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

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

GRE Math Problem 58

A prize of $240 is divided between two persons. If one person receives $180, then what is the difference between the amounts received by the two persons?
(A) $30
(B) $60
(C)$120
(D)$210
(E)$420


Solution for GRE Math Problem 57:

Let’s take three consecutive integers: n-1, n, n+1
Their sum is n-1 + n + n+1 = 3n – is a multiple of 3.
Among the numbers given, only 28 is not the multiple of 3.


Answer:(C) 28

GRE Math Problem 57

Which of the following CANNOT be expressed as the sum of three consecutive integers?
(A) 18
(B) 24
(C) 28
(D) 33
(E) 36

Solution for GRE Math Problem 56:
What was Camille's total score?
75x6=450

What was Camille's total score after her teacher had dropped her lowest grade?
85x5=425

What was her lowest grade?
450-425=25
Answer:(B) 25

Analytical Problem 4

John purchased a toy for his young daughter. The toy included nine colorful plastic pieces in the shape of the numbers from 1 to 9. In examining these pieces he made the following observations:

• One piece is purple and two pieces each are red, green, yellow, and blue.
• The two red pieces are consecutive numbers.
• The number 4 is green.
• The two blue pieces are not consecutive numbers.
• Both the 1 and the 9 are yellow.
• The purple piece is not a number immediately greater than or less than either green piece.

Question I. If one of the red pieces is the number 3, what number is the other red piece?
(A) 2   (B) 4   (C) 5   (D) 6   (E) 7

Question II. If the number 5 is green, each of the following could be true EXCEPT

(A) The purple piece and one of the yellow piece are consecutive numbers.

(B) The number 6 is purple.

(C)  The two blue pieces and the purple piece are three consecutive numbers.

(D) The number 2 is red.

(E) The number 8 is red.

Question III. If the number 6 is green, which of the following could be true?

(A) The number 2 is blue.

(B) The number 3 is purple.

(C) The number 5 is red.

(D)The number 5 is purple.

(E) The number 7 is blue.

Question IV.Which of the following, if true, would enable you to determine the color of every number?

(A) The 2 is blue.

(B) The 3 is blue.

(C) The 5 is red.

(D)The 7 is blue.

(E) The 7 is green.

Solution for GRE Math Problem 51:
If we take the second researcher's time as 100%, the first one's time will be 75%. Therefore, the ratio is 75:100=3:4. So, we should divide 35 into the ratio of 3:4. We'll get 15 and 20, their difference is 20-15=5<7
Answer:B

GRE Math Problem 49

m and n are positive integers mn = 25

Column A	Column B
m	n

Solution for GRE Math Problem 48:
Putting r=0, t=0 we have 0=0. Putting r=1, t=1 we have 10 > 6
Answer: D, unable to determine

GRE Math Problem 38

A number is a palindrome if it reads exactly the same from right to left as it does from left to right.
For example, 959 and 24742 are palindromes.

Column A	Column B
The probability that a three-digit number chosen at random is a palindrome	0.1

Solution for GRE Math Problem 37:
If all 16 shirts had costed $10, total amount of money spent would have been $160. It is $27 less, than the actual amount. Since Each $13 shirt is $3 more expensive than $10 shirt, there were 9 $13 shirts and 7 $10 shirts
Answer: A

GRE Math Problem 37

A man buys 16 shirts. Some of them cost $13 each, while the remainder cost $10 each. The cost of all 16 shirts is $187.

Column A	Column B
The number of $13 shifts purchased	The number of $10 shirts purchased

Solution for GRE Math Problem 36:

If all the tickets hab been sold for $25, the total sum would be 25x11=$275, which is 275-227=.$48 less, than the actual amount of money. Since the $13 ticket is 25-13=$12 cheaper, theree were = 4 tickets for $13 and 11-4=7 tickets for $25.
 Answer: Column A is bigger

GRE Math Problem 36

Tickets to a concert cost $25 and $13. An agent sells 11 tickets for a total price of $227.

Column A	Column B
The number of $25 tickets sold	The number of $13 tickets sold

Solution for GRE Math Problem 35:
If a²=b, then .
Now, if a=1, then Column A = 2 > Column B = 1.
And if  a=2, then Column A = 1 < Column B = 4.
Answer: D, unable to define.

GRE Math Problem 31

Compare two values:

Column A	Column B
The remainder when 10100 is divided by 575	The remainder when 10100 is divided by 755

Solution for GRE Math Problem 30:
The average of three digits is 2 means, that their sum is 6. There are 7 tripples of digits with the sum of 6. they are: 6,0,0; 5,1,0; 4,2,0; 4,1,1; 3,3,0; 3,2,1; 2,2,2.
6,0,0 gives only one 3-digit number: 600.
For 5,1,0 we have 510, 501, 150, 105.
For 4,2,0: 420, 402, 240, 204.
For 4,1,1: 411, 141, 114.
For 3,3,0: 330, 303.
For 3,2,1: 321, 312, 231, 213, 123, 132.
For 2,2,2: 222.
Total 21 numbers.
Answer: Column A is greater.

GRE Math Problem 30

Compare two values:

Column A	Column B
The number of positive three-digit numbers for which the average (arithmetic mean) of the three digits is equal to 2	20

Solution for GRE Math Problem 29:

Answer:(E)

GRE Math Problem 26

The sum of four consecutive odd positive integers is always
(A) an odd number
(B) divisible by 4
(C) a prime number
(D) a multiple of 3
(E) greater than 24

Solution for GRE Math Problem 25:

And again we have the right triangle with legs of 180 and 240. And again we don't have to get square root from 180²+240². We can notice, that 180=3x60 and 240=4x60. So, it is an Egyptian triangle with hypotenuse of 5x60=300 yards.
Answer:(B) 300 yards

GRE Math Problem 22

Compare two values:

Column A	Column B
The units digit of 1414	The units digit of 1616

Solution for GRE Math Problem 21:

The area of the circle with the radius of R is R². To cet the are of the ring, we should substract the area of the inner circle from the area of the outer circle. So, the are of the smallest circle is . The area of the whole figure is x4²=16. The area aof the yellow resion is x3²-x1²=8. It is exactly the half of the whole figure's area. It means, that the area of the shaded resion equals the area of the yellow region.
Answer:C

GRE Math Problem 19

Jordan has taken 5 math tests so far this semester. If he gets a 70 on his next test, it will lower the average (arithmetic mean) of his test scores by 4 points. What is his average now?
(A) 74 (B) 85 (C) 86 (D) 90 (E) 94

Solution for GRE Math Problem 18:

According to the trianglae unequation, if a triangle has sides of a and (a > =b), it is true for the third side, c, that a-b < c < a+b. Then, among the possible values for c, only (II) 11 fits the tringle unequality: 10-9 < 11 < 10+11.

Answer:(C) II only

GRE Math Problem 16

Compare two values:

Column A	Column B
2468	8+6x10+4x102+2x103

Solution for GRE Math Problem 15:

If b books can be purchased for d dollars, then of a book can be purchased for 1 dollar. Then, for m dollars, books can be purchased.



Answer:A

Math Problem 8

A person is standing on a staircase. He walks down 4 steps, up 3 steps, down 6 steps, up 2 steps, up 9 steps, and down 2 steps. Where is he standing in relation to the step on which he started?
(A) 2 steps above
(B) 1 step above
(C) the same place
(D) 1 step below
(E) 2 steps below

Solution for GRE Math Problem 7:

Let's abstract from costs, profits and other financial terms. We just have two equations:
p=4
c+p=20
c - ?
Then,
c=20-p
c=16

Answer:(C)$16

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