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by GMATGuruNY » Fri Dec 19, 2014 7:49 am
buoyant wrote:If 6x=8y=14z, then what is a possible sum of positive integers x, y, and z?

a)52
b)58
c)84
d)122
e)168
If we divide the equations by 2, we get:
3x = 4y = 7z.

Set the equations equal to 84, the LCM of 3, 4 and 7:
3x = 4y = 7z = 84.
Here, x=28, y=21 and z=12.
Implication:
The smallest possible value for x+y+z = 28+21+12 = 61.

The correct answer must be a multiple of 61.
Only D is viable:
122 = 2*61.

The correct answer is D.
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by buoyant » Fri Dec 19, 2014 10:22 am
GMATGuruNY wrote:
buoyant wrote:If 6x=8y=14z, then what is a possible sum of positive integers x, y, and z?

a)52
b)58
c)84
d)122
e)168
If we divide the equations by 2, we get:
3x = 4y = 7z.

Set the equations equal to 84, the LCM of 3, 4 and 7:
3x = 4y = 7z = 84.
Here, x=28, y=21 and z=12.
Implication:
The smallest possible value for x+y+z = 28+21+12 = 61.

The correct answer must be a multiple of 61.
Only D is viable:
122 = 2*61.

The correct answer is D.

we should take the same multiple for each variable in this case 3x = 4y = 7z
right?
that is we cannot take x= 28*2 , y=21 , z= 12*4
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by MartyMurray » Fri Dec 19, 2014 10:55 am
buoyant wrote:we should take the same multiple for each variable in this case 3x = 4y = 7z
right?
that is we cannot take x= 28*2 , y=21 , z= 12*4
You could see it this way.

Start by dividing 6x=8y=14z by 2 to get 3x=4y=7z.

Then since 3, 4 and 7 do not share any factors, we know that x must be divisible by 4 and 7, y must be divisible by 3 and 7 and z must be divisible by 4 and 3.

So x is a multiple of 28, y is a multiple of 21 and z is a multiple of 12.

To be clear on the ratios of x, y and z we can do the following.

x is a * 28. Y is b * 21 and z is c * 12.

So 3(a)(28) = 4(b)(21) = 7(c)(12) and 84(a) = 84(b) = 84(c)

So a, b and c are equal and any combination of x, y and z will be a multiple of 28 + 21 + 12 = 61.
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by Matt@VeritasPrep » Mon Dec 22, 2014 12:13 pm
buoyant wrote:we should take the same multiple for each variable in this case 3x = 4y = 7z
right?
that is we cannot take x= 28*2 , y=21 , z= 12*4
Yeah, one way of thinking about this is that 3x = something, 4y = (the same thing), and 7z = (the same thing). Since 3x, 4y, and 7z are each different ways of representing the same number, whatever that number is, it must be a multiple of 3, 4, and 7.

Algebraically, you could represent it this way:

3x = k
4y = k
7z = k

You can pick any k you want. An easy one is 3*4*7, or 84. Now we have

3x = 84
4y = 84
7z = 84

and we solve from there as shown above.
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by Jeff@TargetTestPrep » Wed Dec 13, 2017 10:34 am
buoyant wrote:If 6x=8y=14z, then what is a possible sum of positive integers x, y, and z?

a)52
b)58
c)84
d)122
e)168
We can divide the given equation by 2 and we have:

3x = 4y = 7z

In order for each term to be equal, x could be 4(7) = 28, y could be 3(7) = 21, and z could be 3(4) = 12.

Thus, we see that the smallest value of x + y + z is 28 + 21 + 12 = 61. However, since that is not an answer choice, the next highest value is 2 x 61 = 122.

Answer: D

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