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DS on inequalities

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Source: — Data Sufficiency |
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Re: DS on inequalities

by life is a test » Tue Nov 03, 2009 9:06 pm
r_walid wrote:IF material A cost 3$/kg and B cost 5$/kg.
If 10 kg of material K conssit of x kg of A and y kg of B, is x>y?

1- y>4

2- the cost of 10kg of K is less than $40.
IMO D

the question can be rephrased as is x>5?

1. if y>4 then x = 5, 4,3, 2, 1. in all cases it is not more than y hence suff.

2. 3x + 5(10-x) < 40 -> 10<2x -> x<5 -> suff

hope that helps.
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by venubab » Wed Nov 04, 2009 1:10 am
Answer B

1. y>4

Say, y=4.1, x=5.9 x>y
Say, y=6, x=4 x<y

Not sufficient

B is sufficient
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by life is a test » Fri Nov 06, 2009 11:10 pm
r_walid can you please post the OA.

thanks.
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by okigbo » Tue Nov 17, 2009 12:01 pm
life is a test: how can A be sufficient if we dont know the total cost? I think it is B - two equations / two unknowns
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by enniguy » Tue Nov 17, 2009 4:19 pm
life is a test wrote:
r_walid wrote:IF material A cost 3$/kg and B cost 5$/kg.
If 10 kg of material K conssit of x kg of A and y kg of B, is x>y?

1- y>4

2- the cost of 10kg of K is less than $40.
IMO D

the question can be rephrased as is x>5?

1. if y>4 then x = 5, 4,3, 2, 1. in all cases it is not more than y hence suff.

2. 3x + 5(10-x) < 40 -> 10<2x -> x<5 -> suff

hope that helps.
Statement 1 says y>4. So, y can be 4.2. Then x will be 5.8. So, x>y. Now, y can also be 8, since 8 is greater than 4.
Then x will be 2. So here y > x.
Hence statement 1 is not sufficient.

Statement 2 as you all know is very sufficient.
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by palvarez » Tue Nov 17, 2009 10:51 pm
okigbo wrote:life is a test: how can A be sufficient if we dont know the total cost? I think it is B - two equations / two unknowns

Master this problem. You gonna see similar problems on GMAT: at least two or three.


It is a weighted avg problem.

A's average: 3, B's average: 5
The mixture's average < 4


When you combine A and B equally, the average gonna be 4.

When the average < 4. Then we can say that A's quantity < B's quantity.


Look at a generalization: You are given a list of n numbers.

The lowest in that list <= average <= the largest.

In these questions, don't waste your time setting up all algebraic equations, etc. If you notice the pattern, answering this question would take a couple of seconds.
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by beatthegmat2910 » Thu Nov 19, 2009 4:19 pm
IMO D.
Condition 1 - Y > 4; y=5,6,7,8,9
x=5,4,3,2,1

x=y for one set ,hence Not sufficient.

Condition 2-
3x+5y < 40 ; This is also not sufficient , since 3x+5y could be anything under 40.

Combining 1 and 2 , the only value that 3X+5Y can hold so that Condition 1 is true is 39.
Solving the two equations, you get the answer.
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by Birgit Anne » Mon Nov 30, 2009 11:17 am
Answer B (Statement 2 alone is sufficient):

3x + 5y < 40
We also know that:
x+y = 10

Let`s pick numbers
3*5 + 5*5 =40 not in line with statement 2
3*5,1 + 5*4,9 = 39,8 --> in line with statement 2
Therefore x>y --> Statement 2 alone is sufficient

x must be greater than 5 and y must be less than 5.
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by ssuarezo » Wed Dec 02, 2009 8:35 am
palvarez wrote:
It is a weighted avg problem.

A's average: 3, B's average: 5
The mixture's average < 4 WHERE DIS U GET THIS FROM? IT'S NOT IN THE STEM QUESTION.

When you combine A and B equally, the average gonna be 4. OK .

When the average < 4. Then we can say that A's quantity < B's quantity. OK .

Look at a generalization: You are given a list of n numbers. WE HAVE DIFFERENT NUMBERS, NOT NECESARILY RELATED

The lowest in that list <= average <= the largest. ARE YOU TALKING ABOUT 3,5,10, 4, 40????.

In these questions, don't waste your time setting up all algebraic equations, etc. If you notice the pattern, answering this question would take a couple of seconds. COULD YOU SOLVE THE PROBLEM USING YOUR APPROACH
Hi PAlvarez:

I'm interested in your approach, could you answer my questions in underlined black upper case?

Thank you.
Silvia
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