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rajatmeh

2nd Level
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Everything posted by rajatmeh

  1. What are you doing? A is insufficient, so now move forward and test (B). How come just by using (A) you say (A).
  2. © Assume a is the price of one copy and b is the number of copies planned to be sold without applying new strategy. ab=100 --(i) (a-5)(b+10)=100--(ii) solving (i) and (ii) gives 2a=b+10--(iii) solving (ii) and (iii) a^2-5a-50=0 (a-10)(a+5)=0 ==>a=10 or a=-5 total copies sold later = b+10 = 2a = 2*10=20 as copies sold cannot be negative. Hence ©.
  3. (B). Using (A) ====== x-y = 22k ==> x= y + 22k ==> x+y = 2y+22k = 2(y+11k) 2(y+11k) is multiple of y only when y=11 else not. Using (B) ====== x=10a+a=11a always multiple of 11 y=11b+b=11b always multiple of 11 ==> x+y=11(a+b) always multiple of 11. Hence (B).
  4. Yes looks like ©. but can we be sure that just by plugging in n = 3,6,9 no other value of n will produce a prime number.
  5. Well I don't understand that if the room has same length and width = 10 and the cylinder is kept in the center then how come the farthest horizontal distance be 10. Even if you take the distance from corner to center of room it will be 5sqrt(2). Also since height of cylinder is not provided we cannot calculate the volume of cylinder. IMO (E). Was not very confident solving this. Guys help.
  6. agree that xa can be any thing but for that anything if present then anything + 5 and anything-5 is also present. Question mentions that if 5 is that anything then are all multiple of 5 present.
  7. IMO ©. individually each statement is not sufficient. Combining both statements ==> 2 c=3 ==> b>3 since x = (b!)* c ==> x has at least TWO 3, one from b! and one from c. Hence sufficient.
  8. No that's not the case. Imagine X = 5 and Y=4 X 5 Y ==> 5>4. If it had been 2X
  9. Using (1) ====== 3X+2Y ==> X Using (2) ====== 4X + 3Y ==> X Answer (B).
  10. IMO (D). Cracked within 1 min. 450=2*3^2*5^2 ==> 450x to be a cube x has to contain 2^2*3*5 = 60.
  11. IMO ©. Interest from X is r1*i1 Interest from Y is r2*i2 Assume r1*i1 > r2*i2 ==> (r1/r2)*(i1/i2) > 1 --(i) since r1 > r2 ==> r1/r2>1 Also since i1/i2>r1/r2 ==> i1/i2 > 1 ==> (i) is true ==> X is more profitable ==> © is sufficient.
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