gmat prep
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Let the smallest be x, then the series is x,x+1,.....,x+9,x+10.
i. The average of first 9 is 7, so sum of them is 9*7 = 63. First 9 from the series mean x,x+1,...,x+8. So we have a 9 number AP with the smallest number as X and the difference between each as 1.
Sum of n series AP (S) = (2a+(n-1)*d)*n/2, where a = smallest number in series and d is difference.
Equating S to 63, we can solve for x. Hence (i.) is sufficient and same for (ii). Hence Either of them.
i. The average of first 9 is 7, so sum of them is 9*7 = 63. First 9 from the series mean x,x+1,...,x+8. So we have a 9 number AP with the smallest number as X and the difference between each as 1.
Sum of n series AP (S) = (2a+(n-1)*d)*n/2, where a = smallest number in series and d is difference.
Equating S to 63, we can solve for x. Hence (i.) is sufficient and same for (ii). Hence Either of them.
- cubicle_bound_misfit
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IMO D
in a series when odd number of consecutive terms are arranged from least to geratest
AM = median
so stmt 1 :
Average of first 9 int is 7.
hence series is 3 4 5 6 7 8 9 10 11 ---> complete series 12 13
hence Average 8 ---Suff
Stmt 2
average of last 9 int is 9. same way the series can be determined ---SUFF
ans is D.
let me know if this is correct.
regards,
in a series when odd number of consecutive terms are arranged from least to geratest
AM = median
so stmt 1 :
Average of first 9 int is 7.
hence series is 3 4 5 6 7 8 9 10 11 ---> complete series 12 13
hence Average 8 ---Suff
Stmt 2
average of last 9 int is 9. same way the series can be determined ---SUFF
ans is D.
let me know if this is correct.
regards,
Cubicle Bound Misfit