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Re: Is 4^(x+y)=8^(10) ?
[#permalink]
25 Aug 2016, 15:07
Expert Reply
Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?
(1) x - y = 9 (2) y/x = 1/4
We need to determine whether 4^(x+y) = 8^10. We start by breaking down our two bases into prime factors.
4^(x+y) = (2^2)^(x+y) = 2^(2x+2y)
8^10 = (2^3)^10 = 2^30
We can now rephrase the question as:
Is 2^(2x+2y) = 2^30 ?
Because the bases are the same, we can drop them and set the exponents equal to each other. The question becomes:
Is 2x+2y = 30 ?
Is x + y = 15 ?
After simplifying the equation, we see that we need to determine whether the sum of x and y is equal to 15.
Statement One Alone:
x – y = 9
Knowing the difference of x and y is not the same as knowing the sum of x and y; thus, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.
Statement Two Alone:
y/x = ¼
When we cross multiply obtain:
4y = x
4y = x is not enough information to determine the value of x + y. Statement two alone is not sufficient. We can eliminate answer choice B.
Statements One and Two Together:
Using statements one and two we know the following:
x – y = 9 and 4y = x
Since 4y = x, we can substitute 4y for x into the equation x – y = 9 and we have:
4y – y = 9
3y = 9
y = 3
Since y = 3, x = 4(3) = 12.
Thus, x + y = 12 + 3 = 15. We can answer yes to the question. Both statements together are sufficient.
Re: Is 4^(x+y)=8^(10) ?
[#permalink]
17 Jul 2017, 06:24
1
Is \(4^{x+y}=8^{10}\) ?
\(4^{x+y}=8^{10}\)
\(2^{2(x+y)}=2^{30}\)
\(2(x+y) = 30\)
\(x+y = 15\)
So, we will find if \(x+y = 15\)?
(1) \(x - y = 9\)
Clearly not sufficient as we can have multiple values for x and y
Hence, (1) ===== is NOT SUFFICIENT
(2) \frac{y}{x = 1/4}
\(x = 4y\)
Again this is not sufficient as we can get multiple values of x and y
Hence, (2) ===== is NOT SUFFICIENT
Combining (1) & (2)
When we substitute \(x = 4y\) in equation \(x - y = 9\)
we get,
\(4y - y = 9\)
\(3y = 9\)
\(y = 3\)
\(x = 4y\)
\(x = 4 * 3\)
\(x = 12\)
\(x + y = 12 + 3 = 15\)
Hence, (1) & (2) ===== is SUFFICIENT
Hence, Answer is C
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Is 4^(x+y)=8^(10) ?
[#permalink]
Updated on: 15 Jan 2018, 08:00
1
Top Contributor
Expert Reply
Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?
(1) x - y = 9 (2) y/x = 1/4
Target question:Is 4^(x + y) = 8^10? This is a good candidate for rephrasing the target question.
Given equation: 4^(x + y) = 8^10 Rewrite 4 and 8 as powers of 2 to get: (2²)^(x + y) = (2³)^10 Apply power of a power law to get: 2^(2x +2y) = 2^30 This means that: 2x + 2y = 30 Divide both sides by 2 to get: x + y = 15 In other words, asking whether 4^(x + y) = 8^10 is the SAME as asking whether x + y = 15 REPHRASED target question:Is x + y = 15?
Statement 1: x - y = 9 Is this enough information to answer the REPHRASED target question? No. Consider these two CONFLICTING cases: Case a: x = 12 and y = 3. In this case, x + y = 12 + 3 = 15. So, x + y DOES equal 15 Case b: x = 10 and y = 1. In this case, x + y = 10 + 1 = 11. So, x + y does NOT equal 15 Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y/x = 1/4 Is this enough information to answer the REPHRASED target question? No. Consider these two CONFLICTING cases: Case a: x = 12 and y = 3. In this case, x + y = 12 + 3 = 15. So, x + y DOES equal 15 Case b: x = 8 and y = 2. In this case, x + y = 8 + 2 = 10. So, x + y does NOT equal 15 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 x - y = 9 Statement 2 tells us that y/x = 1/4 Since we have 2 different linear equations with 2 variables, we COULD solve the system for the individual values of x and y, which means we COULD answer the REPHRASED target question with certainty. Of course, we wouldn't waste precious time performing such calculations, since our sole goal is to determine the sufficiency of the combined statements. Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
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Re: Is 4^(x+y)=8^(10) ?
[#permalink]
07 Apr 2018, 15:14
I understand that statement one alone is not sufficient, however when we plug in for statement 2 we end up with 4y=x. So wouldn’t it be as follows, 4y+y= 15 5y= 15 Y= 3 and then we can solve for x Isn’t the answer B instead?
Re: Is 4^(x+y)=8^(10) ?
[#permalink]
08 Apr 2018, 03:00
Ekjotb wrote:
I understand that statement one alone is not sufficient, however when we plug in for statement 2 we end up with 4y=x. So wouldn’t it be as follows, 4y+y= 15 5y= 15 Y= 3 and then we can solve for x Isn’t the answer B instead?
Hello
It is NOT given that 4^(x+y) = 8^10. We have to determine whether this is actually true or not.
Re: Is 4^(x+y)=8^(10) ?
[#permalink]
28 Sep 2018, 09:13
Hey Guys,
Quick question from my side. Why are we allowed in Statement B to make this move:
y/x=1/4 -> 4y = x
I am confused, because we don`t know for sure that either x or y are really positive and hence we would not be allowed to divide or multiple a variable.
Re: Is 4^(x+y)=8^(10) ?
[#permalink]
28 Sep 2018, 09:17
1
Expert Reply
HHPreparation wrote:
Hey Guys,
Quick question from my side. Why are we allowed in Statement B to make this move:
y/x=1/4 -> 4y = x
I am confused, because we don`t know for sure that either x or y are really positive and hence we would not be allowed to divide or multiple a variable.
Thank you for your help!
We are concerned about the sign when dealing with inequalities: we should keep the sign if we multiply by a positive value and flip the sign when we multiply by a negative value.
For equations we can multiply/divide by a variable regardless of its sign (providing it's not 0).
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Re: Is 4^(x+y)=8^(10) ?
[#permalink]
30 Dec 2018, 07:00
Bunuel wrote:
Is \(4^{x+y}=8^{10}\) ?
(1) x - y = 9 (2) y/x = 1/4
Practice Questions Question: 41 Page: 278 Difficulty: 600
From the question, 2^2(x+y) = 2^3x10 =>v2x + 2y = 30 => x+y = 15 ?? That's what it asks basically ;
Now, from (1), x-y=9 ; We can say there are 2 variables & 1 equation so there are many values possible for which x-y = 9 But, only the value 12-3=9 & 12+3=15 for all other values we can say whatever maybe the value of x or y such that x-y=9 it won't be equal to 15 ; Since, we don't know what are their values so (1) is clearly INSUFFICIENT.
From, (2), 4y=x ; Same here we can't say what are the value of x & y ; But, here only & only if x=3 & y=4x3=12 then only we can get 12+3=15 ; But, we are not sure as per (2) & hence, (2) is INSUFFICIENT.
From, (1) & (2), we can determine x & y and after solving we get x=12 & y=3 => x+y = 15 => That's what the question asks => SUFFICIENT => Answer option C ; _________________
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Re: Is 4^(x+y)=8^(10) ?
[#permalink]
29 Sep 2019, 03:52
Hi Bunuel, Can I ask you a question? In statements such as (1) x - y = 9 and (2) y/x = 1/4, would it be safe to assume that since there are two variables x and y can take multiple values?
Plugging in numbers in both the statements does led you to the right answer but it's a little time consuming.
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