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ISEE Upper Level Quantitative : Equations

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Example Questions

Example Question #81 : How To Find The Solution To An Equation

$b =- a+\frac{1}{2 }$

2a + 2b = 1

Which is the greater quantity?

(A)

(B)

Possible Answers:

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(A) is greater

Correct answer:

It is impossible to determine which is greater from the information given

Explanation:

The two equations are actually equivalent, as is proved here:

$b =- a+\frac{1}{2 }$

$a + b =a - a+\frac{1}{2 }$

$a + b = \frac{1}{2 }$

$2 \cdot \left (a + b \right )= 2 \cdot \frac{1}{2 }$

2a + 2b = 1

Therefore, we need only test the first equation. However, it can be shown that it is possible for either of the two to be greater or both to be equal; as can be determined from that third equation $a + b = \frac{1}{2 }$ , any two values of and that add up to $\frac{1}{2}$ will solve the system, such as $\left ( 0, \frac{1}{2}\right )$ , $\left ( \frac{1}{4}, \frac{1}{4} \right )$ , or $\left ( \frac{1}{2}, 0\right )$ .

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Example Question #82 : How To Find The Solution To An Equation

$-\frac{3}{4} M =- \frac{6}{7}$

$-\frac{4}{5} N =- \frac{5}{6}$

Which is the greater quantity?

(A)

(B)

Possible Answers:

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(B) is greater

(A) is greater

Correct answer:

(A) is greater

Explanation:

$-\frac{3}{4} M =- \frac{6}{7}$

$\left ( - \frac{4}{3}\right ) \cdot\left ( -\frac{3}{4} M \right )=\left ( - \frac{4}{3}\right ) \cdot\left ( - \frac{6}{7} \right )$

$M = \frac{24}{21} = \frac{8}{7} = 1 \frac{1}{7}$

$-\frac{4}{5} N =- \frac{5}{6}$

$\left (-\frac{5} {4} \right ) \cdot \left (-\frac{4}{5} N \right )= \left (-\frac{5} {4} \right ) \cdot \left (- \frac{5}{6} \right )$

$N = \frac{25}{24} = 1\frac{1}{24}$

$1 \frac{1}{7} > 1 \frac{1}{24}$ , so M > N , making (A) greater.

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Example Question #81 : Algebraic Concepts

Give the -coordinate of the point on the graph of the equation $\frac{3}{x} + \frac{2}{y} = 5$ that has -coordinate $-\frac{1}{4}$ .

Possible Answers:

$-\frac{2}{7}$

$\frac{17}{2}$

$\frac{2} {17}$

No such point exists.

$-\frac{7}{2}$

Correct answer:

$\frac{2} {17}$

Explanation:

The point $\left ( - \frac{1}{4}, y\right )$ is on the graph of the equation $\frac{3}{x} + \frac{2}{y} = 5$ . Finding the -coordinate of this point is the same as evaluating for $x= -\frac{1}{4}$ . Substitute, and we get:

$\frac{3}{x} + \frac{2}{y} = 5$

$\frac{3}{ -\frac{1}{4}} + \frac{2}{y} = 5$

$-12+ \frac{2}{y} = 5$

$-12+ \frac{2}{y} + 12 = 5 + 12$

$\frac{2}{y} = 17$

$\frac{2}{y}\cdot \frac{y}{17} = 17 \cdot \frac{y}{17}$

$\frac{2} {17} = y$

The -coordinate is therefore $\frac{2} {17}$ .

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Example Question #83 : How To Find The Solution To An Equation

Give the -coordinate of the point on the graph of the equation $2\sqrt{x} + \sqrt{y} = 25$ that has -coordinate 64.

Possible Answers:

No such point exists.

Correct answer:

Explanation:

The point $\left ( 64, y\right )$ is on the graph of the equation $2\sqrt{x} + \sqrt{y} = 25$ . Finding the -coordinate of this point is the same as evaluating for x=64 . Substitute, and we get:

$2\sqrt{x} + \sqrt{y} = 25$

$2\sqrt{64} + \sqrt{y} = 25$

$2 \cdot 8 + \sqrt{y} = 25$

$16+ \sqrt{y} = 25$

$16+ \sqrt{y} - 16 = 25 - 16$

$\sqrt{y} = 9$

$\left (\sqrt{y} \right )^{2} = 9 ^{2}$

y =81

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Example Question #81 : Equations

Give the -coordinate of the point on the graph of the equation $3\sqrt{x} +2 \sqrt{y} = 23$ that has -coordinate 64.

Possible Answers:

No such point exists.

$\frac{1}{4}$

$-\frac{1}{4}$

Correct answer:

No such point exists.

Explanation:

The point $\left ( 64, y\right )$ is on the graph of the equation $3\sqrt{x} +2 \sqrt{y} = 23$ . Finding the -coordinate of this point is the same as evaluating for x=64 . Substitute, and we get:

$3\sqrt{x} +2 \sqrt{y} = 23$

$3\sqrt{64} +2 \sqrt{y} = 23$

$3 \cdot8 +2 \sqrt{y} = 23$

$24 +2 \sqrt{y} = 23$

$24 +2 \sqrt{y} -24 = 23 -24$

$2 \sqrt{y} = -1$

$2 \sqrt{y} \div 2 = -1 \div 2$

$\sqrt{y} = - \frac{1}{2}$

Since the square root of a number must be positive, there is no solution. Therefore, there is no point on this graph with -coordinate 64.

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Example Question #84 : Algebraic Concepts

Give the -coordinate of the point on the graph of the equation $\frac{4}{x} + \frac{3}{y} = 12$ that has -coordinate $\frac{1}{3}$ .

Possible Answers:

$-\frac{1}{4}$

No such point exists.

$\frac{1}{4}$

Correct answer:

No such point exists.

Explanation:

The point $\left ( \frac{1}{3}, y\right )$ is on the graph of the equation $\frac{4}{x} + \frac{3}{y} = 12$ . Finding the -coordinate of this point is the same as evaluating for $x= \frac{1}{3}$ . Substitute, and we get:

$\frac{4}{x} + \frac{3}{y} = 12$

$\frac{4}{\frac{1}{3}} + \frac{3}{y} = 12$

$12+ \frac{3}{y} = 12$

$12+ \frac{3}{y}- 12 = 12 - 12$

$\frac{3}{y} = 0$

However, there is no number that can be divided into 3 to yield a quotient of 0, so there is no solution. Therefore, there is no point on this graph with -coordinate $\frac{1}{3}$ .

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Example Question #84 : Equations

$y = \frac{1}{3} x + \frac{4}{3}$

2x - 3y = 4

What is ?

Possible Answers:

$\frac{8}{3}$

$\frac{4}{3}$

Correct answer:

Explanation:

Substitute $\frac{1}{3} x + \frac{4}{3}$ for in the second equation:

2x - 3y = 4

$2x - 3 \left ( \frac{1}{3} x + \frac{4}{3} \right ) = 4$

2x - x -4 = 4

x -4 = 4

x -4+ 4 = 4 + 4

x = 8

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Example Question #85 : Algebraic Concepts

$y = \frac{1}{3} x + \frac{4}{3}$

2x - 3y = 4

What is ?

Possible Answers:

$\frac{4}{3}$

$\frac{8}{3}$

Correct answer:

Explanation:

Solve for in the top equation:

$y = \frac{1}{3} x + \frac{4}{3}$

$3 \cdot y = 3 \cdot \left ( \frac{1}{3} x + \frac{4}{3} \right )$

3 y = x +4

3 y-4 = x +4 -4

x = 3 y-4

Substitute 3 y-4 for in the second equation:

2x - 3y = 4

$2 \left ( 3 y-4 \right )- 3y = 4$

6 y-8- 3y = 4

3 y-8 = 4

3 y-8 + 8 = 4 + 8

3 y = 12

$3 y \div 3 = 12 \div 3$

y = 4

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Example Question #81 : Algebraic Concepts

If $\frac{x}{4}+2y=16$ , then what is an expression for x in terms of y?

Possible Answers:

x=64-8y

x=24-2y

x=8y-64

x=4y

x=16-2y

Correct answer:

x=64-8y

Explanation:

To solve this problem, isolate for x. First, move the y term over to the left side. This gives you $\frac{x}{4}=16-2y$ . Then, multiply both sides by 4. This gives you x=(16-2y)4 . Then, distribute the four to the terms inside the parantheses. This gives you a final answer of x=64-8y .

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Example Question #86 : Equations

y = 0.5x + 1.4

1.5x-0.5y = 9.8

Evaluate .

Possible Answers:

x = 8.4

x = 7.28

x = 5.6

x = 5.2

The answer cannot be determined from the information given.

Correct answer:

x = 8.4

Explanation:

Substitute 0.5x + 1.4 for in the second equation as follows:

1.5x-0.5y = 9.8

$1.5x-0.5 \left ( 0.5x + 1.4 \right ) = 9.8$

1.5x-0.25 x - 0.7 = 9.8

1.25 x - 0.7 = 9.8

1.25 x - 0.7 + 0.7= 9.8 + 0.7

1.25 x =10.5

$1.25 x\div 1.25 =10.5 \div 1.25$

x = 8.4

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Equations

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #81 : How To Find The Solution To An Equation

Example Question #82 : How To Find The Solution To An Equation

Example Question #81 : Algebraic Concepts

Example Question #83 : How To Find The Solution To An Equation

Example Question #81 : Equations

Example Question #84 : Algebraic Concepts

Example Question #84 : Equations

Example Question #85 : Algebraic Concepts

Example Question #81 : Algebraic Concepts

Example Question #86 : Equations

All ISEE Upper Level Quantitative Resources

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