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

Agree with A. Let the old wages be 'j' and 'm'; the new wages would be 1.06*j and 1.06*m. We need to find the value of (1.06*j - 1.06*m), which can be re-prashed as 1.06(j - m) St1] Before the wage increases, Jack’s hourly wage was 5 USD per hour more than Mark’s This means (j-m) = 5 hence our answer is 1.06*5 Suff. St2] Before the wage increases, the ration of Jack’s hourly wage to Mark’s hourly wage was 4 to 3. This is only a ratio and hence the real values could have innumerable possibilities.

For any value of 'x' complying to the inequalities x>2 or x 0, 'a' could be 0 while the inequality still remains true. If 'a' could be 0 our answer to the question 'is a positive' would be no, and as rightly presented by you above the answer to the same question is yes. This according to me makes st1. insuff.

We need to maximize on the number of $20 chips lost, as greater the number of $20 chips lost out of 16 lost chips, greater is going to be the amount cashed in. Assume the x = $20 chips lost hence $100 chips lost = x-2 x+x-2 = 16 x = 9 9*20 + 7*100 = 880 and 3000 - 880 = 2120.

I choose C, though the contstruction of st1 is quite confusing. We need to know whether 4 or fewer distinct keys of the TW function. St1] Though we know that the monkey presses distinct keys for each 4 letter sequence, we do not know whether any of those keys function. St2] Though we know that the monkey is able to create 24 4-letter words, we do not know whether these words are distinct. Together we know that 4! = 24 and hence to construct 24 4-letter words by typing distinct key, atleast 4 keys need to be functioning. What's the OA?

Agree with E. Is a>0? 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. Insuff. St2] ax^2 + 1 > 0 ax^2 > -1 If x=0 then 'a' could be any number- positive, negative or 0. If x^2 is non-zero i.e. positive then 'a' could be a negative number, making a*x^2 a negative fraction greater than -1 or 'a' could be any positive number. Insuff. Together, If x>2 or x or if x=0 or x=2, then 'a' could be a positive value and the answer to our main question is yes.

Agree with clock. Good one. The new member figure for the 12th week = 5*x^11 If x is a multiple of 5, then the power of 5 will be 12 and the power of the remainder of x will be 11. Otherwise we must see and option identical to 5*x^11. Only D fits in.

'A' by all means.

You can't assume this (bolded part above) as it is not mentioned that 'y' is an integer. e.g. (4th root of 5)^4 * 10^3 = 5000

With st1] we know that 'p' could be a factor of either n! or (1 + ((n+1) * (n+2)) With st2] we know that 'p' could be either a factor of (n+1) or (n+2) (if either one is prime) or a prime no. such that 1 Combining the inferences based on both the statements: ( 1+ ((n+1) * (n+2)), since it is 1 added to a multiple of both (n+1) and (n+2) can never be a multiple of either and hence p must be a factor of n! itself. Let me know if you have any more doubts.

Thanks for taking the time; appreciate the effort. I'll certainly call them to enquire about the gurantee that they offer. And yes, If i decide to join, will keep you posted via PM. Thanks again.