If 0<x<y, is y-x < 0.00005

There are two ways to deal with it.

Method 1:

Is y-x < 0.00005?

We can see that both statements alone are not sufficient.

(1) x>1/60,000
(2) y<1/15,000

We know that we can add inequalities when they have the same sign ie.
a < b
c < d
then, a+c < b+d

Also, when we multiply an inequality by -1, the inequality sign flips.
x>1/60,000 implies -x < -1/60,000

You can add these two inequalities: -x < -1/60,000 and y<1/15,000 to get y-x < 1/15000 - 1/60,000
which is y-x < 1/20,000 i.e. y-x < 0.00005

Another method is to see this on the number line. Draw a number line to understand this.

0<x<y implies that x and y are both positive and x is to the left of y on the number line.
Is y-x < 0.00005 means is the distance between x and y less than .00005?

(1) x>1/60,000
means x lies to the right of 1/60,000

(2) y<1/15,000
means y lies to the left of 4/60,000

So the distance between them must be less than 4/60,000 - 1/60,000 = 3/60,000 = .00005

(1) Insufficient. We know nothing about \(y\).
(2) Insufficient. We know nothing about \(x\).

(1)+(2) Sufficient.
We know that \(y<\frac{1}{15,000}\) and \(-x<-\frac{1}{60,000}\). If we add this two inequalities we will get:
\(y-x<\frac{1}{15,000}-\frac{1}{60,000}=\frac{1}{20,000}=0.00005\)


The correct answer is C.

ADDING/SUBTRACTING INEQUALITIES

1. You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

2. You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

Check for more the links below:
Inequalities Made Easy!

dave13:
I want to make the fractions comparable and want to be able to perform operations on them. When are two fractions comparable? They are intuitive to compare and we can perform operations on them when they have the same denominator.

One fraction given is 1/60,000.

Another fraction given is 1/15,000. So I multiply and divide it by 4 to get 4/60,000. Since I multiply and divide by the same number, the fraction does not change.
1/15,000 = 4/60,000

Now I can easily perform operations on 1/60,000 and 4/60,000.

If 0<x<y, is y-x < 0.00005

(1) x>1/60,000
(2) y<1/15,000


Solution: (Answer is C)

What do we know?

X is positive and Y is greater than X.

What do we need to know?

Is Y is less than 0.00005 + X?

Whenever you face a Data Sufficiency question asking Yes, No. Simply substitute and try to disprove the statement.

Statement(1):

X is greater than 1/60,000 = 0.00001666

Which does not tell any relation between X and Y

Hence it is insufficient.

Statement (2) is also insufficient as it only tells that Y is less than 0.000066
(It is very important to know the importance of converting fractions to percentage)

If we combine both the statements, we get that X is greater than 0.000016 and Y is less than 0.000066

Now the question is asking us that y-x<0.00005, to try to disprove that we need to maximize y-x and for that let us get the maximum value of y and minimum value of x.
Let us say y = 0.000065 and x = 0.000017
So the maximum difference is = 0.000065 - 0.000017 = 0.000048

Hence combining both the statements we can say that y-x will always be less than 0.000048. Hence answer is (C)

for 2 ineq to subtract they have to be with opposit direction , one of them is bigger than and 2nd is smaller than and what u do is keep the sign ( direction in terms of bigger than or smaller than) of the ineq from which u subtract the 2nd ....


Another way of seeing it is as follows

if we subtract 1 from 2

is like flipping the sign of 1 and adding it to the 2nd , thus


x>1/60,000 becomes -x<-1/60,000...............1 after changing direction ( flipping the sign)

now add 2 to 1

y>1/15,000........2

y+ (-x) > 1/15,000 + (-1/60,000).................. simplify

y-x > 1/20,000

\(\frac{1}{20000}=\frac{1}{2}*\frac{1}{10000}=0.5*10^{-4}=5*10^{-5}\)

ahh thank you, I was multiplying .5 by 10,000 instead of 1/10,000 and knew it wasnt possible

Elite097 Draw 0, x, and y from left to right as described in the free info.
The question is about the distance between x and y. We know that y sits to the right of x, and the question is whether the distance between them is "microscopic." 0.00005 = 5/100,000 = 1/20,000. Neither statement can possibly be sufficient on its own, because we're interested in the distance between x and y. Combining the statements may be useful, as we're given a bottom limit for x and an upper limit for y, forcing the distance between them to be less than some quantity. What quantity, exactly? Well, (1/15,000 - 1/60,000). Let's expand by a factor of 60,000 and infer that y-x < 3.
Rephrase the question, expanding by the same factor of 60,000 and get: Is y-x < 3?
So the statements, combined, give us a definitive YES.
To your second question: Given two inequalities, we can reason that the sum of the bigger sides of the two inequalities must be greater than the sum of the two smaller sides of the two inequalities. If you want to subtract one inequality from the other, you can expand/reduce by a factor of (-1) the inequality that you wish to subtract (remembering to flip the sign, since the action you took flipped everything to the other side of zero, the mirror image), and then "add" the inequalities.

I have one question is this step possible for this question ( I have re written the equations in this way)

1/60000 < X
Y< 4/60000

Add both equations and then subtract you reach back to the original question and can prove sufficiency.

I took off the difficulty, but GMAT Prep did rate this medium level.

Anyway, I understand why the both statements are insufficient but I do not know how you combines them? What did you when you 'subtracted 2 from 1'?

When you subtract 1 from 2, you get the value of y-x. However, since we know only one sided limits of these values, let's consider those values.

y-x=(1/15000)-(1/60,000)

Taking L.CM. y-x=1/20,000

y-x=0.00005

However, this just gives us the limit of the difference. Since y<1/15,000 and x>1/60,000, a bigger number on the L.H.S is being subtracted from a smaller number and hence, the actual difference will be less than 1/20,000. This is by applying concept. Let us test values for better understanding.

For e.g. the value of y could be y=1/20,000(the greater the denominator, the smaller the number and hence y>1/15000) and x=1/40,000(by similar idea)

y-x=1/20,000-1/40,000=1/40,000
1/40,000<1/20,000. Hence proved.

Hope that helps!

In relation to this question, I have a basic scientific notation question, how would you write the sci notation of 1 / 20,000?

How do you know what sign the combined inequality takes when combining two inequalities with different signs?

This question deals with advanced knowledge of decimals and fractions.

X < Y → First Condition

Statement 1 - X is greater than 1/60000 → 1/6 = 0.16
1/60 → 0.016
1/600 → 0.0016
1/6000 → 0.00016
1/60000 → 0.000016

But we don’t have the value of “y”

Statement 2 - y < 1/15000

1/15 = ½ of 1/7 → 0.07
1/150 → 0.007
1/1500 → 0.00007
1/15000 → 0.000007

If the value of Y is to be greater than 0.007 then the value can be 0.7, 0.1, 1 upto infinitely positive numbers

But we still don’t have the value of “x”

Combining 1 + 2 we can easily conclude that the difference will be greater/lesser

WHAT IF . . .
y= 0.000067 and x= 0.000017
y-x=0.00005
That gives answer as NO

0<x<y means both + and y>x

y-x<0.00005?
y-x < 0.5*10^-4?

1) x>1/60,000
x>1/(6*10^4)
x > 1/6 * 10^-4
No info about y, NS

2) y<1/15,000
y < 1/(1.5*10^4)
y < 1/1.5 * 10^-4
y < 2/3 * 10^-4
No info about x, NS

3)
LT (2/3 * 10^-4) - GT (1/6 * 10^-4) < 0.5*10^-4 ?
LT 2/3 - GT 1/6 < 1/2 ?
2/3 - 1/6 is 4/6 - 1/6 --> 3/6 = exactly 1/2
But since our actual number is less than 2/3 and other number is greater than 1/6 the difference will be smaller than 1/2.
So it's sufficient.