Hi guys,
Please suggest what should be the answer to this question.I think we can not answer this question with 1 & 2 and the answer should be E;but correct answer is C.
Is the surface area of a rectangular pool larger than 1000 square feet?
(1) The pool measures fifty feet diagonally.
(2) One side of the pool measures twenty five feet.
Thanks,
Amit
I am not sure if anyone is following this thread.
If xyz not equal 0,if x(y+z) >= 0.
1) y+z = |y| + |z
2) x+y| = |x| + y
someone answered C,
I think,we have been asked to figure out if x(y+z) = 0 or x(y+z) > 0 .if we can answer either of them,we are good to say if the statement is true or false
from St 1...it proves y+z is pstive and x not being equal to 0,it always says x(y+z) cant be equal to 0...so this is enough to answer the query.
St 2 is useful,if someone wants to know if x(y+z) > 0 only
Correct me If I am wrong...
Thanks
Soumik
Hi guys,
Please suggest what should be the answer to this question.I think we can not answer this question with 1 & 2 and the answer should be E;but correct answer is C.
Is the surface area of a rectangular pool larger than 1000 square feet?
(1) The pool measures fifty feet diagonally.
(2) One side of the pool measures twenty five feet.
Thanks,
Amit
Amit C should be correct.
break the rectangle into two triangles...
so area triangle 1 = area triangle 2
triangle 1 ; let take ABC is the trangle.
AC = diagonal = 50
any one side is 25.
now if we divide rect..into two traingle..each will be right angled triangle...so by putting pythagors theorem...we would know the other side...of the triangle..which will come arnd ~43..so now we know the area of trng ABC....multiply by 2..you know rectangle area...and hence if its greater or smaller..than 1000 sq ft
i hope this helps...
I am not sure if anyone is following this thread.
If xyz not equal 0,if x(y+z) >= 0.
1) y+z = |y| + |z
2) x+y| = |x| + y
someone answered C,
I think,we have been asked to figure out if x(y+z) = 0 or x(y+z) > 0 .if we can answer either of them,we are good to say if the statement is true or false
from St 1...it proves y+z is pstive and x not being equal to 0,it always says x(y+z) cant be equal to 0...so this is enough to answer the query.
St 2 is useful,if someone wants to know if x(y+z) > 0 only
Correct me If I am wrong...
Thanks
Soumik
from statement 1, we can infer that y+z is +ve. so the value of x(y+z) will depend on whether x is positive or negative. if x is positive, then x(y+z)>0, and is x is negative, x(y+z)
Quote:
Originally Posted by soumikc View Post
I am not sure if anyone is following this thread.
If xyz not equal 0,if x(y+z) >= 0.
1) y+z = |y| + |z
2) x+y| = |x| + y
someone answered C,
I think,we have been asked to figure out if x(y+z) = 0 or x(y+z) > 0 .if we can answer either of them,we are good to say if the statement is true or false
from St 1...it proves y+z is pstive and x not being equal to 0,it always says x(y+z) cant be equal to 0...so this is enough to answer the query.
St 2 is useful,if someone wants to know if x(y+z) > 0 only
Correct me If I am wrong...
Thanks
Soumik
From Statement 1 We can not predict that (y+z) will be positive;
for example Y=-7 AND Z= -5 here Y+Z = |Y| + Z = +12
BUT VALUE OF (Y+Z) will be -12
please correct me if I am wrong
Friends, Take a look at this DS Problem :-
Is n an integer greater than 4?
(1) 3n is a positive integer.
(2) n/3 is a positive integer.
Answer is E
I am accepting answer as E, But would like to know various approaches, since such questions are more frequent ....
Is n an integer greater than 4?
(1) 3n is a positive integer.
(2) n/3 is a positive integer.
Answer is E
I am accepting answer as E, But would like to know various approaches, since such questions are more frequent ....
Friends, Take a look at this DS Problem :-
Is n an integer greater than 4?
(1) 3n is a positive integer.
(2) n/3 is a positive integer.
Answer is E
I am accepting answer as E, But would like to know various approaches, since such questions are more frequent ....
Answer has to be E.here's my approach:
Statement 1-
3n is a positive integer.
3n can be 3,6,9,12, 15,21....etc
that implies n can be 1,2,3,4,5,6,7...etc
so we cannot conclude n is greater than 4.
Statement 1 is insufficient.
Statement 2-
n/3 is positive integer.
n/3 can be 1,2,4,...etc
that gives n as 3,6,12..etc
so we again cannot conclude that n is greater than 4.
Statement 2 is insufficient.
Using both 1 and 2 we get n can be n can 3,6,12 etc..
so we cannot conclude n is greater than 4.
Hence answer is E.
Hope this helps.
Regards,
Neha
hi neha
Q. is complete as far as it is given in the book
this is from tmh gmat book
actually this weekend i was starting quant n was looking for some good option to get started, so for the time being i started with tmh.
i scored 44 in gmat prep test 1 quant sec without any preparation so i guess quant is not a very big hurdle
so could u plz suggest me some good strategy for quant
i mean i need a decent book/ material which covers all the concepts.
@ Target..the question is definitely not complete...from whr does 4/5 comes onlt TMH can clarify further
in case you have been able to get further explanation/clarification...please share
Not sure if this has been asked before...
If x and y are integers, is xy>2?
1) x+y = 3
2) x-y = 1
Not sure if this has been asked before...
If x and y are integers, is xy>2?
1) x+y = 3
2) x-y = 1
I think answer is E.
Since it is given that x and y are integers, it could be positive or negative. so we got to consider both cases.
Statement 1-
x+y=3
1) x and y both are positive
x=1, y=2, x+y=2+1=3
in this case yes xy>2
2) x postive and y negative
x=7, y= -4, x+y=7+(-4)=3
in this case xyand same for x negative and y positive
so Statement 1 is insufficient.
Statement2-
x-y=1
1) x and y both are positive
x=6, y=5, x-y= 6-5=1
in this case, yes xy>2
2) x negative and y negative
x= -6, y= -7, x-y= -6-(-7)=1
in this case, yes xy>2
3) x=1 and y=0
then x-y=1-0=1
and xy=0 in this case, no xy
So Statement 2 is Insufficient.
Now consider Statement 1 and 2 together-
x+y=3
x-y=1
now we'l get specific values of x and y. on solving the above two equations we get x=2 and y=1
So xy=2*1=2 that is equal to 2.
we are asked is xy>2?
and our ques is answered xy is not greater to 2.
Hence answer is E.
Regards,
Neha
Hey please go through the solution you have posted again...
Statement 1-
x+y=3
1) x and y both are positive
x=1, y=2, x+y=2+1=3
in this case yes xy>2. - 2*1 = 2 so xy=2. Hence no i feel.
I feel the answer should be C. Both are sufficient. Please let me know if there is any mistake in my approach...
I think answer is B.
Since it is given that x and y are integers, it could be positive or negative. so we got to consider both cases.
Statement 1-
x+y=3
1) x and y both are positive
x=1, y=2, x+y=2+1=3
in this case yes xy>2
2) x postive and y negative
x=7, y= -4, x+y=7+(-4)=3
in this case xy
and same for x negative and y positive
so Statement 1 is insufficient.
Hey please go through the solution you have posted again...
Statement 1-
x+y=3
1) x and y both are positive
x=1, y=2, x+y=2+1=3
in this case yes xy>2. - 2*1 = 2 so xy=2. Hence no i feel.
I feel the answer should be C. Both are sufficient. Please let me know if there is any mistake in my approach...
Arnav, two things:
1) ple read the bold statements in my post above.
I said x+y=1+2=3, in this case yes xy>2
but if take one of x and y as a negative integer and positive integer then xythat's why Statement 1 is insufficient.
In the ques, it is said x an y are integers, where is it given that x and y can only be positive integers? we can't assume that x and y are positive so we have to consider the negative integers also.
2) this is not only for this ques.
If both statements are individually/alone sufficient to answer the ques then answer is D.
and if both statements are together sufficient to answer the ques then answer is C.
I think answer is B.
Since it is given that x and y are integers, it could be positive or negative. so we got to consider both cases.
Statement 1-
x+y=3
1) x and y both are positive
x=1, y=2, x+y=2+1=3
in this case yes xy>2
2) x postive and y negative
x=7, y= -4, x+y=7+(-4)=3
in this case xyand same for x negative and y positive
so Statement 1 is insufficient.
Statement2-
x-y=1
1) x and y both are positive
x=6, y=5, x-y= 6-5=1
in this case, yes xy>2
2) x negative and y negative
x= -6, y= -7, x-y= -6-(-7)=1
in this case, yes xy>2
So Statement 2 is Sufficient.
and i think this condition -- x-y=1 is not valid for x negative and y positive or vice-versa.
Hence Answer B.
Do tell the OA and if there is any mistake in my solution.
Regards,
Neha
----->u must consider cases for 0 .-->x - y = 1 ...Could be x = 1 and y = 0 so xy = 0
rohan_vus Says----->u must consider cases for 0 .-->x - y = 1 ...Could be x = 1 and y = 0 so xy = 0
Thanks Rohan.
when i did the ques, i knew even this statement is insufficient but couldn't get hold of the value for which it will be invalid. forgot to consider this value. was only checking positive and negative values.
So yes Statement 2 is insufficient too.
Now consider Statement 1 and 2 toegether-
x+y=3
x-y=1
now we'l get specific values of x and y. x=2 and y=1
So xy=2*1=2 that is equal to 2.
we are asked is xy>2?
and our ques is answered xy is not greater to 2.
Hence answer is E.
will edit my solution above also.
thanks :)
Regards,
Neha
Hey, I think answer shud be A (Statement 1 alone is sufficient).
Option1: x+y=3
case1: x>0, y >0: then xy is ALWAYS = 2 (which isn't > 2)
case2: anyone is = 0; then xy = 0 (again not > 2)
case 3: anyone 2)
Hence we can say that given x & y are integers and x+y = 3 then xy is not > 2.
Thus 1 is sufficient.
Option2: x-y=1
x=1, y=0 then xyx=3, y=2 then xy>2
Thus 2 is insufficient.
I cudn't find any value for x & y which satisfies x+y=3 and have xy >2.
Hey, I think answer shud be A (Statement 1 alone is sufficient).
Option1: x+y=3
case1: x>0, y >0: then xy is ALWAYS = 2 (which isn't > 2)
case2: anyone is = 0; then xy = 0 (again not > 2)
case 3: anyone 2)
Hence we can say that given x & y are integers and x+y = 3 then xy is not > 2.
Thus 1 is sufficient.
Option2: x-y=1
x=1, y=0 then xyx=3, y=2 then xy>2
Thus 2 is insufficient.
I cudn't find any value for x & y which satisfies x+y=3 and have xy >2.
Yes, that's the right answer.
Shouldn't the answer be A.
Since we'l get specific values of x and y. x=2 and y=1
So xy=2*1=2 that is equal to 2.
and we are asked is xy>2?
and our ques is answered xy
let us try with one positive and one negative value also:
x = -5
y = 8
x+y = 3
and xy=-40 which again implies xy
Hence solution should be A.
Thanks Rohan.
when i did the ques, i knew even this statement is insufficient but couldn't get hold of the value for which it will be invalid. forgot to consider this value. was only checking positive and negative values.
So yes Statement 2 is insufficient too.
Now consider Statement 1 and 2 toegether-
x+y=3
x-y=1
now we'l get specific values of x and y. x=2 and y=1
So xy=2*1=2 that is equal to 2.
we are asked is xy>2?
and our ques is answered xy is not greater to 2.
Hence answer is E.
will edit my solution above also.
thanks :)
Regards,
Neha
Set S consists of 5 consecutive integers and Set T consists of 7 consecutive integers. Is the median of the numbers in set S equal to the median of the numbers in set T?
1) The median of the numbers in set S is 0
2) Sum of numbers in set S is equal to sum of numbers in set T.
Can some one tell me what should be the answer?
My take :
Statement 2 alone is sufficient but not 1.
Set S consists of 5 consecutive integers and Set T consists of 7 consecutive integers. Is the median of the numbers in set S equal to the median of the numbers in set T?
1) The median of the numbers in set S is 0
2) Sum of numbers in set S is equal to sum of numbers in set T.
Can some one tell me what should be the answer?
My take :
Statement 2 alone is sufficient but not 1.
well i'm not sure what the answer should be. but i also don't understand why 2 is sufficient. i'm not able to get any set of 5 consecutive and set of 7 consecutive integers whose sum is same.
Shouldn't the answer be A.
Since we'l get specific values of x and y. x=2 and y=1
So xy=2*1=2 that is equal to 2.
and we are asked is xy>2?
and our ques is answered xy
let us try with one positive and one negative value also:
x = -5
y = 8
x+y = 3
and xy=-40 which again implies xy
Hence solution should be A.
My take is A too... same as ur explanation..
And statement 2 is insufficient as you can get eighter way with diff values...
Hence i feel A alone is sufficient..