Hi Rich,
thanks for reverting. It says fact 1 - 9th term is 81 so

term shd be some constant on term 9 is 81 where as in your example you state the second term to be 81 itself 81. I do get the logic of number increase or decrease but its difficult when you show in fact 1 and two. can you please show in examples ....

Fact 1: the 9th term = 81

If the terms are 9, 81, 729, etc., then ALL 9 terms are greater than 1.
If the terms are 1, 81, 81(81), etc. then 8 of the germs are greater than 1.
Fact 1 is INSUFFICIENT

Fact 2: the 5th term = 1

Since the sequence either increases or decreases, we'd have...
4 numbers less than 1, 1, 4 numbers greater than 1
or
4 numbers greater than 1, 1, 4 numbers less than 1

Regardless of which option, we end up with exactly 4 terms.
Fact 2 is SUFFICIENT

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

The next number in a certain sequence is defined by multiplying the previous term by some positive constant k, where k ≠ 1. How many of the first nine terms in this sequence are greater than 1?

(1) The ninth term in this sequence is 81.

(2) The fifth term in this sequence is 1.

There are 2 variables in the original condition (the initial value and variable k). Therefore in order to match the number of equations and variables, we need 2 equations. Since there is 1 each in 1) and 2) C has high probability of being the answer. Using both 1) & 2) together, 9-th=k^4=81 gives us k=3 and therefore 5-th=3, 6-th=9, 7-th=27, 8-th=81. Since 4 is larger than 1, the conditions are sufficient and therefore the answer is C. However in tip 4, we can't have a trivial case as our answer and therefore we should try using 1) and 2) separately.

In case of 1), we don't know the value of k. therefore the condition is not sufficient.
In case of 2), when k>1, 1/k^4, 1/k^3, 1/k^2, 1/k,1,k,k^2,k^3,k^4 and therefore the 4 latter terms are greater than 1. When 0<k<1, the former 4 terms are greater than 1, therefore the answer is unique. Thus the answer is B.


Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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