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2 Quant problems

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2 Quant problems

by vaibhav.iit2002 » Sun Sep 13, 2009 8:42 am
Can anyone suggest the best way to solve theses problem.

1. Nth term, T(n) = (-1)power(k) X (2)power(-k)
Which one is correct regarding sum of 1 to 10 terms:
a. T>2
b. 1<T<2
c. 1/2<T<1
d. 1/4 < T < 1/2
e. T < 1/4

2. 450y=(n)power(3) , where y and n are integers. Which of the following are integers
i. y/(3 X 2power2 X 5)
ii. y/(3power2 X 2 X 5)
iii. y/(3 X 2 X 5power2)

a. none
b. i only
c. ii only
d. iii only
e. i,ii,iii


Answers coming soon. :D
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by prindaroy » Tue Sep 15, 2009 6:45 pm
1) The best way to solve this would not be through using the geometric progression formula. Too tedious. Best way to solve it is through intuition. First number is -1/2, next is 1/4, and then -1/8 and so on and so forth, so essentially when you add the first and second term you get -1/4, so the answer has to be in negative realm. So the summation is essentially less than 1/4 for when t(n) = (-1^k)*(2^-k) = (-1/2)^k

2) There is a slight problem with the question; it has to state MUST or COULD be integers...because if you factor out 450 = 2*3^2*5^2, so y should be of any such form; 2^2*3*5, so it can also be 2^5*3^4*5^4 and so on so forth. Now 2^5*3^4*5^4 is divisible by all three of them provided y is greater than 2*3^2*5^2 which is the smallest such form. All you need to do is create 450y such that it can be cube rooted. So there are infinitely many such forms all of which are divisible by the three of them if the question was; "which of the following could be integers?". If the question was "which of the following must be integers" then clearly the answer is 1 only.
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