Algebra - GMAT Quantitative

Card 0 of 1488

Question

True or false: $\left { a_{n} \right }$ . is an arithmetic sequence.

Statement 1: $a_{4} - a_{3} = a_{2} -a_{1}$

Statement 2: $a_{6}- a_{5} = a_{8} - a_{7}$

Answer

An arithmetic sequence is one in which the difference of each term in the sequence and the one preceding is constant.

The two statements together demonstrate that two such differences are equal to each other, and that two other such differences are equal to each other. No information, however, is given about any other differences, of which there are infinitely many. Therefore, the question of whether the sequence is arithmetic or not is unresolved.

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Question

True or false:

Statement 1:

Statement 2: $x^{2}< 81$

Answer

Assume Statement 1 alone. can be rewritten as .

Assume Statement 2 alone. It can be rewritten as

$x^{2}< 81$

the solution set of which is

From either statement alone, it follows that .

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Question

What is the value of z?

Statement 1: $\dpi{100} \small x+y+z=4$

Statement 2: $\dpi{100} \small 2x+y^{2}+z=17$

Answer

To solve for three variables, you must have three equations. Statements 1 and 2 together only give two equations, so the statements together are not sufficient.

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Question

Is the equation linear?

Statement 1:

Statement 2: is a constant

Answer

If we only look at statement 1, we might think the equation is not linear because of the term. But statement 2 tells us the is a constant. Then the equation is linear. We need both statements to answer this question.

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Question

Data sufficiency question- do not actually solve the question

Solve for :

$\small 2x+3xy+4y=7$

1. $\small x=1$

2. $\small x+4y=13$

Answer

When solving an equation with 2 variables, a second equation or the solution of 1 variable is necessary to solve.

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Question

Data Sufficiency Question

Solve for and .

1.

2. Both and are positive integers

Answer

Using statement 1 we can set up a series of equations and solve for both and .

Additionally, the information in statement 2 indicates that there is only one possible solution that satisfies the requirement that both are positive integers.

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Question

Solve the following for x:

4x+7y = 169

1. x > y

2. x - y = 12

Answer

To solve with 2 unknowns, we must create a system of equations with at least 2 equations. Using statement 2 as a second equation we can easily get our answer. Solve statement 2 for x or y, and plug in for the corresponding variable in the equation given by the problem.

So, solving statement 2 for x, we get x=12+y. Replacing x in the equation from the problem, we get 4(12+y) + 7y=169. We can distribute the 4, and combine terms to find 48+11y=169. Subtract 48 from both sides, we get 11y=121. So y=11. Reusing either equation and plugging in our y value gives our x value. So x - 11=12, or x=23. This shows that x > y, and statement 1 is true. But even though it's true, it is completely unneccessary information. Therefore the answer is that we only need the information from statement 2, and statement 1 is not needed.

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Question

How many solutions does this system of equations have: one, none, or infinitely many?

Statement 1:

Statement 2:

Answer

If the slopes of the lines are not equal, then the lines intersect at one solution; if they are equal, then they do not intersect, or the lines are the same line. Write each equation in slope-intercept form, :

$4y \div 4 = \left (-Ax + 12 \right )\div 4$

$y = - \frac{A}{4} x + 3$

$By \div B =\left ( - 20x + C \right )\div B$

$y = - \frac{20}{B}x + \frac{C}{B}$

The slopes of the lines are $- \frac{A}{4} , - \frac{20}{B}$ .

We need to know both and in order to determine their equality or inequality, and only if they are unequal can we answer the question.

Set and .

$- \frac{A}{4} = - \frac{10}{4} = - \frac{5}{4}$

$- \frac{20}{B} = - \frac{20}{5} = -4$

The slopes are unequal, so the lines intersect at one point; the system has exactly one solution.

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Question

Given that both $A,B \neq 0$ , how many solutions does this system of equations have: one, none, or infinitely many?

Statement 1:

Statement 2:

Answer

If the slopes of the lines are not equal, then the lines intersect at one solution. If the slopes are equal, then there are two possibilties: either they do not intersect or they are the same line. Write each equation in slope-intercept form:

$By \div B = \left (-Ax + 12 \right ) \div B$

$y = \frac{-A}{B}x + \frac{12}{B}$

$5y \div 5 = \left (- 20x + C \right )\div 5$

$y = -4x + \frac{C}{5}$

The slopes of these lines are $\frac{-A}{B}, -4$ .

If Statement 1 is true, then we can rewrite the first slope as $\frac{-3B}{B} = -3$ , meaning that the lines have unequal slopes, and that there is only one solution. Statement 2 tells us the value of , which is irrelevant.

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Question

Data Sufficiency Question

Solve for and :

1.

2.

Answer

In order to solve an equation set, one requires a number of equations equal to the number of variables. Therefore, either of the statements allow the problem to be solved.

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Question

Data Sufficiency Question

Solve for , , and :

1.

2.

Answer

In order to solve an equation set, one requires a number of equations equal to the number of variables. Therefore, three equations are needed and both statements are required to solve the problem.

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Question

Given that , evaluate .

Statement 1:

Statement 2:

Answer

Solve for in each statement.

Statement 1:

$4z \div 4= 8 \div 4$

Statement 2:

From either statement alone, it can be deduced that .

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Question

A toy store sells dolls for dollars each and trucks for dollars each. How many dolls did the store sell last week?

(1) Last week, the store sold twice as many trucks as dolls.

(2) Last week, the store made , dollars from selling trucks and dolls.

Answer

Let:

t: the number of trucks sold last week

d: the number of dolls sold last week - d is the value we are looking to find

To evaluate the statements, we translate the word problems into equations

(1): or

(2):

Each statement provides a single equation with two unknowns which is unsolvable, so each statement alone is not enough.

The two statements taken together give us a system of two equations with two unknowns, which we can solve.

Therefore the right answer is C.

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Question

is an integer. Is there a real number such that $x^{A}=B$ ?

Statement 1: is negative

Statement 2: is even

Answer

The equivalent question is "does have a real $A \textrm{th }$ root?"

If you know only that is negative, you need to know whether is even or odd; negative numbers have real odd-numbered roots, but not real even-numbered roots.

If you know only that is even, you need to know whether is negative or nonnegative; negative numbers do not have real even-numbered roots, but nonnegative numbers do.

If you know both, however, then you know that the answer is no, since as stated before, negative numbers do not have real even-numbered roots.

Therefore, the answer is that both statements together are sufficient to answer the question, but neither statement alone is sufficient to answer the question.

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Question

Which is the greater quantity, $3^{x}$ or $9^{y}$ - or are they equal?

Statement 1:

Statement 2:

Answer

From Statement 1 alone,

$3^{x} = 3^{2y+2} = 3^{2y} \cdot 3^{2} =\left ( 3^{2} \right ) ^{y}\cdot 9 = 9^{y} \cdot 9>9^{y}$

Now assume Statement 2 alone. We show that this is insufficient with two cases:

Case 1:

$3^{3} = 27$ ; $9^{1} = 9$ ; therefore, $3^{x} >9^{y}$

Case 1:

$3^{-3} = \frac{1}{27}$ ; $9^{-1} = \frac{1}{9}$ ; therefore, $3^{x}<9^{y}$

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Question

Does $\log_{2} \left ( MN \right )$ exist?

Statement 1: and are both negative.

Statement 2: divided by 2 yields an integer.

Answer

A logarithm can be taken of a number if and only if the number is positive. If Statement 1 alone is true, then , being the product of two negative numbers, must be positive, and $\log_{2} \left ( MN \right )$ exists.

Statement 2 is irrelevant; 4 and both yield integers when divided by 2, but $\log_{2} 4 = 2$ and $\log_{2}\left ( -4 \right )$ does not exist.

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Question

Johnny was assigned to write a number in scientific notation by filling the circle and the square in the pattern below with two numbers.

$\bigcirc \times 10 ^{\square }$

Johnny filled in both shapes with numbers. Did he succeed?

Statement 1: He filled in the circle with the number "10".

Statement 2: He filled in the square with a negative integer.

Answer

The number $a \times 10 ^{n}$ is a number written in scientific notation if and only of two conditions are true:

$1 \leq \left | a \right | < 10$

is an integer

By Statement 1 Johnny filled in the circle incorrectly, since it makes .

By Statement 2, Johnny filled in the square correctly, but the statement says nothing about how he filled in the circle; Statement 2 leaves the question open.

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Question

Solve the following rational expression:

$\frac{x^{m-n}10^{n-1}}{10^{(\frac{m}{2}-2)}x^{n}}$

(1)

(2)

Answer

When replacing m=5 in the expression we get:

$\frac{x^{5-n}10^{n-1}}{10^{\frac{1}{2}}x^{n}}$

Therefore statement (1) ALONE is not sufficient.

When replacing m=2n in the expression we get:

$\frac{x^{2n-n}10^{n-1}}{10^{(\frac{2n}{2}-2)}x^{n}}=\frac{x^{n}10^{n-1}}{10^{n-2}x^{n}}=\frac{10^{n-1}}{10^{n-2}}=10$

Therefore statement (2) ALONE is sufficient.

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Question

Is $x^{2} < x$ ?

(1)

(2)

Answer

$x^{2} < x \Leftrightarrow x^{2}-x<0 \Leftrightarrow x(x-1)<0$

From statement 1 we get that and .

So the first term is positive and the second term is negative, which means that is negative; therefore the statement 1 alone allows us to answer the question.

Statement 2 tells us that . If , we have $x^{2}=0.25$ which is less than . Therefore in this case $x^{2}<x$ .

For , we have $x^{2} = 4$ which is greater than . So in this case $x^{2} > x$ .

So statement 2 is insufficient.

Therefore the correct answer is A.

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Question

is a number not in the set $\left{-1, 0, 1 \right}$ .

Of the elements $\left { x^{4}, x^{5}, x^{6}\right }$ , which is the greatest?

Statement 1: is a negative number.

Statement 2: $\left | x\right | > 1$

Answer

Statement 1 alone is inconclusive, as can be demonstrated by examining two negative values of other than .

Case 1: .

Then

$x^{4} = (-2)^{4} = 16$

$x^{5} = (-2)^{5} = -32$

$x^{6} = (-2)^{6} = 64$

$x^{6}$ is the greatest of these values.

Case 2: $x = -\frac{1}{2}$

Then

$x ^{4}=\left ( -\frac{1}{2} \right )^{4} = \frac{1}{16}$

$x ^{5}=\left ( -\frac{1}{2} \right )^{5} = -\frac{1}{32}$

$x ^{6}=\left ( -\frac{1}{2} \right )^{6} = \frac{1}{64}$

$x^{4}$ is the greatest of these values.

Now assume Statement 2 alone. Either or .

Case 1: .

Then $x \cdot x^{4} > 1\cdot x^{4}$ , so $x^{5} > x^{4}$ ; similarly, $x^{6} > x^{5}$ .

$x^{6}$ is the greatest of the three.

Case 2: .

Odd power $x^{5}$ is negative, and even powers $x^{4}$ and $x^{6}$ are positive, so one of the latter two is the greatest. Since , it follows that $x^{2} > 1$ . It then follows that $x^{2} \cdot x^{4} > 1 \cdot x^{4}$ , or $x^{6} > x^{4}$ .

Again, $x^{6}$ is the greatest of the three.

Statement 2 alone is sufficient, but not Statement 1.

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