1. r is what percentage of s ? Option 1: r/s = 3/4 => r is 75% of s. Sufficient. Option 2: r + s = 75/100. Can't be sure. Not sufficient. Answer should be A.
2. Is r/s = integer ? Option 1: Not sufficient. Every not all the factors of s is a factor of r. Option 2: Every prime factor of s is also a factor of r. Sufficient as denominator will become 1 on solving. Answer should be B.
3. z^n = 1. What is the value of z ? Option 1: N is non zero integer. z can be 1 or -1. Non sufficient. Option 2: z>0. Many combinations of z and n - (2,0) (3.0). Not sufficient. Option 1 and 2: Now z can only be 1. Sufficient. Answer should be C.
4. If x != y, is (x-y)/(x+y) > 1 Option 1: x>0. Can't know as no info about y is given. Option 2: yOption 1 and 2. Again can't be sure as not sure which is greater between x and y. Answer should be E.
What are the OAs ?
OAs are D , A , C , E.
For 4th question can you please elaborate. Using both the options we know x>0 and yy ...Ohh I understood .... whether x > y or not is not known hence it can not be deduced if (x-y)/(x+y) > 1.
For 3rd question I missed -1 as one possible value of z. Thanks :)
I am totally clueless about 1st. For 2nd I thought D as answer.
For 4th question can you please elaborate. Using both the options we know x>0 and yy ...Ohh I understood .... whether x > y or not is not known hence it can not be deduced if (x-y)/(x+y) > 1.
For 3rd question I missed -1 as one possible value of z. Thanks :)
I am totally clueless about 1st. For 2nd I thought D as answer.
Oh, 2nd one striked me now I think - Option 1 says every factor of s is a factor r i.e. suppose s = 2x2x3x7x7x11, then r will be having all these factors. So, r/s will be integer.
But as per option 2, every prime factor of s is a factor of r Ex - s = 2x2x3x7x7x11, then r = 2x3x7x11. Observe that in this case every prime factor of s is also a factor of r but r/s is not an integer.
Thats why A will be the answer. Its easier to explain the solution when correct answer is known. :p
You are correct about 4th. We can get +ve or -ve answer in different cases satisfying both option 1 and 2.
Guys I need help with the solution of following questions from OG 11.
1. If r and s are positive numbers, r is what percentage of s? a. r/s = b. r+s = 75/100
2. If r and s are positive integers, is r/s an integer? a. Every factor of s is also a factor of r. b. Every prime factor of s is also a prime factor of r.
3. If z^n =1, what is the value of z? a. N is a nonzero integer. b. Z>0
4. If x != y, is (x-y)/(x+y) > 1 a. X>0 b. Y
My Picks:
1. 'A' r/s = 3/4 => we can get the %'s -> sufficient r+s = 75/100, i.e r+s = 3/4... given that r and s are (+) ve numbers.. there are infinte solutions -> no sufficent to get the %s.
2. 'A' stmt1: every factor .. so r/s must be an integer => sufficient. stmt 2:every prime factor: there is a possibility of repetition.. and hence can't be sure if r/s yields an integer...=> not sufficient.
3. 'C' stmt 1: if n=1,2,3,4,,,, z=1; if n=2,4,6,8,... z = 1 or -1...=> not sufficient. stmt 2: Z>0...when n=0, Z can be any value for the equality.. => not sufficienet.
combining=> Z=1 ...
4. 'E' x != y, is (x-y)/(x+y) > 1 - or x >y stmt 1: x>0... when x=1, x=y, when x=3, x>y.. not sufficient stmt 2: yy... not sufficent..
combining too.. we cannot tell anything abt inequality..
For 4th question can you please elaborate. Using both the options we know x>0 and yy ...Ohh I understood .... whether x > y or not is not known hence it can not be deduced if (x-y)/(x+y) > 1.
For 3rd question I missed -1 as one possible value of z. Thanks :)
I am totally clueless about 1st. For 2nd I thought D as answer.
Hi- jus looked at the OA's... Can't figure out why it is 'C' for the first question.. can you pls recheck the question?
Originally Posted by amitrverma View Post Guys I need help with the solution of following questions from OG 11.
1. If r and s are positive numbers, r is what percentage of s? a. r/s = b. r+s = 75/100
Hello amitrverma, Here is my firstestest post as I attempt to answer Q1! :)
1. Ans.
Option 1. r/s = --> Clearly r is 75% of s (No doubt!) --> Sufficient Option 2. r+s = 75/100 --> Not sufficient. By no means can we say anything about r/s just by this data, here (r,s) values can be infi, example: (0.30,0.45) or (0.25,0.50) etc etc. which all add up to 0.75. Hence,r/s ratios are infinite as well. Hence Not suff. And OA is WRONG!!
Oh, 2nd one striked me now I think - Option 1 says every factor of s is a factor r i.e. suppose s = 2x2x3x7x7x11, then r will be having all these factors. So, r/s will be integer.
But as per option 2, every prime factor of s is a factor of r Ex - s = 2x2x3x7x7x11, then r = 2x3x7x11. Observe that in this case every prime factor of s is also a factor of r but r/s is not an integer.
Thanks Dopa and others. Even I dont understand how with r+s = 75/100 one can find the value of r/s but OG11 OA indeed is D. Now is there any chance of OA being wrong? Or are all of us missing something here?
Hello amitrverma, Here is my firstestest post as I attempt to answer Q1! :)
1. Ans.
Option 1. r/s = --> Clearly r is 75% of s (No doubt!) --> Sufficient Option 2. r+s = 75/100 --> Not sufficient. By no means can we say anything about r/s just by this data, here (r,s) values can be infi, example: (0.30,0.45) or (0.25,0.50) etc etc. which all add up to 0.75. Hence,r/s ratios are infinite as well. Hence Not suff. And OA is WRONG!!
3. If n is a positive integer, is n^3-n divisible by 4? a. N=2k+1, where k is an integer b. N^2 + n is divisible by 6
Hi again,
Answer should be: 1. C 2. C 3. By n^3-n, did you mean (n^3)-n OR n^(3-n) !!? Also I suppose, n & N over here are both the same, correct me if I'm wrong! :idea: Assuming it to be (n^3)-n, answer should be A
Answer should be: 1. C 2. C 3. By n^3-n, did you mean (n^3)-n OR n^(3-n) !!? Also I suppose, n & N over here are both the same, correct me if I'm wrong! :idea: Assuming it to be (n^3)-n, answer should be A
Hi,
1. C
2. can we solve this way...
1/p>p/(p^2+1)
(p^2+1)> p^2
which is always true...
so the answer will be A.....
wat say ???
3. A is right answer .. considering is 0 divisible by everynumber .....
3. A is right answer .. considering is 0 divisible by everynumber .....
Well, what if p, in which case you can't carry p from 1/p to the RHS as you've done without changing the inequality sign. I guess you're getting my point. What are the OAs ?
Answer should be: 1. C 2. C 3. By n^3-n, did you mean (n^3)-n OR n^(3-n) !!? Also I suppose, n & N over here are both the same, correct me if I'm wrong! :idea: Assuming it to be (n^3)-n, answer should be A
I kindda understood.' The mistake I had been making was cross multiplying in case of inequalities. Which is wrong unless you know the sign, as in case of negative numbers equality sign would reverse if we cross multiply. This explains 1st and 2nd question.
Regarding third , n^2+n = n(n+1) .. divisible by 6 --> so 2 and 3 are factors of n(n+1) but this dont help to deduce that it would be divisible by 4 also :).
Thanks Dopa and others. Even I dont understand how with r+s = 75/100 one can find the value of r/s but OG11 OA indeed is D. Now is there any chance of OA being wrong? Or are all of us missing something here?
Well....it was a typo !!! Not in OG, but in your post Amit !
Option B was r divided by s (sign: - with a dot on top and bottom). You misread/mistyped it as + !
Well....it was a typo !!! Not in OG, but in your post Amit !
Option B was r divided by s (sign: - with a dot on top and bottom). You misread/mistyped it as + !
OG sab ka baap hai biddu
Oh boy!! I have soft copy of OG11 and perhaps the scanner could not put those dots properly and I ended up in reading that as + instead division sign. And I have been breaking my head on this problem.
Oh boy!! I have soft copy of OG11 and perhaps the scanner could not put those dots properly and I ended up in reading that as + instead division sign. And I have been breaking my head on this problem.
That explains it all. Thanks KingCat :)
Hey amit, I've been searching for OG to facts-inference-judgment, If you have the soft copy of the same, I would be thankful to you if you could mail it to me.
