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GRE Math Challenge #29-five consecutive integers is t
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Updated on: 09 May 2018, 09:52
Updated on: 09 May 2018, 09:52
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If the sum of five consecutive integers is t, then, in terms of t, what is the greatest of these integers?
A. \(\frac{(t -20)}{5}\)
B. \(\frac{(t - 10)}{5}\)
C. \((\frac{t}{5})\)
D. \(\frac{(t + 10)}{5}\)
E. \(\frac{(t + 20)}{5}\)
A. \(\frac{(t -20)}{5}\)
B. \(\frac{(t - 10)}{5}\)
C. \((\frac{t}{5})\)
D. \(\frac{(t + 10)}{5}\)
E. \(\frac{(t + 20)}{5}\)
Re: GRE Math Challenge #29
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10 Sep 2014, 02:23
10 Sep 2014, 02:23
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
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
Re: GRE Math Challenge #29-five consecutive integers is t
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29 Mar 2018, 19:14
29 Mar 2018, 19:14
1
sandy wrote:
If the sum of five consecutive integers is t, then, in terms of t, what is the greatest of these integers?
A. (t -20)/5
B. (t - 10)/5
C. (t/5)
D. (t + 10)/5
E. (t + 20)/5
A. (t -20)/5
B. (t - 10)/5
C. (t/5)
D. (t + 10)/5
E. (t + 20)/5
the ans should be D not E, please check it out.
the mean of the 5 consecutive integer is t/5= median and the greatest is (t/5)+ 2 Which is equal to D (t + 10)/5.
