so answer is either of the statement alone is sufficient
When it says six times more often, i think it should be interpreted as "when he throws an odd number "x" times, he throws an even number "6x" times"..what do u say naga?
For eg; if he throws an odd number 2 times..he throws an even number 12 times and so on.. There is a difference between "six times more" and "six more"
Just to note that both answers do not have to be same. Even if the final answer is different, both would still independently be able to solve the question!
from 1)
x + x + 6 = 30 ..
so odd game is 12 and even game is 18 ..
from 2)
4(x+6) + 2x = 96
6x = 72 ..
odd game is 12 and even game is 18
both gives same conclusion ...
so answer is either of the statement alone is sufficient
When it says six times more often, i think it should be interpreted as "when he throws an odd number "x" times, he throws an even number "6x" times"..what do u say naga?
For eg; if he throws an odd number 2 times..he throws an even number 12 times and so on.. There is a difference between "six times more" and "six more"
i would stick on to ma interpretation .. as u say the word 'times' might confuse .. IMO, here times meant Trials 😃
A certain game requires a player to roll a standard six-sided die. If a player rolls even numbers six times more often than he rolls odd numbers, how many times did he roll even numbers?
(1) The player rolled the die a total of 30 times.
(2) The player received 4 points each time he threw an even number, and 2 points each time he threw an odd number. His final score was 96.
Please tell the correct approach for above problem???
Bad diction. The intended meaning is not clear. 6 times definitely means ODD x; EVEN 6x. Maybe the author is trying to say "rolled 6 more even numbers than odd numbers." What is the source?
Bad diction. The intended meaning is not clear. 6 times definitely means ODD x; EVEN 6x. Maybe the author is trying to say "rolled 6 more even numbers than odd numbers." What is the source?
This problem is from Grockit.com. I agree that it's poorly worded.
From (1). p divides n! + n!(n+1)(n+2) => n!(1+ (n+1)(n+2)) Plug-in numbers Case 1: n=4 4!(1+5*6) = 4*3*2*1(31) - Since P divides this and is a prime number P could be 2,3 or even 31. But 31 does not divide 4!. We don't Know Case 2: n=3 3!(1+4*5) = 3*2(3*7) - Since P divides this and is a prime number P could be 2,3 or even 7. But 7 does not divide 4!. So We don't Know. Insufficient.
Case (2): (n + 2)! / n ! = (n+1)(n+2). Plug-in numbers: n=1, so 2*3=6. P divides 6 so it can be 2 or 3. 2 divides 2! but 3 does not. So we don't know. Insufficient.
Combining (1) & (2): Case 1: n = 3 P divides : 3!(1+ (4)(5)) = 3*2*3*7 and also 4*5 = 2*2*5. The only value that P can take is 2 So P divides n!=3! Case 1: n = 4 P divides : 4!(1+ (5)(6)) = 4*3*2*31 and also 5*6 = 2*3*5. The only values that P can take are 2,3 so P divides n!=4! in both cases. Hence 1 & 2 together are sufficient. Choice C
From 1: X can be -1.9,...-1,0,....1.5,....1.9 (Insufficient) From 2: Case1 : X = -X => -x=3x-2 => x=0.5 ....(This does not satisfy |0.5 = 3*.5-2 = -0.5...RHS has to be +ve because LHS is +ve) Case2 : X = X => x=3x-2 => x=1 , (sufficient) Answer choice B.
From 1: X can be -1.9,...-1,0,....1.5,....1.9 (Insufficient) From 2: Case1 : X = -X => -x=3x-2 => x=0.5 Case2 : X = X => x=3x-2 => x=1 , (Insufficient) (1) & (2) : 0.5 & 1 (Insufficient) Answer choice E.
Hi Gail, For statement 2, x= 0.5 will not make the eqn correct. x = 3x-2
When x =0.5, LHS = 1/2 RHS = 3/2 - 2 = -0.5 LHS is not equal to RHS. So x = 0.5 will not solve the eqn.
Hi Gail, For statement 2, x= 0.5 will not make the eqn correct. x = 3x-2
When x =0.5, LHS = 1/2 RHS = 3/2 - 2 = -0.5 LHS is not equal to RHS. So x = 0.5 will not solve the eqn.
-Deepak.
he has wriiten -x = 3x-2 (so LHS is -0.5)
since x = 3x-2
either x = 2-3x (which gives x=0.5)
or
x=3x-2 (which gives us x=1)
my take - E
option B
if x=0.5 , 3x-2 would be -ve. So x =1 is da only soln
Yes option B is the correct choice. I have corrected my original post. I want to rather note here that whenever we find a solution for a modulus question we should always substitute that in the original equation and check if its valid. That's the only way how test makers make a Moduli question tricky. This will pretty much happen in all(90-95%) the 600-800 level modulus questions.
