If x, y, and k are positive and x is less than y, then (x + k)/(y + k)

let x=5 ,y=6 and k =1
solve
\(\frac{x + k}{y + k}\) = 6/7
and x/y=5/6
we see that 6/7>5/6
IMO B

x<y
So, x/y= less than 1
If x=1
y=2
x/y=1/2=.5

Now, if we increase both numerator and denominator by same number then result must be increased.
1+1/2+1=2/3=.67

Answer is B

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Let's compare (x + k)/(y + k) and x/y:

Is (x + k)/(y + k) > x/y?

Is (x + k)y > (y + k)x?

Is xy + ky > xy + kx?

Is ky > kx?

Is y > x? Yes!

Since y > x, all the inequalities above hold. So, (x + k)/(y + k) is greater than x/y.

Notice that we multiplied each side of the inequalities by constants several times, but all of x, y and k are positive; so we never had to worry about reversing the direction of the inequality.

One way to solve this question is to know that if you add the same constant to the numerator and denominator of a positive fraction, then the value of the fraction increases and will always tend to 1. But even if we did not know this, we can easily solve the question by using the answer choices. Remember this is a PS question, so there have to be only ONE possible answer.

We have x, y and k to be positive and x < y. The question asks us for (x + k)/(y + k)

A. If (x + k)/(y + k) = 1, then x = y. This is not possible since x < y
B. If (x + k)/(y + k) > x/y, cross multiplying (since x and y are positive), we get xy + yk > xy + xk. Cancelling xy on both sides, we get y > x or x < y. So B works.

Hope this helps!

Aditya
CrackVerbal Quant Expert

let x= 2, y= 4, and y = 1
Then 2/4 = 0.5
(2+1)/(4+1) = 0.6

so the value is greater than x/y.

The answer is B.

Obviously there is no reason why the answer would be equal to 1 - so let us ignore that.
We basically need to compare the new fraction (x+k)/(y+k) with x/y

Method 1
Given x, y and k are positive and x < y => x/y is a fraction less than 1
For example: Let us look at 2/3 (= 0.67)
If we add 1 to both numerator and denominator, we have 3/4 (= 0.75)
Thus, the new fraction is greater than the initial one.
The reason for this is as follows:

Since x < y, when we add the same constant k to x and to y, the percent increase in the value of x is more than that for y
Thus, the numerator increases more in comparison to the denominator.
Thus, the new fraction increases in value

Thus, if x, y, k are positive and x < y, we have: (x+k)/(y+k) > x/y

Method 2
We need to compare (x+k)/(y+k) and x/y
The symbol between them would be '>' or '<' or '='

Let the symbol be #

Thus: (x+k)/(y+k) # x/y

Since all terms are positive, we can easily cross-multiply:
=> xy + yk # xy + xk
=> yk # xk

Cancelling k from both sides (since k is positive):
y # x

But, we know that y > x

Thus: '#' represents '>'

Thus: (x+k)/(y+k) > x/y


Answer B

Asked: If x, y, and k are positive and x is less than y, then \(\frac{x + k}{y + k}\) is

x<y
x/y < 1
For example
x = 2; y=3; k = 1
x =2 < y=3
x/y = 2/3
(x+k)/(y+k) = 3/4 > 2/3 = x/y

IMO B

Can the method in which you used simplification and # and to represent =, >, or < be used in any situation to find the relationship between two functions/equations?

Assume x/y as 2/3. this we know is equal to 0.67.
now take k as 7. (2+7)/(3+7)=9/10. This is equal to 0.9.

Hence, x/y<(x+k)/(y+k).

Using concepts I learnt from TTP,

If x is less than y, then x/y is a positive proper fraction.

Positive Proper fractions have a value between 0 and 1.

K, a positive integer, when added to the numerator and the denominator of a fraction brings the value of a fraction closer to 1.

In this case, as x/y is a fraction that has a value between 0 and 1, (x+k)/(y+k) will be closer to 1 and greater than x/y.

Therefore, (x/y) < [(x+k)/(y+k)]

Ans. B

We are adding the same thing to both the numerator and denominator, so the ratio of those additive quantities is 1:1. Therefore, the starting ratio will move along the number line toward 1 (since 1:1 is 1).
Since the starting ratio (x/y) was between 0 and 1 (x and y are positive, x is less than y), moving toward 1 means it is increasing.

BrentGMATPrepNow / chetan2u, sir can you elaborate further on the properties of fractions? For instance what happens when a constant is subtracted/added from/to numerator(N) and denominator(d) when N<D & N>D etc.?

Happy to help.
Please note that the following only applies to positive fractions.

If we have a fraction in the form N/D, where N > D, then N/D > 1
So, (N - k)/(D - k) will be greater than N/D (for positive values of k)
For example, (10 - 4)/(7 - 4) > 10/7

Conversely, if we have a fraction in the form N/D, where N < D, then N/D < 1
So, (N - k)/(D - k) will be less than than N/D (for positive values of k)
For example, (3 - 2)/(7 - 2) < 3/7

Please check fractions below. It may help you.

ScottTargetTestPrep
Thank you for this helpful explanation. I have two questions:
Question 1: Why do you put y(x+k) on the left-hand side of > vs x(y+k)... is this the order of operations for cross-multiplying fractions in inequalities?

Question 2: If you had x>y as an end result, what would be the next step when looking at the answer choices? You just know that (x+k)/(y+k) is not greater than x/y but is it equal to 1 (choice A), equal to x/y (choice C) etc. Would you repeat the process setting (x+k)/(y+k)=1 and if that doesn't work (x+k)/(y+k)=x/y

I just reasoned through the question per Avi Gutman's approach knowing that adding 1 brings me closer to one but not quite at 1. However, I was curious about the algebra approach here.

Thank you

Hey woohoo921

Let me try to elaborate the points from Scott’s crisp solution. I will answer both your questions one by one. 😊

Your First Question: “Why do you put y(x+k) on the left-hand side of > vs x(y+k)”
This is due to the principle of cross-multiplication. Let’s take a closer look at the process of cross-multiplication in the context of inequalities. It is actually composed of two different operations that just happen at the same time.
Let me try to explain the same to you through a simple example in which I will show you both these operations separately first and then together.

Example:
Consider ¾ > ½. ----------(I)
We will try to remove the denominators from both fractions one by one.
Step 1: Let’s start with the denominator 2 from ½. Multiplying inequality-I above throughout by 2, we get:

• ¾ × 2 > ½ × 2
• ¾ × 2 > 1 --------(II)

Note: Since 2 is a positive number, multiplying both sides of the inequality by 2 will not affect the sign of inequality. Observe how the denominator from the right-hand side (2 of 1/2) went to get multiplied with ¾ on the left-hand side. The numerator 3 did not change its place.
Step 2: Next, let’s try to remove the denominator 4 from ¾. Multiplying inequality-II above throughout by 4, we get:

• ¾ × 2 × 4 > 1 × 4
• 3 × 2 > 1 × 4
• 6 > 4

Note: Since 4 is a positive number, multiplying it to both sides of the inequality does not impact the sign of the inequality. Observe how the denominator from the left-hand side (4 of ¾) went to get multiplied with 1 on the right-hand side. The numerator 1 did not change its place.

SUMMARY:

1. In both these steps, we took the positive denominators from each side of the inequality and multiplied them to the numerator on the other side without changing the inequality sign. The product in each case was written on the same side as the numerator.
2. Everything we did in two steps above could also have been done in a single step. (This is what we call cross-multiplication).
   
   1. We could straight go from ¾ > ½ to 3 x 2 > 1 x 4.
3. This is precisely what Scott did. He went from (x + k)/(y + k) > x/y to y(x + k) > x(y + k).

Your Second Question: “What if we came to “Is x > y” at the end?”
Let’s try to create this scenario. Let’s suppose Scott had started with “Is (x + k)/(y + k) < x/y?” Then, we would have:

Is (x + k)/(y + k) < x/y?
Is (x + k)y < (y + k)x?
Is xy + ky < xy + kx?
Is ky < kx?
Is y < x?
Now, we are given that y > x. Thus, the answer to the last question (Is y < x?) is NO!
And since the last inequality does not hold true, all the above inequalities would not hold true as well. So, (x + k)/(y + k) is not less than x/y.

Now, since < is rejected, we have two possibilities left: > or =. But how do we decide which one?
Well, for now, let’s just put a “?” between the two fractions being compared. This “?” will eventually be either “>” or “=”.

Here we start: (x + k)/(y + k) ? x/y
Since, all terms are positive, we can cross-multiply without affecting the sign of the inequality. Let’s do that.
y(x + k) ? x(y + k)
xy + ky ? xy + kx
ky ? kx
y ? x
But we already know that y > x! So “?” must actually represent “>”.
Thus, (x + k)/(y + k) > x/y.

And we’re done!

Takeaways:

1. Cross-multiplying fractions in an inequality involves keeping each numerator as is, and bringing the denominators from the other side for multiplication.
   
   1. The inequality sign remains the same if the denominator being moved is positive. For example, 1/3 > ¼. When we cross-multiply, the sign remains the same and we get 1 x 4 > 1 x 3.
2. For questions like this, where you need to compare two entities, the easiest approach is to start with a “?” between the two entities. As you would simplify this (using normal rules of equations and inequalities), you would be able to use some given information to infer what “?” stands for out of >, <, or =.

Hope this helps!

Best Regards,
Ashish
Quant Expert, e-GMAT

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