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Re: GRE Math Challenge #29 [#permalink]
1
13divyasha wrote:
assuming the consecutive nos. as n,n+1,n+2,n+3,n+4
we have => n+n+1+n+2+n+3+n+4 = t
=> 5n+10 = t => n = (t-10)/5
from the series assumed, the greatest no. is n+4 = {(t-10)/5}+4 = (t-10+20)/5 = (t+10)/5
the answer is D



I too got the answer as D.
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Re: GRE Math Challenge #29-five consecutive integers is t [#permalink]
Another option for solving for those of us who don't like algebra:

Pick easy numbers for the consecutive series, (I chose 1, 2, 3, 4, 5). Sum of all 5 will equal n*(n+1)/2 where n=5, so sum = 15, therefore t = 15.

Now check the answer choices to see which one gives you 5 as an answer. This is similar to asking: "Which of the following CAN be equal to 5 where t = 15?"

A) <0, cross it out
B) =1, cross it out
C) =3, cross it out
D) =5, this is the answer

if you want to check E for good measure...
E) =7, cross it out
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Re: GRE Math Challenge #29-five consecutive integers is t [#permalink]
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