Column A | Column B |
The average (arithmetic mean) of all the positive multiples of 5 less than 26 | The average (arithmetic mea: of all the positive multiples of 7 less than 26 |
Friday, April 30, 2010
GRE Math Problem 72
Compare two values:
Tuesday, April 27, 2010
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
Monday, April 26, 2010
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.
Monday, April 19, 2010
Analytical Problem 5
The planning committee of an academic
conference is planning a series of panels using eight professors, M, N, Q, R, S,
T, U, and V. Each panel must be put together in accordance with the following conditions:
N, T, and U cannot all appear on the
same panel.
M. N, and R cannot all appear on the
same panel.
Q and V cannot appear on the same panel.
If V appears on a panel, at least two
professors of the trio M, S, and U must also appear on the panel.
Neither R nor Q can appear on a panel
unless the other also appears on the panel.
If S appears on a panel, both N and
V must also appear on that panel.
Question I
Which of the following CANNOT appear
on a panel with R?
(A) M
(B) N
(C) Q
(D) S
(E) T
Question II
Exactly how many of the professors can
appear on a panel alone?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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.
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
Question V
Which of the following is an acceptable
group of professors for a panel?
(A) M, N, Q, R
(B) M, Q, R,T
(C) M, R, T, U
(D) M, S, U, V
(E) N,R,T,U
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
(B) N,Q,M
(C) Q,R, S
(D) Q,R,V
(E) V, R, N
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
Solution for GRE Math Problem 71:
Peter's
grade > Victor's grade
Georgette’s
grade > Victor's grade
We can’t
determine, which of these grades is higher
Answer:D
Tuesday, April 13, 2010
GRE Math Problem 71
Peter's grade was higher than that of Victor.and Victor's grade was less than that of Georgette.
Solution for GRE Math Problem 70:
Area=length x width
New Area=length x (1.25 width) = 1.25 (length x width) = 1.25 Area
Area will increase by 25%.
Answer:(A) 25%
Column A | Column B |
Georgette's grade | Peter's grade |
Solution for GRE Math Problem 70:
Area=length x width
New Area=length x (1.25 width) = 1.25 (length x width) = 1.25 Area
Area will increase by 25%.
Answer:(A) 25%
Monday, April 12, 2010
GRE Math Problem 70
If the width of a rectangle is increased by 25% while the length remains constant, the resulting area is what percent of the original area?
(A) 25%
(B)75%
(C) 125%
(D)225%
(E) Cannot be determined from the information given.
Solution for GRE Math Problem 69:
At what rate is the water filling the tank?
800-300=500 cubic feet
How many minutes will it take to completely fill a water tank?
3,750:500 = 7.5 minutes = 7 min. 30 sec.
Answer:(C) 7 min. 30 sec.
Sunday, April 11, 2010
GRE Math Problem 69
How many minutes will it take to completely fill a water tank with a capacity of 3,750 cubic feet if the water is being pumped into the tank at the rate of 800 cubic feet per minute and is being drained out of the tank at the rate of 300 cubic feet per minute?
(A) 3 min. 36 sec.
(B) 6 minutes
(C) 7 min. 30 sec.
(D) 8 minutes
(E) 1,875 minutes
Solution for GRE Math Problem 68:
The legs of triangle BCE equal to the sides of the square ABCD. So, its area is the half of the area of ABCD. Thus, the area of ABCD is 16
Answer:(C) 16
Saturday, April 10, 2010
GRE Math Problem 68
If the area of the triangle BCE is 8, what is the area of the square ABCD?
(A) 4
(B) 8
(C) 16
(D) 22
(E) 82
Solution for GRE Math Problem 67:
As and , then PQRS is a parallelogram. And as PQ=QR, it is a rhomb. Its obtuse angle equals 120 degrees, and the acute angle is 60 degrees. Then, triangle PQR is equilateral and PQ=QR=QP=3
Answer: (C) 3
Friday, April 9, 2010
GRE Math Problem 67
Here and .
If PQ=3 and QR=3, then what is the length of PR?
(A)
(B)
(C) 3
(D)
(E)
Solution for GRE Math Problem 66:
As the y-coordinates of A and B are equal, the distance between them is 2-(-4)=6. That’ll be the diameter of the circle. The radius of the circle is 3. Its area will be
Answer:(C) 9
Thursday, April 8, 2010
GRE Math Problem 66
If A is the point (-4, 1) and B is the point (2, 1), what is the area of the circle which has AB as a diameter?
(A) 3
(B) 6
(C) 9
(D) 12
(E) 36
Solution for GRE Math Problem 65:
1 < ab < 2
2 < a+b < 4
So, ab < 2 < a+b
Answer:B
(A) 3
(B) 6
(C) 9
(D) 12
(E) 36
Solution for GRE Math Problem 65:
1 < ab < 2
2 < a+b < 4
So, ab < 2 < a+b
Answer:B
Wednesday, April 7, 2010
GRE Math Problem 65
a < b
Solution for GRE Math Problem 64:
As we can see, the circle whose diameter is a is located within the square whose side is a for any value of a. So, the square’s area is greater.
Answer: A
Column A | Column B |
ab | a+b |
Solution for GRE Math Problem 64:
As we can see, the circle whose diameter is a is located within the square whose side is a for any value of a. So, the square’s area is greater.
Answer: A
Tuesday, April 6, 2010
GRE Math Problem 64
a<1
Solution for GRE Math Problem 63:
As
then
Let’s find the difference
Then c
Answer:B
Column A | Column B |
The area of a square whose side is a | The area of a circle whose diameter is a |
Solution for GRE Math Problem 63:
As
then
Let’s find the difference
Then c
Answer:B
Monday, April 5, 2010
GRE Math Problem 63
c and d are positive
Solution for GRE Math Problem 62:
Point (a,b) lays in the quarter where x<0 and y>0. So, a<0 and b>0 and b>a
Answer:B
Column A | Column B |
c | d |
Solution for GRE Math Problem 62:
Point (a,b) lays in the quarter where x<0 and y>0. So, a<0 and b>0 and b>a
Answer:B
Sunday, April 4, 2010
GRE Math Problem 62
Column A | Column B |
a | b |
Solution for GRE Math Problem 61:
326(31)-326(19) =326(31-19)=326(12) – here we need not make any calculations, because there is only one answer which is more than 10 times greater than 326
Answer:(A)3,912
Saturday, April 3, 2010
GRE Math Problem 61
326(31)-326(19) =
(A)3,912
(B)704
(C)100
(D)326
(E)10
Solution for GRE Math Problem 60:
for x>0
Answer:B
Friday, April 2, 2010
GRE Math Problem 60
x>0
Solution for GRE Math Problem 59:
The perimeter of the square is
Answer:(C) 3x + 4
Column A | Column B |
Solution for GRE Math Problem 59:
The perimeter of the square is
Answer:(C) 3x + 4
Thursday, April 1, 2010
GRE Math Problem 59
The length of each side of a square is
What is the perimeter of the square?
(A) x + 1
(B) 3x + 1
(C) 3x + 4
(D)
(E) It cannot be determined from the information given.
Solution for GRE Math Problem 58:
How much, in dollars, will receive the second one?
240-180 = 60
What is the difference between the amounts received by the two persons?
180-60=120
Answer: (C)$120
What is the perimeter of the square?
(A) x + 1
(B) 3x + 1
(C) 3x + 4
(D)
(E) It cannot be determined from the information given.
Solution for GRE Math Problem 58:
How much, in dollars, will receive the second one?
240-180 = 60
What is the difference between the amounts received by the two persons?
180-60=120
Answer: (C)$120
Subscribe to:
Posts (Atom)