12 Days of Christmas GMAT Competition - Day 4: If p and q are

If you square and solve the 2 equations, it leads to x<11. I think is part is straight-forward.
Given the question talks about the 2 prime solutions, they have to be positive.
This means p = 2 (smallest prime) and q = 7 (largest prime less than 11). So p-q = 2-7=-5

Simplifying

√(x^2−4x+4)<√(x−20)^2
√(x-2)^2< √(x-20)^2
|x-2|<|x-20|

Solving the inequality leads to
x<11

Prime numbers less than 11 are 2,3,5 and 7

So, p=2, q=7

p-q=-5

LHS is (x-2)^2 .It can also be written as |x-2| as it is under root.
Similarly RHS is |x-20| .
If x is 11 then LHS = RHS . So x=7 is max (q) and x=2(p) is least prime that satisfies the equation .
therfore p-1 =2-7 =-5

Ans if C. If we square both the sides, we get x^2 - 4x + 4 < ( x - 20)^2.

If we solve this inequality, we get x < 11, and the prime numbers which are < 11 are 2, 3 , 5 and 7.

So p is 2 and q is 7, and p - q = -5

The inequality becomes:

(x2 - 4x + 4)^(1/2) < |x - 20|

Square both sides (valid since both sides are non-negative):
x2 - 4x + 4 < (x - 20)2
x2 - 4x + 4 < x2 - 40x + 400
-4x + 4 < -40x + 400
36x < 396
x < 11

So, x must be less than 11.

Prime numbers less than 11: 2, 3, 5, 7
Therefore:
p = 2 (least prime solution)
q = 7 (greatest prime solution)
p - q = 2 - 7 = -5

The answer is C. -5

Lest prime is 2, greatest prime is 7
P-q=-5

Case 1: x<2x < 2
∣x−2∣=2−xand∣x−20∣=20−x
The inequality becomes:
2−x<20−x
Simplify:
2<20
This is always true. Thus, all x<2 satisfy the inequality.
Case 2: 2≤x≤20
∣x−2∣=x−2and∣x−20∣=20−x
The inequality becomes:
x−2<20−x
Simplify:
2x < 22 = x<11
Thus, for this case, 2≤x<11
Case 3: x>20x > 20
∣x−2∣=x−2and∣x−20∣=x−20
The inequality becomes:
x−2<x−20
Simplify:
−2<−20-2 < -20
This is never true. Thus, no x>20 satisfies the inequality.
Step 3: Combine results
The solutions to the inequality are:
x<2or2≤x<11
This simplifies to:
x<11

Step 4: Find prime numbers
The prime numbers less than 11 are:
2,3,5,7

• The smallest prime (p) is 2.
• The largest prime (q) is 7.

Step 5: Calculate p−q
p−q=2−7=−5
Final Answer:
−5\boxed{-5}

\(\\
\sqrt{x^2 - 4x + 4} < \sqrt{(x-20)^2}\\
\)

We can square both sides to eliminate the square root signs
which gives us

\(\\
x^2 - 4x + 4 < (x-20)^2\\
\)

Open the brackets on the RHS (Right-Hand-Side)

\(\\
x^2 - 4x + 4 < x^2 - 40x +400\\
\)

Collect Like terms

\(\\
x^2 - x^2 - 4x + 40x < 400 - 4\\
\\
36x < 396\\
\\
x < 11\\
\)

The least Prime Number less than 11 is 2 and the greatest is 7

\(\\
p = 2 \\
q = 7\\
\)

Therefore
\(\\
p - q = 2 - 7 = -5\\
\)

Answer is C

sqrt(x^2 - 4x + 4) < sqrt((x-20)^2)
x^2 - 4x + 4 < x^2 - 40x + 400
36x < 396 x < 11
p = 2, q = 7
p - q = -5

IMO C

If we simplify the equation, The inequality becomes:

∣x−2∣<∣x−20∣

There are multiple ways to solve this, We can go by traditional method of dividing this equation into multiple cases..

But as we know the x has to be prime, Lowest prime is 2 and that satisfy the equation hence p=2.

The other point would be x-2 =-(x-20) => x=11.

So at 11 this will be equality. At 7 this will satisfy. Hence q=7

p-q = -5

IMO C

We know \(\sqrt{x^2}\)=|x|

Also, x^2 - 4x + 4 = (x-2)^2
Hence the given equation becomes:
|x-2|<|x-20|

Critical points are 2,20

1. assume x>20 Then opening modulus sign we get
x-2 < x-20
No solution

2. assume 2<x<20
(x-2) < -(x-20)
i.e. x<11

So greatest prime no' solution = 7 {cant take 11}

3. least prime number is 2, lets test if its a valid solution

|2-2| < |2-20|
Yes 0<18
So greatest prime no' solution = 2

Hence the required answer is 2-7=(-5)

We can rewrite the equation as:

sqrt[(x-2)^2] < sqrt[(x-20)^2]
=> |x-2| < |x-20| .................................(1)

Now we know p is the least prime that is a solution, let's try and substitute 2 in equation and check if it holds valid => we get 0 < 18, which is valid. Hence, we confirm that p = 2

Looking at the answer choices, we know now that q can be 11, 7, 5, or 3

Testing from greatest by substituting in (1),
For 11 => we get 9 < 9 which is invalid.
For 7 => we get 5 < 13 which is valid. Hence, we confirm that q = 7

Now, p-q = 2-7 = -5

Answer C

If p and q are, respectively, the least and greatest prime number solutions to \(\sqrt{x^2 - 4x + 4} < \sqrt{(x-20)^2}\), what is the value of p - q?

\(\sqrt{x^2 - 4x + 4} < \sqrt{(x-20)^2}\)
\(\sqrt{(x-2)^2} < \sqrt{(x-20)^2}\)

|x-2| < |x-20|

Case 1: x = 2; |2-2|=0 < |2-20| = 18
Case 2: x = 3; |3-2|=1 < |3-20| = 17
Case 3: x = 5; |5-2|=3 < |5-20| = 15
Case 4: x = 7; |7-2|=5 < |7-20| = 13
Case 5: x = 11; |11-2|=9 = |11-20| = 9


p = 2; since |2-2|=0 < |2-20| = 18
q = 7; since |7-2|=5 < |7-20|=13

p - q = 2 - 7 = -5

IMO C

First thing to do is to convert the left side of the equation into something that is comparable to the right side.

The left side then becomes the (X-2)^1/2 which needs to be less than (X-20)^1/2

The smallest prime number is 2, we plug 2 into both sides and receiver 0 < 324 so we know that works and can serve as our p value.

Then we find a prime number that is much larger like 11 plug that in and see that both sides equal 81 so therefore the q value HAS to equal the next lowest prime number which would be 7.

In terms of finding the higher value to test, we know that the value cannot be very high because if we are relying on the right side of the equation to be positive before we square it, the equation as a whole will never work. So it has to result in the squaring of a negative value that is larger than the left side of the equation can produce.

P=2 and Q=7 therefore p-q is equal to -5

Final Quality is this

|x−2∣<∣x−20∣

So we wil have the 3 Cases
1. X<=2
2. 2<X<=20
3. X>20


1. For 1st, (X<=2)
-X+2<-x+20

we get 2<20 (Always True)

So, in the X<=2, we have only 2 as Prime Number

2. For 2nd, (2<X<=20)

x-2<-x+20
x<11

in this we get X = (2,11). so, we have the Prime number 3,5,7

3. for 3rd case (x>20)
We get -2<-20 (False)


SO prime numbers as the solution we get (2, 3, 5, 7)

So Min-Max = -5

\sqrt{x^2 - 4x + 4} < \sqrt{(x-20)^2}[/m]
\sqrt{(x-2)^2} < \sqrt{(x-20)^2}[/m]
|x-2|<|x-20|

If 2 < x < 20
|x-2| become x-2
|x-20| become 20-x

Formula become x-2 < 20-x
2x < 22
x <11

If x < 11, the least prime number p is 2 and the maximum prime number q is 7
p-q = 2-7 = -5
Therefore the answer is C

The answer is -5.

from the given we have |x-2| < |x-20|

when we find the limits of x, we can find p and q.

Solving the inequality we have,
when x>20,
x-2<x-20
-2<-20 which is not possible

When 2≤x<20
we have x-2<20-x
2x<22 x<11

When x<2
we have 2-x<20-x
2<20

so we have 2≤x<11 and the prime numbers are 2,3,5,7 --> 2-7 = -5