2, but x > y. Now such cases are not possible.
3. x < y^2 --> x is definitely less than y when x is a fraction. If we check St2 we had a case in which x^2 < y, but x = y. Such cases are not possible now.
Using both the statements we can say that x is definitely less than y irrespective of the nature of x (whether a fraction or an integer)
Hope I am clear now.