Hi,
Can someone help and tell me the way to solve this one?
If k not equal to 0,1 or -1 ,is 1/k>0?
1>1/k-1 >0
2>1/k+1 >0
OA is A.Thanks
Hi,
Can someone help and tell me the way to solve this one?
If k not equal to 0,1 or -1 ,is 1/k>0?
1>1/k-1 >0
2>1/k+1 >0
OA is A.Thanks
I assume.
1> 1/(k-1) > 0 not (1/k)-1 > 0
2> 1/(k+1) > 0
1 => (k-1) > 0 => k>1 => 1/k is always greater than 0..answer is always YES.
2 => (k+1) > 0 => k>-1 => k= -0.5, -0.4, 0.5, 3...so 1/k can be >0 &
So A.
Hi,
I did not understand how first condition gives you k-1>0 or how second gives you k+2>0.please explain.
Thanks.
Hi,
I did not understand how first condition gives you k-1>0 or how second gives you k+2>0.please explain.
Thanks.
For 1/(k-1) to be positive, (k-1) should be positive..so (k-1)>0
if (k-1) is 1/(k-1) is
same applies to second statement..
Hope this clarifies.
Hi,
Can someone help and tell me the way to solve this one?
If k not equal to 0,1 or -1 ,is 1/k>0?
1>1/k-1 >0
2>1/k+1 >0
OA is A.Thanks
Using St.1 alone,
1/(k-1)>0...taking reciprocals, k-1hence, k
st.1 alone does not give a definite answer.
Using St.2 alone,
1/(k+1)>0...taking reciprocals, k+1hence, k
st.2 alone gives a definite answer i.e. k is always LESS than 0..
hence, st.2 alone is sufficient to answer the question..
answer is A..
Hi,
Can someone help me with the way to pick values and solve this problem?
If zt
1> z2> t
Thanks.OA is E
Hi,
Can someone help me with the way to pick values and solve this problem?
If zt
1> z2> t
Thanks.OA is E
Hi suruchee,
from next time please dont give OA and let the people come up with their answers .................
for this probs
Given that Zt
lets take first statement
it says Z
Now lets take second statement
It says tin this case for every value of z i.e. z czn be less than or greater than 4 so we can not say anything by second statement as well
even by taking both together we wont be able to tell if z
hope i am making some sense..........
do correct me if i am not........
Hi suruchee,
from next time please dont give OA and let the people come up with their answers .................
for this probs
Given that Zt
lets take first statement
it says Z
Now lets take second statement
It says tin this case for every value of z
i.e. z czn be less than or greater than 4 so we can not say anything by second statement as well
even by taking both together we wont be able to tell if z
hope i am making some sense..........
do correct me if i am not........
hey...small typo error ..statement in bold is incorrect...if t is -ve, z is +ve, but value cud be less than or more than 4...not conclusive...
combined statements..
t= -6 and z= 1
Or t=-6 and z= 6......not conclusive that z is below 4...Ans E
Hi,
Here is a question from GMATPREP:
is x^4 + y^4 >z^4>
1>x^2 + y^2 > z^2
2>x+y > z
Please discuss how you solved.
Thanks.
Hi,
Here is a question from GMATPREP:
is x^4 + y^4 >z^4>
1>x^2 + y^2 > z^2
2>x+y > z
Please discuss how you solved.
Thanks.
IMO: A
Pl post the OA
Hi,
Here is a question from GMATPREP:
is x^4 + y^4 >z^4>
1>x^2 + y^2 > z^2
2>x+y > z
Please discuss how you solved.
Thanks.
Where have you gone suruchee?? Post the OA's for the Q's..
can SOme body help me with this... 
5 = 5, |-5| = 5, so |x| = x, if x is positive or 0 and |x = -x if x is negative.
x| = x, if x is positive or 0 and |x = -x if x is negative.
HOw is this possible
that |x= -X if x is negative.. i mean i did not get the rule..
:-x happy
can SOme body help me with this...
5 = 5, |-5| = 5, so |x| = x, if x is positive or 0 and |x = -x if x is negative.
x| = x, if x is positive or 0 and |x = -x if x is negative.
HOw is this possible
that x= -X if x is negative.. i mean i did not get the rule..
:-x happy
Hey obstaclez,
Let, y=x (signifies y takes the absolute of x irrespective of whether x is +ve or -ve)
It means that, when x is positive y is also positive
so whenever x>=0 the above equation translates to y=x.
When x is negative y is still positive because that's what absolute value means. So when x
eg: if x=4 then , y=x becomes y=x. hence y=4
but if x=-4 then to make y +ve we need to multiply x with a -ve sign.
Hence y=-x,and so y=-(-4)=4
That's why we get the V-shaped graph of the function y=|x !!
Hope this solves your doubt.....:smile:
I have one doubt on below DS from 1000DS q's
22. If 1 d d equal to 9?
(1) d + 0.01 (2) d + 0.05 > 2
My answer is E ( Neither Alone nor taken together is sufficient to answer)
But the actual answer states B (Statement 2 alone is sufficient).
How is it possible?
I have one doubt on below DS from 1000DS q's
22. If 1 d d equal to 9?
(1) d + 0.01 (2) d + 0.05 > 2
My answer is E ( Neither Alone nor taken together is sufficient to answer)
But the actual answer states B (Statement 2 alone is sufficient).
How is it possible?
here,d=1.xxxx
so,
statement 1:d+0.01so,dcannot say
statement 2:d+0.05>2
d>1.95
so,it is sufficient.
so,statement 2 alone is sufficient.
hope that helps.
here,d=1.xxxx
so,
statement 1:d+0.01so,dcannot say
statement 2:d+0.05>2
d>1.95
so,it is sufficient.
so,statement 2 alone is sufficient.
hope that helps.
can you pl throw more light on your explanation.
Try plugging some values
Statement 1
if d= 1.91
then d + 0.01 = 1.92 In this case tenths digit is 9, so try some other number.
if d= 1.3
then d + 0.01 = 1.31 In this case tenths digit is not 9.
Conclusion: Statement 1 alone is not sufficient.
Statement 2
Since If 1 d Statement 2 i.e. d+0.05>2 actually means that 1.95 This clearly shows that the tenths digit will always be "9". SO statement II alone is sufficient, hence answer is "B".
hey guys a friends showed me this question can you guys explain:
If Fran jumps straight up off the floor and lands on her feet T seconds later, her feet will reach a maximum height of 1.22t^2 meters above the floor. On one such jump, was Fran off the floor for less than 1 second?
1) On her jump Fran's feet reached a maximum height of 1 meter above the floor.
2) On her jump Fran's spent more than 1/4 second ascending.
I think the answer should be "A".
rajatmeh SaysI think the answer should be "A".
I don't think I'm understanding the question right, could you explain your answer?
Thanks