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ISEE Middle Level Math : Algebraic Concepts

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Example Questions

Example Question #71 : Variables

Simplify the following statement

(14b^2t^5)(2b^3t^9)

Possible Answers:

$14b^5t^{14}$

$28b^5t^{14}$

$28b^4t^{4}$

$28t^5b^{14}$

Correct answer:

$28b^5t^{14}$

Explanation:

Simplify the following statement

(14b^2t^5)(2b^3t^9)

To multiply these terms, first multiply the 2 and the 14

28(b^2t^5)(b^3t^9)

Next, we can combine the b's and the t's by adding together their exponents.

$28(b^2t^5)(b^3t^9)=28b^5t^{14}$

So our answer is:

$28b^5t^{14}$

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Example Question #71 : Variables

Combine the following terms:

$5q^3*8q^4*9q^{11}$

Possible Answers:

$360q^{18}$

$360q^{7}$

$36q^{18}$

$360q^{12}$

Correct answer:

$360q^{18}$

Explanation:

Combine the following terms:

$5q^3*8q^4*9q^{11}$

Let's begin by multiplying the coefficients (numbers out in front)

5*8*9=40*9=360

So our answer must have 360 in front.

Next, let's multiply our q's. To do so, we need to add up our exponents.

$q^3*q^4*q^{11}=q^{3+4+11}=q^{18}$

So, put it together to get:

$360q^{18}$

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Example Question #72 : Variables

$x^{2} = 64$ . Evaluate $x^{4}$ .

Possible Answers:

512

4,096

Correct answer:

4,096

Explanation:

$x^{2} = 64$ , so either $x = \sqrt{64} = 8$ or $x = - \sqrt{64} =- 8$ .

If the former is true,

$x^{4} = 8^{4} = 8 \cdot 8 \cdot 8 \cdot 8 = 4,096$ .

If the latter is true, then, since an even-numbered power of any number is positive,

$x^{4} =( - 8)^{4} = 8^{4} = 4,096$ .

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Example Question #73 : Variables

Simplify the following expression

$54x^3*8x^2*\frac{x^2}{9}$

Possible Answers:

48x^7

$1244x^{12}$

540x^7

$48x^{12}$

Correct answer:

48x^7

Explanation:

Simplify the following expression

$54x^3*8x^2*\frac{x^2}{9}$

Let's begin by grouping our x's and our coefficients

$(54*8*\frac{1}{9})(x^3*x^2*x^2)$

We can combine our coefficents by multiplying them, and we can combine the x's by adding the exponents

$(54*\frac{1}{9}*8)(x^7)=(6*8)x^7=48x^7$

So, our answer is:

48x^7

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Example Question #74 : Variables

2a = 3b = 66

Evaluate .

Possible Answers:

121

726

676

Correct answer:

726

Explanation:

First, use the given equations to solve for both the individual variables.

2a = 66

$2a \div 2 = 66 \div 2$

a = 33

3b = 66

$3b \div 66 = 66\div 3$

b = 22

Now that both variables have been solved for, find their product.

$ab =22 \cdot 33 = 726$

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Example Question #21 : Algebra

$3\frac{2}{5} \div 1 \frac{1}{5}=$

Possible Answers:

$4\frac{2}{5}$

$2\frac{5}{6}$

$\frac{102}{25}$

Correct answer:

$2\frac{5}{6}$

Explanation:

First convert each mixed number into and improper fraction

$\frac{17}{5} \div \frac{6}{5}$

Then convert the operation to multiplication and flip the second fraction

$\frac{17}{5} \ast \frac{5}{6}=$

Reduce where possible and multiply to solve:

$\frac{17}{6}=2\frac{5}{6}$

The answer is $2\frac{5}{6}$

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Example Question #22 : Algebra

Simplify:

$\frac{8x^3-6x^2+2x}{2x}$

Possible Answers:

4x^3-3x^2+x

4x^3+3x-x

4x^2-3x+1

4x^2+3x-1

Correct answer:

4x^2-3x+1

Explanation:

$\frac{8x^3-6x^2+2x}{2x}=\frac{8x^3}{2x}-\frac{6x^2}{2x}+\frac{2x}{2x}=4x^2-3x+1$

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Example Question #23 : Algebra

Solve for B

AX+B(2Y)=C

Possible Answers:

$2B=\frac{C-AX}{Y}$

$\frac{C-2Y}{AX}=B$

$B=\frac{C-AX}{2Y}$

Not enough information provided.

B=4

Correct answer:

$B=\frac{C-AX}{2Y}$

Explanation:

This problem looks strange since it mostly contains variables, in particular unknown variables. Still remember the order of operations: parentheses, exponents, multiplication, division, addition, and subtraction. We still use those operations in this case. All we must do is move the variables around to solve for .

In this case we can subtract (additive inverse) to the other side first.

AX-AX+B(2Y)=C-AX

Rewrite our problem.

B(2Y)=C-AX

Stop for a moment and look at the progress. Go back to the original question, solve for . Our problem shows about to be multiplied to , but we do not necessarily have to multiply because we can divide by to both sides of the equation, which would save us a step as well as have B by itself.

$\frac{B(2Y)}{2Y}=\frac{C-AX}{2Y}$

$B=\frac{C-AX}{2Y}$

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Example Question #24 : Algebra

Suppose you know the values of , , and , and you want to evaluate the expression below.

$A + B^{2}\div C$

Which of the following is the first step you must complete?

Possible Answers:

Evaluate A^2 .

Evaluate $B\div C$ .

Evaluate B^2 .

Evaluate $A\div C$ .

Evaluate A+B .

Correct answer:

Evaluate B^2 .

Explanation:

Use the order of operations, PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction

In our expression, there are no parentheses, so square first.

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Example Question #5 : How To Divide Variables

Divide:

$\frac{8x^2-6x+4}{-2}$

Possible Answers:

$-4x^2+3x-2^{2}$

4x^2-3x+2

-4x^2+3x-2

4x^2+3x-2

-4x^2-3x+2

Correct answer:

-4x^2+3x-2

Explanation:

$\frac{8x^2-6x+4}{-2}=\frac{8x^2}{-2}-\frac{6x}{-2}+\frac{4}{-2}=-4x^2+3x-2$

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ISEE Middle Level Math : Algebraic Concepts

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

All ISEE Middle Level Math Resources

Example Questions

Example Question #71 : Variables

Example Question #71 : Variables

Example Question #72 : Variables

Example Question #73 : Variables

Example Question #74 : Variables

Example Question #21 : Algebra

Example Question #22 : Algebra

Example Question #23 : Algebra

Example Question #24 : Algebra

Example Question #5 : How To Divide Variables

All ISEE Middle Level Math Resources

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