Data Sufficiency

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.

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.

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

r_walid can you please post the OA.

thanks.

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

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.

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.

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.

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.

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