There are the real gems among the flash games. You can play great free online games, read their reviews and wathc walkthroughs on my new site GameGems.org. Welcome!

Saturday, August 20, 2011
Sunday, July 17, 2011
Smart Flash Games
The whole first half of 2011 I was involved in the creation of interesting flash games. So, I uploaded them to the web and created my portfolio blog, where you can play them and have fun.
The games, besides being entertaining, hepl the players train their memory, attention and reaction, which is significant for successful GRE exams.
Welcome to Smart Flash Games!
Sky Control
5 Brain Teasers
I Wanna Win!
Saturday, July 10, 2010
GRE Math Problem 78
In the preceding figure, if l1||l2, what is the value
of x?
(A) 36 (B)54 (C)90 (D)144 (E) 154
As the angle between the base of the triangle and the vertical line is not marked, the relation between x and y could be of any value. Answer:D
Wednesday, May 26, 2010
GRE Math Problem 77
Compare two values:
| Column A | Column B |
| x | y |
Solution for GRE Math Problem 76:
Then, angle b is two thirds of and angle between "11" and "12", then, it is two thirds of 30 degrees, which equals to 20 degrees.
Total 120+20=140 degrees
Answer: C
Thursday, May 13, 2010
GRE Math Problem 76
Compare two values:
Solution for GRE Math Problem 75:
As he made a 25% profit on one painting, he sold it for 125% of the original price.
$2000=125%
$1600=100% - its original price was $1600
As he had a 25% loss on the second painting, he sold it for 75% of the original price.
$2000=75%
$2666.(6)=100% - its original price was $2666.(6)
Last year Leo paid for both paintings $1600+$2666.(6)=$4266 and this year he got $4000 for selling them. So, he lost more than $100
Answer:(C) He lost more than $100.
| Column A | Column B |
| The measure, in degrees, of the smaller angle formed by the hour hand and the minute hand of a clock at 11:20 | 140 |
Solution for GRE Math Problem 75:
As he made a 25% profit on one painting, he sold it for 125% of the original price.
$2000=125%
$1600=100% - its original price was $1600
As he had a 25% loss on the second painting, he sold it for 75% of the original price.
$2000=75%
$2666.(6)=100% - its original price was $2666.(6)
Last year Leo paid for both paintings $1600+$2666.(6)=$4266 and this year he got $4000 for selling them. So, he lost more than $100
Answer:(C) He lost more than $100.
Thursday, May 6, 2010
GRE Math Problem 75
Last year Leo bought two paintings. This year he sold them for ">2000 each. On one, he made a 25% profit, and on the other he had a 25% loss. What was his net loss or profit?
(A) He broke even.
(B) He lost less than $100.
(C) He lost more than "$100.
(D) He earned less than $100.
(E) He earned more than $100.
Solution for GRE Math Problem 74:
By what fraction is
greater than
?
So,
of the number is 7. Therefore, the whole number is 12.
But we were asked about the
of this number, so, the answer is 12*3/5=20. Thank you for pointing this in comment :)
Answer:(D)20
(A) He broke even.
(B) He lost less than $100.
(C) He lost more than "$100.
(D) He earned less than $100.
(E) He earned more than $100.
Solution for GRE Math Problem 74:
By what fraction is
So,
But we were asked about the
Answer:(D)20
Tuesday, May 4, 2010
GRE Math Problem 74
If
of a number is 7 more then
of the number. What is
of the number?
(A) 12
(B) 15
(C) 18
(D) 20
(E) 24
Solution for GRE Math Problem 73:
In order to determine the total number of players we have to multiply the number of players in one team by the number of teams in one division by the number of divisions. We’ll get dtp
Answer: (B) dtp
(A) 12
(B) 15
(C) 18
(D) 20
(E) 24
Solution for GRE Math Problem 73:
In order to determine the total number of players we have to multiply the number of players in one team by the number of teams in one division by the number of divisions. We’ll get dtp
Answer: (B) dtp
Sunday, May 2, 2010
GRE Math Problem 73
The Center City Little League is divided into d divisions. Each division has t teams, and each team has p players. How many players are there in the entire league?
(A) d+t+p
(B) dtp
(C)
(D)
(E)
Solution for GRE Math Problem 72:
All the positive multiples of 5 less than 26 are: 5, 10, 15, 20, 25. Their average will be:
=1+2+3+4+5=15
All the positive multiples of 7 less than 26 are: 7, 14, 21. Their average will be:
=14
Answer:A
(A) d+t+p
(B) dtp
(C)
(D)
(E)
Solution for GRE Math Problem 72:
All the positive multiples of 5 less than 26 are: 5, 10, 15, 20, 25. Their average will be:
All the positive multiples of 7 less than 26 are: 7, 14, 21. Their average will be:
Answer:A
Friday, April 30, 2010
GRE Math Problem 72
Compare two values:
| 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 |
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
Answer: (C) 3
Friday, April 9, 2010
GRE Math Problem 67
Here
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
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