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

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

GRE Math Problem 15

If b books can be purchased for d dollars, how many books can be purchased for m dollars?

1. 
2. bdm
3. 
4. 
5. 

Solution for GRE Math Problem 14:

By opening the brackets we get x+y+7 in Column A, which is bigger than x+y-7 in Column B.


Answer:A

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 12

Given: The average (arithmetic mean) of the measures of two angles of a quadrilateral is 60°.

Compare two values:

Column A	Column B
The average of the measures of the other two angles	120°

Solution for GRE Math Problem 11:

Average of x and y lies exactly in the middle between x and y. Average of x, y and y lies between x and y, but closer to y. As y is bigger than x, columb B is greater.
Answer:B

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 10

In 1980, the cost of p pounds of potatoes was d dollars. In 1990, the cost of 2p pounds of potatoes was dollars. By what percent did the price of potatoes decrease from 1980 to 1990?

(A) 25% (B) 50% (C) 75% (D) 100% (E) 400%

Solution for GRE Analytical Problem 2:

We’ll name countries for their first letters: B, C, D, E, F, G. Now we shall make some preliminary conclusion from the conditions.

1. At least one film will be shown each day.
2. No more than three films will be shown on any two consecutive days.

 

Having 6 films and 4 days, we’ve got a math problem here:
In how many ways can we get 6 out of 4 positive integers, with any two consecutive terms not bigger than 3?
If the first number is 2, then the second must be 1. The third and the forth numbers can be 1+2 or 2+1.
So, we have two ways for now: 2+1+1+2 and 2+1+2+1.
And if the first number is 1, the second number can’t 1 (otherwise the sum of the third and the fourths numbers would be 4). So, only one possibility is there: 1+2+1+2

Now we have three possible patterns for showing films during the festival:
1+2+1+2
2+1+1+2
2+1+2+1
They’ll be useful during our further solution

3. The Belgian film must be shown on Saturday.

 

It’s time to start filling the table:

Thu	Fri	Sat	Sun
		B

4. The Canadian film must be shown on the same day as another film.

C+X

5. The French film must be shown on a day before the German film is shown.

F – G

6. The Danish film must be shown on a day after the English film is shown.

E – D

IMPORTANT Note: phrase “on a day after” shows that any number of days can pass between English and Danish films. To indicate two consecutive days phrases like “on a day immediately after”, “on a next day” etc.

And one more to do in the preparatory phase is to make a restriction table:

Thu	Fri	Sat	Sun
G D			F E

Now we may proceed to answering questions.

Question I

B cant’ be shown on Thu, for it has already been scheduled to Sat. C must be accompanied by another film, so, it can’t be shown on Thu, either. D and G are in our restriction table for Thu. So, only F remaining.

Answer (D) France

Question II

If F is on Sat, the G is on Sun. The only pattern with 2 films on Sat is 2+1+2+1, so, C must be on Thu, so must E. And В will be shown on Fri

Thu	Fri	Sat	Sun
C E	D	B F	G

Now we can see, that (B) The film from Denmark will be shown on Friday is the only possible answer.

Answer (B)

Question III

With two films on Thu and two on Sun, we have 2+1+1+2 pattern. So, C can’t be shown on Fri. Neither does B, because it is shown on Sat. And any of other four films may be shown on Friday.
D:

Thu	Fri	Sat	Sun
E F	D	B	G C

E:

Thu	Fri	Sat	Sun
F C	E	B	D G

F:

Thu	Fri	Sat	Sun
E C	F	B	D G

G:

Thu	Fri	Sat	Sun
F E	G	B	D C

Answer (D) 4

Question IV

If the English and German films are shown on the same day, we have

F – E+G –D

Thus, E+G can’t be shown neither on Thu, nor on Sun. These films can’t be shown on Sat, because in that case we would have three films then. So, E+G must be on Fri and we have 1+2+1+2 pattern.

Thu	Fri	Sat	Sun
	E G	B

F must be on Thu, D and C – on Sun

Thu	Fri	Sat	Sun
F	E G	B	C D

That’s why answer (E) “he film from Canada will be shown on Sunday” is the only true possibility.

Answer (E)

Question V

If the film from England is shown on Saturday, we have 2+1+2+1 pattern.

Thu	Fri	Sat	Sun
		B E

Sunday slot must be taken by D.

Thu	Fri	Sat	Sun
		B E	D

C must be shown on Thursday:

Thu	Fri	Sat	Sun
C		B E	D

At last, F and G will take Thu and Fri slots.

Thu	Fri	Sat	Sun
C F	G	B E	D

Now we can see, that the film from Germany cannot be shown on the same day as any other film.

Answer (E) The film from Germany.

Analytical problem 2

The Fairfield Foreign Film Festival plans to show six films—one each from Belgium, Canada, Denmark, England, France, and Germany. The festival will open on a Thursday and close three days later on a Sunday. Each film will be shown once during the four days of the festival.

The order in which the films will be shown must adhere to the following conditions:

1. At least one film will be shown each day.
2. No more than three films will be shown on any two consecutive days.
3. The Belgian film must be shown on Saturday.
4. The Canadian film must be shown on the same day as another film.
5. The French film must be shown on a day before the German film is shown.
6. The Danish film must be shown on a day after the English film is shown.

Question I
If only one film is shown on Thursday, it could be the entry from which of the following countries?
(A) Belgium
(B) Canada
(C) Denmark
(D) France
(E) Germany

Question II
If the Belgian and French films are shown on the same day, which of the following must be true?
(A) The film from Denmark will be shown on
Thursday.
(B) The film from Denmark will be shown on
Friday.
(C) The film from Germany will be shown on
Thursday.
(D) The film from Canada will be shown on
Sunday.
(E) The film from England will be shown on
Sunday.

Question III
If the director of the festival decides to show two films on Thursday and two on Sunday, how many films would be eligible to be shown on Friday?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

Question IV
If the English and German films are shown on the same day, which of the following must be true?
(A) The film from England will be shown on
Thursday.
(B) The film from France will be shown on Friday.
(C) The film from Canada will be shown on
Saturday.
(D) The film from Germany will be shown on
Saturday.
(E) The film from Canada will be shown on
Sunday.

Question V
If the film from England is shown on Saturday, which of the following films cannot be shown on the same day as any other film?
(A) The film from Belgium.
(B) The film from Canada.
(C) The film from England.
(D) The film from France.
(E) The film from Germany.

Solution for GRE Math Problem 9:

"ratio of the number of insects in a given population having characteristic X to the number of insects in the population not having characteristic X is 5:3" means that of population has characteristics X, and of the population has not characteristics X.

To find, what proportion of the total insect population are male insects having the characteristic X, we must multiply the fraction of X-insects in the whole population by the fraction of males in X-insects.

Answer: D

Math Problem 9

A student conducts an experiment in biology lab and discovers that the ratio of the number of insects in a given population having characteristic X to the number of insects in the population not having characteristic X is 5:3, and that of the insects having characteristic X are male insects. What proportion of the total insect population are male insects having the characteristic X?

(A)1
(B)
(C)
(D)
(E)

Solution for GRE Math Problem 8:

Total steps down: 4+6+2=12
Total steps up: 3+2+9=14
So, he is standing by 14-12=2 steps above

Answer:(A) 2 steps above

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 7

If the profit on an item is $4 and the sum of the cost and the profit is $20, what is the cost of the item?
(A)$24
(B)$20
(C)$16
(D)$12
(E) Cannot be determined from the information given.

Solution for GRE Math Problem 6:

From 6:15 to 7:15 the minute hand will drew a complete circle, 360°. Then, from 7:15 to 7:45, it drew a half of a circle, 180°. Total: 360+180=540

Answer:(D) 540°

Math Problem 6

From the time 6:15 p.m. to the time 7:45 p.m. of the same day, the minute hand of a standard clock describes an arc of
(A) 30°
(B) 90°
(C) 180°
(D) 540°
(E) 910°

Solution for GRE Math Problem 5:

Value in column A is 5x3=15. Value in column B is 2+4x4=18. therefore, column B is greater.

Answer:B

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

Analytical problem 1

Dr. William Weinberg is a psychiatrist who has to schedule one therapy session for each of six patients— Bernstein, Cox, Dawson, Edson, Friedman, and Grant. Each session will take place on the same day. The sessions will be scheduled at 9:00, 10:00, 11:00, 1:00, 2:00, and 3:00, with a lunch break at noon.

In making up the schedule for the session, Dr. Weinberg will conform to the following conditions:

• Friedman will have the first session in the afternoon.
• Dawson's session must be earlier in the day than Edson's.
• Bernstein's session must be earlier in the day than either Edson's or Friedman's.
• Cox's session must immediately precede Grant's session.

Question I. If Bernstein is only available at 10:00, in how many ways can the sessions be scheduled?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 6

Question II. If Friedman switches to a morning session, and Cox takes the 1:00 spot vacated by Friedman, which of the following must be true?

(A) Bernstein has the first session of the day.

(B) Bernstein or Friedman has the second session

of the day.

(C) Bernstein or Dawson has the second session of

the day.

(D) Dawson or Friedman has the third session of

the day.

(E) Dawson or Grant has the last session of the

day.

Question III. If Grant has the last session of the day, each of the following could be true EXCEPT

(A) Bernstein has the first session of the day.

(B) Dawson has the first session of the day.

(C) Dawson has his session at 10:00.

(D) Edson has her session at 10:00.

(E) Cox has his session at 2:00.

Question IV. If Cox's session is earlier in the day than Friedman's, which of the following must be true?

(A) Bernstein has the first session of the day.

(B) Cox has the first session of the day.

(C) Bernstein has a session at 11:00.

(D) Grant has a session at 11:00.

(E) Dawson has a session at 2:00.

Solution for GRE Math Problem 4:
Number 2n has all the prive factors, which number n has. Additionaly, as number n is odd, prime number 2 is not among its prime factors, but number 2n has 2 among it prime factors. So, the number of prime factors of 2n is bigger than the number of prime factors of n.

Answer: B

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

Math Problem 1

Compare two quantities.

Column A	Column B
The sum of the positive divisors of 19	The product of the positive divisors of 19

For the questions like this the correct answer is
A if the quantity in Column A is greater

B if the quantity in Column B is greater

C if the two quantities are equal

D if it is impossible to determine which quantity is greater