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 21

The figure below consists of four circles with the same center. The radii of the four circles are 1, 2, 3, and 4.
Compare two values:

Column A	Column B
The area of the shaded region	The area of the yellow region

Solution for GRE Math Problem 20:
Since x + y = a, y + z = b, and x + z = c, a+b+c=2x+2y+2z=2(x+y+z). So, x+y+z=. And the average of x, y and x is
Answer:(D)

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 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 18

If the lengths of two of the sides of a triangle are 9 and 10, which of the following could be the length of the third side?

I. 1 II. 11 III. 21

(A) None

(B) I only

(C) II only

(D) I and II only

(E) I, II, and III

Solution for GRE Math Problem 17:

The amount of money, which employee earned per week is M=40x. As it left the same after the reduction of hours, the new amount of money earned per hour is
Answer:D

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.

Analytical Problem 3

A farm is divided into four fields designated as the North Field, the East Field, the South Field, and the West Field. For the new growing season, each of the fields will be planted with exactly one of four crops—corn, beans, barley, or oats— according to the following plan:

• The West Field must be planted with either barley or beans.
• At least one field must be planted with corn.
• If, in the previous year, a field was planted with beans, then it must be planted with beans again.
• If, in the previous year, a field was planted with either wheat or alfalfa, then it must be planted with oats.

 

Question I. Which of the following could be the crops planted?

	North	East	South	West
A	beans	corn	barley	oats
B	beans	barley	oats	beans
C	wheat	barley	oats	beans
D	oats	barley	corn	beans
E	alfalfa	corn	beans	beans

Question II. If, in one year, the East Field is planted with wheat and the South Field is planted with beans, which of the following must be true of the year immediately following?

(A) The North Field is planted with barley.

(B) The North Field is planted with corn.

(C) The West Field is planted with barley.

(D) The South Field is planted with oats.

(E) The East Field is planted with beans.

Question III. In the previous growing season, the West Field could have been planted with any of the following crops EXCEPT

(A) alfalfa

(B) beans

(C) barley

(D)oats

(E) corn

Question IV. If, in the previous growing season, the West Field was planted with corn and the other three fields were planted with oats, what is the minimum number of fields that must be planted with crops that are different from those planted there for the previous growing season?

(A) None

(B) One

(C) Two

(D) Three

(E) Four

Question V. If three of the four fields were planted with beans for the previous growing season, which of the following could also have been true of the previous growing season?

(A) Barley was planted in the West Field.

(B) Corn was planted in the West Field.

(C) The fourth field was planted with oats.

(D) The fourth field was planted with wheat.

(E) The fourth field was planted with beans.

Question VI. If, for the previous growing season, the North Field was planted with oats, the East Field was planted with beans, and the South Field was planted with alfalfa, which of the following must be true of the new growing season?

(A) Corn is planted in the North Field.

(B) Oats are planted in the North Field.

(C) Barley is planted in the East Field.

(D) Barley is planted in the West Field.

(E) Beans are planted in the West Field.

Solution for GRE Math Problem 16:

This problem doesn't need much calculation. Knowing that 2468=2000+400+60+8=2x³+4x10²+6x10++8 we can say that the columns are equal


Answer:C