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ISEE Upper Level Math : ISEE Upper Level (grades 9-12) Mathematics Achievement

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Example Questions

Example Question #701 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Find the value of x.

7,x, 343, 2401,16807

Possible Answers:

Correct answer:

Explanation:

Find the value of x.

7,x, 343, 2401,16807

We have a geometric series here. In other words, each term is a consistent multiple of the number before it. To find our common multiple, we need to take one term and divide it by the term before it. Let's try using the 3rd and 4th terms.

$CM=\frac{2401}{343}=7$

So, to move from one term to the next, we need to multiply by seven.

To find x, we need to multiply our first term (7), by 7.

7*7=49

So, our answer is 49.

We can check this by multiplying 49 by 7

49*7=343

Doing so yields 343, which is in fact the next term in our sequence.

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Example Question #1 : Equations

Solve for x:

$\dpi{100} 3x^{2}+4=31$

Possible Answers:

$\dpi{100} 9$

$\dpi{100} \pm 1$

$\dpi{100} \pm 9$

$\dpi{100} \pm 3$

Correct answer:

$\dpi{100} \pm 3$

Explanation:

First, subtract 4 from both sides:

$\dpi{100} 3x^{2}+4-4=31-4$

$\dpi{100} 3x^{2}=27$

Next, divide both sides by 3:

$\dpi{100} \frac{3x^{2}}{3}=\frac{27}{3}$

$\dpi{100} x^{2}=9$

Now take the square root of both sides:

$\dpi{100} \sqrt{x^{2}}=\sqrt{9}$

$\dpi{100} x=\pm 3$

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Example Question #2 : Equations

Solve for $\dpi{100} x$ :

$\dpi{100} \frac{x}{3}+22 = 45$

Possible Answers:

$\dpi{100} 69$

$\dpi{100} 3$

$\dpi{100} 135$

$\dpi{100} 66$

Correct answer:

$\dpi{100} 69$

Explanation:

$\dpi{100} \frac{x}{3}+22 = 45$

$\dpi{100} \frac{x}{3}+22-22 = 45-22$

$\dpi{100} \frac{x}{3} = 23$

$\dpi{100} (3)\cdot \frac{x}{3} = 23\cdot (3)$

$\dpi{100} x=69$

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Example Question #3 : Equations

Solve for :

| 3x + 9| = 24

Possible Answers:

$x = -5 \textrm{ or }x = 11$

$x = -11 \textrm{ or }x = 11$

$x = -5\textrm{ or }x = 5$

$x = 5 \textrm{ or }x = 11$

$x = -11 \textrm{ or }x = 5$

Correct answer:

$x = -11 \textrm{ or }x = 5$

Explanation:

Rewrite this as a compound statement and solve each separately:

| 3x + 9| = 24

$3x + 9 = - 24 \textrm{ or }3x + 9 = 24$

3x + 9 = - 24

3x + 9 -9= - 24-9

3x = - 33

$3x\div 3 = - 33 \div 3$

x = -11

3x + 9 = 24

3x + 9 -9= 24-9

3x = 15

$3x\div 3 = 15 \div 3$

x = 5

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Example Question #4 : Equations

Solve for :

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

Possible Answers:

$x = 1\textrm{ or }x = 7$

x = 12

x = 7

$x = 3\textrm{ or }x = 4$

$x = 1\textrm{ or }x = 4$

Correct answer:

x = 7

Explanation:

FOIL each of the two expressions, then solve:

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

$x ^{2} + 3 \cdot x -4\cdot x -4 \cdot 3 = x^{2} -1 \cdot x - 2 \cdot x + 2\cdot1$

$x ^{2} + 3 x -4 x -12 = x^{2} -1 x - 2x + 2$

$x ^{2} - x -12 = x^{2} - 3x + 2$

$x ^{2} - x -12 - x^{2} = x^{2} - 3x + 2 - x^{2}$

- x -12= - 3x + 2

Solve the resulting linear equation:

- x -12 +3x + 12 = - 3x + 2+3x+12

2x =14

$2x\div 2 =14 \div 2$

x = 7

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Example Question #5 : Equations

Solve for :

$x^{2} -10x = 24$

Possible Answers:

$x=-4\textrm{ or }x=6$

$x=-2 \textrm{ or }x=12$

$x=-6\textrm{ or }x=4$

$x=-12 \textrm{ or }x=2$

$x=-6\textrm{ or }x=-4$

Correct answer:

$x=-2 \textrm{ or }x=12$

Explanation:

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

$x^{2} -10x = 24$

$x^{2} -10x -24= 24-24$

$x^{2} -10x -24= 0$

Now factor the quadratic expression $x^{2} -10x -24$ into two binomial factors (x + ?)(x + ?) , replacing the question marks with two integers whose product is and whose sum is . These numbers are -12,2 , so:

(x -12)(x + 2) = 0

$x-12 = 0 \Rightarrow x = 12$

$x+2 = 0 \Rightarrow x = -2$

The solution set is $\left \{ -2,12\right \}$ .

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Example Question #6 : Equations

Solve for :

$\left ( x-3\right )(2x-1) = 18$

Possible Answers:

$x= -5 \textrm{ or } x= 3$

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

$x= -3 \textrm{ or } x= 5$

$x= - 5\textrm{ or } x= 1 \frac{1}{2}$

$x = 9\frac{1}{2} \textrm{ or }x=19$

Correct answer:

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

Explanation:

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

$\left ( x-3\right )(2x-1) = 18$

$x\cdot 2x -x\cdot 1 -3 \cdot 2x +3\cdot 1= 18$

$2x^{2} -x -6x +3= 18$

$2x^{2} -7x +3= 18$

$2x^{2} -7x +3-18= 18 -18$

$2x^{2} -7x -15= 0$

Use the method to factor the quadratic expression $2x^{2} -7x -15$ ; we are looking to split the linear term by finding two integers whose sum is and whose product is 2 (-15) = -30 . These integers are -10,3 , so:

$(2x^{2} -10x)+(3x -15)= 0$

2x (x -5)+3(x -5)= 0

$\left (2x +3 \right) \left ( x-5 \right )= 0$

Set each expression equal to 0 and solve:

2x +3= 0

2x = -3

$x = -\frac{3}{2} = -1\frac{1}{2}$

x-5= 0

x=5

The solution set is $\left \{ - 1 \frac{1}{2}, 5\right \}$ .

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Example Question #7 : Equations

Solve for :

$x^{2} + 12 = 8x$

Give all solutions.

Possible Answers:

$x = -6 \textrm{ or }x =- 2$

$x = 2 \textrm{ or }x = 6$

$x = -8 \textrm{ or }x = 8$

$x = -6 \textrm{ or }x =6$

$x = 6 \textrm{ or }x = 8$

Correct answer:

$x = 2 \textrm{ or }x = 6$

Explanation:

Rewrite this quadratic equation in standard form:

$x^{2} + 12 = 8x$

$x^{2} -8x + 12 = 8x-8x$

$x^{2} -8x + 12 = 0$

Factor the expression on the left. We want two integers whose sum is and whose product is . These numbers are -6,-2 , so the equation becomes

(x-6)(x-2)=0 .

Set each factor equal to 0 separately, then solve:

$x - 6 = 0 \Rightarrow x = 6$

$x - 2 = 0 \Rightarrow x = 2$

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Example Question #8 : Equations

Solve for :

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

Possible Answers:

x = 0

The equation has the set of all real numbers as its solution set.

The equation has no solution.

x =4

x = 2

Correct answer:

x = 0

Explanation:

Simplify both sides, then solve:

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

$4 \cdot x + 4 \cdot 6 - 3 \cdot x - 3 \cdot 8 = 2 \cdot x + 2 \cdot 4 +2 \cdot x - 2 \cdot 4$

4x + 24 - 3 x - 24 = 2x +8 +2 x -8

4x - 3 x+ 24 - 24 = 2x +2 x +8-8

$\left ( 4 - 3 \right ) x= \left ( 2 +2 \right ) x$

x= 4 x

x - 4x = 4 x - 4x

$\left (1 - 4 \right ) x = 0$

-3 x = 0

x = 0

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Example Question #9 : Equations

3 (x + 2) + 4x = 5x + 2 (x +3)

Possible Answers:

The equation has the set of all real numbers as its solution set.

x = 0

The equation has no solution.

x = -2

x = 6

Correct answer:

The equation has the set of all real numbers as its solution set.

Explanation:

Simplify both sides, then solve:

3 (x + 2) + 4x = 5x + 2 (x +3)

$3 \cdot x + 3 \cdot 2 + 4x = 5x + 2 \cdot x + 2 \cdot 3$

3x + 6 + 4x = 5x + 2x + 6

3x + 4x + 6 = 5x + 2x + 6

$\left ( 3 + 4 \right ) x + 6 = \left ( 5 + 2 \right ) x + 6$

7x + 6 =7 x + 6

This is an identically true statement, so the original equation has the set of all real numbers as its solution set.

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : ISEE Upper Level (grades 9-12) Mathematics Achievement

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #701 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #1 : Equations

Example Question #2 : Equations

Example Question #3 : Equations

Example Question #4 : Equations

Example Question #5 : Equations

Example Question #6 : Equations

Example Question #7 : Equations

Example Question #8 : Equations

Example Question #9 : Equations

All ISEE Upper Level Math Resources

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