If n is an integer and 1/(n+1)

Question

If n is an integer and  \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n} , then what is the value of n?

A) 9
B) 10
C) 11
D) 12
E) 13

Bunuel · Accepted Answer

Given:  \frac{1}{n+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{1}{n} .

Now, obviously  3*(\frac{1}{33})<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<3*(\frac{1}{31}) , as {3 times the least #} < {given sum} < {3 times the largest #}:

\frac{3}{33}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31} ;

\frac{1}{11}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{3}{30} ;

\frac{1}{10+1}<\frac{1}{31}+\frac{1}{32}+\frac{1}{33}<\frac{3}{31}<\frac{1}{10} ;

n=10 .

Answer: B.

gmacforjyoab · Answer

Did it on similar grounds as Bunuel 1/(n+1) < ( 1/31 + 1/32 + 1/33) < 1/n Substitute ( 1/31 + 1/32 + 1/33) to be 1/a 1/(n+1) < ( 1/a) < 1/n ......... ...................... hence n+1 > a > n --------------------------------- eq 1 1/a > 3/33 ( i.e 1/11) ... Hence a<11 from eq 1 --- n+1 >a>11 ................ n10 from eq 1 --- n+1>a>10 .... hence n+1>10 ... n> 9 Ans n=10

vy3rgc · Answer

Approximate 1/31 + 1/32 + 1/33 = 0.09+

Now POE. B fits the inequality.

0.09 (1/11) < 0.09+ < 0.1 (1/10)

atalpanditgmat · Answer

hi gmacforjyoab,

I guess i am lacking some mathematics in the highlighted part. Could you please throw some light. It would be great help.

Regards
Atal Pandit

gmacforjyoab · Answer

Since (1/n+1) < 1/a < 1/n , we can say that n+1 >a > n ( when u take the reciprocal of two numbers in an Inequality , the inequality flips ) Consider this ---- 1/4<1/3<1/2 , which would mean 4>3>2 ... Oh and lets say - all the numbers were 1/33 , then the sum would be 3/33 , but all the numbers are not 1/33 , the other two numbers are 1/32 and 1/31 . and these two numbers are greater than 1/33 , hence the sum of 1/31 +1/32 + 1/33 would also be grater than 3/33 hence , 1/a > 3/33 i.e 1/11 HTH Jyothi

tritringuyen · Answer

1/n> 1/31+1/32+1/33> 1/33+ 1/33 + 1/33 = 3/33 = 1/11 ====> n<11 1/(n+1)< 1/31 + 1/32+ 1/33 < 1/31 + 1/31 + 1/31 = 3/31 ====> n>9,3 Then, n=10. B.

Asifpirlo · Answer

...............

Amazing solution..... glad to learn this.....

mvictor · Answer

solved it the other way...and probably the fastest way...
suppose we have 1/33 + 1/33 + 1/33
we have 3/33 or 1/11
since we have 1/31 and 1/32, logically, the result would be slightly more than 1/11.
10 works just fine...
we have 1/n+1 => 1/11, and we have 1/10

10 works just fine!

Sana87 · Answer

Is the formula always 3 times the least number and 3 times the largest number? Or did we take 3 as there are 3 fractions in the middle? So if there were 4 it would have been 4 times? Or it's always 3 times regardless the number of fractions in the middle?

Bunuel · Answer

The factor (3 in this case) corresponds to the number of fractions being summed. Since there are 3 fractions in this problem, we use 3 times the smallest and 3 times the largest value as bounds.

If there were 4 fractions in the sum, we would use 4 times the smallest and 4 times the largest number instead.

bumpbot · Answer

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If n is an integer and 1/(n+1)<1/31+1/32+1/33<1/n, then what

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