Mixture Problem

The ans should be "A" here.

According to question :

0.10X+0.02Y=0.05(X+Y)

The q asks for X...

To know that we need to know at least one of the variables..

From stem 1 : Y=10 so we can get the value of X.Sufficient.

From 2: Z=16 which is nothing but X+Y=Z according to question.
But we need to know the individual value/s of one of the variables not the sum.So insufficient.

Amit

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hijja: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Mon Mar 10, 2008 4:25 pm; Thanked: 1 times

by hijja » Wed Jul 30, 2008 5:16 am

IMO D
From the question, we can come with 2 equations i.e. x+y = z and 10x+2y=5z.

From statement 1, we have y =10, i.e. z=x+10 , so
10x +20 = 5x + 50 i.e. 5x = 30 , so x = 6

From statement 2, we have z =16, i.e. 16=x+y , so
10x +2y = 80 ..solve these to get x = 6.

Hence both the statements are sufficient to answer this question.

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amitansu: Master | Next Rank: 500 Posts; Posts: 294; Joined: Tue Feb 26, 2008 9:05 pm; Thanked: 13 times; Followed by:1 members

by amitansu » Wed Jul 30, 2008 7:01 am

Yes.. you are right .
I missed it.

Amit

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olika: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Sat Nov 10, 2007 10:35 am; Thanked: 1 times

by olika » Wed Jul 30, 2008 3:54 pm

hijja wrote:
From statement 2, we have z =16, i.e. 16=x+y , so
10x +2y = 80 ..solve these to get x = 6.

x and y might be non-integers, so you can't solve for x.

I go with ans A.

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parallel_chase: Legendary Member; Posts: 1153; Joined: Wed Jun 20, 2007 6:21 am; Thanked: 146 times; Followed by:2 members

by parallel_chase » Wed Jul 30, 2008 4:14 pm

olika wrote:
hijja wrote:
From statement 2, we have z =16, i.e. 16=x+y , so
10x +2y = 80 ..solve these to get x = 6.
x and y might be non-integers, so you can't solve for x.

I go with ans A.

Well it does not matter weather x and y are integers or not, here we are talking about tons of mixture. x could be 6.0 or y=5.67, it does not matter.

Dont confuse number properties with mixtures and ratios.