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

Question

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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?

A. -9
B. -7
C. -5
D. -3
E. -1

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

Simplify:

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

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

Recall that  a^2 = |a|

|x - 2| < |x - 20|

This implies that x is closer to 2 than it is to 20. The halfway point between 2 and 20 is (2 + 20)/2 = 11. Thus, x must be somewhere in the green interval:

----------------(2)---------------- (11)----------------(20)---------------- 
 
Thus, x < 11.

Since p and q are, respectively, the least and greatest prime numbers in this range, p = 2 and q = 7. Therefore, p - q = 2 - 7 = -5.

Answer: C.

Functionx · Answer

\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

hr1212 · Answer

|x-2|<|x-20|

Case 1 : x < 2

2 - x < 20 - x
2 < 20

True for all values of x

Case 2 : 2 <= x < 20

x - 2 < 20 - x
2x < 22
x < 11

True for values  ∞, 11)

So smallest prime value is p = 2 and largest prime value is q = 7

p - q = 2 - 7 = -5

Answer: C

Krunaal · Answer

We can rewrite the equation as: sqrt < sqrt => |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

Kinshook · Answer

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

HansikaSachdeva · Answer

The equation can be rewritten as: 
√(x^2 − 4x + 4) < √(x^2 − 40x + 400)

Squaring both sides to remove the square root,

= x^2 − 4x + 4 < x^2 − 40x + 400
= − 4x + 4 < − 40x + 400
= 36x < 396 
= x < 11

Prime numbers less than 11 are 2, 3, 5, 7
Greatest, q = 7 
Least, p = 2

Answer: 2 - 7 = -5

TheLegacy · Answer

The correct answer is C) -5 :

Let's do the sqrt of both sides, knowing that we must have two abs quantities under the roots.

|x-2| < |x-20|

Now we must "discuss" 4 different scenarios: + + , + - , - + , - - .

++:

x>=2 and x>=20 -> x>=20 always

It will give: x-2 <= x-20
Hence: -2 <= -20 which is ALWAYS FALSE. NO SOLUTIONS.

+ - :

x >= 2 and x<20 -> 2 =< x < 20

Hence: x < 11 --> 2 <=x < 11

Here our least prime number is 2 and the greatest is 7.
Hence the solution is 2 - 7 = -5

- + :

x<2 and x>20

FALSE -> x cannot be less than 2 and more than 20 at the same time!

- - :

x<2

We can discuss it, but it will give me negative numbers, and honestly, I am not interested in them since we are talking about prime numbers.

The solution is then -5.

dhaphuong · Answer

\sqrt{x^2 - 4x + 4} < \sqrt{(x-20)^2} --> < --> |x-2| < |x-20| Case 1: x<2 |x-2| < |x-20| --> 2 - x < 20 - x --> 2 < 20 (always true) Case 2: x>=20 |x-2| < |x-20| --> x - 2 < x - 20 --> - 2 < - 20 (always wrong) Case 3: 2<= x <20 |x-2| < |x-20| --> x - 2 < 20 - x --> 2x < 22 --> x < 11 Combining the condition of the case: 2 <= x < 11. Prime numbers in the range: 2, 3, 5, 7 p = 2 q = 5 --> p - q = -5 --> ANSWER C

Prakhar9802 · Answer

Reducing the question stem to simplest form -

\sqrt{(x^2 - 4x + 4)} = \sqrt{(x-2)^2}
= |x-2|

\sqrt{(x-20)^2} = |x-20|

Question becomes-

|x-2| < |x-20|

Least prime number 2

Put in 2

|2-2| < |2-20|
0<18

Hence, P = 2

For Q, we know that mod is basically the distance between 2 numbers on the number line. Therefore, to satisfy this inequality, we need the distance of our number Q from 20 to be greater than the distance of Q from 2.

If we plug in 11, we can see that it is equidistant from both the numbers,

|11-2| < |11-20|
9 < 9

This inequality does not hold.
Therefore, we need to plug in a smaller prime number than 11.

Try 7,

|7-2| < |7-20|
5 < 13

Therefore, Q = 7

P-Q = 2-7 = -5

FINAL ANSWER = Option C

Lizaza · Answer

It's clear that we can change the left hand side to: 
  \sqrt{(x-2)^2} < \sqrt{(x-20)^2}

Then we can get rid of the root:
 |x-2| < |x-20| , or  |x| < |x-18|.

As   x   is prime, we're subtracting 1 from a positive number - so to get this equation, the difference between   x   and 18 has to be bigger than   x   itself.
The numbers that fit are:  2, 3, 5, 7.  From 11, we already get   |11-18|=7,  which is too big.

As  p=2  and  q=7 , then  p-q=2-7=-5 .   The answer is C.

aarati17 · Answer

Ans. C. -5 Since (x^2-4x+4) = (x-2)^2 The equation can be written as- |x-2|<|x-20| For x<2 |x-2| will be written as 2-x and |x-20| will be written as 20-x =>2-x < 20-x =>2<20 Hence, for all values of x<2, the equation will be satisfied Prime number cannot be negative so we do not have any prime number for when x>2 For 2<=x<20 |x-2| will be written as x-2 and |x-20| will be written as 20-x =>x-2 < 20-x =>2x<22 =>x<11 So, for 2<=x<20, we have x<11 which means values of x is between For x>=20 |x-2| will be written as x-2 and |x-20| will be written as x-20 =>x-2 < x-20 =>2>20 This is not possible so we can eliminate this Hence, From 2<=x<20, we get the least prime number p=2 and the greatest prime number q=7 p-q = 2-7 = -5

NilayMaheshwari · Answer

|(x-2)| < |(x-20)|
|x-2| - |x-20| < 0
 We'll check for three ranges,

1st range- x<2,

-x+2+x-20 <0
-18<0
which is always true, therefore this whole range is solution.

2nd range- 2=<x<20

x-2+x-20<0
2x-22<0
2(x-11)<0
x-11<0
x<11.
Therefore, the range of solutions is 2=<x<11.

3rd range- x>=20

x-2-x+20<0
18<0
which is false, therefore this whole range can't be a solution.

Now the solution is x<11.
In this range, smallest prime number is 2 and greatest prime number is 7.
2-7=-5.
Therefore, C

minnieminnoon · Answer

Hi! Quick question, since I’m in the middle of reviewing error logs from last year.
Do we still get this type of questions in GMAT Focus Edition?

Bunuel · Answer

Yes, this type of question is perfectly aligned with the GMAT Focus Edition syllabus. It tests core quantitative concepts such as roots, inequalities, absolute value, and number properties, all of which remain fully relevant on the current exam.

DonBosco7 · Answer

For this question if you remember a simple property of absolute values, the question is solvable in a minute.
 property:- if |a| = |b|
then, a=b or a = -b
 
apply this to the question

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

then,  \sqrt{(x-2)^2} < \sqrt{(x-20)^2}

this is written as |x-2| < |x-20|........................ apply the above property

1] x-2 < x-20
2] x-2 < -(x-20)

least prime => p = 2
greatest prime from above x = 7

hence p-q = 2-7 = -5

Answer C

Demonicgod · Answer

thanks man it helped a lot.

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