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SAT Math : SAT Mathematics

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Example Questions

Example Question #1124 : Algebra

The population of a city will decrease by 15 percent every 50 years and the population starts at 120,000 people. Construct a function that describes this situation.

Possible Answers:

P(t)=120,000(0.85)^t

$P(t)=120,000(0.15)^{\frac{t}{50}}$

$P(t)=120,000(0.80)^{\frac{t}{50}}$

$P(t)=120,000(0.85)^{50t}$

$P(t)=120,000(0.85)^{\frac{t}{50}}$

Correct answer:

$P(t)=120,000(0.85)^{\frac{t}{50}}$

Explanation:

To construct a function that describes this situation first identify what is known.

Since this particular situation is talking about population decrease, the function will be an exponential decay.

Recall that an exponential decay function is in the form,

$P(t)=I(1-r)^{t}$

where,

$\\P(t)=\text{ Finial population} \\I=\text{ Initial population} \\r=\text{ Rate} \\t=\text{Time}$

Since the statements says that the population decreases every 50 years we can rewrite the general form to,

$P(t)=I(1-r)^{\frac{t}{50}}$

Now substituting in the known values, the function can be written.

$P(t)=120,000(0.85)^{\frac{t}{50}}$

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Example Question #1125 : Algebra

Screen shot 2016 02 18 at 2.50.12 pm

The above graph shows supply and demand for a particular Product. What is the equation for the demand of this product?

Possible Answers:

y=-4x+50

y=-2x+50

y=4x+50

y=2x+50

y=-4x

Correct answer:

y=-4x+50

Explanation:

We can determine the demand equation by using point slope form.

Point slope form is y-y_0=m(x-x_0) , where (x_0, y_0) , and is the slope, where $m=\frac{y_2-y_1}{x_2-x_1}$ .

Let y_2=50 , y_1=10 , x_2=0 , and x_1=10 .

$m=\frac{50-10}{0-10}=-4$

Now we have

y-y_0=-4(x-x_0)

Choose a point (0,50) ,

y-50=-4(x-0)

y=-4x+50

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Example Question #1126 : Algebra

Screen shot 2016 02 18 at 2.50.12 pm

If the equation of the demand line is y=-4x+50 , and the equation for supply is y=5x , determine the point where supply and demand is the same.

Possible Answers:

$x=\frac{50}{9}$

$x=\frac{14}{9}$

$x=\frac{500}{9}$

$x=\frac{5}{9}$

x=6

Correct answer:

$x=\frac{50}{9}$

Explanation:

To solve this, all we need to do is set the equations equal to each other.

-4x+50=5x

Now solve for

50=9x

$x=\frac{50}{9}$

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Example Question #1127 : Algebra

Amanda has 5000 ants in an ant farm and their population grows $20\%$ annually. How many ants will be in Amanda's ant farm in 6 years?

Possible Answers:

15000

14700

14930

14920

14900

Correct answer:

14930

Explanation:

This is an exponential growth problem, so let's recall the equation for exponential growth.

P(t)=P(1+r)^t , where is the starting amount of ants, is the growth rate, and is the time in years.

First step is to convert $20\%$ into a decimal. $20\%=0.20$

P(t)=5000(1+0.20)^t

P(t)=5000(1.20)^t

P(6)=5000(1.20)^6=14930

So in 6 years, Amanda will have 14930 ants.

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Example Question #1128 : Algebra

The equation for the universal gravitation is $F_g=\frac{Gm_1m_2}{r^2}$ , $m_1=\text{mass of object 1}$ , $m_2=\text{mass of object 2}$ , $r=\text{radius between masses}$ , and is the universal gravitational constant. If $m_1=10 \ kg$ , $m_2=20 \ kg$ , $F_g=1.34\times10^{-12} N$ and $G=6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2}$ , what is the radius between the two masses? Round to the nearest tenth.

Hint: $\frac{\text{kg\:m}}{s^2}=1 \text{Newton}$

Possible Answers:

$r=99.60\ m$

$r=99.90\ m$

$r=100\ m$

$r=99.70\ m$

$r=99.80\ m$

Correct answer:

$r=99.80\ m$

Explanation:

The first step is to plug in all the values into the equation.

$1.34\times10^{-12} N=\frac{6.67408\times10^{-11}m^3kg^{-1}s^{-2}\cdot 10kg\cdot 20kg}{r^2}$

Now we will solve for .

$r^2=-\frac{6.67408\times10^{-11}m^3kg^{-1}s^{-2}\cdot 10kg\cdot 20kg}{1.34\times10^{-12} N}$

$r^2=\frac{1.334816\times10^{-8}m^3kg^{-1}s^{-2} kg^2}{1.34\times10^{-12} N}$

$r^2=9961.313\:m^2$

Take the square root on each side

$r=\sqrt{9961.313\:m^2}=99.80\ m$

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Example Question #1129 : Algebra

If g(x) = 3x+2 , what is the value of g(-2) ?

Possible Answers:

Correct answer:

Explanation:

g(x) = 3x+2

g(-2) = 3(-2)+2=-6+2=-4

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Example Question #221 : New Sat

If f(x) = 4x + 2(x+1) , then f(8x) =?

Possible Answers:

48x + 8

16x + 4

48x + 2

32x + 1

Correct answer:

48x + 2

Explanation:

f(x) = 4x + 2(x+1) ,

f(8x) = 4(8x) + 2(8x+1) = 32x + 16x + 2 = 48x + 2

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Example Question #162 : How To Find F(X)

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate $(g \circ f ) (10)$ .

Possible Answers:

$\textup{None of the other choices gives the correct response}$

$-12 \frac{1}{2}$

$24 \frac{1}{2}$

$2 \frac{3}{4}$

Correct answer:

$24 \frac{1}{2}$

Explanation:

By definition, $(g \circ f ) (10) = g [f(10)]$ .

First, evaluate f(10) by setting x = 10 in the definition of f(x) :

f(x) = 4x- 17

$f(10) = 4 \cdot 10 - 17= 40 - 17 = 23$

$(g \circ f ) (10) = g [f(10)] = g(23)$ , so evaluate g(23) similarly:

$g(x) = \frac{1}{2}x+ 13$

$g(23 ) = \frac{1}{2} \cdot 23 + 13 = 11 \frac{1}{2} + 13 = 24 \frac{1}{2}$

$(g \circ f ) (10) = 24 \frac{1}{2}$

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Example Question #202 : Algebraic Functions

Define $f(x) = \sqrt{2x+ 3}$ , restricting the domain of the function to the interval [11, 39 ] .

Give the range of the function.

Possible Answers:

[-9, 9 ]

$(-\infty, \infty )$

[5, 9 ]

None of these

$[-9, -5 ] \cup [5, 9 ]$

Correct answer:

[5, 9 ]

Explanation:

If $11 \le x \le 39$ , then, by way of the properties of inequality, we can multiply all expressions by 2:

$2 \cdot 11 \le 2 \cdot x \le 2 \cdot 39$

$22 \le 2 x \le 78$

then add 3 to all expressions:

$22+ 3 \le 2 x+ 3 \le 78 + 3$

$25 \le 2 x+ 3 \le 81$

Taking the square root of all expressions, we get

$\sqrt{25 }\le \sqrt{2 x+ 3} \le \sqrt{81}$

$5 \le \sqrt{2 x+ 3} \le 9$

$5 \le f(x) \le 9$ .

The correct range is [5, 9 ] .

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Example Question #162 : How To Find F(X)

Define two functions as follows:

f(x) = 4x- 17

$g(x) = \frac{1}{2}x+ 13$

Evaluate:

$(f \circ g ) (10)$

Possible Answers:

$24 \frac{1}{2}$

None of these

$-12 \frac{1}{2}$

$2 \frac{3}{4}$

Correct answer:

Explanation:

By definition, $(f \circ g ) (10) = f [g(10)]$ .

First, evaluate g(10) by setting x = 10 in the definition of g(x) :

$g(x) = \frac{1}{2}x+ 13$

$g(10) = \frac{1}{2} \cdot 10 + 13 = 5 + 13 = 18$

$(f \circ g ) (10) = f [g(10)] = f(18)$ , so evaluate f(18) similarly:

f(x) = 4x- 17

$f(18) = 4 \cdot 18 - 17= 72 - 17 = 55$

$(f \circ g ) (10) = 55$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1124 : Algebra

Example Question #1125 : Algebra

Example Question #1126 : Algebra

Example Question #1127 : Algebra

Example Question #1128 : Algebra

Example Question #1129 : Algebra

Example Question #221 : New Sat

Example Question #162 : How To Find F(X)

Example Question #202 : Algebraic Functions

Example Question #162 : How To Find F(X)

All SAT Math Resources

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