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

GRE Math Problem 65

a < b

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

GRE Math Problem 48

Compare:

Column A	Column B
5(r +t)	5r + t

Solution for GRE Math Problem 47:
x(x + y) + y(x + y) = (x+y)(x+y) = (x + y)²
Answer:C, they are equal

GRE Math Problem 47

Compare:

Column A	Column B
(x + y)2	x(x + y) + y(x + y)

Solution for GRE Math Problem 46:
(a + 2)(b + 3)=ab+3a+2b+6
(a + 3)(b + 2)=ab+2a+3b+6

Let's find the difference:
(a + 2)(b + 3)–(a + 3)(b + 2) = ab+3a+2b+6 – (ab+2a+3b+6) = a–b

 Since there are no restrictions for a and b, their difference can have any sign, then, answer is D, unable to determine
Answer:D

GRE Math Problem 46

Compare:

Column A	Column B
(a + 2)(b + 3)	(a + 3)(b + 2)

Solution for GRE Math Problem 45:
Just make calculations: 2*3+3*2=12
Answer: (C) 12

GRE Math Problem 45

6. If x = 3 and y = 2, then 2x + 3y =
(A) 5
(B)10
(C)12
(D)14
(E)15

Solution for GRE Math Problem 44:

The fraction , is bigger than if a > 2b and less than otherwise. The only fraction here which is less than is (E)
Answer: E

GRE Math Problem 43

If 3x-4y=5 and then what is x?
(A)-5y
(B)-5x
(C)1
(D)3
(E) 4

Solution for GRE Math Problem 42:

bd=3^ax3^c=3^a+c
Answer:(B)

GRE Math Problem 35

Given:
a²=b,
a>0

Compare two values:

Column A	Column B

Solution for GRE Math Problem 34:
To divide a number by a power of 10 we should remove as amany zeroes, as the respected power of ten has. So, the first fraction can be reduced by dividing the nominator and denominator by 10, and the second one - by 1000. The both will equal to
Answer: C, they are equal.

GRE Math Problem 28

Because her test turned out to be more difficult than she intended it to be, a teacher decided to adjust the grades by deducting only half the number of points a student missed. For example, if a student missed 10 points, she received a 95 instead of a 90. Before the grades were adjusted the class average was A. What was the average after the adjustment?
(A)
(B)
(C)
(D)
(E)A+25

Solution for GRE Math Problem 27:
The sum of all vertical segments in the path P-R-T equals to the vertical distance between P and T. The sum of all horizontal segments in the path P-R-T equals to the horizontal distance between P and T. The same for the path P-Q-T. So, these paths are equal.
Answer:C

GRE Math Problem 20

If x + y = a, y + z = b, and x + z = c. what is the average (arithmetic mean) of x, y, and z?
(A)
(B)
(C)
(D)
(E) a+b+c

Solution for GRE Math Problem 19:

Let X be Jordan's average of test scores after 5 tests. Then, he got total 5X points after 5 tests. If he gets 70 points after the next test, he'll have 5X+70 points. His average will be points. But it is given, that his average will be lower by 4 points. We have an equation:

5X+70=6X-24
X=94

Answer:(E) 94

GRE Math Problem 17

A business firm reduces the number of hours its employees work from 40 hours per week to 36 hours per week while continuing to pay the same amount of money. If an employee earned x dollars per hour before the reduction in hours, how much does he earn per hour under the new system?



A

B

C

D

E 9x

Solution for Analytical Problem 3:

Let’s fix the conditions:

The West Field must be planted with either barley or beans:

W: Ba OR Be

At least one field must be planted with corn.

Co: N OR E OR S

If, in the previous year, a field was planted with beans, then it must be planted with beans again.

Be –> Be

If, in the previous year, a field was planted with either wheat or alfalfa, then it must be planted with oats.

Wh –> Oa

Al –> Oa

To answer Question I we should first use condition 1 to cross out answer A (oats of the field for beans or barney). Then, we can see, that answers B and C have no place for corn. Then, in answer E one filed is planned for Alfalfa, which is not mentioned among four crops for this yesr. So, the answer is D.

Answer: D oats, barley, corn, beans.

To solve Question II, let’s draw a table:

	N	E	S	W
Year 1		Wh	Be
Year 2

We can determine the crops for Year 2 in E and S:

	N	E	S	W
Year 1		Wh	Be
Year 2		Oa	Be

The only place for corn left in the Year 2 is the North field, so, answer B must be true.

Answer: B The North Field is planted with corn.

Among the answers Question III we must look for one, which violates conditions for Western field. And we can see that if the Western field had been planted with alfalfa, then we’d have to plant neither barney, nor beans, but only wheat there the following year.

Answer: A alfalfa.

We’ll need a table again to arrange data from Question IV:

	N	E	S	W
Year 1	Oa	Oa	Oa	Co
Year 2

So, the crop in the Western must be changed. Besides, the crop in one of the other fields must be changed to corn. That’s why no less that two fields must be planted with crops that are different from those planted there for the previous year.

Answer: C two.

In Question V we should remember about the field for corn. The following year three fields will be planted with beans and the forth one – with corn. But the fourth one can’t be the Western field, because the Western field will be planted with beans or barney. The fourth field can’t be planted neither with wheat or beans, because it will be planted either with oats, or with beans the following year. So, we can see, that the fourth field was planted with oats.

Answer: C the fourth field was planted with oats.

Question VI can be solved with the help of the table:

	N	E	S	W
Year 1	Oa	Be	Al
Year 2

Using the initial conditions. We can determine the crops for Year 2:

	N	E	S	W
Year 1	Oa	Be	Al
Year 2	Co	Be	Oa	Ba OR Be

Answer: A Corn is planted in the North Field.

GRE Math Problem 14

Compare two values:

Column A	Column B
	x+y-7

Solution for GRE Math Problem 13:

Let's use formula for the squares of sum and difference:

(a+b)²=2²+2ab+b²

(a-b)²=a²-2ab+b²

It seems that column A is bigger by value of 4ab. But wait! What if ab<0? Then Column B is bigger. So, it is impossible to determine, which column is greater. Answer:D

GRE Math Problem 13

Given:

Compare two values:

Column A	Column B
(a+b)2	(a-b)2

Solution for GRE Math Problem 12:

The sum of those two angles of quadrilateral is 2x60=120^o. Since the sum of all four angles of quadrilateral is 360^o, the sum of other two angles is 360-120=240^o and their average is 240/2=120. So, the columns are equal.
Answer:C

GRE Math Problem 11

Given: x<y
Compare two values:

Column A	Column B
The average (arithmetic mean) of x and y	The average (arithmetic mean) of x, y, and y

Solution for GRE Math Problem 10:

In 1990 for twice less money we could buy twice more potatoes. So, the cost became 2x2=4 times less than in 1980. So, the price in 1990 was a quarter or 25% of the price in 1980. But the question is “By what percent did the price of potatoes decrease from 1980 to 1990?” Assuming price in 1980 as 100%, we get 100%–25%=75%
Answer:C

Math Problem 4

n is an odd positive integer

Compare two values:

Column A	Column B
The number of prime factors of n	The number of prime factors of 2n

Solution for GRE Math Problem 3:

Lets consider three possibilities: numbers x and y can be bot positive, negative and positive and 0 and 1. (They both can't be negative, because their sum is x+y=1>0).

In the first case, as x>0, y>0, x+y=1, we can say that x<1 and y<1, so their product will be less than 1.

In the second case, the product of positive and negative values is negative, so, less than one.

And for the pair (1;0), xy=0<1. That's why column B is always bigger.

Answer: B

Math Problem 3

Given: x+y=1

Compare two values:

Column A	Column B
xy	1

Solution for GRE Math Problem 2:

This question is a bit tricky. Of course, we can see, that 65% of $100 is $65, and 2/3 of $100 is $66.(6). But if a=0 then both values become equal. So, as no restrictions for a are given, in general we can't define which column is bigger.

Answer: D, it is impossible to determine which quantity is greater

Math Problem 2

Compare two values:

Column A	Column B
65% of a

Solution for GRE Math Problem 1:

Number 19 has only two positive divisors. They are numbers 1 and 19. So, their sum is 1+19=20, and their product is 1x19=19. Column A is bigger.

Answer: A