If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

Question

If  r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}  and  s = 1 + \frac{1}{3}r , then  s  exceeds  r  by

(A)  \frac{1}{3} 
(B)  \frac{1}{6} 
(C)  \frac{1}{9} 
(D)  \frac{1}{27} 
(E)  \frac{1}{81}

MahmoudFawzy · Accepted Answer

we can rewrite R and S as the following:
 R = 3^0 + 3^{-1} + 3^{-2} + 3^{-3}

S = 3^0 + 3^{-1}R 
 S = 3^0 + 3^{-1}(3^0 + 3^{-1} + 3^{-2} + 3^{-3}) = 3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}

upon subtraction,
 S-R  =  (3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})   -(3^0 - 3^{-1} - 3^{-2} - 3^{-3})   = 3^{-4}

which is E

dave13 · Answer

r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}  = \frac{40}{27}

s = 1 + (\frac{1}{3} * \frac{40}{27})  =  \frac{121}{81}

\frac{121}{81}  -  \frac{40}{27}  =  \frac{1}{81}

E

AmanAngrish · Answer

r= 1 + 1/3 + 1/9 + 1/27
Apply G.P summation formula, sum1 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum1 = 13/9
s= 1 + 1/3(r)
i.e s = 1 + 1/3 + 1/9 + 1/27 + 1/81
Apply G.P summation formula, sum2 = (a*(1-r^n))/(1-r)
where a = 1; r = 1/3; n = 3
Therefore sum2 = 40/27
Do sum1 - sum2 = 1/27

Option (D)

MahmoudFawzy · Answer

your concept is very interesting and correct,
however, I think there is a small mistake in the application.

I think n=4 for sum1 (as 4 elements are added),
and n=5 for sum 2 (as 5 elements are added).

sum1 would be  40/27 
sum2 would be  121/81 
and by subtraction, the result would be  1/81

KanishkM · Answer

So i thought of some other method, it did not work

r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}

s = 1 + \frac{1}{3}r

Just substitute r in the second expression

s = 1 + 1/3 *  1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}

So it will give us 1/81

E

anujpal · Answer

If  r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}  and  s = 1 + \frac{1}{3}r , then  s  exceeds  r  by

(A)  \frac{1}{3} 
(B)  \frac{1}{6} 
(C)  \frac{1}{9} 
(D)  \frac{1}{27} 
(E)  \frac{1}{81}

s = 1 + \frac{1}{3}r  =  1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}

Now, s-r =  1+ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}  - ( 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} ) =  \frac{1}{81}

BrentGMATPrepNow · Answer

Given:  s = 1 + \frac{1}{3}r

Replace  r  with  1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}  to get:  s = 1 + \frac{1}{3}(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27})

Expand to get:  s= 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}

Since we're told  r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} , we can see that  s  exceeds  r  by  \frac{1}{81}

Answer: E

Jason256 · Answer

To determine how much s exceeds r, we can start by calculating the values of r and s using the given formulas.

Given:
r = 1 + 1/3 + 1/9 + 1/27
s = 1 + 1/3(r)

Let's simplify r first:
r = 1 + 1/3 + 1/9 + 1/27
r = (27/27) + (9/27) + (3/27) + (1/27)
r = (27 + 9 + 3 + 1)/27
r = 40/27

Now, substitute r into the formula for s:
s = 1 + 1/3(r)
s = 1 + 1/3(40/27)
s = 1 + (40/81)
s = (81/81) + (40/81)
s = 121/81

To find how much s exceeds r, we subtract r from s:
s - r = (121/81) - (40/27)
s - r = (12127 - 4081)/(81*27)
s - r = (3267 - 3240)/(2187)
s - r = 27/2187

Therefore, s exceeds r by 27/2187.

TargetMBA007 · Answer

KarishmaB  - Wonder, if there's a more efficient, logic-based strategy that can be used here?

KarishmaB · Answer

This is how one can do it:

r = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}  ............................. (I)

s = 1 + \frac{1}{3}r = 1 + \frac{1}{3}(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27})

s = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81}  ...................(II)

Compare r and s in (I) and (II). 
s is simply 1/81 more than r.

Answer (E)

Why would I do this? Because I see that r is a GP with 1/3 as common ratio. When I multiply it again by 1/3 in s, I will get further terms of the GP. So only one term will be extra. So I shouldn't try to calculate anything here but just write down the terms as it is.

TargetMBA007 · Answer

Thanks Karishma,

Absolutely love how you are able to almost eliminate all calculations in your solutions. Always so elegant!

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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

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If r= 1 + 1/3 + 1/9 + 1/27 and s= 1 + 1/3(r), then s exceeds r by

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