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HSPT Math : Algebra

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6 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #31 : How To Multiply Variables

Simplify:

$(2)(15y^{2})(x^{3})(10)$

Possible Answers:

$300x^{3}y^{2}$

$27 x^{4}y^{2}$

$27x^{3}y^{2}$

$30x^{31}y^{2}$

$20x^{4}y^{30}$

Correct answer:

$300x^{3}y^{2}$

Explanation:

Begin by moving all of the related variables (and constants) next to each other. You can group these in parentheses to make it clear. This is allowed because of the associative rule for multiplication.

$(2 * 15 * 10)(x^{3})(y^{2})$

There are no variables to combine. We only need to combine the numerical coefficient. This gets you:

$(300)(x^{3})(y^{2})$

This is the same as:

$300x^{3}y^{2}$

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Example Question #671 : Concepts

Simplify:

$x * y * (x^{3}) * (xy)$

Possible Answers:

$6x^{3}y$

$3x^{5}y^{2}$

10xy

$x^{3}y$

$x^{5}y^{2}$

Correct answer:

$x^{5}y^{2}$

Explanation:

$(x * x* x^{3})(y * y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(x^{1+1+3})(y^{1+1})$

$(x^{5})(y^{2})$

This is the same as:

$x^{5}y^{2}$

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Example Question #1267 : Hspt Mathematics

Simplify:

$x * a^{3} * y^{4} * x^{2} * x*a^{2}$

Possible Answers:

$a^{6}xy^{2}$

$a^{6}x^{2}y^{4}$

80axy

$a^{5}x^{3}y^{4}$

$a^{5}x^{4} y^{4}$

Correct answer:

$a^{5}x^{4} y^{4}$

Explanation:

$(a^{3} * a^{2})(x * x^{2} * x)(y^{4})$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(a^{3+2})(x^{1+2+1})(y^4)$

$(a^{5})(x^{4})(y^{4})$

This is the same as:

$a^{5}x^{4} y^{4}$

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Example Question #32 : How To Multiply Variables

Simplify:

$xy(x^{2} + 2aby)$

Possible Answers:

$3abx^{3}y^{2}$

$x^{2}y + 2abxy$

$3abx^{3}y$

$2x^{2}y + 4abxy$

$x^{3}y + 2abxy^{2}$

Correct answer:

$x^{3}y + 2abxy^{2}$

Explanation:

$xy(x^{2} + 2aby)$

Distribute the outside term into the parentheses. Multiply each term in parentheses by :

$(xy)(x^{2}) + (xy)(2aby)$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

$(x*x^{2})(y) + (2)(a)(b)(x)(y*y)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(x^{1+2})(y)+(2)(a)(b)(x)(y^{1+1})$

$(x^{3})(y) + (2)(a)(b)(x)(y^2)$

This is the same as:

$x^{3}y + 2abxy^{2}$

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Example Question #31 : Operations

Simplify:

2ab(5xy - 3ax + 4by)

Possible Answers:

$12a^{4}b^{4}x^{2}y^{2}$

$7abxy – 5a^{2}bx + 6ab^{2}y$

$10abxy - 6a^{2}bx + 8ab^{2}y$

12abxy

6abxy - 6abx + 5aby

Correct answer:

$10abxy - 6a^{2}bx + 8ab^{2}y$

Explanation:

2ab(5xy - 3ax + 4by)

Distribute the outside term into the parentheses. Multiply each term in parentheses by 2ab :

(2ab)(5xy) - (2ab)(3ax)+ (2ab)(4by)

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

(2*5)(abxy) - (2*3)(a*a)(bx)+ (2*4)(a)(b*b)(y)

When multiplying variables of the same type, you add their exponents together. This gets you:

$(10)(abxy)-(6)(a^{1+1})(bx)+(8)(a)(b^{1+1})(y)$

$(10)(abxy) - (6)(a^{2})(bx)+ (2*4)(a)(b^{2})(y)$

This is the same as:

$10abxy - 6a^{2}bx + 8ab^{2}y$

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Example Question #32 : Operations

Simplify:

$5bv(vx^{2} - 10bx - 5vb^{6})$

Possible Answers:

$5bv^{2}x^{2} + 50b^{2}v^{2}x – 25b^{7}$

$6bv^{2}x^{2} – 5b^{2}v^{2}x$

$10bv^{2}x^{2} – 50bv^{2}x – 10b^{5}v^{2}$

$5bv^{2}x^{2} - 50b^{2}vx - 25b^{7}v^{2}$

Correct answer:

$5bv^{2}x^{2} - 50b^{2}vx - 25b^{7}v^{2}$

Explanation:

$5bv(vx^{2} - 10bx - 5vb^{6})$

Distribute the outside term into the parentheses. Multiply each term in parentheses by 5bv :

$(5bv)(vx^{2}) - (5bv)(10bx) - (5bv)(5vb^{6})$

Now, for each member, move the similar variables into separate groups. You can do this because of the associative rule for multiplication:

$(5)(b)(v * v)(x^{2}) - (5 * 10)(b *b)(vx) - (5*5)(b *b^{6})(v*v)$

When multiplying variables of the same type, you add their exponents together. This gets you:

$(5)(b)(v^{1+1})(x^2)-(50)(b^{1+1})(vx)-(25)(b^{1+6})(v^{1+1})$

$(5)(b)(v^{2})(x^{2}) - (50)(b^{2})(vx) - (25)(b^{7})(v^{2})$

This is the same as:

$5bv^{2}x^{2} - 50b^{2}vx - 25b^{7}v^{2}$

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Example Question #61 : Algebra

$x^{2}* x^{5}=$

Possible Answers:

$x^{7}$

$x^{3}$

Cannot be simplified further

Correct answer:

$x^{7}$

Explanation:

To simplify, add the exponents:

$x^{2}* x^{5}=x^{7}$

Answer: $x^{7}$

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Example Question #62 : Algebra

$y^{5}* y^{9}=$

Possible Answers:

$y^{4}$

Cannot be simplified further

$y^{59}$

$y^{14}$

Correct answer:

$y^{14}$

Explanation:

Because the two terms have the same base, they can be multiplied together by adding the exponents:

$y^{5}* y^{9}=y^{14}$

Answer: $y^{14}$

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Example Question #63 : Algebra

$z^{3}* z^{6}=$

Possible Answers:

Cannot be simplified further

$z^{6}$

$z^{3}$

$z^{9}$

Correct answer:

$z^{9}$

Explanation:

To simplify the expression, add the exponents and keep the base unchanged:

$z^{3}* z^{6}=z^{9}$

Answer: $z^{9}$

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Example Question #64 : Algebra

$w^{10}* w^{10}=$

Possible Answers:

w+20

Cannot be simplified further

20w

$w^{100}$

$w^{20}$

Correct answer:

$w^{20}$

Explanation:

To multiply two terms with the same base and different exponents, add the exponents and leave the base unchanged:

$w^{10}* w^{10}=w^{20}$

Answer: $w^{20}$

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HSPT Math : Algebra

Study concepts, example questions & explanations for HSPT Math

All HSPT Math Resources

Example Questions

Example Question #31 : How To Multiply Variables

Example Question #671 : Concepts

Example Question #1267 : Hspt Mathematics

Example Question #32 : How To Multiply Variables

Example Question #31 : Operations

Example Question #32 : Operations

Example Question #61 : Algebra

Example Question #62 : Algebra

Example Question #63 : Algebra

Example Question #64 : Algebra

All HSPT Math Resources

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