help me with these questions

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help me with these questions

by kimmie2k » Tue Oct 06, 2009 11:28 pm
1) if a>b>c>0, c=3?
1. 1/c > 1/3
2. 1/a + 1/b + 1/c = 1

Answer is D, Why?

2) John deposits $1000 in an account with annual rate x percent (compound quarterly), no other activity for this account, is account balance over $1050 after one year?
1. (1+x/200)^2 > 1.05
2. (1+x/400)^2 > 1.025

Answer is D, Why?


3) if |2x| > |3y|, is x>y?
1. x>0
2. y>0

Answer is A, why?

4) A(a,b) is the coodinate, is it on the line y=x?
1. a=2
2. b=a+2

Answer is B, Why?
Source: — Data Sufficiency |

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by fiercepoint » Wed Oct 07, 2009 4:12 am
First problem.

1) Plug the value of c into the inequality, 1/3 > 1/3. This is not right so it is sufficient to say that c isn't 3.

2) Again plug the value into the inequality, 1/a + 1/b + 1/3 = 1. This means 1/a + 1/b would have to equal 2/3 but a>b>c>0 so 1/a and 1/b will both be smaller than 1/3. Adding two smaller fractions to 1/3 will not equal 1 so again this statement is sufficient to say that c isn't 3.

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by fiercepoint » Wed Oct 07, 2009 4:29 am
I'll take the third one as well.

The absolute value of x is obviously higher than y but the problem is that x might be a larger negative number. For example if x = -2 and y = 1 then |2x| = 4 and |3y| = 3. All you need to know to solve this is whether x is positive or negative.

1) Tells you that x is positive. Sufficient.

2) Doesn't help at all, regardless of value of y, x could still be a larger negative number.

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by kimmie2k » Wed Oct 07, 2009 10:24 pm
Thanks fiercepoint.

I understood your explaination on Q3.
and For Q1, I actually made a typo. Original question should be 1) if a>b>c>0, c<3?

Any idea?