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A contractor combined -GMATPrep [#permalink]
30 Jan 2009, 11:10

A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2% gravel G, by weight, to produce z tons of a mixture that was 5% gravel G, by weight. What is the value of x?

Re: A contractor combined -GMATPrep [#permalink]
30 Jan 2009, 11:27

IMO D.

here is my approach.

It is given that 0.1x + 0.02y = 0.05z. Eq-----> Eq1 Also x + y = z ----> Eq 2

Two eqautions and 3 unknowns. To solve for x, we need to have atleast 3 distinct linear equations.

Stmt1---> y = 10. from this we have 3 equations two from the question stem and third from the statment. Three equations and 3 unknowns. Sifficient. Stmt2 ---> z = 16 from this we have 3 equations two from the question stem and third from the statment. Three equations and 3 unknowns. Sifficient.

Hence D is the ans.

Last edited by mrsmarthi on 30 Jan 2009, 12:46, edited 1 time in total.

Re: A contractor combined -GMATPrep [#permalink]
30 Jan 2009, 11:50

2

This post received KUDOS

Expert's post

D

This is a C-trap.

1. Fast 10sec approach (if we don't have enough time):

- if we have two mixtures with different % and prepare a new mixture, it is obvious that final % will depend only on ratio between two start mixtures. Therefore, if we know all % and weight of any mixture (z or y), we can find x.

2. Usual approach.

We have system of equations: 0.1x+0.02y=0.05z x+y=z

Now, we can write out the system separately for two conditions and will get a system with two equations and two variables - each condition is sufficient:

Re: A contractor combined -GMATPrep [#permalink]
30 Jan 2009, 12:44

walker wrote:

D

This is a C-trap.

1. Fast 10sec approach (if we don't have enough time):

- if we have two mixtures with different % and prepare a new mixture, it is obvious that final % will depend only on ratio between two start mixtures. Therefore, if we know all % and weight of any mixture (z or y), we can find x.

2. Usual approach.

We have system of equations: 0.1x+0.02y=0.05z x+y=z

Now, we can write out the system separately for two conditions and will get a system with two equations and two variables - each condition is sufficient:

1) y=10 0.1x+0.2=0.05z x+10=z

x=6

2) z=16 0.1x+0.02y=0.8 x+y=16

x=6

Rightly said. And looks obvoise that that I am frequent victim of that trap w.r.t DS. BTW, I have corrected my previous post. I missed that x + y = z fact.

Re: A contractor combined -GMATPrep [#permalink]
31 Jan 2009, 12:24

walker wrote:

D

This is a C-trap.

1. Fast 10sec approach (if we don't have enough time):

- if we have two mixtures with different % and prepare a new mixture, it is obvious that final % will depend only on ratio between two start mixtures. Therefore, if we know all % and weight of any mixture (z or y), we can find x.

2. Usual approach.

We have system of equations: 0.1x+0.02y=0.05z x+y=z

Now, we can write out the system separately for two conditions and will get a system with two equations and two variables - each condition is sufficient:

1) y=10 0.1x+0.2=0.05z x+10=z

x=6

2) z=16 0.1x+0.02y=0.8 x+y=16

x=6

Hi walker,

can you explain me the 10 second approach in detail.

Re: A contractor combined -GMATPrep [#permalink]
31 Jan 2009, 13:43

Expert's post

Let's consider another similar situation: we have two paints, yellow and blue. In order to get green color we have to mix blue and yellow paints at a special ratio. For example, 1 liter of yellow paint and 2 liters of blue paint give us necessary green paint (1:2 ratio). In other words, we can get green paint only if we follow strictly the ratio, otherwise resulted paint will contain more blue or more yellow. In our example, if we have 20 litters of yellow paint, we should get 40 litters of blue paint, not more or less. If we make 60 litters of green paint, we should take 20 liters of yellow and 40 litters of blue.

The same reasoning in the problem. We know all concentrations. Therefore, we have a special ratio of initial mixtures in order to get resulted mixture. knowing amount only one mixture is enough to find others using the ration.

By the way, this problem 7-t75289 uses the same idea. So, you can solve it very quickly.
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