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lets say p=3/2,5/2 and 2.
therefore p2= 9/4,25/4 and 4
2p=3,5,4
therefore in case of 3/2:
p2<2p
incase of 5/2:
p2>2p
and, incase of 2:
p2=2p
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This is could be true (or false) question. As long as some of the answers are within the condition 1<p<3, the answer is valid.

I. p^2 < 2p --> 0<p<2
Some of 0<p<2 are within 1<p<3
CBT (Could Be True)

II. p^2 = 2p
p=0 or p=2. p=2 is within 1<p<3
CBT

III. p^2 > 2p --> p<0 or p>2
Some p>2 are within 1<p<3
CBT

FINAL ANSWER IS (E)
I, II, III

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test value of p with 3/2,2 & 5/2
we get
I. p^2 < 2p
II. p^2 = 2p
III. p^2 > 2p
to be valid and true ; since its could be true type
IMO E is correct

If 1 < p < 3, then which of the following could be true?

I. p^2 < 2p
II. p^2 = 2p
III. p^2 > 2p

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
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Here p = 3/2, 2, 5/2
I. When, P = 3/2, then, p^2 = 9/4 and 2p =3/2*2 = 3, Ok
II. When p = 2, then, p^2 = 4, and 2p = 4, Ok.
III. When, p = 5/2, then p^2 = 25/4, 2p = 2*(5/2) = 5, OK
Answer. E
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If 1 < p < 3,then which of the following could be true?

Could be true = yes and no
Must be true = consistently yes
Cannot be true = consistently no
Constraint: 1<p<3
.: p= 3/2 or 2 or 5/2

I. p^2 < 2p
When p= 3/2
(3/2)^2 < 2(3/2) —> 9/4 < 3? Yes
When p= 5/2
(5/2)^2 < 2(5/2) —> 25/4 < 5? No
(I) could be true ,Eliminate B and C

II. p^2 = 2p
When p= 3/2
(3/2)^2 =2(3/2) —> 9/4 =3? No
When p=2
2^2=2(2) —> 4=4? Yes
(II) could be true, Eliminate A

III. p^2 > 2p
When p=3/2, 9/4 > 3? No
When p= 5/2 , 25/4 >5? Yes
Eliminate D ,we left with E
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

Hit that E ;)

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If 1 < p < 3, then which of the following could be true?

I. \(p^2 < 2p\) .. Let's consider \(P=1.1\)
\(1.1^2 < 2*1.1\)
\(1.21<2.2\) .. True

II. \(p^2 = 2p\) .. let's consider \(P=2\)
\(2^2 = 2*2\)
\(4 = 4\) .. True

III. \(p^2 > 2p\) .. let's consider \(P=2.5\)
\(2.5^2 > 2*2.5\)
\(6.25 > 5\) .. True

Option 'E' is the winner
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Given, \(1 < p < 3\)

I. \(p^2 < 2p\)
--> \(p^2 - 2p < 0\)
--> \(p(p - 2) < 0\)
--> \(0 < p < 2\) --> Could be True (as 1 < p < 2 satisfies the given range for p)

II. \(p^2 = 2p\)
--> \(p^2 - 2p = 0\)
--> \(p(p - 2) = 0 \)
--> \(p = 0\) or \(2\) --> Could be True (as p = 2 falls given range for p)

III. \(p^2 > 2p\)
--> \(p^2 - 2p > 0\)
--> \(p(p - 2) > 0\)
--> \(p < 0\) or \(p > 2\) --> Could be True (as p > 2 satisfies given range for p)

Option E
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Given that 1<p<3, we are to determine which of the following could be true.

It is worth noting that could be true means we only need to prove that there exists a number within the given range that satisfies the given condition.

I. p^2 < 2p
Let p=1.5, 2p=3, and p^2=2.25. Since p^2(2.25) < 2p(3) I could be true.

II. p^2=2p
Let p=2, then 2p=4 and p^2=4. Since p^2=2p=4, II could be true.

III. p^2>2p
Let p=2.5, then 2p=5 and p^2=6.25. Since p^2(6.25)>2p(5), III could be true.

The answer is E.
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If 1 < p < 3, then which of the following could be true?

I. \(p^2 < 2p\)
If p = 1.5, 2p = 3
\(1.5^2 = 2.25 \)TRUE

II. \(p^2 = 2p\)
If p = 2, 2p = 4
\(2^2 = 4\) TRUE

III. \(p^2 > 2p\)
If p = 2.5, 2p = 5
\(2.5^2 = 6.25\) TRUE

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

ANSWER E.
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If 1 < p < 3, then which of the following could be true?

I. \(p^{2} < 2p\)
—> \(p^{2} —2p = p(p—2) <0\)
0 <p <2
—> p is in 1< p <2 (could be true)

II. \(p^{2} = 2p\)
p *(p—2)= 0
p= 2 and p= 0
—> p =2 and p is in 1 <p <3 (could be true)

III. \(p^{2} > 2p\)
—> p*(p—2) <0
p <0 and p >2
—> p is in 2 <p <3 (could be true)

The answer is E.

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Quote:
If 1 < p < 3, then which of the following could be true?

I. p^2 < 2p
II. p^2 = 2p
III. p^2 > 2p

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

I. p^2-2p<0, p(p-2)<0; p>0 then p-2<0, p<2 (1<p<3) true;

II. p^2 = 2p, p^2-2p=0, p(p-2)=0; p=0, or p=2, but p≠0 (1<p<3) true;

III. p^2 > 2p, p^2-2p>0, p(p-2)>0, p>0 then p-2>0 p>2 (1<p<3) true;

Ans (E)
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Solving with values

I. p^2 < 2p
When p = 1.1

II. p^2 = 2p
When p =2

III. p^2 > 2p
When p=2.1

I,II,III could be true
Hence, E is correct
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Ans: E

1)p2<2p
p(p-2)<0..so, 0<p<2..so if P can be 1.5 which is in the mentioned range 1<p<3

2)p2=2p
p(p-2)=0, so p=0 or p=2..so when p=2, it's in the mentioned range 1<p<3

3)p2>2p
p(p-2)>0..so, p>2 or p<0..so when P=2.5, it's in the mentioned range 1<p<3

so, all three could be true.
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Bunuel

Competition Mode Question



If 1 < p < 3, then which of the following could be true?

I. p^2 < 2p
II. p^2 = 2p
III. p^2 > 2p

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

If p = 1.5, we see that p^2 = 2.25 < 2p = 3. So I could be true.

If p = 2, we see that p^2 = 4 = 2p. So II could be true.

If p = 2.5, we see that p^2 = 6.25 > 2p = 5. So III could be true.

Answer: E
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