Situaciones en inglés

1) If a = (1/4)b and c = 7a, then which of the following represents the average (arithmetic mean) of a, b, and c, in terms of a ?

(A) a + 4

(B) (11/3)a

(C) 4a

(D) (4 1/7)a

(E) (7 1/4)a

ANS C

2) Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 12, 0.13, and 4.068 are three terminating decimals. If j and k are positive integers and the ratio j/k is expressed as a decimal, is j/k a terminating decimal?

(1) k = 3

(2) j is an odd multiple of 3.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

ANS C

3)Data Sufficiency
What is the value of a?
(1) a - b = 6
(2) b = a - 6

ANS E Why?

4) A sporting goods store purchased a number of baseballs and a number of basketballs. If the cost of each baseball was $2.50 and the cost of each basketball was $5.50, what was the total cost of the baseballs and basketballs purchased by the store?

(1) The number of baseballs purchased by the store was 1/3 the total number of baseballs and basketballs purchased by the store.

(2) The number of basketballs purchased by the store was twice as many as the number of baseballs purchased by the store.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

ANS E

5) For all numbers a and b, the operation @ is defined by a @ b = a^b + b. If x @ 3 = 2, then x =
(A) -3
(B) -2
(C) -1
(D) 1
(E) 2

ANS C

6) Is xy > 7 ?
(1) x + y = 7
(2) 1 ≤ x ≤ 3 and 2 ≤ y ≤ 4

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

ANS C

7) x bags of marbles are to be created, each consisting of y red marbles, z blue marbles, and nothing else. If the marbles are to be drawn from a total of 84 red marbles and 30 blue marbles and all marbles must be placed in one of the bags, what is the maximum number of bags that can be created?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

ANS D

8) Is the perimeter of square S greater than the circumference of circle C ?
(1) S is inscribed in circle C.
(2) The ratio of the area of S to the area of C is 2:pi.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

ANS D

9) Which of the following CANNOT yield an integer when divided by 7 ?
(A) The sum of three consecutive integers
(B) An integer with only even prime factors
(C) The product of two odd integers
(D) An integer divisible by 8
(E) An even integer

ANS B

10) What is the value of t^3 - m^3 ?
(1) t^2 - m^2 = 18
(2) t - m = 2

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

ANS C

CLASES PARTICULARES ONLINE GMAT QUANT GRE QUANT SAT QUANT
En clasesgmat.es impartimos clases y cursos OnLine para preparar tu GMAT QUANT, tu GRE QUANT and tu SAT QUANT, en ESPAÑA y el mundo.
Toma tus clases OnLine GMAT QUANT, GRE QUANT, SAT QUANT, desde cualquier parte de España:

Madrid, Valencia, Corcega, Barcelona, Sevilla, Mallorca, Tenerife, Ibiza, Granada, Málaga, Lanzarote, Palma de Mallorca, Gran Canaria, San Sebastian, Bilbao, Alicante, Marbella, Córdoba, Fuenteventura, Santiago de Compostela, Islas Baleares, Girona, Ronda, Costa Brava, Cádiz, Costa del Sol, Toledo, Benidorm, Las Palmas de Gran Canaria, Formentera, Jerez de La Frontera, Salamanca, Segovia, Zaragosa, La Gomera, Sitges, Tarragona, Murcia, Asturias, Pamplona, La Coruña, La Palma, Santa Cruz de Tenerife, Almería, Banalmadena, Nerja, Alcúdia.

Experimenta nuestro Open your Mind! Reserva tus clases OnLine hoy.
Las clases particulares OnLine para preparar tu GMAT QUANT, GRE QUANT y SAT QUANT, se emiten desde Chile para España y el mundo.

Las clases son en español con material en españo y en inglés (este úñtimo con material oficial, GMAT, GRE o SAT).

Solicita tus clases OnLine GMAT MATH, GRE MATH y SAT MATH al WhatsApp +569 455 17215. No esperes más y reserva tus clases ya!

Clases y curso OnLine GMAT parte matemática, GRE parte matemática y SAT parte matemática para España y el mundo, son impartidas por:

Claudio Hurtado Maturana
Ex docente de la Pontificia Universidad Católica de Chile.
Consultor razonamiento matemático
GMAT QUANT
GRE QUANT
SAT QUANT
Correo Electrónico clasesgmatchile@gmail.com
Sitio Web https://clasesgmat.es/
WhatsApp +569 455 17215
Dirección Avenida Hernando de Aguirre 128 Of 904, Providencia Santiago Chile.

EXPLANATION

1) Answer: C
The average of the three variables is (a + b + c)/3. However, we need to solve in terms of a, which means we must convert b and c into something in terms of a.

We’re told that a = (1/4)b, which is equivalent to b = 4a. We can plug that in and simplify the average to: (a + 4a + c)/3

We also know that c = 7a, which we can plug directly into the average expression:

(a + 4a + 7a)/3

= 12a/3 = 4a, choice (C).

2) Answer: C
Statement (1) is insufficient. If the denominator of the fraction is 3, the decimal would be terminating if the numerator is a multiple of 3. For instance, 6/3 = 2, a terminating decimal. However, if the numerator is not a multiple of 3, it will not be terminating, as in 7/3 = 2.33.

Statement (2) is also insufficient. The important factor in determining whether a fraction is equivalent to a terminating decimal is the denominator. If j = 9, the fraction could be 9/3 (terminating) or 9/7 (not terminating).

Taken together, the statements are sufficient. j/k is equal to (3(integer))/3 = integer. An integer is, as defined in the question itself, a terminating decimal. Choice (C) is correct.

3)

4) Answer: E
Call the number of baseballs s, the number of basketballs k, and the total cost of the balls t. Ths, the total cost of the balls purchased is:
2.5s + 5.5k = t
Statement (1) is insufficient. It tells us that s = (1/3)(s + k). Combined with the equation given in the question, we have two equations and three variables. That’s not enough to solve.

Statement (2) is also insufficient. It says that k = 2s. Again, combined with the equation in the question, we have two equations and three varaibles.

Taken together, the statements are still insufficient. It may look like we have three equations now, but the equations given by the two statements turn out to be equivalent. Simplify the equation from (1):
s = (1/3)(s+k)
3s = s + k
2s = k
That’s the same as the equation in (2). Thus, we only have two equations, but there are still three variables (s, k, and t). Choice (E) is correct.

5) Answer: C
Since a @ b = a^b + b, x @ 3 = x^3 + 3. Now we can turn to more traditional algebra:

x^3 + 3 = 2

x^3 = -1

x = -1, choice (C).

6) Answer: C
Statement (1) is insufficient: if x = 1 and y = 6, xy is less than 7; if x = 3 and y = 4, xy is greater than 7.

Statement (2) is also insufficient: if you take the smallest possible numbers, x = 1 and y = 2, xy is less than 7; if you take the largest possible numbers, x = 3 and y = 4, xy is greater than 7.

Taken together, the statements are sufficient. Given the ranges for x and y in (2), the only possible pair of numbers that sums to 7 is x = 3 and y = 4. If you choose a smaller value for x, y would have to be bigger, and (2) precludes that possibility. The same is true if you try a smaller value for y. Thus, there’s only one possible pair of numbers for x and y, and the product is greater than 7. Choice (C) is correct.

7) Answer: D
Important to note is that all marbles must be used. The question gives variables for the numbers of red and blue marbles in each bag to indicate that the number of each color marble in each bag is equal, so if one bag has, say, 7 blue marbles, every bag has 7 blue marbles.

This drastically reduces our options. For instance, the number of bags must be a factor of 84, limiting us to 1, 2, 3, 4, 6, 7, 12, 21, 28, 42, and 84. That’s still quite a few options, but of course not all of these will work, since we also have to consider the 30 blue marbles. The number of bags also must be a factor of 30: 1, 2, 3, 5, 6, 10, 15, or 30.

The number of bags, then, must be a member of both of those lists: 1, 2, 3, or 6. We’re looking for the maximum number of bags that could be created, so we take the largest number from the list, 6, choice (D).

8) Answer: D
Statement (1) is sufficient: a square inscribed in a circle always has the same relationship with the circle. The diagonal of the square is the diameter of the circle, so you can work out the exact relationship and determine the ratio between the sizes of the figures, which allows you to answer the question.

Statement (2) is also sufficient: if you have the ratio of the areas, you can determine the ratio of the side of the square to the radius of the circle, from which you could compare the perimeter and the circumference of the figures. Choice (D) is correct.

9) Answer: B
Consider each choice. (A) is possible; for instance, 6 + 7 + 8 = 21. (B) is impossible, so it is correct. An integer with only even prime factors cannot have 7 as a prime factor; if it doesn’t have 7 as a prime factor, it can’t yield an integer when divided by 7. (C) is possible: If either of the odd integers is 7 or divisible by 7, the result is also divisible by 7. (D) is possible: for example, 56 is divisible by both 7 and 8. Finally, many even integers are divisible by 7 (28, 42, etc.), so (E) is possible as well. (B) is the correct choice.

10) Answer: C
Within the scope of GMAT math, there is no way to simplify t^3 - m^3, so in order to answer the question, we’ll need the values of both t and m.

Statement (1) is insufficient. We can factor and find that (t + m)(t - m) = 18, but that doesn’t give us what we need.

Statement (2) is also insufficient. Two variables and one equation isn’t enough to solve for the variables.

Taken together, the statements are sufficient. If t - m = 2, we can substitute that into the factored version of (1):
(t + m)(2) = 18
t + m = 9
Now we have two equations and two variables:
t + m = 9
t - m = 2
Add the equations:
2t = 11
t = 5.5
From there, we can find m and answer the question. No need to do the rest of the math, and the answer is (C).