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PSAT Math : PSAT Mathematics

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Example Questions

Example Question #1 : How To Find Compound Interest

Donna wants to deposit money into a certificate of deposit so that in exactly ten years, her investment will be worth $100,000. The interest rate of the CD is 7.885%, compounded monthly.

What should Donna's initial investment be, at minimum?

Possible Answers:

$\$45,802.62$

$\$ 46,815.46$

$\$45,569.99$

$\$55,912.78$

More information is needed to answer the question.

Correct answer:

$\$45,569.99$

Explanation:

The formula for compound interest is

$A = A_{0}\left ( 1+ \frac{r}{N} \right )^{Nt}$

where $A_{0}$ is the initial investment, is the interest rate expressed as the decimal equivalent, is the number of periods per year the interest is compounded, is the number of years, and is the final value of the investment.

Set A = 100,000, r = 0.07885, N = 12 (monthly = 12 periods), and t = 10 , and evaluate $A_{0}$ :

$100,000 = A_{0} \left ( 1+ \frac{0.07885}{12} \right )^{12 \cdot 10}$

$100,000 \approx A_{0} \left ( 1+0.0065708\right )^{120}$

$100,000 \approx A_{0} \left ( 1 .0065708\right )^{120}$

$100,000 \approx A_{0} \cdot 2.1944$

$A_{0} \approx 100,000 \div 2.1944$

$A_{0} \approx 45,569.99$

The correct response is $45,569.99.

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Example Question #1 : How To Find Compound Interest

Tom invests $\$15$ , 000 in a savings account with an annual interest rate of $6\%$ . If his investment is compounded semiannually, how much interest does he earn after years?

Possible Answers:

$\$1390.91$

$\$1942.35$

$\$1758.45$

$\$1882.63$

$\$1841.31$

Correct answer:

$\$1882.63$

Explanation:

In order to find the interest earned, used the compound interest formula

$\text{Final Balance} = \text{Principal} \cdot (1+\frac{IR}{C})^{(\text{Time})(\text{C})}$

where represents the number of times the account is compounded each year, and represents the interest rate expressed as a decimal.

$=15,000\cdot(1+\frac{.06}{2})^{(2)(2)}$

$=15,000\cdot(1+.03)^{4}$

$=15,000\cdot(1.03)^{4}$

=16882.63

The account is worth $16882.63 after two years. Therefore Tom earns $1882.63 in interest.

16882.63-15000=1882.63

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Example Question #2 : Pattern Behaviors In Exponents

If a^x·a⁴ = a¹² and (b^y)³ = b¹⁵, what is the value of x - y?

Possible Answers:

-4

-9

-2

Correct answer:

Explanation:

Multiplying like bases means add the exponents, so x+4 = 12, or x = 8.

Raising a power to a power means multiply the exponents, so 3y = 15, or y = 5.

x - y = 8 - 5 = 3.

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Example Question #1 : How To Find Patterns In Exponents

If p and q are positive integrers and 27^p= 9^q, then what is the value of q in terms of p?

Possible Answers:

(3/2)p

(2/3)p

Correct answer:

(3/2)p

Explanation:

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 3^3p= 3^2q. So then 3p = 2q, and q = (3/2)p is our answer.

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Example Question #2 : How To Find Patterns In Exponents

Simplify 27^2/3.

Possible Answers:

125

729

Correct answer:

Explanation:

27^2/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations.

27^2/3 = (27²)^1/3 = 729^1/3 OR

27^2/3 = (27^1/3)² = 3²

Obviously 3² is much easier. Either 3² or 729^1/3 will give us the correct answer of 9, but with 3² it is readily apparent.

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Example Question #1 : How To Find Patterns In Exponents

If and are integers and

$\left ( \frac{1}{3} \right )^{a}=27^{b}$

what is the value of $a\div b$ ?

Possible Answers:

$-\frac{1}{3}$

$\frac{1}{3}$

Correct answer:

Explanation:

To solve this problem, we will have to take the log of both sides to bring down our exponents. By doing this, we will get $\dpi{100} \small a\ast log\left (\frac{1}{3} \right )= b\ast log\left ( 27 \right )$ .

To solve for $\dpi{100} \small \frac{a}{b}$ we will have to divide both sides of our equation by $\dpi{100} \small log\frac{1}{3}$ to get $\dpi{100} \small \frac{a}{b}=\frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ .

$\dpi{100} \small \frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ will give you the answer of –3.

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Example Question #4 : How To Find Patterns In Exponents

If $\log 2=0.301$ and $\log 3=0.477$ , then what is $\log 12$ ?

Possible Answers:

1.592

1.116

1.255

1.079

1.346

Correct answer:

1.079

Explanation:

We use two properties of logarithms:

$log(xy) = log (x) + log (y)$

$log(x^{n}) = nlog (x)$

So $\log 12=2 \log2+\log3$

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Example Question #571 : Algebra

Evaluate:

$x^{-3}x^{6}$

Possible Answers:

$x^{6}$

$x^{9}$

$x^{-3}$

$x^{3}$

$x^{-18}$

Correct answer:

$x^{3}$

Explanation:

$x^{m}\ast x^{n} = x^{m + n}$ , here m=-3 and n=6 , hence -3+6=3 .

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Example Question #5 : How To Find Patterns In Exponents

Solve for

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$

Possible Answers:

None of the above

Correct answer:

Explanation:

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$ = $\left ( \frac{3}{2} \right )^{3} = \left ( \frac{2}{3} \right )^{-3}$

x+1=-3 which means x=-4

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Example Question #6 : How To Find Patterns In Exponents

Which of the following statements is the same as:

$\dfrac{4^{x+2}}{2^y}$

$I)\ 2^{2x+4-y} \quad II)\ 2^{x+2} / 4^y \quad III)\ (4^x)( 4^2 )(2^{-y} )$

Possible Answers:

$I \mbox{ and } II \mbox{ only}$

$I \mbox{ only}$

$I \mbox{ and } III \mbox{ only}$

$I, II \mbox{ and } III$

$II \mbox{ and } III \mbox{ only}$

Correct answer:

$I \mbox{ and } III \mbox{ only}$

Explanation:

Remember the laws of exponents. In particular, when the base is nonzero:

$b^{m+n} = b^m b^n \qquad (b^m)^n = b^{mn} \\ (b \times c)^n = b^n c^n \qquad b^{-n} = 1/b^n$

An effective way to compare these statements, is to convert them all into exponents with base 2. The original statement becomes:

$4^{x+2} / 2^y = 4^{x+2} (2^{-y}) = (2^2)^{x+2} 2^{-y} = 2^{2(x+2) } 2^{-y} = 2^{2x+4-y}$

This is identical to statement I. Now consider statement II:

$2^{x+2} / 4^y = 2^{x+2} ( 4^{-y} ) = 2^{x+2} ( 2^2 )^{-y} = 2^{x+2} 2^{-2y} = 2^{x+2-2y} \neq 2^{2x+4-y}$

Therefore, statement II is not identical to the original statement. Finally, consider statement III:

$(4^x)(4^2 ) (2^{-y}) = (2^2)^x (2^2)^2 (2^{-y}) = (2^{2x} )( 2^4 ) (2^{-y}) = 2^{2x+4-y}$

which is also identical to the original statement. As a result, only I and III are the same as the original statement.

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PSAT Math : PSAT Mathematics

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : How To Find Compound Interest

Example Question #1 : How To Find Compound Interest

Example Question #2 : Pattern Behaviors In Exponents

Example Question #1 : How To Find Patterns In Exponents

Example Question #2 : How To Find Patterns In Exponents

Example Question #1 : How To Find Patterns In Exponents

Example Question #4 : How To Find Patterns In Exponents

Example Question #571 : Algebra

Example Question #5 : How To Find Patterns In Exponents

Example Question #6 : How To Find Patterns In Exponents

All PSAT Math Resources

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