Solution for Questions GRE Analytical Problem 5, Part 2

Question IV

If S appears on a panel, that panel must consist of at least how many professors?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

C6: S => (NV)

And there are no more conditions that require more members.

Answer (A) 3

Question V

Which of the following is an acceptable group of professors for a panel?

(A) M, N, Q, R – violates C2: !(MNR)

(B) M, Q, R,T – is quite acceptable

(C) M, R, T, U  – violates C5: R <=> Q

(D) M, S, U, V – violates C6: S => (NV)

(E) N,R,T,U – violates C5: R <=> Q

Answer (A) M, N, Q, R

Question VI

Which of the following groups of professors can form an acceptable panel by doing nothing more than adding one more professor to the group?

(A) M, R, T

C5:R <=> Q

We must add Q and we’ll get an acceptable panel

(B) N,Q,M

C5:R <=> Q

We must add R/ But that panel would violate

C2: !(MNR)

(C) Q,R, S

According to

C6: S => (NV)

We must add both N and V

(D) Q,R,V

C4: V=>(MS) or (MU) or (SU) or (MSU)

We must add at least two members

(E) V, R, N

C4: V=>(MS) or (MU) or (SU) or (MSU)

We must add at least two members

Answer (A) M, R, T

Question VII

Of the group N, S, T, U, V, which professor will have to be removed to form an acceptable panel?

(A) N

(B) S

(C) T

(D) U

(E) V

This panel violates

C1:!(NTU)

So, we must remove one of them.

But the presence of N and V is required by S

C6: S => (NV)

Then, we must remove T

Answer (C) T

Solution for Questions GRE Analytical Problem 5, Part 1

Solution for Questions GRE Analytical Problem 5:

Let’s fix the conditions:

C1: N, T, and U cannot all appear on the same panel.

!(NTU)

In another form it can be written:

(NT)=>!V

(NU)=>!T

(TU)=>!N

C2: M. N, and R cannot all appear on the same panel.

!(MNR)

In another form it can be written:

(MN)=>!R

(MR)=>!N

(NR)=>!M

C3: Q and V cannot appear on the same panel.

!(QV)

In another form it can be written:

Q <=> !V

V <=> !Q

C4: If V appears on a panel, at least two professors of the trio M, S, and U must also appear on the panel.

V=>(MS) or (MU) or (SU) or (MSU)

Or, equivalent to this:

(!M!S) or (!M!U) or (!S!U)=>!V

C5: Neither R nor Q can appear on a panel unless the other also appears on the panel.

R <=> Q

From this follows:

!R <=> !Q

C6: If S appears on a panel, both N and V must also appear on that panel.

S => (NV)

Then,

!N or !V => !S

Now we can proceed to question solving.

Question I

Which of the following CANNOT appear on a panel with R?

(A) M

(B) N

(C) Q

(D) S

(E) T

C5: R <=> Q

C4: Q <=> !V

C6: !N or !V => !S

So, professor S CANNOT appear on a panel with R

Answer (D) S

Question II

Exactly how many of the professors can appear on a panel alone?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

C4: V=>(MS) or (MU) or (SU) or (MSU)

Professor V can’t appear on a panel alone.

C5: R <=> Q

Professor’s R and Q can’t appear on a panel alone.

C6: S => (NV)

Professor S can’t appear on a panel alone.

So, only four of them can.

Answer (D) 4

Question III

If neither V nor N appears on a panel, then which of the following must be true?

(A) M appears on the panel.

(B) Q appears on the panel.

(C) T appears on the panel.

(D) S does not appear on the panel.

(E) U does not appear on the panel.

C6: !N or !V => !S

Answer (D) S does not appear on the panel.

Math Problem 5

An apartment building has 5 floors, one of which has only 2 apartments. Each of the other floors has 4 apartments.

Compare two values:

Column A	Column B
3 times the number of floors in the building	The number of apartments in the building

Solution for GRE Analytical Problem 1:

Let’s name the patients B, C, D, E, F, G for their first letters. L will be used for lunch.

Let’s draw a table for sessions and fill it with known information:

• a lunch break at noon.
• Friedman will have the first session in the afternoon.

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	F

Then we’ll make some notes to fix other pieces of information:

• Dawson's session must be earlier in the day than Edson's.

D – E

• Bernstein's session must be earlier in the day than either Edson's or Friedman's.

B – E

B – F

• Cox's session must immediately precede Grant's session.

CG

What preliminary conclusion cam we make of this?

1. We need two adjacent sessions for C and G
2. E can’t be neither on the first, nor on the second session (because both B and D must be before him)
3. B can’t be neither on the last, nor on the penultimate session (because he must be followed by both E and F)
4. D can’t be the last, for he must be followed by E
5. F can’t be the first, for he must be preceded by B

Restrictions 2-5 can be shown graphically:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
E F	E				B	B D

 

Now, we can proceed to Question I.

“Bernstein is only available at 10:00” means, that B will in the second slot

9:00	10:00	11:00	12:00	1:00	2:00	3:00
	B		L	F

Now, looking at the table, we can see only one pair of adjacent sessions for CG: at 2:00 and 3:00

9:00	10:00	11:00	12:00	1:00	2:00	3:00
	B		L	F	C	G

Then, as E can take the session at 9:00, he’ll take 11:00 session:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
	B	E	L	F	C	G

And the session at 9:00 is for D

9:00	10:00	11:00	12:00	1:00	2:00	3:00
D	B	E	L	F	C	G

So, we have only one way to schedule sessions if B is on 10:00.

Answer A

Note: we have not to check if D at 9:00 fits the requirements, because there isn’t 0 among possible answers. Otherwise, to judge between 0 and 1 we’d had to prove that D at 9:00 doesn’t violate any requirement.

Question II alters initial data:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	C

Having place for C, we can see he place for G:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	C	G

Session at 3:00 can’t be taken by F (because he switched to the morning), by B or by D. That’s why the last session must be taken by E.

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	C	G	E

Since F can’t be the first, we have two possibilities for him:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
	F		L	C	G	E
		F

Then, we’ll place B before F

9:00	10:00	11:00	12:00	1:00	2:00	3:00
B	F		L	C	G	E
B		F
	B

And in every arrangement the place for D is clearly determined

9:00	10:00	11:00	12:00	1:00	2:00	3:00
B	F	D	L	C	G	E
B	D	F
D	B

Now we can see that in every case Dawson or Friedman has the third session of the day.

Answer D

We shall start solving Question III from our initial table with G added to 3:00 slot

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	F		G

The place for C is determined:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
			L	F	C	G

Now, looking to restriction table we find place for E:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
		E	L	F	C	G

And we have two variants for B and D

9:00	10:00	11:00	12:00	1:00	2:00	3:00
B	D	E	L	F	C	G
D	B

Now we can see, that answer “(D) Edson has her session at 10:00” can’t be true;

Answer: D

Note: We could find this answer merely by looking at our restriction table, thus, saving precious time.

In Question IV there are two places for C (and for G, then):

9:00	10:00	11:00	12:00	1:00	2:00	3:00
C	G		L	F
	C	G

Then, place for B is determined:

9:00	10:00	11:00	12:00	1:00	2:00	3:00
C	G	B	L	F
B	C	G

Now we can see, that possibilities (A)-(D) are not necessary true. Then, correct answer if “(E) Dawson has a session at 2:00”. Really, as 3:00 slot is restricted for D, he can be placed only to 2:00 slot, and 3:00 session must be for E

9:00	10:00	11:00	12:00	1:00	2:00	3:00
C	G	B	L	F	D	E
B	C	G

Answer: E