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Is 4^(x+y)=8^(10) ?

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Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 02:19
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 02:20
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SOLUTION

Is 4^{x+y}=8^{10} ?

Work with the same base: is 4^{x+y}=8^{10} ? --> is 2^{2(x+y)}=2^{30} ? --> is 2(x+y)=30? is x+y=15?

(1) x - y = 9. Not sufficient.
(2) y/x = 1/4 --> x=4y. Not sufficient.

(1)+(2) We have two distinct linear equation with two unknowns (x - y = 9 and x=4y), hence we can solve for both of them and get whether x+y=15 is true. Sufficient.

Answer: C.
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 04:20
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Is 4^{x+y}=8^{10} ?
(1) x - y = 9
(2) y/x = 1/4


4^{x+y}=8^{10} ---> 2^2{x+y}=2^3{10}----->2^{x+y}=2^{15}
The question can be restated as Is x+y = 15?

1) x-y = 9 --->Since x & y can assume any value---->Insufficient
2) x= 4y ---> No Absolute value is given for x & y, Only ratio is given-----> Insufficient
1+2) 4y-y = 9 ----> 3y=9---->Sufficient

Thus Answer C
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 05:49
Bunuel wrote:

Is 4^{x+y}=8^{10} ?

(1) x - y = 9
(2) y/x = 1/4




the given eqn can be solved as:
2^2(x+y) = 2^3(10)

=> x+y = 15

Stmt 1) x - y = 9 can take many values for x and y and satisfy the eqn
Stmt 2) y/x = 1/4, here also x and y can take many values

combining both 1) and 2) the eqn can be solved to get values of x and y and can be verified if these x and y satisfies the given eqn

Hence C
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 09:13
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Bunuel wrote:
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Is 4^{x+y}=8^{10} ?

(1) x - y = 9
(2) y/x = 1/4

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From the question is X + Y = 15?

1st Option X - Y = 9 it can be any thing and any no. not necessarly the nos whose sum is equal to 15. Not Sufficient.
2nd Option: Y/x = 1/4 again Y and X can be any thing not necessarly the nos. whose summ is equal to 15. Not sufficient.

Combining Both options:
X - X/4 = 9
X = 12
therefore Y = 3.

so Answer is "C".

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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 18 Sep 2012, 09:17
first i calculated E... then after luking at above solution..i wud say..its C..
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 19 Sep 2012, 23:39
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The question is simply is x + y = 15?
All we need is to know the value of the sum of x and y to get sufficiency.

(1) x -y= 9

10 - 1 is 9
11 - 2 is 9
24 - 9 is 15

INSUFFICIENT

(2) y/x = 1/4 or 4y = x many possibilities , INSUFFICENT

X - y = 9
(4y) - Y = 9
y = 3
x = 12

answer: c
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 21 Sep 2012, 02:46
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SOLUTION

Is 4^{x+y}=8^{10} ?

Work with the same base: is 4^{x+y}=8^{10} ? --> is 2^{2(x+y)}=2^{30} ? --> is 2(x+y)=30? is x+y=15?

(1) x - y = 9. Not sufficient.
(2) y/x = 1/4 --> x=4y. Not sufficient.

(1)+(2) We have two distinct linear equation with two unknowns (x - y = 9 and x=4y), hence we can solve for both of them and get whether x+y=15 is true. Sufficient.

Answer: C.

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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 21 Sep 2012, 02:57
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Ans:C
4^(x+y)=8^(10)
=>2^2(x+y)=2^3(10)
=>x+y=15?
Statement1:x-y=9----Insufficent
Statement2:y/x=1/4=>x=4y---Insufficent
By combining St1&st2
y=3 and x=12 which gives x+y=15---Sufficient
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 03 Mar 2013, 21:21
Hi, could you please help clarify? I'm a little bit confused.

Isn't the answer D?

(1) X-Y=9 ----> X=9+Y
We can just put "X=Y+9" in the initial equation X+Y=15 to solve the equation without any help from the 2nd equation, y/x = 1/4.

The same goes to the second equation
(2) y/x = 1/4 -------> X=4Y
Just put substitute the X in the initial equation, X+Y=15, with the 4Y and then solve the equation for both X and Y.

So, we can actually get the answer using only 1 of statement. So why is the answer C.

Billion thanks in advance!
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 04 Mar 2013, 02:27
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wutthiyo wrote:
Hi, could you please help clarify? I'm a little bit confused.

Isn't the answer D?

(1) X-Y=9 ----> X=9+Y
We can just put "X=Y+9" in the initial equation X+Y=15 to solve the equation without any help from the 2nd equation, y/x = 1/4.

The same goes to the second equation
(2) y/x = 1/4 -------> X=4Y
Just put substitute the X in the initial equation, X+Y=15, with the 4Y and then solve the equation for both X and Y.

So, we can actually get the answer using only 1 of statement. So why is the answer C.

Billion thanks in advance!


Welcome to GMAT Club!

Actually the correct answer is C, not D.

From the stem the question became: is x + y = 15?

Now, (2) says that x - y = 9. Can we tell from this whether x + y = 15? No! Consider x = 10 and y = 1 for a NO answer and x = 12 and y = 3 for an YES answer. Hence, the first statement is NOT sufficient.

The same for the second statement.

The problem with your solution is that you assumed that we have two equations for each statement, whereas we have just one: x - y = 9 for (1) and y/x = 1/4 for (2). The second equation, x + y = 15 is not given to be true, we are asked to find whether it's true.

Hope it's clear.
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 05 Mar 2013, 22:50
Bunuel wrote:

Welcome to GMAT Club!

Actually the correct answer is C, not D.

From the stem the question became: is x + y = 15?

Now, (2) says that x - y = 9. Can we tell from this whether x + y = 15? No! Consider x = 10 and y = 1 for a NO answer and x = 12 and y = 3 for an YES answer. Hence, the first statement is NOT sufficient.

The same for the second statement.

The problem with your solution is that you assumed that we have two equations for each statement, whereas we have just one: x - y = 9 for (1) and y/x = 1/4 for (2). The second equation, x + y = 15 is not given to be true, we are asked to find whether it's true.

Hope it's clear.


Oh! Thanks a lot! That's clear.

So when a question ask whether an equation is true or not, we cannot use it in solving the problem. Am I right? ^___^
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 06 Mar 2013, 01:35
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wutthiyo wrote:
Bunuel wrote:

Welcome to GMAT Club!

Actually the correct answer is C, not D.

From the stem the question became: is x + y = 15?

Now, (2) says that x - y = 9. Can we tell from this whether x + y = 15? No! Consider x = 10 and y = 1 for a NO answer and x = 12 and y = 3 for an YES answer. Hence, the first statement is NOT sufficient.

The same for the second statement.

The problem with your solution is that you assumed that we have two equations for each statement, whereas we have just one: x - y = 9 for (1) and y/x = 1/4 for (2). The second equation, x + y = 15 is not given to be true, we are asked to find whether it's true.

Hope it's clear.


Oh! Thanks a lot! That's clear.

So when a question ask whether an equation is true or not, we cannot use it in solving the problem. Am I right? ^___^


Yes, you cannot use it as a given.
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 07 Jun 2013, 04:13
Hey guys,

Why do you have to reduce the base to 2? Why cant we just change the base of 8 to 4 so thats it reads X+y=20? That would seem to be easier but mathematically Im not coming out to the correct answer. Can someone help?
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 07 Jun 2013, 04:32
AlphaMan21 wrote:
Hey guys,

Why do you have to reduce the base to 2? Why cant we just change the base of 8 to 4 so thats it reads X+y=20? That would seem to be easier but mathematically Im not coming out to the correct answer. Can someone help?


8^{10} cannot be written as 4^{20}, 4^{20}=4^{2*10}=16^{10}\neq{8^{10}}.

Hope it's clear
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 08 Jun 2013, 05:12
Thanks a lot - found the root of my error there
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 08 Jun 2013, 05:12
No pun intended! Ha!
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 27 Oct 2013, 07:20
Bunuel wrote:
SOLUTION

Is 4^{x+y}=8^{10} ?

Work with the same base: is 4^{x+y}=8^{10} ? --> is 2^{2(x+y)}=2^{30} ? --> is 2(x+y)=30? is x+y=15?

(1) x - y = 9. Not sufficient.
(2) y/x = 1/4 --> x=4y. Not sufficient.

(1)+(2) We have two distinct linear equation with two unknowns (x - y = 9 and x=4y), hence we can solve for both of them and get whether x+y=15 is true. Sufficient.

Answer: C.


Here's the approach I did, it took me roughly 5 minutes to complete, would you be able to tell me how to get faster, and if i plugged Y into the correct question.

From the Q.Stem I simplified the Question down to, does "x+y=15" Yes/No.

Statement 1) x-y=9
x+y=9+y
X=9+Y

I used substation, plugged in X with the original equation, and I got 9+2y=15, reduced down to y=3.

So the question is reduced to "does Y=3, yes or no" from what I know, I'm Not sure. So therefore it's not sufficient.

Statement 2) y/x=1/4 is simplified to 4Y=X,

I Plugged in to the original equation X+Y=15
So now I have 4Y+Y=15
5Y=15, The question is now rephrased as does y equal 3? again, it's not sufficient.

(C) X+Y=15
From Statement 1 & Statement 2, Y=3, I plugged in the following

X+Y=15
X-3=-3
X=12

Banuel, If I went wrong anywhere, can you tell me where I went wrong. Should I test the answer choice to see if it's insufficient, instead of labeling the answer not sufficient?

For statement one, I found myself answering, yes it's sufficient to solve, but X is not given, so I should not assume that the number is automatically "x=12" correct? The simplification is reducing the question still" Does Y=3? and that changes it all, because I don't know if Y=3. So is that enough information to move on? Same reasoning on the second statement. Threw me off for a bit.
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 27 Oct 2013, 23:35
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selfishmofo wrote:
Bunuel wrote:
SOLUTION

Is 4^{x+y}=8^{10} ?

Work with the same base: is 4^{x+y}=8^{10} ? --> is 2^{2(x+y)}=2^{30} ? --> is 2(x+y)=30? is x+y=15?

(1) x - y = 9. Not sufficient.
(2) y/x = 1/4 --> x=4y. Not sufficient.

(1)+(2) We have two distinct linear equation with two unknowns (x - y = 9 and x=4y), hence we can solve for both of them and get whether x+y=15 is true. Sufficient.

Answer: C.


Here's the approach I did, it took me roughly 5 minutes to complete, would you be able to tell me how to get faster, and if i plugged Y into the correct question.

From the Q.Stem I simplified the Question down to, does "x+y=15" Yes/No.

Statement 1) x-y=9
x+y=9+y
X=9+Y

I used substation, plugged in X with the original equation, and I got 9+2y=15, reduced down to y=3.

So the question is reduced to "does Y=3, yes or no" from what I know, I'm Not sure. So therefore it's not sufficient.

Statement 2) y/x=1/4 is simplified to 4Y=X,

I Plugged in to the original equation X+Y=15
So now I have 4Y+Y=15
5Y=15, The question is now rephrased as does y equal 3? again, it's not sufficient.

(C) X+Y=15
From Statement 1 & Statement 2, Y=3, I plugged in the following

X+Y=15
X-3=-3
X=12

Banuel, If I went wrong anywhere, can you tell me where I went wrong. Should I test the answer choice to see if it's insufficient, instead of labeling the answer not sufficient?

For statement one, I found myself answering, yes it's sufficient to solve, but X is not given, so I should not assume that the number is automatically "x=12" correct? The simplification is reducing the question still" Does Y=3? and that changes it all, because I don't know if Y=3. So is that enough information to move on? Same reasoning on the second statement. Threw me off for a bit.


5 minutes is too much for this problem.

The question boils down to whether x+y=15.

(1) says x - y = 9. Can we answer whether x+y=15? No, because infinitely many pairs of (x, y) satisfy x - y = 9, and only one of them yields the sum of 15, namely x=12 and y=3.

(2) says x=4y. Basically the same here: can we answer whether x+y=15? No, because infinitely many pairs of (x, y) satisfy x=4y, and only one of them yields the sum of 15.

When we combine the statements we have x-y=9 and x=4y. So, we have two two distinct linear equation with two unknowns, hence we can solve for both of them and get whether x+y=15 is true.

As you can see we don't need to solve anything for this question to get the answer.

Hope it helps.
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Re: Is 4^(x+y)=8^(10) ? [#permalink] New post 28 Oct 2013, 14:09
Thanks Banuel, It really helps. Yeah, 5 Minutes is way too long. I'm having to brush up on linear equations as you can tell.
The second time I worked the problem it made more sense, and cut my timing in half.
Re: Is 4^(x+y)=8^(10) ?   [#permalink] 28 Oct 2013, 14:09
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