Search found 9 matches
Remember as a general thumb rule: If you know the parity (negative or positive) of the product of any number of variables, then the same will be the parity of any combination of those variables multiplied in numerator or denominator. Reason is pretty simple the parity of 'a' is always the same as th...
- by vineet.nitd
Tue Oct 25, 2016 7:41 am- Forum: Problem Solving
- Topic: Positive / Negatives
- Replies: 6
- Views: 3577
You may multiply each side of the inequality by |x| without flipping the sign and then rearrange.
x(|x|-1)>0
x>0 AND |x|>1 => x>1
OR
x<0 AND |x|<1 => -1<x<0
From the two range of values above, we know for sure that x>-1.
Hence choice B is correct.
- by vineet.nitd
Tue Oct 25, 2016 2:16 am- Forum: Problem Solving
- Topic: Absolute values
- Replies: 7
- Views: 4029
Remainder when 32^(32^32) OR (28+4)^(32^32) is divided by 7 is the same as remainder when 4^(32^32) is divided by 7. This is because all the terms in the expansion will be multiples of 7 except the last one, which is 4^(32^32). Also 32^32 when divided by 3 gives remainder (-1)^32 OR 1=>32^32 = 3k+1 ...
- by vineet.nitd
Tue Oct 25, 2016 1:59 am- Forum: Problem Solving
- Topic: Tough Reminder Question
- Replies: 3
- Views: 2457
The problem with your approach is that you are recounting stuff. When you have 4 distinct items to be sent to 4 distinct places, each item has 4 choices to go. (Box 1 has 4 choices) AND (Box 2 has 4 choices) AND (Box 3 has 4 choices) AND (Box 4 has 4 choices) = 4*4*4*4 Now truck 1 could have any num...
- by vineet.nitd
Tue Oct 25, 2016 1:37 am- Forum: Problem Solving
- Topic: Grouping and selecting
- Replies: 9
- Views: 4772
What is the implication of ab + c being odd? It implies that one of the two components (ab OR c) is even, while the other is odd. If that is true their product is also going to be even, since one component is even for sure. Hence, 3 must be true. Statements 1 and 2 have symmetry as “a� AND “b...
- by vineet.nitd
Tue Oct 25, 2016 1:14 am- Forum: Problem Solving
- Topic: Even Odd Problem
- Replies: 9
- Views: 4750
- by vineet.nitd
Wed Jul 15, 2015 9:34 pm- Forum: Problem Solving
- Topic: ratio of the surface area of a cube
- Replies: 4
- Views: 1855
In such problems, one should try to satisfy the condition first. The condition here is that at least one marble of each type should be in the case. So, we give one marble of each type to the case. Now we are left with suit that can accommodate two more marbles out of the remaining marbles: 4Cs, 4Ss,...
- by vineet.nitd
Wed Jul 15, 2015 9:25 pm- Forum: Problem Solving
- Topic: P & C
- Replies: 9
- Views: 3823
Solve the quadratic in ab to get the values of ab as -3 and 1. However, as per problem ab>0, hence ab=-3 can be discarded. So, ab = 1. Now, since there is no restriction on a or b belonging only to set of integers, we can assign any value to 'a' (except 0), such that we always have its multiplicativ...
- by vineet.nitd
Wed Jul 15, 2015 8:19 pm- Forum: Problem Solving
- Topic: If ab>0 and (a^2)*(b^2) + 2ab – 3 = 0
- Replies: 11
- Views: 3848
In such problems where distinct objects have to be assigned or distributed across distinct places, think from the perspective of 'what' is being distributed and 'where' is it being distributed. 'What' is people and 'where' is office here! The first person has two choices as there are two offices. So...
- by vineet.nitd
Wed Jul 15, 2015 5:14 am- Forum: Problem Solving
- Topic: Assign employees
- Replies: 16
- Views: 12713