IMO B
ac/bd = a/d * c/b > c/b
therefore a/d > 1. since a,b,c,d are postive integers therefore the expression must be greater than zero. in statment 2 a > d therefore a/d must be greater than 1 Hence statement II is sufficient.
Hence B
IMO D
I got the answer. there was a misprint in the question. read the equation as
Root {(2*root 63) + 2/(8 + (3*root 7))}.. the mistake is the positive sign. the question has mistakently been written as negative. the correct question would be with the positive sign
here is how i solved it ....eventually
realize that 3* root 7 = root 63
now let root 63 be A then the equation would read
root {2*{A+1/(8+A)} ....taking 2 common
root {2*{[A*(8+A)+1]/(8+A)}
root {2*{[8A+A^2 +1]/(8+A)
now put A = 3*root 7
then 8A+A^2-1 = 8*3*root 7 + 9*7 +1
= 8*3*root 7 + 63 +1 = 8*3*root7 +64
take 8 common 8(8+ 3* root 7)
put 3*root 7 = A
then = 8(8+A)
put the expression back in the main equation we get
root {2*{8*[8+A]/[8+A]}}
= root 8*2 = root 16 = 4 hence D is the answer
Hi all
just saw this thread. would appreciate it if we could also add Sets 1 to 20 in the same thread also. I have recently started doing sets and have reached set 7.
Also i only do Quants section in the sets because in the verbal section there are too many mistakes and the questions are also wrong most of the time. would appreciate it if those who are going through verbal part of the sets could elaborate if they are worth doing.
thanks.
Stuck on this one. would appreciate it if someone can assist
Qs. Root {(2*root 63)-2/(8 + (3*root 7))}
A. 8 + (3* root 7)
B. 4 + (3* root 7)
C. 8
D. 4
E. root 7
Thanks
IMO C
Statemetn I root x > y. now two possibilities could exist that x is less than y but both are fractions therefore its root is more than y for example let x = 1/4 and y = 1/3. now root x = 1/2 which is more than y.
another possibility is that x is larger than y and is so large that even its root is greater than y for example let x = 16 and y = 2 therefore root 16 = 4 therefore it is greater than y.
Statement II clarifies that point. It says that x is greater than y. This rules out the first possibility becasuse the premise of the possbility is that x is less than y and both are fractions. Hence that leaves us with possibility that x > y therefore x^3 must be greater than y. BTW it has already been pointed out that both x and y are positive integers.
This is a Thread Started to discuss the Sets. I am sure that this would help us all so here is the first installment. :tup:
BTW answers are given at the end of the set but would appreciate it if someone can solve it along with explanations. Will try it myself and will try to post the solutions.