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SAT II Math II : SAT Subject Test in Math II

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

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All SAT II Math II Resources

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

Example Questions

Example Question #1 : Mathematical Relationships

Add the numbers: 64+67+88

Possible Answers:

109

229

129

219

209

Correct answer:

219

Explanation:

Add the ones digits.

4+7+8=19

Add the tens digits using the one as the carryover.

6+6+8 +(1)= 21

Combine this number with the ones digit of the first number solved.

The answer is: 219

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Example Question #1 : Mathematical Relationships

Add the numbers: 84+86+36

Possible Answers:

216

194

196

206

294

Correct answer:

206

Explanation:

Add the ones digits.

4+6+6 = 16

Carry the one and add the tens digits.

8+8+3+(1) = 20

Combine this number with the ones digit of the first number.

The answer is: 206

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Example Question #31 : Sat Subject Test In Math Ii

Add the following numbers: 23+64+97

Possible Answers:

274

284

184

164

174

Correct answer:

184

Explanation:

Add the ones digits.

3+4+7= 14

Carry the one and add the tens digits.

2+6+9+(1) = 18

Combine this number with the ones digit of the first number.

The answer is: 184

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Example Question #32 : Sat Subject Test In Math Ii

Add the numbers: 35+36+88

Possible Answers:

149

249

139

259

159

Correct answer:

159

Explanation:

Add the ones digits.

5+6+8 = 19

Add the tens digits with the one as the carryover.

3+3+8+(1)=15

The answer is: 159

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Example Question #33 : Sat Subject Test In Math Ii

Add the numbers: 34+64+87

Possible Answers:

205

195

175

185

165

Correct answer:

185

Explanation:

Add the ones digits.

4+4+7= 15

Add the tens digit with the carryover.

3+6+8+(1) = 18

The answer is: 185

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Example Question #1 : Exponents And Logarithms

Solve for :

$5^{x} = 432$

Give the solution to the nearest hundredth.

Possible Answers:

$x \approx 3.77$

$x\approx 3.37$

$x \approx 1.94$

$x \approx 2.44$

$x \approx 2.86$

Correct answer:

$x \approx 3.77$

Explanation:

One way is to take the common logarithm of both sides and solve:

$5^{x} = 432$

$\log \left (5^{x} \right )= \log 432$

$x \log 5= \log 432$

$x = \frac{\log 432}{\log 5} \approx \frac{2.6355}{0.6990} \approx 3.77$

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Example Question #1 : Exponents And Logarithms

Solve for :

$\left (e+1 \right ) ^{k} = 100$

Give your answer to the nearest hundredth.

Possible Answers:

$k \approx 3.61$

$k \approx 1.43$

$k \approx 3.51$ '

$k \approx 1.89$

$k \approx 1.69$

Correct answer:

$k \approx 3.51$ '

Explanation:

Take the common logarithm of both sides and solve for :

$\left (e+1 \right ) ^{k} = 100$

$\log \left (e+1 \right ) ^{k} = \log 100$

$k \log \left (e+1 \right ) = 2$

$k= \frac{2}{ \log \left (e+1 \right ) }$

$k= \frac{2}{ \log \left (2.7183+1 \right ) } \approx \frac{2}{ \log 3.7183 } \approx \frac{2}{1.3133} \approx 3.51$

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Example Question #1 : Exponents And Logarithms

Solve for :

$10^{t} + 1 = e$

Give your answer to the nearest hundredth.

Possible Answers:

$t \approx 0.46$

$t \approx 0.42$

$t \approx 0.24$

$t \approx -0.57$

The equation has no solution.

Correct answer:

$t \approx 0.24$

Explanation:

$10^{t} + 1 = e$

$10^{t} = e - 1$

Take the common logarithm of both sides and solve for :

$\log 10^{t} = \log \left (e - 1 \right )$

$t = \log (e - 1) \approx \log (2.718-1) \approx \log 1.718 \approx 0.24$

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Example Question #1 : Exponents And Logarithms

Solve for :

$e^{x} = \pi$

Give your answer to the nearest hundredth.

Possible Answers:

$x \approx 0.87$

$x \approx 1.14$

The equation has no solution.

$x \approx 2.01$

$x \approx 0.50$

Correct answer:

$x \approx 1.14$

Explanation:

Take the natural logarithm of both sides and solve for :

$e^{x} = \pi$

$\ln e^{x} = \ln \pi$

$x = \ln \pi \approx \ln 3.1416 \approx 1.14$

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Example Question #1 : Exponents And Logarithms

To the nearest hundredth, solve for :

$2^{x+4} = 5^{x}$

Possible Answers:

$x \approx 1.20$

$x \approx -1.68$

$x \approx 6.32$

$x \approx 3.03$

$x \approx 4.48$

Correct answer:

$x \approx 3.03$

Explanation:

Take the common logarithm of both sides, then solve the resulting linear equation.

$2^{x+4} = 5^{x}$

$\log 2^{x+4} = \log 5^{x}$

$(x + 4) \log 2 = x \log 5$

$x\log 2 + 4\log 2 = x \log 5$

$4\log 2 = x \log 5 - x\log 2$

$4\log 2 = x\left ( \log 5 - \log 2 \right )$

$x = \frac{4\log 2 }{ \log 5 - \log 2 }$

$x \approx \frac{4 \cdot 0.3010}{0.6990-0.3010} \approx 3.03$

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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Mathematical Relationships

Example Question #1 : Mathematical Relationships

Example Question #31 : Sat Subject Test In Math Ii

Example Question #32 : Sat Subject Test In Math Ii

Example Question #33 : Sat Subject Test In Math Ii

Example Question #1 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

Example Question #1 : Exponents And Logarithms

All SAT II Math II Resources

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