Hi PGMAT:

The expression is simplified as follows:
For some random number a
sqrt(a^2) = |a|

So sqrt((y-4)^2) = |y-4| = |4-y|

In this question we are asked whether |4-y| = (4-y)
Or in other words, is 4-y >=0 (|a| = a only if x>=0)
4-y>=0 => y<=4. So if we can prove sufficiently that y is less than or greater than 4 we are good.

Now let's look at the options.

1. |y-3|<=1
=> -1 <= y-3 <=1
=> -2 <= y-4 <=0 (adding -1 to all parts of the expression).
So y-4 <= 0 means that y<=4. This is what we set out to find. Sufficient.

2. y.|y|>0
Since |y| >=0, so |y| can't be negative, hence y*|y| >0 means that y>0.
This does not tell us whether y<=4. Insufficient.

So the final answer is A
For more information regarding moduli, refer to my post.
https://www.beatthegmat.com/squares-squa ... tml#476551

Let me know if this helps

PGMAT wrote:Is sqrt((y-4)^2) = 4-y?
1. |y-3|<=1
2. y|y|>0

Can some one explain how to simplify the expression in the question? Thanks.

Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.

Thus, √(y-4)² = |y-4| = the distance between y and 4.

Question rephrased: Is |y-4| = 4-y?
In other words:
Is the DISTANCE between y and 4 equal to the DIFFERENCE between 4 and y?

A DIFFERENCE can be negative, 0, or positive.
A DISTANCE must be greater than or equal to 0.
For the DIFFERENCE between two values to be equal to the DISTANCE between the two values, the DIFFERENCE -- like the DISTANCE -- must be greater than or equal to 0:
4-y≥0
y≤4.

Question rephrased: Is y≤4?

Statement 1: |y-3|≤ 1.
-1 ≤ y-3 ≤1
2 ≤ y ≤ 4
Since y must be between 2 and 4, inclusive, it must be less than or equal to 4.
SUFFICIENT.

Statement 2: y|y|>0 .
Since |y| cannot be negative, the left-hand side must be (+)(+).
Thus, y > 0.
Since it's possible that y<4, y=4, or y>4, INSUFFICIENT.

The correct answer is A.

Mitch, thanks for excellent explanation.
I have a related question. Please help me understand if I am solving correctly. Thanks.

If x^3<16x; what are the possible solutions of x?

x^3<16x
x^3-16x<0
x(x^2-16)<0
x<0 or x^2<16 ==> |x|<4 ==> -4<x<4

PGMAT wrote:Mitch, thanks for excellent explanation.
I have a related question. Please help me understand if I am solving correctly. Thanks.

If x^3<16x; what are the possible solutions of x?

x^3<16x
x^3-16x<0
x(x^2-16)<0
x<0 or x^2<16 ==> |x|<4 ==> -4<x<4

Not quite.
I recommend that you use the CRITICAL POINT approach.

Step 1: Move all of the variables to one side of the inequality and factor.
x³ < 16x
x³ - 16x < 0
x(x²-16) < 0
x(x+4)(x-4) < 0.

Step 2: Identify the CRITICAL POINTS.
The CRITICAL POINTS are where the lefthand side is EQUAL to 0:
x=-4, x=0, x=4.

Step 3: To determine the valid ranges of x, test one value to the left and right of each critical point.

x<-4:
Test x=-10 in x³ < 16x:
(-10)³ < 16(-10)
-1000 < -160.
This works: x<-4 is a valid range.

-4 < x < 0:
Test x=-1 in x³ < 16x:
(-1)³ < 16(-1)
-1 < -16.
Doesn't work: -4<x<0 is NOT a valid range.

0 < x < 4:
Test x=1 in x³ < 16x:
(1)³ < 16(1)
1< 16.
This works: 0<x<4 is a valid range.

x>4:
Test x=10 in x³ < 16x:
(10)³ < 16(10)
1000 < 160.
Doesn't work: x>4 is NOT a valid range.

Thus, the two ranges that satisfy x³ < 16x are x < -4 and 0<x<4.

If you type "critical points" and "gmatguruny" into the search bar, you'll find other problems that I've solved with the CRITICAL POINT approach.