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Of the following sums, which is greatest ?

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by uptowngirl92 » Wed Jun 24, 2009 1:52 am
Of the following sums, which is greatest ?

(A) 1/root 2 + 1/root 3+ 1/root 4+ 1/root 5

(B) 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2

(C) 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5

(D)1-1/2+1/3-1/4

(E) 1/2+1/3+1/4 +1/5

OA:A

Which is the easiest and fastest way to solve this problem??
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Source: — Problem Solving |

uptowngirl92 wrote:Of the following sums, which is greatest ?

(A) 1/root 2 + 1/root 3+ 1/root 4+ 1/root 5

(B) 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2

(C) 1/2^2 + 1/2^3 + 1/2^4 + 1/2^5

(D)1-1/2+1/3-1/4

(E) 1/2+1/3+1/4 +1/5

OA:A

Which is the easiest and fastest way to solve this problem??
Rationalize the denominator in A and just compare choices and answer should follow. Comparing A and E you will see that all numerators in E is 1 but denominators are larger where as 3 denominators in A are 2 and the numerators are different from 1. For the last terms respectively root 5 is greater than one so this shows that A > than E. The rest can triavially be eliminated.
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by niksraza » Wed Jun 24, 2009 7:42 am
This is a good fractions problem.

Firstly start by knowing that the greater the denominator the smaller the fraction.

That eliminates B,C,E (left with A and D)

Secondly understand that 1/sqrt of a no. is always > 1/ a number so at the very least A represents

A>1/2+1/3+1/4+1/5 or A>E. Hence A is the answer.
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by ssuarezo » Wed Jun 24, 2009 1:22 pm
niksraza wrote:This is a good fractions problem.

Firstly start by knowing that the greater the denominator the smaller the fraction.

That eliminates B,C,E (left with A and D)

Secondly understand that 1/sqrt of a no. is always > 1/ a number so at the very least A represents

A>1/2+1/3+1/4+1/5 or A>E. Hence A is the answer.
---- Direct and clear niksraza... thank you
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