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Senior | Next Rank: 100 Posts
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by nehakhas1 » Fri Jan 30, 2009 4:15 am
x + 2y + z = 8;
2x + y + z = 7
What are the values of x, y, z?
I. x, y, z are positive integers. II. x, y, z are distinct numbers

We can find the ans without these given conditions as well .Then should our answer be E?
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Source: — Data Sufficiency |
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Just trying

by Alara533 » Fri Jan 30, 2009 9:48 am
x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
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Re: Just trying

by ajmoney09 » Fri Jan 30, 2009 12:24 pm
Alara533 wrote:x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
I dont agree....

You nee three distinct equations to solve for the three different values.. you are only given two.

What is the OA?
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by Alara533 » Fri Jan 30, 2009 2:38 pm
We need three equations to solve three unknowns when we don't have any other information about to the unknown. But in this case we have additional information.

For eg:- x + y + z = 3 can be solved with only this one equation, assuming we know x, y and z are all positive integers.
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Re: Just trying

by coffee5251 » Fri Jan 30, 2009 3:37 pm
Alara533 wrote:x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
Why can't x = 2 and z = 0?
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by masuarezdl » Fri Jan 30, 2009 3:47 pm
Alara533 wrote:We need three equations to solve three unknowns when we don't have any other information about to the unknown. But in this case we have additional information.

For eg:- x + y + z = 3 can be solved with only this one equation, assuming we know x, y and z are all positive integers.
AJmoney09, the thing here is to substract one equation from the other, that way you will be left with two unknowns and only one set of numbers does the job. Let me try to illustrate it:

We subsract equation 2 from equation 1...
x + 2y + z = 8
- 2x - y - z = -7

The remaining equation is -x + y = 1

Then we find the value of y = x + 1, and substitute in any of the two equations given:

x + 2(x+1) + z = 8
x + 2x + 2 + z = 8
3x + z = 6

In this last equation, if all numbers are required to be positive, then x=1 and z=3, only those numbers satisfy the condition. Afterwards, you will only have to substitute to find the value of y.

Coffee5251, the number 0 is neither positive nor negative. Statement 1 indicates that x, y and z are all positive numbers.

Hope it helps.
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