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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.

Let's denote as A(p, q) and B(p-2, q+c). Given y=2x+5 the slope is 2. Now if we let x=p, y=q, x1=p-2 and y1=q+c, then the formula of a slope of a straight line is given as: (y1-y)/(x1-x) or substituting the above values into this formula you should get: [(q+c-q)/(p-2-p)]=2 or c=-4. Hence, A.

Alpha, Yes, I do think that both statements 1 and 2 refer back to Henry.

Alpha, 1) The regular price of the most expensive item was $50 and the regular price of the next most expensive item was $20. I assume that three items are ranked as: most expensive, next most expensive and least expensive. But, even if Y=Z=20, then 1.5 + 0.1Z > 0.15Z will hold true.

Let's denote as X, Y and Z the regular prices of the most to the least expensive items. The question stem reveals the following information about the discounts that Henry received: 0.2X+0.1Y+0.1Z. Now, we need to find out if: 0.2X+0.1Y+0.1Z > 0.15(X+Y+Z) *** Statement 1: We know that X=$50 and Y=$20. Substituting these values into *** we get: 12+0.1Z> 10.5 + 0.15Z or 1.5 + 0.1Z > 0.15Z. This inequality will always hold true for all positive (assuming price is not zero) values of Z which are less than $20. - SUFFICIENT Statement 2: We know that Z=$15 - INSUFFICIENT Hence, (IMO) A.

Yes, I agree with you guys. I stopped at the point where XY=6 or -6. It's important to substitute the values back into equation and see if other equations hold true with these values. The answer should be D.

I see now that D is a way to go.

(X^3 * Y^3 – X^2 * Y^2)=[XY(XY[XY-1])] *** Statement 1: XY*XY=36, (X^3 * Y) =24, Solving we get: XY=6 or -6. But, XY=-6 will not satisfy (X^3 * Y) = 24, as X^2*XY=X^2*(-6) should give us 24 and it is not true for XY=-6. Therefore, XY=6. Substituting XY=6 into *** we can solve the given equation. -SUFFICIENT Statement 2: We know that: X^3 * Y^2 = 72 and that (X^3 * Y) = 24. Now we have:[ X^3 * Y^2]=(X^3 * Y)*Y or 24Y=72. Y=3. The actual values of X and Y are: X=2 and Y=3. - SUFFICIENT Hence, D.

Md Minuddin, Thanks for noting it. Of course, it should be: T= y/x+(80-y)/1.25x. Why X=4 and Y=20? Well, many other combinations would also do. I looked for numbers that will be divisible by 1.25. For instance, (80-y)=60 in my example, and it is divisible by 1.25.

Statement 1: Given that (X^2)>9 we have to possibilities: 1) X>3 2) X Since, the question stem says that X is negative, then we should only accept the inequality X

Apoorva.srivastva, Statement 2 is not sufficient because if X=-4, say, then we have: (-4)^3