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I think we should worry about wherher or not the set S contains all multiples of 5. Other numbers do not play a significant role. The stem tells us that 5 is, indeed, in S. (IMO) C.

Statement 1: Here b will have 3 as one of its multiples. For x to be a multiple of 9 we need another 3. If c is 3 or any multiple of 3, then x will be a multiple of 9. Otherwise x will not necessarily be a multiple of 9. - INSUFFICIENT Statement 2: If c is less than 4, c may or may not be a multiple of 9. For instance, if c=-9, then x will be a multiple of 9, otherwise not necessarily. - INSUFFICIENT Combining both statements together, we can find out that x is a multiple of 9. Hence, C.

OK, clear now. Thanks!

Sure, the answer is 288. What I calculated earlier applies only to numbers that end in 5. Thanks to everyone for all your comments. Special thanks to Md Minuddin!

All integers (positive, negative and zero), which are divisible by two are even. The rest of the integers is odd.

I think statement 2 is sufficient. If we take the lowest possible limit w=1, then it will give us: w=1, W=w+5=1+5=6. The required percentage is: (1/6)*100=16.666... This is more than 10 percent. If we take a higher value of w, the percentage will be higher than 16.666. So, we can say that statement 2 is sufficient. Hence, B. Please post your comments if you think differently.

Here's my approach: We have numbers from 101 to 999. Their difference is: 999-101=898 numbers not including the last number 999. Out of 898 numbers, half are odd and half are even. So, we have: 449 odd numbers plus one last number 999, or 450 odd numbers. Cycle of occurence of 5 within the odd numbers is 5. That means in 450 we have (450/5)=90 numbers that end in 5. The required total number of three digit odd numbers that do not contain five should be: 450-90=360.

If you are choosing D, then you should have a sign that will satisfy the requirement stated in Statement 1. What is your sign?

If we employ the minus sign, then we will get: k-l-m=2k-l-m k=2k, which will be true for k=0. That means, k@(l+m) = (k@l) + (k@m) is NOT true for all numbers when we use the minus sign.

The answer should be B. We have: y=mx+b. Now assuming that b=0 and m=2, we have y=2x. If we rotate this line 90 degrees, the equation of this new line will be y=(-1/2)*x. Here, the slope is (-1/m). Hence, B.

mkx2000, Statement 1 says that k@1 does not equal to1@k for some numbers. It doesn't say that k@1 equals to 1@k for ALL numbers. What is your number k, so that for some numbers, k times 1 won't be equal to 1 times k? In my opinion, there is no such real number. There is a possibility of k being equal to 1 for cases of subtraction and division, so that for these cases only k-1=1-k=0 and (k/1)=(1/k)=1 But can we settle at one sign?

Statement 2: is not sufficient. Consider X=86 and Y=57. Their difference is 29, which is not divisible by 9.

Let's denote intensity as I. We are required to find the value of: [i(n+1)]/[i(n)]. If we assume that I(n)=2, say, then we have: I(3)=2 10*I(3)=2*10=20=I(4) 10*I(4)=20*10=200=I(5) 10*I(5)=200*10=2000=I(6) 10*I(6)=2000*10=20000=I(7) 10*I(7)=20000*10=200000=I(8) The required ratio is: [i(n+1)]/[i(n)]=[(2*(10^5))/2]=10^5. Hence, C.

Statement 1: We know that the sign @ is not + nor *. - INSUFFICIENT Statement 2: We know that the sign @ represents munius. - SUFFICIENT. Hence, (IMO) B.

My guess is 6 colors. We have 6 unique colors and (5+4+3+2+1) unique combinations. Hence, (IMO) B.