Ans (C)
Statement 1 alone is not sufficient, take 2,11 as examples, one gives a +ve integer and other doesn't.
Stat 2 alone is also not sufficient as 2Q could be (-)ve.
But both the statements taken together can give the ans, since prime nos are only +ve and 2Q is divisible by 11 therefore 2*2Q/11 will be an even +ve integer.
Is 4Q/11 a positive integer?
1. Q is a prime number
2. 2Qis divisible by 11
Ans (C)
Statement 1 alone is not sufficient, take 2,11 as examples, one gives a +ve integer and other doesn't.
Stat 2 alone is also not sufficient as 2Q could be (-)ve.
But both the statements taken together can give the ans, since prime nos are only +ve and 2Q is divisible by 11 therefore 2*2Q/11 will be an even +ve integer.
Is 4Q/11 a positive integer?
1. Q is a prime number
2. 2Qis divisible by 11
1 alone as Prime numbers are nothing but positive . . .But with 2 alone we cannot get the answer. . .
Is 4Q/11 a positive integer?
1. Q is a prime number
2. 2Qis divisible by 11
Solution: C
Opt I : Q is a prime number so we can take prime numbers ranging from 2- 11. with Q = 2,
4*Q = 8 and 4Q/11 = 8/11 which is not an integer.
now lets say Q = 11 then 4Q/11 = 4 , an integer. Hence this isn't suffcient.
OPTII: lets say Q = 11/2 then 2Q/11 = 1 an integer. Thus we cannot conclude that Q is an integer using 2 alone.
OPT I nd II: since Q is a prime number and 2Q is divisible by 11, it implies that the only possible value of Q is 11. Hence sufficient to say tht 4Q/11 is an integer.
My Doubt in your reasoning for the 1st statement:
Say ABC=60deg, then the triangle becomes equilateral, and a=1,
then this solution becomes incorrect......
Anything wrong in this reasoning?? I am unsure.....
Second stmt, same reasoning, does not give a definite answer.
That is why i had chosen E.
OA Please......
A for me.
Area of triangle = 1/2 * 2/a * a^2 = a. Hence, we need to find out whether a > 1/
From stmt1: AB^2 + BC^2 > AC^2 or 2*AB^2 > AC^2
or 2*(a^4 + 1/a^2) > 4/a^2
or, (a^6 + 1) > 2 or, a^6 > 1 that means a > 1. Hence, sufficient.
(a could be -2 also but since since a is a measurement of length of sides, a cannot be -ve. therefore a has some +ve value.)
Simplifying stmt2 will give a > 0...insufficient.
Perimeter = AB + BC + AC = 2AB + AC = 2*sqrt((1/a^2)+ a^4) + 2/a
= 2*(sqrt(1+a^6))/a + 2/a > 4/a
or, sqrt(1+a^6) + 1 > 2
or, sqrt(1+a^6) > 1 or a > 0.
Hope it clears the air.
Ok My take..
area of the triangle = a
Option A
---------
Here lets assume ABC is 90
then our condition boils down to .. a = 1
(a^6 = 1 .. hence only possible solution a=1)
now it goes on to show that : a^6 > 1 only when a>1and a^6
Option B
---------
Here the equation boils down to .. a^4 > 0 .. which actually tells nothing about the case in question.
Hence we can only get the answer from A alone ..
1. An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?
(1) The tens digit of n is a factor of the units digit of n.
(2) The tens digit of n is 2.
1. An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?
(1) The tens digit of n is a factor of the units digit of n.
(2) The tens digit of n is 2.
IMO A
given n is int, n>20,
1) Sufficient
let the no is xy
y= ax..
xy= 10x +y ==> 10x + ax ==> x(10 +a), clearly the no gonna be a multiple of some integer..
2) Not sufficient clearly for ex 23
Hence A
Try this...
1. Is z even? (1) 5z is even. (2) 3z is even.
My choice is D, each option is individually sufficient
Try this...
1. Is z even? (1) 5z is even. (2) 3z is even.
Ans. C.
Explanation: 5*z is even ==> z ={2,4,6....} or {2/5,4/5...}
hence 1. is insufficient.
likewise 3z alone being even is insufficient as the sample set for z includes
z= {2,4,6....} or {2/3,4/3...}
Taking both together,
z={2,4,6...} Hence z is even. Sufficient.
For any integers x and y, min(x,y) and max(x,y) denote the minimum and maximum of x and y, respectively. For example, min(5,2)=2 and max(5,2)=5. For the integer w, what is the value of min(10,w)?
(1) w=max(20,z) for some integer z.
(2) w=max(10,w)
For any integers x and y, min(x,y) and max(x,y) denote the minimum and maximum of x and y, respectively. For example, min(5,2)=2 and max(5,2)=5. For the integer w, what is the value of min(10,w)?
(1) w=max(20,z) for some integer z.
(2) w=max(10,w)
IMHO....
(2) alone gives us the answer, as it shows that w is greater than 10.
(1) alone can also give us the answer, as, if z=11, w=20. If z=32, w=32. In any case, w will be greater than 10!(and so min(10,w)=10).
Either statement alone is sufficient.
So, (D).
Ans. C.
Explanation: 5*z is even ==> z ={2,4,6....} or {2/5,4/5...}
hence 1. is insufficient.
likewise 3z alone being even is insufficient as the sample set for z includes
z= {2,4,6....} or {2/3,4/3...}
Taking both together,
z={2,4,6...} Hence z is even. Sufficient.
C is correct...
hey just kidding but must say, i think u still remember that concept which u missed
when u answered my last ds question...rem?? u assumed Z as an integer unnecessarily in that
question....
This time u made no error..C is correct
For any integers x and y, min(x,y) and max(x,y) denote the minimum and maximum of x and y, respectively. For example, min(5,2)=2 and max(5,2)=5. For the integer w, what is the value of min(10,w)?
(1) w=max(20,z) for some integer z.
(2) w=max(10,w)
Ans should be op D...same explanation as given by the crab365
For any integers x and y, min(x,y) and max(x,y) denote the minimum and maximum of x and y, respectively. For example, min(5,2)=2 and max(5,2)=5. For the integer w, what is the value of min(10,w)?
(1) w=max(20,z) for some integer z.
(2) w=max(10,w)
I am a little confused by this problem. Consider the second option.
w=max(10,w)
now, conventional logic states that LHS is the output of RHS. In this case RHS is a function which works on it's inputs to produce the result. let's evaluate the above statement in this light.
11 = max(10,11) is perfectly valid because you as well as the function can evaluate 11>10.
however we know the same function will not have a solution in case of
9 = max(10,9) which is an impossible solution.
However since w is a variable, this is a valid equation too. Pls can someone shed some light into this seeming dichotomy
...
I am a little confused by this problem. Consider the second option.
w=max(10,w)
now, conventional logic states that LHS is the output of RHS. In this case RHS is a function which works on it's inputs to produce the result. let's evaluate the above statement in this light.
11 = max(10,11) is perfectly valid because you as well as the function can evaluate 11>10.
however we know the same function will not have a solution in case of
9 = max(10,9) which is an impossible solution.
However since w is a variable, this is a valid equation too. Pls can someone shed some light into this seeming dichotomy...
Buddy, "w" is a variable and in fact "w=max(10,w)" has multiple solutions, the solution set contain int values between (11,infinity)...so c w is a variable and it is taking variable values...
Moreover,
as in DS the given options has to be correct, means u consider only those cases in which the value is valid or possible
Hi.. friends...do you have soft copy of OG12 or OG11 ....or any link..please share
please send me a PM i will send u the soft copy of OG 10 and 11 😃
In the rectangular co-ordinate system, are the points (r,s) and (u,v) equidistant from the origin?
(1) r+s=1
(2)u=1-r and v=1-s