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If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 01:04
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If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z? A. 15 B. 20 C. 25 D. 40 E. 47
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 01:39
Given problem : If x=2y=4z for positive integers x, y, and z, find the average of x,y,3z? Since x=2y=4z, assume x=4x,y=2x and z=x So the average of x,y,3z will be \(\frac{(4x+2x+3x)}{3} = 3x\) Of the answer options, only 15(Option A) is a multiple of 3 and is our correct answer.
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 01:50
MathRevolution wrote: If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?
A. 15 B. 20 C. 25 D. 40 E. 47 If x=2y=4z, x + y + 3z = 4z + 2z + 3z = 9z Average of 3 numbers =\(\frac{9z}{3}\) = 3z Since the average is multiple of 3, answer has to be a multiple of 3 as well.
Only option A is a multiple of 3. Hence Answer A
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 02:16
MathRevolution wrote: If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?
A. 15 B. 20 C. 25 D. 40 E. 47 x = 2y = 4z x, y, 3z Average= Sum/Count Average = (4z+2z+3z)/3 = 9z/3 = 3z As we can see the answer should be multiple of 3. Answer is A
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If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 09:35
MathRevolution wrote: If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?
A. 15 B. 20 C. 25 D. 40 E. 47 Given \(x=2y=4z\) \(x = 4z\)  (i) \(2y = 4z\) \(y = 2z\)  (ii) Average of \(x, y\) and \(3z = \frac{x + y + 3z}{3}\) Substituting the values of \(x\) and \(y\) from (i) and (ii) respectively, we get; \(\frac{4z + 2z + 3z}{3} = \frac{9z}{3} = 3z\) Average of \(x, y\) and \(3z\) must be a multiple of \(3\). Among the options Only (Option A) \(15\) is multiple of \(3\). Answer (A)...



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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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30 Jun 2017, 10:42
MathRevolution wrote: If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?
A. 15 B. 20 C. 25 D. 40 E. 47 Let, x=2y=4z = 4 So, x = 4 , y = 2 & z = 1 Arithmetic Mean of x, y & 3z is \(\frac{4 + 2 + 3}{3} = 3\) So, The number must be a multiple of 3, only (A) satisfies, thus answer will be (A) 15
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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02 Jul 2017, 18:13
==> You get x=4z and y=2z, and the average=\(\frac{(x+y+3z)}{3}=\frac{(4z+2z+3z)}{3}=3z\), so it always needs to be the multiple of 3. Therefore, the answer is A. Answer: A
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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23 Jul 2017, 07:27
Hi Bunuel, Vyshak, If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z? x=2y ==> y=x/2 x=4z ==> z=x/4 arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3 =10x/12 ==>5x/6 Can somebody please tell me where am'I going wrong?
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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23 Jul 2017, 08:16
NandishSS wrote: Hi Bunuel, Vyshak, If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z? x=2y ==> y=x/2 x=4z ==> z=x/4 arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3 =10x/12 ==>5x/6 Can somebody please tell me where am'I going wrong? \(\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}\), not 5x/6.
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If x=2y=4z for positive integers x, y, and z, which of the following c
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23 Jul 2017, 10:55
MathRevolution wrote: If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z?
A. 15 B. 20 C. 25 D. 40 E. 47 y=x/2 z=x/4 x+x/2+3*x/4=9x/4 mean=(9x/4)/3=3x/4 only 15 is a multiple of 3 A



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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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24 Jul 2017, 06:58
Bunuel wrote: NandishSS wrote: Hi Bunuel, Vyshak, If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z? x=2y ==> y=x/2 x=4z ==> z=x/4 arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3 =10x/12 ==>5x/6 Can somebody please tell me where am'I going wrong? \(\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}\), not 5x/6. Tagging VeritasPrepKarishma, Bunuel, I'm Sorry for very silly mistake.But 3x/4, does ans should be multiple of 3. Can you please explain bit.
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Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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07 Aug 2017, 10:12
NandishSS wrote: Bunuel wrote: NandishSS wrote: Hi Bunuel, Vyshak, If x=2y=4z for positive integers x, y, and z, which of the following can be the average (arithmetic mean) of x, y and 3z? x=2y ==> y=x/2 x=4z ==> z=x/4 arithmetic mean of x, y and 3z = [x+(x/2)+(3x/4)]/3 =10x/12 ==>5x/6 Can somebody please tell me where am'I going wrong? \(\frac{x+\frac{x}{2}+\frac{3x}{4}}{3}=\frac{\frac{4x+2x+3x}{4}}{3}=\frac{3x}{4}\), not 5x/6. Tagging VeritasPrepKarishma, Bunuel, I'm Sorry for very silly mistake.But 3x/4, does ans should be multiple of 3. Can you please explain bit. Since the given options are all integers, it needs to be assumed that \(\frac{3x}{4}\) will be an integer. Therefore, the resultant will 3 * (some number). In the given options, the only option that is a multiple of 3 is Option A. That is why it is the right option. Kapish?




Re: If x=2y=4z for positive integers x, y, and z, which of the following c
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