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SAT II Math I : SAT Subject Test in Math I

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Example Questions

Example Question #7 : Simplifying Expressions

Decrease 9x + 4y by 20%. Which of the following will this be equal to?

Possible Answers:

7.2x + 3.2y

9x + 3.2y

1.8x+4y

1.8x+0.8y

7.2x + 4y

Correct answer:

7.2x + 3.2y

Explanation:

A number decreased by 20% is equivalent to 100% of the number minus 20% of the number. This is taking 80% of the number, or, equivalently, multiplying it by 0.8.

Therefore, 9x + 4y decreased by 20% is 0.8 times this, or

$0.8 \left (9x + 4y \right ) = 0.8 \cdot 9x + 0.8 \cdot 4y = 7.2x + 3.2y$

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Example Question #2 : Simplifying Expressions

Divide:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

Possible Answers:

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

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

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

$\frac{ 3 }{ 2 }x ^{2} + 2x + 8 + \frac{1}{x}$

$\frac{ 3 }{ 2 }x ^{3} + 2x^{2} + 8x + 1$

Correct answer:

$\frac{ 3 }{ 2 }x ^{2} + 2x + 8 + \frac{1}{x}$

Explanation:

Divide termwise:

$\left ( 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x \right ) \div 8x^{2}$

$=\frac{ 12 x ^{4} + 16x^{3}+ 64x^{2}+8 x }{ 8x^{2}}$

$=\frac{ 12 x ^{4} }{ 8x^{2}} + \frac{ 16x^{3} }{ 8x^{2}} + \frac{ 64x^{2} }{ 8x^{2}} +\frac{ 8 x }{ 8x^{2}}$

$=\frac{ 12 }{ 8 }x ^{4-2} + \frac{ 16 }{ 8}x^{3-2} + \frac{ 64 }{ 8 } +\frac{ 8 }{ 8} \cdot \frac{1}{x^{2-1}}$

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

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Example Question #9 : Simplifying Expressions

Simplify the expression:

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

Possible Answers:

x-4

x+2

x-2

x+4

Correct answer:

x-2

Explanation:

To solve this problem, we first need to factor the numerator. We are looking for two numbers that multiply to equal -8 and sum to equal 2.

4*(-2)=-8

4+(-2)=2

Now, we can write out our expression in fraction form.

$\frac{(x+4)(x-2)}{x+4}$

Since we have the like term x+4 in the numerator and denominator, we can cancel them out of our expression.

$\frac{(x+4)(x-2)}{x+4}= x-2$

Thus, our answer is x-2 .

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Example Question #3 : Simplifying Expressions

Simplify:

$\frac{(x^5y^2z^3)(x^2yz)}{x^4y^{-3}z^3}$

Possible Answers:

x^3y^5z

x^4y^1z

x^6y^5

x^3y^6z

x^3z

Correct answer:

x^3y^6z

Explanation:

To simplify, we begin by simplifying the numerator. When muliplying like bases with different exponents, their exponents are added.

For x: $\small 4+3=7$

For y: $\small 2+1=3$

For z: $\small 3+1=4$

The numerator is now x^7y^3z^4 .

When dividing like bases, their exponents are subtracted.

For x: $\small 7-4=3$

For y: $\small 3-(-3)=6$

For z: $\small 4-3=1$

Thus, our answer is x^3y^6z .

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Example Question #111 : Single Variable Algebra

Decrease 6y + 17 by 30%. Which of the following will this be equal to?

Possible Answers:

1.8y +5.1

4.2y + 17

4.2y + 11.9

6y-13

The correct answer is not among the other responses.

Correct answer:

4.2y + 11.9

Explanation:

A number decreased by 30% is equivalent to 100% of the number minus 30% of the number. This is taking 70% of the number, or, equivalently, multiplying it by 0.7.

Therefore, 6y + 17 decreased by 30% is 0.7 times this, or

$0.7 \left (6y + 17 \right ) = 0.7 \cdot 6y +0.7 \cdot 17 = 4.2y + 11.9$

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Example Question #12 : Simplifying Expressions

The polynomial $x^{4} - 27x^{2}+ 5x - A$ is divisible by the linear binomial x - 6 . Evaluate .

Possible Answers:

A = 354

A = - 354

None of the other choices gives the correct answer.

A =-294

A =294

Correct answer:

A = 354

Explanation:

By the factor theorem, a polynomial p (x) is divisible by the linear binomial x - c if and only if p(c) = 0 . Therefore, we want the value of that makes the polynomial equal to 0 when evaluated at x = 6 .

$x^{4} - 27x^{2}+ 5x - A = 0$

$6^{4} - 27 \cdot 6^{2}+ 5 \cdot 6 - A = 0$

$1,296- 27 \cdot 36+ 30- A = 0$

1,296- 972+ 30- A = 0

354- A = 0

354=A

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Example Question #13 : Simplifying Expressions

Factor:

$125T^{3}+729$

Possible Answers:

$(5T+9)(5T-9)^{2}$

$(5T+9)(25T^{2}+81)$

$(5T+9)(25T^{2}-45T +81)$

$(5T+9)^{3}$

$(5T+9)(25T^{2}+45T +81)$

Correct answer:

$(5T+9)(25T^{2}-45T +81)$

Explanation:

$125T^{3}+729$ can be rewritten as $(5T)^{3} + 9^{3}$ and is therefore the sum of two cubes. As such, it can be factored using the pattern

$A^{3}+B^{3} =(A+B) (A^{2}-AB +B^{2})$

where A = 5T, B = 9 .

$125T^{3}+729$

$=(5T)^{3} + 9^{3}$

$= (5T+9)[(5T)^{2}-5T \cdot 9 +9^{2}]$

$= (5T+9)(25T^{2}-45T +81)$

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Example Question #271 : Sat Subject Test In Math I

Factor completely:

$t^{3} +4t^{2}-9t -36$

Possible Answers:

$\left (t ^{2}+6 \right ) \left (t-6} \right )$

$\left (t ^{2}-6 \right ) \left (t+6} \right )$

$\left (t +3 \right )^{2} \left (t -4} \right )$

$\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )$

$\left (t ^{2}+9 \right ) \left (t-4} \right )$

Correct answer:

$\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )$

Explanation:

The grouping technique works here:

$t^{3} +4t^{2}-9t -36$

$=\left (t^{3} +4t^{2} \right )-\left (9t +36 \right )$

$=\left (t ^{2} \cdot t +t^{2} \cdot 4} \right )-\left (9 \cdot t +9 \cdot 4 \right )$

$=t ^{2} \left (t + 4} \right )-9 \left ( t + 4 \right )$

$=\left (t ^{2}-9 \right ) \left (t + 4} \right )$

The first factor is the difference of squares and can be factored further accordingly:

$\left (t ^{2}-9 \right ) \left (t + 4} \right )$

$=\left (t ^{2}- 3^{2} \right ) \left (t + 4} \right )$

$=\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )$

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Example Question #272 : Sat Subject Test In Math I

Factor completely:

$t^{3} - 243$

Possible Answers:

$(t - 3)\left ( t^{2}+ 3t + 81\right )$

The polynomial is prime.

$(t - 3)(t-9) ^{2}$

$(t - 3)\left ( t^{2} + 81\right )$

$(t - 3)\left ( t^{2}+9t + 81\right )$

Correct answer:

The polynomial is prime.

Explanation:

Since the first term is a perfect cube, the factoring pattern we are looking to take advantage of is the difference of cubes pattern. However, 243 is not a perfect cube of an integer $(6^{3} = 216 < 243 < 7^{3} = 343)$ , so the factoring pattern cannot be applied. No other pattern fits, so the polynomial is a prime.

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Example Question #271 : Sat Subject Test In Math I

Exponentiate:

$\left (x^{2} + x + 7 \right )^{2}$

Possible Answers:

$x^{4} + x^{3} +8x^{2} + 14x+ 14$

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

$x^{4} + 2x^{3} +8x^{2} + 14x+ 14$

$x^{4} + x^{2} + 49$

$x^{4} + x^{3} +15x^{2} + 14x+ 49$

Correct answer:

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

Explanation:

$\left (x^{2} + x + 7 \right )^{2} = \left (x^{2} + x + 7 \right )\left (x^{2} + x + 7 \right )$

Vertical multiplication is perhaps the easiest way to multiply trinomials.

$x^{2} + x + 7$

$\underline{x^{2} + x + 7}$

$7x^{2} + 7x + 49$

$x ^{3 } + x^{2} + 7x$

$\underline{x ^{4 } + x^{3 } + 7x^{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; }$

$x^{4} + 2x^{3} +15x^{2} + 14x+ 49$

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SAT II Math I : SAT Subject Test in Math I

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #7 : Simplifying Expressions

Example Question #2 : Simplifying Expressions

Example Question #9 : Simplifying Expressions

Example Question #3 : Simplifying Expressions

Example Question #111 : Single Variable Algebra

Example Question #12 : Simplifying Expressions

Example Question #13 : Simplifying Expressions

Example Question #271 : Sat Subject Test In Math I

Example Question #272 : Sat Subject Test In Math I

Example Question #271 : Sat Subject Test In Math I

All SAT II Math I Resources

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