guys,
look at this one. seems to me a simple straightforward (d). but was surprised to see the reported answer as (e). I really doubt if this (reported answer) is correct.
Is there a chance that I am missing something here?
If x and y are positive, is x3 > y?
(1) (x^0.5) > y
(2) x > y
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
thanks,
Jamal
guys,
look at this one. seems to me a simple straightforward (d). but was surprised to see the reported answer as (e). I really doubt if this (reported answer) is correct.
Is there a chance that I am missing something here?
If x and y are positive, is x3 > y?
(1) x > y
(2) x > y
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
thanks,
Jamal
both statements u written are same??:shocked:
hi madnikhil,
typo happened when I pasted the Q from a pdf. have corrected it now.
thanks for pointing out.
guys,
look at this one. seems to me a simple straightforward (d). but was surprised to see the reported answer as (e). I really doubt if this (reported answer) is correct.
Is there a chance that I am missing something here?
If x and y are positive, is x3 > y?
(1) (x^0.5) > y
(2) x > y
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
thanks,
Jamal
correct answer would be D only..
e in any case is not possible.. because both statements says clearly that x>y, so x3>y ..
Hence D..
may be oa is wrong...
If x and y are positive, is x3 > y?
(1) (x^0.5) > y
(2) x > y
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.
Let x = 0.81 and y = 0.78
From Statement 1, root(x) > y ..Agreed
0.9 > 0.78
But (0.81)^3 is not Greater than 0.78 ....So we cannot determine from this statement alone....However we get x^3 > y..when x,y > 1
Similarly from Statement II we cannot determine....
Combining both..we cannot determine again...
Hence..Option (e)....
Let x = 0.81 and y = 0.78
From Statement 1, root(x) > y ..Agreed
0.9 > 0.78
But (0.81)^3 is not Greater than 0.78 ....So we cannot determine from this statement alone....However we get x^3 > y..when x,y > 1
Similarly from Statement II we cannot determine....
Combining both..we cannot determine again...
Hence..Option (e)....
is it x^3
or x3????
is it x^3
or x3????
I think..we have seen loads of Typo Errors By Fellow Puys...So better discern on your own ..Yes it would be x^3....rather than x3..what is x3..either it would have been 3x....or x^3...so i did from x^3..
guys,
look at this one. seems to me a simple straightforward (d). but was surprised to see the reported answer as (e). I really doubt if this (reported answer) is correct.
Is there a chance that I am missing something here?
If x and y are positive, is x3 > y?
(1) (x^0.5) > y
(2) x > y
thanks,
Jamal
for 0
I) if you take a value for y such that it lies b/w ( x^3 , x^2) it violates
II) if you take a vlue for y such that it lies b/w ( x^3 , x ) it violates
Hence even using both , you cannot determine the given inequality ...
Can any1 elucidate me rgarding d Chinese Remainder Theorem nd Fermat's little Theorem with an example each???
or else post me a link where i can study them...thnks n advance
Can any1 elucidate me rgarding d Chinese Remainder Theorem nd Fermat's little Theorem with an example each???
or else post me a link where i can study them...thnks n advance
hi,
have a look at this post...it will definitely help you...
http://www.pagalguy.com/forum/cat-and-related-discussion/43743-the-shout-boxers-team-09-a-24.html#post1680098
also, plz try to search the thread before posting any query...
there is a 90% chance that the query has been already answered...
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
plz can anyone tellme the approach to solve this question ??
Quote:
Originally Posted by barclays_boss View Post
how many 3 digit nos. are there , the cubes of which end with "56" ?
1. 30
2. 27
3. 18
4. 20
I think answer should be 18 as numer shud be of type AB6 which on expanding binomially give .(...+ 3*36*AB0 + 216 ) .You have to just find what value of 'B' on multiplication of 8 give last digit as 4.Therefore B can be 8 or 3.So 'B' can take value 3 or 8. A can take value 1 to 9.henece 18 values...
"In a number system the product of 122 and 41 is 5442. What is the base of this number system?"
plz can anyone tellme the approach to solve this question ??
"In a number system the product of 122 and 41 is 5442. What is the base of this number system?"
plz can anyone tellme the approach to solve this question ??
Getting 6.........
Let the base be x then ............
*(4x + 1) = 5(x^3) + 4(x^2) + 4x + 2 or
(x^3) - 5(x^2) - 6x = 0.......which gives
x = 0,-1,6 where 0 and -1 are not possible hence ans 6........
Puys, Please solve this:
2^495 mod 625 = ?
Puys, Please solve this:
2^495 mod 625 = ?
Getting 293.........
By Inverse Euler's.........
E(625) = 500 so we have to find
2^(-5) mod 625 multiplying numerator by 626^5 we get.....
Now this ques becomes lengthy.......
313^5 mod 625 = 293...........
Wait for shorter approach by someone........
method pls
Puys, Please solve this:
2^495 mod 625 = ?
2^495 mod 625
(2^10)^49 * 2^5 mod 625
(24^49 + 24^49 * 1000) * 32 mod 625
32*224 mod 625
293
can u please tell me wats euler's rule
2^495 mod 625
2^495 mod 5 => 3
2^495 mod 125 = 43
125a+43 = 5b+3
Answer = 293
125 & 5 are not coprime or am I missing something?
my take:-
2^495 mod 5^4 =
32.(1000+24)^49 mod 5^4 =
32.(24^49 + 24^49 *1000) mod 5^4 =
32(599-375) mod 5^4 =
32*224 mod 5^4 = 293
OR
2^500 mod 5^4 = 1,
2^495 mod 5^4 = p,==>
32p = 625k +1
p = 19k + (17k+1)/32,==>k = 15
p = 293