In the triangle whose sides are a,b and hypoteneous is c, is x>90 (x is the angle between a and b)?
(1) a^2 + b^2
(2) c >4
Thanks,
Anurag...
In the triangle whose sides are a,b and hypoteneous is c, is x>90 (x is the angle between a and b)?
(1) a^2 + b^2
(2) c >4
Thanks,
Anurag...
Do you mean "hypotenuse is c" ? If yes, X should be 90. Am I missing something here?
Gail.Wynand SaysDo you mean "hypotenuse is c" ? If yes, X should be 90. Am I missing something here?
Ok, I think I framed the question in a wrong way. I did not have the question in front of me while posting it here. Apologies...
a and b are the two adjacent sides and c is the longest side of the triangle. x is the angle between a and b.
So, the DS question is, Is x >90 degree.
(1) a^2 + b^2
(2) c >4
Thanks,
Anurag...
Hi
I had a simple question.Say we pick a no like 5 .Is 0 considered a multiple of 5
as I was doing the following problem and I marked the answer as B but actually it was E
If x is a prime no what is the value of x
(1) x (2) (x-2) is a multiple of 5
The problem is from the Kaplan premier book.
Also wanted to ask you is there any good book / any best practices for permutations/combinations/probablity as I am weak in this area.
Hi
Any ideas for this one .The OA is E
(1) Not sufficient as 2,3,5,7 etc can al be possible values
(2) (x-2) a multiple of 5 .The OA says that both 2 and 7 are possiblities ie 0 and 5 can be multiples of 5
hence answer is E
Hi
Any ideas for this one .The OA is E
(1) Not sufficient as 2,3,5,7 etc can al be possible values
(2) (x-2) a multiple of 5 .The OA says that both 2 and 7 are possiblities ie 0 and 5 can be multiples of 5
hence answer is E
Actually after reading your post I wanted to check before commenting.
Here goes the explanation:
Let x and y be integers. Then x is a multiple of y if there exists
another integer z such that x = y*z.
That is the definition. Now let x = zero and let y be an arbitrary
integer. Can we find an integer z such that 0 = y*z? I think
you'll see that choosing z = zero will do the trick every time. So
zero is a multiple of every integer.
source : Is Zero a Multiple of Every Number?
I don't agree with the above since 0 is not an integer. Its neither a +ve integer nor a -ve integer. A 0 is zero.
Although wikipedia also supports the same wiki/Multiple_(mathematics)
Moreover a logic just popped in my mind:
If X is a multiple of Y => X = K*Y. where Y is NON-zero.
If we say that X, a multiple is 0 then 0=K*Y => K = 0 => Y=0/0 (undefined).
This is the reason why 0 as a multiple does not make sense to me.
Actually after reading your post I wanted to check before commenting.
Here goes the explanation:
Let x and y be integers. Then x is a multiple of y if there exists
another integer z such that x = y*z.
That is the definition. Now let x = zero and let y be an arbitrary
integer. Can we find an integer z such that 0 = y*z? I think
you'll see that choosing z = zero will do the trick every time. So
zero is a multiple of every integer.
source : Is Zero a Multiple of Every Number?
I don't agree with the above since 0 is not an integer. Its neither a +ve integer nor a -ve integer. A 0 is zero.
Although wikipedia also supports the same wiki/Multiple_(mathematics)
Moreover a logic just popped in my mind:
If X is a multiple of Y => X = K*Y. where Y is NON-zero.
If we say that X, a multiple is 0 then 0=K*Y => K = 0 => Y=0/0 (undefined).
This is the reason why 0 as a multiple does not make sense to me.
Yes I totally agree with your explanation . 0 is neither +ve no -ve .BTW this problem is from the Kaplan Premier book.Thanks for the solid explanation .
Actually after reading your post I wanted to check before commenting.
Here goes the explanation:
Let x and y be integers. Then x is a multiple of y if there exists
another integer z such that x = y*z.
That is the definition. Now let x = zero and let y be an arbitrary
integer. Can we find an integer z such that 0 = y*z? I think
you'll see that choosing z = zero will do the trick every time. So
zero is a multiple of every integer.
source : Is Zero a Multiple of Every Number?
I don't agree with the above since 0 is not an integer. Its neither a +ve integer nor a -ve integer. A 0 is zero.
Although wikipedia also supports the same wiki/Multiple_(mathematics)
Moreover a logic just popped in my mind:
If X is a multiple of Y => X = K*Y. where Y is NON-zero.
If we say that X, a multiple is 0 then 0=K*Y => K = 0 => Y=0/0 (undefined).
This is the reason why 0 as a multiple does not make sense to me.
Yes I agree with you . 0 is neither positive nor negative. BTW this problem is from the Kaplan Premier book .Thanks for the solid explanation.
ajayreddy Sayssee, both the options are insufficient to solve the question when used individually...but when used together, we knw that the lower limit of population was 5,500,000,000 and the lower limit of income was 330,000...per capita income is calculated as income/population...so this comes out to be 16,666..this means that the per capita is atleast 16,666 since this is the lower limit..hence we can conclude that the per capita income is greater than 16,500....answer is C
Little late for reply
Was just going through the questions. To calculate Per capita income: we should have lower limit for income and upper limit of population.
Answer should be E.
ohh boy... i posted my approach to get it checked. Thanks for helping me to find the catch in my approach. refer to the blue bold part. I think i made a mistake there. Modulus of any value yield negative result only when the entire value has a negative sign before it, which in our case is (x-5). x-5| = 5-x only when x
x-5 = 5-x only when x-5Any value less than 5 should satisfy the equation. this is true for previous equation as well.
@deepakram, can you comment on this approach? is this ok?
@Mission , the expr |x-5 = 5-x only when x-5
-Deepak.
1. If x is an integer, does x have a factor n such that 1
(1) x > 3!
(2) 15! + 2 x 15! + 15
2. Is n an integer?
(1) n/3 is an integer.
(2) 3n is an integer.
1. Is ab > ba ? (Both a and b are nonzero integers.)
(1) a = b3
(2) b
2. Is n a whole number if n is a positive whole number?
(1) (9n) is an integer.
(2) (6n) is not an integer.
3. What is the value of s in the fraction r/s if r and s are positive integers?
(1) The least common denominator of r/s and 1/4 is 12.
(2) r = 2
2. Is n an integer?
(1) n/3 is an integer.
(2) 3n is an integer.
Statement 1 : n/3 is an integer then n has to be an integer .So statement 1 is suff.
Statement 2 : 3n is an integer..Then n need not be an integer . n = 1/3.So insuff..
I will go with option A
3. What is the value of s in the fraction r/s if r and s are positive integers?
(1) The least common denominator of r/s and 1/4 is 12.
(2) r = 2
Statement 1 states that LCD of r/s & 1/4 is 12.We can simply say s=3 .But there is value of R which can impact the value of s.So this statement is insuff.
Statement 2 is insuff since it doesn't talk anyting about value of S.
combining both we can say the value of S = 3 .
So I will go with option C
2. Is n a whole number if n is a positive whole number?
(1) (9n) is an integer.
(2) (6n) is not an integer.
Statement 1 can be modified as 3* sqrt(n) = k where k is an int.So n has to be a perfect square to make k int.So sqrt (n) will be a whole number .Suff
Statement 2 sqrt (6n) is not an int. This doesn't say anything bcos n=9 or n =1 then sqrt (6n) will not be an int.Also for n=3 this will yield an int.So this is insuff.
I will go with option A
1. If x is an integer, does x have a factor n such that 1
(1) x > 3!
(2) 15! + 2 x 15! + 15
A prime number will not qualify : 1 X is not prime
(1) Alone not sufficient since for X>3! there are prime numbers. eg. 7
(2) Alone sufficient. Can be written as: 2(15!/2 + 1) x 15(14!+1).
Clearly the LHS and RHS (of the inequality) satisfies.
Possible values of X are : 15!+2, 15!+3,15!+4,15!+5....,15!+15. Now all these have a 2,3,4,...,15 common in the 15! so will not be a prime. So the answer will always be a yes.
Answer choice B
2. Is n an integer?
(1) n/3 is an integer.
(2) 3n is an integer.
(1) Alone sufficient. since N=3K where K = Integer.
(2) Not sufficient. Since N=K/3 where K is an integer. so N can both be integer and non integer.
Answer choice A
1. Is ab > ba ? (Both a and b are nonzero integers.)
(1) a = b3
(2) b
Ok now this must be a case where a,b represent the digits of a number.
using 1 : We know that for a to be a single digit : b
Using 2: b can be 1,2. A can be 1,2,...,9. So both Yes and no can be answered.
Combining (1) & (2) : b=1,2 implies Yes and No can be answered so not sufficient. Hence correct choice E.
2. Is n a whole number if n is a positive whole number?
(1) (9n) is an integer.
(2) (6n) is not an integer.
Using (1) alone: (9n) = 3n implies : n has to be a square for 3n to be a whole number. Sufficient.
Using (2) alone: (6n) is not an integer means it can be of the form (6X5)(5) is not an integer. But (6X25) so (25) is an integer. Not sufficient.
Answer choice (A) so
3. What is the value of s in the fraction r/s if r and s are positive integers?
(1) The least common denominator of r/s and 1/4 is 12.
(2) r = 2
Using (1) alone: s can be 3,6,12. => Not sufficient.
Using (2) alone: Obviously not sufficient alone. Since we need the values of s.
Combining (1) & (2) s can be 3,6 12 again. So not sufficient. Option (E)
Guys I have a question from OG 12:
If i and j are integers, is i + j an even integer?
(1) i (2) i = j
What i reasoned was:
Statement 1: I
Statement: i=j, although this statement can be written as i+i = 2i, so the answer is yes I+J is even. But if i and j are negative integers then i+j would be a negative number and even and odd is defined only for positive integers. Since its not given that i and j are positive integers I concluded that this question cannot be answered even with both the statements.
But the explanation in the book says this can be answered with only second statement, explanation in the book does not take -ve integer case into account.
May somebody please tell me if my reasoning is wrong ?
Regards
Abhishek Arohi
Guys I have a question from OG 12:
If i and j are integers, is i + j an even integer?
(1) i (2) i = j
What i reasoned was:
Statement 1: I
Statement: i=j, although this statement can be written as i+i = 2i, so the answer is yes I+J is even. But if i and j are negative integers then i+j would be a negative number and even and odd is defined only for positive integers. Since its not given that i and j are positive integers I concluded that this question cannot be answered even with both the statements.
But the explanation in the book says this can be answered with only second statement, explanation in the book does not take -ve integer case into account.
May somebody please tell me if my reasoning is wrong ?
Regards
Abhishek Arohi
dude, integers are numbers without any fractonal/decimal part .. Includes the counting numbers {1,2,3,}, zero {0}, and the negative of the counting numbers {-1, -2, -3, }
so while taking i=j=-9 , the sum becomes -18 which is an ineteger .. since u have already got the concept why statement 2 was correct for postive nos. i believe your doubts are cleared now ..
Guys I have a question from OG 12:
If i and j are integers, is i + j an even integer?
(1) i (2) i = j
What i reasoned was:
Statement 1: I
Statement: i=j, although this statement can be written as i+i = 2i, so the answer is yes I+J is even. But if i and j are negative integers then i+j would be a negative number and even and odd is defined only for positive integers. Since its not given that i and j are positive integers I concluded that this question cannot be answered even with both the statements.
But the explanation in the book says this can be answered with only second statement, explanation in the book does not take -ve integer case into account.
May somebody please tell me if my reasoning is wrong ?
Regards
Abhishek Arohi
Dude, any integer divisible by 2 is even . for ex { -8, -4, ...., 2 , 6}
Hence, answer#2 is correct.
Dude, any integer divisible by 2 is even . for ex { -8, -4, ...., 2 , 6}
Hence, answer#2 is correct.
Thanks for the explanation, actually one of my teachers said that -ve numbers are not even or odd.....thats y i got ocnfused.