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Does 3^(r + s) = 27^6 ?

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Does 3^(r + s) = 27^6 ?  [#permalink]

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Updated on: 05 Nov 2018, 12:10
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Does 3^(r+s) = $$27^6$$ ?

(1) r – s = 8
(2) 5r = 13s

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Last edited by SajjadAhmad on 05 Nov 2018, 12:10, edited 3 times in total.
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Does 3^(r + s) = 27^6 ?  [#permalink]

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14 Mar 2017, 10:58
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1
Official Explanation

This is a “Yes/No” data sufficiency question. Begin by assessing the question. Start by rewriting the equation in the question stem using common bases, to find that 3^(r + s) = $$(3^3)$$6 = 3^18. Since the bases are the same, the exponent expressions can be set as equal. So the question is really asking “Is $$r + s = 18?$$” Now evaluate the statements to see if the information is sufficient to answer the question.

To evaluate Statement (1), $$r – s = 8$$, plug in numbers for r and s. If $$r = 13$$ and $$s = 5$$, then the statement is satisfied. These numbers also produce a “Yes” answer to the question because $$13 + 5 = 18$$. Now, see if there is a way to get a “No” answer. If $$r = 9$$ and $$s = 1$$, then the statement is still satisfied but the answer to the question “Is $$r + s = 18?”$$ is now $$“No.”$$ Statement (1) is insufficient, so write down BCE as the possible answer choices.

Statement (2) is $$5r = 13s$$. Plug in again. If $$r = 13$$ and $$s = 5$$, then the statement is satisfied and the answer to the question Is $$r + s = 18?$$” is “Yes.” Now, see if there is a way to get a “No” answer. If $$r = 0$$ and $$s = 0$$, then the answer to the question is now “No.” Eliminate choice (B).

Look at both statements together and recycle any values that were already used. If $$r = 13$$ and $$s = 5$$, then both statements are satisfied and the answer to the question “Is $$r + s = 18$$?” is “Yes.” Since there is no other possible combination that satisfies both statements.

Hope it helps

pigi512 wrote:
Sorry, I mean that statements 1 & 2 are sufficient separately.
Is there any further official explanation?

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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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08 Mar 2017, 04:20
4
in order to answer the question, we need:
1# exact value of r and s
2# two equations in r and s
3# any other information to predict the value of r and s
4# the value of the expression r+s

St 1: one equation 2 variable. INSUFFICIENT
St 2: one equation 2 variable. INSUFFICIENT

St 1 & St 2: two equations, two variable. SUFFICIENT

Option C
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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14 Mar 2017, 10:16
Excuse me but I do not see how STATEMENTS 1 & 2 are not sufficient:

analyzing the given equation we can extract the formula: r+s=18
from this point we can solve with both the statements, hence 1 & 2 are sufficients.

Where is my mistake?
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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14 Mar 2017, 10:22
pigi512 wrote:
Excuse me but I do not see how STATEMENTS 1 & 2 are not sufficient:

analyzing the given equation we can extract the formula: r+s=18
from this point we can solve with both the statements, hence 1 & 2 are sufficients.

Where is my mistake?

OA is C which states that BOTH Statement 1 and 2 are required to answer the question
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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14 Mar 2017, 10:48
Sorry, I mean that statements 1 & 2 are sufficient separately.
Is there any further official explanation?
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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12 May 2017, 07:13

The question is 3^(r + s) = 27^6.

We know that that's the same as 3^r * 3^s = 3^18

So you get a formula from the question: r+s=18
In statement 1 you get the formula: r-s=8
Now we have 2 different formula's, with just the first statement. Since we have 2 formula's, we can exactly identify r and s.

So the answer should be A, in my opinion!
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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12 May 2017, 07:21
1
MaximD wrote:

The question is 3^(r + s) = 27^6.

We know that that's the same as 3^r * 3^s = 3^18

So you get a formula from the question: r+s=18
In statement 1 you get the formula: r-s=8
Now we have 2 different formula's, with just the first statement. Since we have 2 formula's, we can exactly identify r and s.

So the answer should be A, in my opinion!

No that's not correct.

Does 3^(r + s) = 27^6 ?

Does $$3^{(r + s)} = 3^{18}$$?

Does r + s = 18.

(1) r – s = 8. Both r + s = 18 and r – s = 8 can be true simultaneously for r = 13 and s = 5 (answer YES) but if say r = 10 and s =2, then r + s does not equal to 18 (answer NO). Not sufficient.

(2) 5r = 13s. Again, both r + s = 18 and 5r = 13s can be true simultaneously for r = 13 and s = 5 (answer YES) but if say r = 0 and s = 0, then r + s does not equal to 18 (answer NO). Not sufficient.

(1)+(2) r – s = 8 and 5r = 13s gives r = 13 and s = 5 --> r + s = 18. Sufficient.

P.S. You can check OA (Official Answer) under the spoiler in the original post.
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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12 May 2017, 07:24
1
MaximD wrote:

The question is 3^(r + s) = 27^6.

We know that that's the same as 3^r * 3^s = 3^18

So you get a formula from the question: r+s=18
In statement 1 you get the formula: r-s=8
Now we have 2 different formula's, with just the first statement. Since we have 2 formula's, we can exactly identify r and s.

So the answer should be A, in my opinion!

Hi,
You do not know that r+s=18... Rather you have to answer IS r+s=18?

So you have to find value of R and s OR r+s to check the equation.
Statement I just gives you one equation...
You don't have second equation.
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Does 3^(r + s) = 27^6 ?  [#permalink]

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Updated on: 07 Nov 2019, 07:50
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Does 3^(r + s) = 27^6 ?

(1) r – s = 8
(2) 5r = 13s

Target question: Does 3^(r + s) = 27^6 ?
This is a good candidate for rephrasing the target question.
To get the same base on both sides, rewrite 27 as 3^3.
We get: 3^(r + s) = (3^3)^6
Apply power of power rule to get: 3^(r + s) = 3^18
Now that the bases are the same, we can conclude that: r + s = 18
So, we can REPHRASE the target question....

REPHRASED target question: Does r + s = 18?

Statement 1: r – s = 8
Does this provide enough information to answer the REPHRASED target question? No.
There are several values of r and s that satisfy statement 1. Here are two:
Case a: r = 8 and s = 0, in which case r + s = 8 + 0 = 8
Case b: r = 13 and s = 5, in which case r + s = 13 + 5 = 18
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 5r = 13s
There are several values of r and s that satisfy statement 2. Here are two:
Case a: r = 0 and s = 0, in which case r + s = 0 + 0 = 0
Case b: r = 13 and s = 5, in which case r + s = 13 + 5 = 18
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that r – s = 8
Statement 2 tells us that 5r = 13s
So, we have a system of two different linear equations with 2 different variables.
Since we COULD solve this system for r and s, we could determine whether or not r + s = 18, which means we COULD answer the REPHRASED target question with certainty.
So,the combined statements are SUFFICIENT

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Last edited by GMATPrepNow on 07 Nov 2019, 07:50, edited 1 time in total.
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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20 May 2017, 08:14
Bunuel wrote:
MaximD wrote:

The question is 3^(r + s) = 27^6.

We know that that's the same as 3^r * 3^s = 3^18

So you get a formula from the question: r+s=18
In statement 1 you get the formula: r-s=8
Now we have 2 different formula's, with just the first statement. Since we have 2 formula's, we can exactly identify r and s.

So the answer should be A, in my opinion!

No that's not correct.

Does 3^(r + s) = 27^6 ?

Does $$3^{(r + s)} = 3^{18}$$?

Does r + s = 18.

(1) r – s = 8. Both r + s = 18 and r – s = 8 can be true simultaneously for r = 13 and s = 5 (answer YES) but if say r = 10 and s =2, then r + s does not equal to 18 (answer NO). Not sufficient.

(2) 5r = 13s. Again, both r + s = 18 and 5r = 13s can be true simultaneously for r = 13 and s = 5 (answer YES) but if say r = 0 and s = 0, then r + s does not equal to 18 (answer NO). Not sufficient.

(1)+(2) r – s = 8 and 5r = 13s gives r = 13 and s = 5 --> r + s = 18. Sufficient.

P.S. You can check OA (Official Answer) under the spoiler in the original post.

Bunuel I think on a deeper level, the mistake that person made when analyzing statement 1 is subconsciously assuming that the solution to r - s = 8 must necessarily equate to r + s =18 - sometimes when you're not careful you assume the truth of the premise you're attempting to falsify or prove correct-
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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20 Mar 2019, 03:31
I don’t understand why can’t we take each statement seperately and tackle them. Cutting down the question, it asks does r+s= 18?

S1: r-s=8, so r=s+8 (equation 1)
Now substitute this equation 1 in the question does r+s=18? We get the answer as yes.

S2: 5r=13s, so r=13/5s (equation 2)
Now same way when substituted in question stem, we get an answer as yes.

Yet the answer is C and not D

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Does 3^(r + s) = 27^6 ?  [#permalink]

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20 Mar 2019, 03:36
Shef08 wrote:
I don’t understand why can’t we take each statement seperately and tackle them. Cutting down the question, it asks does r+s= 18?

S1: r-s=8, so r=s+8 (equation 1)
Now substitute this equation 1 in the question does r+s=18? We get the answer as yes.

S2: 5r=13s, so r=13/5s (equation 2)
Now same way when substituted in question stem, we get an answer as yes.

Yet the answer is C and not D

Posted from my mobile device

The problem with your approach is that we don't know from the stem that r + s = 18. The question asks DOES r + s = 18? So, you cannot use r + s = 18 with r - s = 8 to solve (again because we are not given that r+s=18).
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Does 3^(r + s) = 27^6 ?  [#permalink]

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20 Mar 2019, 03:42
Bunuel wrote:
Shef08 wrote:
I don’t understand why can’t we take each statement seperately and tackle them. Cutting down the question, it asks does r+s= 18?

S1: r-s=8, so r=s+8 (equation 1)
Now substitute this equation 1 in the question does r+s=18? We get the answer as yes.

S2: 5r=13s, so r=13/5s (equation 2)
Now same way when substituted in question stem, we get an answer as yes.

Yet the answer is C and not D

Posted from my mobile device

The problem with your approach is that we don't know from the stem that r + s = 18. The question asks DOES r + s = 18? So, you cannot use r + s = 18 with r - s = 8 to solve (again because we are not given that r+s=18).

Alright, so here is my shortcoming! Thanks Bunuel for making it simple
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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20 Mar 2019, 04:16
pigi512 wrote:
Sorry, I mean that statements 1 & 2 are sufficient separately.
Is there any further official explanation?

The questions states Does 3^(r+s) = 27^6 ?
That means we have to answer in Yes, 3^(r+s) = 27^6 or No, 3^(r+s) = 27^6. We don't have to find the exact value.

Statement 1 says r-s = 8
R and S can be any value with the absolute difference of 8 . example (8,0) (9,1) and so on. So there is no unique value. Hence Statement I alone is insufficient

Statement 2 says 5r = 13s
Again, we can not find any unique value. example R =13 and S = 5 OR R = 26 and S = 10 and so on.

So if you see, some values can satisfy the equations but some can not . So we can't mark option D as the values may or may not satisfy the equation.

But if we combine
We get R=13 and s = 5
Now this is a unique value. This is what we need to answer if 3^(r+s) = 27^6. The answer can be Yes/No.

I hope my explanation helps you.
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Re: Does 3^(r + s) = 27^6 ?  [#permalink]

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20 Mar 2019, 06:47
Does 3^(r+s) = $$27^6$$ ?

(1) r – s = 8
(2) 5r = 13s

3^(r+s) = $$27^6$$
r+s=18

#1
r=8+s
#2
r=13/5 * s
from 1&2
we get
s=5 and r= 13
so IMO C
Re: Does 3^(r + s) = 27^6 ?   [#permalink] 20 Mar 2019, 06:47
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