Fiver

I'd still say that the answer is B. If both variables are +ve, is 3x > 7y? St1] x > y + 4 this means 3x > 3y + 12; now if we prove that 3y + 12 >= 7y, we get our answer in yes. So, is 3y + 12 >= 7y ? This possible only when y Do a quick litmus test with nos keeping in mind the above inequality. Assume y=3; therefore as per st1] the smallest integer value for x must be 8. Is 3x > 7y ? or is 3*8 > 7*3. The answer is Yes. Now assume y=4; therefore as per st1] the smallest integer value for x must be 9. Is 3*9 > 7*4 ? The answer is No. St2 I assume is clear to all. Vijay- What's the source & the official explanation for D as the answer ?

Is |a-c| st1] The sum of any two numbers among a, b and c is greater than the third number. This is one property of any given triangle and for this principle to work each of these 3 nos. must be greater than the difference between the other 2 nos. Hence |a-c| must be greater than b in the same way as |b-c| St2] a, b and c are non-zero numbers. Insuff. This is already clear from st1 and this statement by itself can yield several possibilities. Choose A. What's the OA?

Is x + y st1] x St2] y Clearly neither is suff. Together- the minimum impossible value of x + y = 73/72, which is greater than 1. Hence x + y could be between 1 & 73/72 or less than 1. I choose E.

Thanks atishree. ATB. You mean the quant book or the MGMAT CATs?

Given that the order is x,y & z, if z-y = y-x or if z+x = 2y then it means that x, y & z are in AP with y as the median. In other words if we prove that they are in AP the answer to our question is yes otherwise the answer to our question is no. st1] The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. this means x + y + z e.g. 2,3,4 the answer to the question is yes 1,4,5 the answer to the question is no. Insuff. st2] The median of the set {x, y, z, 4} is less than the median of the set {x, y, z } since the given order is x, y & z, the only 2 median possibilities for the set {x,y,z,4} is (x+y)/2 or (y+4)/2, and since given that both these values are less than y; in both the cases it turns out that y>4. Insuff as the series can be in AP or otherwise. Together, if the series is in AP then we know that y has to be less than 4; however with st2 we know that y is >4; hence the series is not in AP & the answer to our question is no. Choose C. Nice question. Atishree, could please share the source ?

Here you have proved that (n-2) is a multiple of 15, which translates to the fact that n cannot be a multiple of 15. Our question is to prove whether n*t is multiple of 15 and if no what is the remainder. The correct choice is C.

Agree with sj2025. C is also incorrect because it suggests that the 'type II disorder' itself does the job of exhibiting 'highs' while the person sufferring from this disorder actually does the job of exhibiting.

I Choose C. Is a> 0 ? St1] a^3-a a (a^2 - 1) For this to hold good the following 2 conditions have to be met 1] a > 0 while a^2 -1 if a^2 - 1 a 0; possible and our answer to the main question is yes. 2] a 0 Is a^2 -1 > 0 then a1 but we have assumed that a Insuff. St2] 1-a^2 > 0 1> a^2 this means that a is a positive or negative proper fraction. Insuff. Together we know 'a' has to be a positive fraction and our answer to the question is yes. Suff.

I Choose C. Given that n = 3x + 2 t = 5y + 3 Derive nt = 15xy + 9x + 10y + 6 What is the remainder when nt is divided by 15 ? Here if we know that x is a multiple of 5 & y is a multiple of 3 we know that the remainder when nt is divided by 15 is 6. st1] n - 2 is a multiple of 5 since n = 3x + 2 & (n - 2) is a multiple of 5 Substract 2 from both sides and we know that 3x is a multiple of 5; hence x is a multiple of 5. But we do not know whether y is a multiple of 3. Insuff. St2] t is a multiple of 3 5y + 3 is a multiple of 3; hence 5y has to be a multiple of 3; hence y is a multiple of 3. But no info about x. Insuff. Together we know that the remainder is 6.

I would choose D. Given that both a & n > 1 & 8! = n(a^n) Find a = ? 1] a^n = 64 Given 8! = n(a^n); 8! = n * 64 Here we can find the value of n, which is a positive integer This positive integer = a^(same integer) Here since given that a>1; there can be only one value for a, which is between 1 & 2, more towards 1. Sufficient. 2] n = 64 Given 8! = 64 * (a^64) Here again a^64 = some positive integer value Hence in the same manner as above value of 'a' can be calulated. What's the OA?

Anytime my friend. Sharing is the nth step towards building :)

The question only says that for a given value 't', 't+2' is also present in the set; it does not confirm vice-versa. In other words -1 could be the first value of the set. The answer is D; infact i have seen a similar question in one of the Gprep tests.

If x and y are positive is 3x>7y 1)x>y+4 Multiply by 3 on both sides of the inequality: 3x > 3y + 12 Now we know that 3x is greater than this value: 3y + 12. So if we also know that this value: 3y + 12 itself is greater than 7y, then it is logical to say that 3x also has to be greater than 7y. So test: Is 3y + 12>7y ? Is 12> 4y ? Is 3>y ? Now this means for all values of 'y' that satisfy the above inequality the answer is yes. For y=3 both sides are equal. And for y>3 the answer is no. 2)-5x This obviously is suff as we end up concluding that 2.5x>7y and, both values being +ve, 3x is > 7y.

You are right. I missed the key words 'for all x' This means, as you rightly suggested, a free range for x rather than for A. Quoting this bit of my post made earlier: st1] x^2 - 2x + a >0 x * (x-2) > -a Here, if either x=0 or x=2 then the LHS is 0 and 'a' is positive. If (x-2)0 then 0 If both x & (x-2) are either positive or negative then LHS becomes positive and 'a' could be positive, negative or 0. Now for A to match all the different possibilities of x listed above, it needs to be positive. Well done 'sevenzerozero' @ ramiy, good question, actually a breed apart. What's the source?

I agree with Testcracker that the question is seeking the value of the difference between the digits a & b and not between the numbers ab & ba. Here's another approach: As suggested by testcracker the aggregate effect of the decrease in the avg is 1.8*10=18. Assume the original number to be 10b+a and hence the replaced number will be 10a+b, the difference of which is 9(b-a), which is equal to 18. Hence (b-a) = 2