SAT Practice Test - Math - Convex Polygons

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Convex Polygons.

_______________________


SAT Practice Test 2015.

The figure above shows a regular hexagon with sides of length a and a square with sides of length a. If the area of the hexagon is $384\sqrt{3}$ square inches, what is the area, in square inches, of the square?

A) 256
B) 192
C) $64\sqrt{3}$
D) $16\sqrt{3}$

Answer:

The area of an equilateral triangle is given by the formula $A_T=L^2\sqrt{3}/4$, where L is the length of the side of the triangle.

The hexagon is composed by 6 equilateral triangles. Therefore the area of the hexagon is given by the formula $A_H=6(A_T)=6L^2\sqrt{3}/4$.

And it was given that the area of the hexagon is $384\sqrt{3}$. Thus
$6L^2\sqrt{3}/4=384\sqrt{3}$
$6L^2/4=384$
$L^2=384(4)/6$
$L^2=256$. This is the area of the square.

Answer: A

_____________________________


SAT Practice Test 2015. $nA=360$
The measure A, in degrees, of an exterior angle of a regular polygon is related to the number of sides, n, of the polygon by the formula above. If the measure of an exterior angle of a regular polygon is greater than 50°, what is the greatest number of sides it can have?
A) 5
B) 6
C) 7
D) 8

Answer:

$nA=360$
$A=360/n$

Since A>50,
$360/n>50$
$360>50n$
$360/50>n$
$7,2>n$

Answer: C

_____________________________

SAT Practice Test - Math - Factoring

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Factoring.

_______________________


SAT 2015 Test. $9a^4+12a^2b^2+4b^4$
Which of the following is equivalent to the expression shown above?
A) $(3a^2+2b^2)^2$
B) $(3a+2b)^4$
C) $(9a^2+4b^2)^2$
D) $(9a+4b)^4$

Answer:

Let's develop the expression in option A:
$(3a^2+2b^2)^2=$
$9a^4+2(3a^2)(2b^2)+4b^4=$
$9a^4+12a^2b^2+4b^4=$

Answer: A

_____________________________

SAT Practice Test - Math - Interest Rate, and Profit Margin

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Interest Rate, and Profit Margin.

_______________________


SAT Practice Test 2015. The amount of money a performer earns is directly proportional to the number of people attending the performance. The performer earns 120 dollars at a performance where 8 people attend.
The performer uses 43% of the money earned to pay the costs involved in putting on each performance. The rest of the money earned is the performer’s profit. What is the profit (in dollars) the performer makes at a performance where 8 people attend?
A) 51.60
B) 57.00
C) 68.40
D) 77.00

Answer:

If the performer uses 43% of the money to pay the costs, the profit will be 57%. Thus we just have to find what is 57% of 120 dollars:
$Profit=0.57(120)$
$Profit=68.4$

Answer: C

_____________________________


SAT Practice Test 2015. Jessica opened a bank account that earns 2 percent interest compounded annually. Her initial deposit was 100 dollars, and she uses the expression $100(x)^t$ to find the value of the account after t years.
What is the value of x in the expression?

Answer:

x is the interest rate, that is, x=1.02

Answer: 1.02

_____________________________


SAT 2015 Test. Jessica opened a bank account that earns 2 percent interest compounded annually. Her initial deposit was 100 dollars, and she uses the expression $100(x)^t$ to find the value of the account after t years.
Jessica’s friend Tyshaun found an account that earns 2.5 percent interest compounded annually. Tyshaun made an initial deposit of 100 dollars into this account at the same time Jessica made a deposit of 100 dollars into her account. After 10 years, how much more money will Tyshaun’s initial deposit have earned than Jessica’s initial deposit? (Round your answer to the nearest cent and ignore the dollar sign when gridding your response.)

Answer:

Jessica will have: $J=100(1.02)^{10}=121,90$
Tyshaun will have: $T=100(1.025)^{10}=128,01$

Tyshaun(-)Jessicas: $128,01-121,90=6.11$

Answer: 6.11

_____________________________


SAT 2015 Test. Alma bought a laptop computer at a store that gave a 20 percent discount off its original price. The total amount she paid to the cashier was p dollars, including an 8 percent sales tax on the discounted price. Which of the following represents the original price of the computer in terms of p?

A) $0.88p$

B) $\frac{p}{0.88}$

C) $(0.8)(1.08)p$

D) $\frac{p}{(0.8)(1.08)}$

Answer:

Let's call the original price R.
The price with a 20% discount, and before tax, is given by the formula: $R(0.8)$
The total amount she paid to the cashier is this value plus tax:

$p=R(0.8)(1.08)$

$\frac{p}{(0.8)(1.08)}=R$

Answer: D

_____________________________


SAT Practice Test. A company’s manager estimated that the cost C, in dollars, of producing n items is $C=7n+350$. The company sells each item for 12 dollars. The company makes a profit when total income from selling a quantity of items is greater than the total cost of producing that quantity of items. Which of the following inequalities gives all possible values of n for which the manager estimates that the company will make a profit?
A. $n<70$
B. $n<84$
C. $n>70$
D. $n>84$

Answer:

The total income for this company is: $12.n$

The company will make a profit when total income is greater than the total cost. Therefore,

$12n>7n+350$
$12n-7n>350$
$5n>350$
$n>70$

Answer: C

_____________________________

SAT Practice Test - Math - Modulus Function

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Modulus Functions.

Learn about Modulus Function

_______________________


SAT Practice Test 2015. Which of the following equations has a graph in the xy-plane for which y is always greater than or equal to -1?
A) $y=|x|-2$
B) $y=x^2-2$
C) $y=(x-2)^2$
D) $y=x^3-2$

Answer:

Choice A and B are wrong, because for x=0, y=-2
Choice C is correct, because $y\geq0$, for any value of x.
Choice D is wrong, because y can take any value.

Answer: C

_____________________________


SAT Practice Test 2015. For what value of n is $|n-1|+1$ equal to 0?
A) 0
B) 1
C) 2
D) There is no such value of n.

Answer:

$|n-1|+1=0$
$|n-1|=-1$

The modulus of any number gives us the magnitude of that number. Therefore the modulus is always positive.

Answer: D

_____________________________

SAT Practice Test - Math - Data Interpretation from Graphs

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Data Interpretation from Graphs.

_______________________


SAT Practice Test 2015.

Two samples of water of equal mass are heated to 60 degrees Celsius (°C). One sample is poured into an insulated container, and the other sample is poured into a non-insulated container. The samples are then left for 70 minutes to cool in a room having a temperature of 25°C. The graph above shows the temperature of each sample at 10-minute intervals. Which of the following statements correctly compares the average rates at which the temperatures of the two samples change?

A) In every 10-minute interval, the magnitude of the rate of change of temperature of the insulated sample is greater than that of the non-insulated sample.
B) In every 10-minute interval, the magnitude of the rate of change of temperature of the non-insulated sample is greater than that of the insulated sample.
C) In the intervals from 0 to 10 minutes and from 10 to 20 minutes, the rates of change of temperature of the insulated sample are of greater magnitude, whereas in the intervals from 40 to 50 minutes and from 50 to 60 minutes, the rates of change of temperature of the non-insulated sample are of greater magnitude.
D) In the intervals from 0 to 10 minutes and from 10 to 20 minutes, the rates of change of temperature of the non-insulated sample are of greater magnitude, whereas in the intervals from 40 to 50 minutes and from 50 to 60 minutes, the rates of change of temperature of the insulated sample are of greater magnitude.

Answer:

The rate of change of temperature corresponds to the slope of the graph.

In the intervals from 0 to 10 minutes and from 10 to 20 minutes, the slope of the curve of the non-insulated sample is greater than that of the insulated sample, whereas in the intervals from 40 to 50 minutes and from 50 to 60 minutes, the slope of the curve of the insulated sample is greater than that of the non-insulated sample.

Answer: D

_____________________________


SAT Practice Test 2015.

According to the line of best fit in the scatterplot above, which of the following best approximates the year in which the number of miles traveled by air passengers in Country X was estimated to be 550 billion?
A) 1997
B) 2000
C) 2003
D) 2008

Answer:

The horizontal line that corresponds to 550 billion miles intersects the line of best fit at a point between 2000 and 2005, closer to 2005. Therefore, the choice that best approximates the year in which the number of miles traveled was estimated to be 550 billion is 2003.

Answer: C

_____________________________


SAT Test 2015.

The graph above displays the total cost C, in dollars, of renting a boat for h hours.

What does the C-intercept represent in the graph?
A) The initial cost of renting the boat
B) The total number of boats rented
C) The total number of hours the boat is rented
D) The increase in cost to rent the boat for each additional hour

Answer:

The total cost C of renting a boat is the sum of the initial cost to rent the boat plus the product of the cost per hour and the number of hours, h, that the boat is rented. The C-intercept represents the point where h=0, that is, the initial cost of renting the boat.

Answer: A

_____________________________


SAT Test 2015. The following question refers to the graph in the previous question.
Which of the following represents the relationship between h and C?
A) C=5h
B) C=(3/4)h+5
C) C=3h+5
D) h=3C

Answer:

This is the graph of a 1st degree equation: C=ah+5, where 5 is the initial cost, as seen in the previous question.
The graph shows that a 5-hour rent costs 20 dollars. Substituting this in the equation:
$C=ah+5$
$20=a5+5$
$20-5=a5$
$15=a5$
$a=3$

And the equation is $C=3h+5$

Answer: C

_____________________________


SAT Test 2015.

The number of rooftops with solar panel installations in 5 cities (A, B, C, D and E) is shown in the graph above. If the total number of installations is 27,500, what is an appropriate label for the vertical axis of the graph?
A) Number of installations (in tens)
B) Number of installations (in hundreds)
C) Number of installations (in thousands)
D) Number of installations (in tens of thousands)

Answer:

City A has 9(x) installations;
City B has 5(x) installations;
City C has 6(x) installations;
City D has 4(x) installations;
City E has 3.5(x) installations.

The sum of these five cities is: (9+5+6+4+3.5)(x) installations, that is, 27.5(x) installations.
This total should equal to the given 27,500 installations:
27.5(x)=27,500
x=1,000

Answer: C

_____________________________

SAT Practice Test - Math - Angles

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Angles.

_______________________


SAT Test 2015.

In the figure above, lines l and m are parallel and lines s and t are parallel. If the measure of ∠1 is 35°, what is the measure of ∠2 ?
A) 35°
B) 55°
C) 70°
D) 145°

Answer:

If two parallel lines are cut by a transversal, the corresponding angles are congruent. The figure below shows corresponding angles that measure 35°:

Given that Supplementary Angles add up to 180 degrees, ∠2 measures 145°.

Answer: D

_____________________________

SAT Practice Test - Math - Algebra

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Algebra.

_______________________


SAT Practice Test 2015.

$I=\frac{P}{4\pi{r}^2}$

At a large distance r from a radio antenna, the intensity of the radio signal I is related to the power of the signal P by the formula above.

Which of the following expresses the square of the distance from the radio antenna in terms of the intensity of the radio signal and the power of the signal?

A) $r^2=\frac{IP}{4\pi}$

B) $r^2=\frac{P}{4\pi{I}}$

C) $r^2=\frac{4\pi{I}}{P}$

D) $r^2=\frac{I}{4\pi{P}}$

Answer:

$I=\frac{P}{4\pi{r}^2}$

Multiplying both sides of the equation by $r^2$:

$Ir^2=\frac{P}{4\pi}$

Now dividing both sides of the equation by $I$:

$r^2=\frac{P}{4\pi{I}}$

Answer: B

_____________________________


SAT Practice Test 2015. This question refers to the same data used in the previous question:

For the same signal emitted by a radio antenna, Observer A measures its intensity to be 16 times the intensity measured by Observer B. The distance of Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
A) 1/4
B) 1/16
C) 1/64
D) 1/256

Answer:

According to the previous question:

$r^2=\frac{P}{4\pi{I}}$

Thus, for Observer B, we should have: ${r_B}^2=\frac{P}{4\pi{I_B}}$

And for Observer A:

${r_A}^2=\frac{P}{4\pi{I_A}}=\frac{P}{4\pi{16I_B}}$

Therefore:

$\frac{{r_B}^2}{{r_A}^2}=\frac{\frac{P}{4\pi{I_B}}}{\frac{P}{4\pi{16I_B}}}$

$\frac{{r_B}^2}{{r_A}^2}=\frac{4\pi{16I_B}}{4\pi{I_B}}=16$

$\frac{{r_A}^2}{{r_B}^2}=\frac{1}{16}$

$\frac{r_A}{r_B}=\frac{1}{4}$

Answer: A

_____________________________


SAT Practice Test 2015. The expression $\frac{5x-2}{x+3}$ is equivalent to which of the following?

A) $\frac{5-2}{3}$

B) $5-\frac{2}{3}$

C) $5-\frac{2}{x+3}$

D) $5-\frac{17}{x+3}$

Answer:

$\frac{5x-2}{x+3}=$

$\frac{5x+15-17}{x+3}=$

$\frac{5x+15}{x+3}-\frac{17}{x+3}=$

$5-\frac{17}{x+3}=$

Answer: D

_____________________________


SAT Practice Test 2015. $R=\frac{F}{N+F}$

A website uses the formula above to calculate a seller’s rating, R, based on the number of favorable reviews, F, and unfavorable reviews, N. Which of the following expresses the number of favorable reviews in terms of the other variables?

A) $F=\frac{RN}{R-1}$

B) $F=\frac{RN}{1-R}$

C) $F=\frac{N}{1-R}$

D) $F=\frac{N}{R-1}$

Answer:

$R=\frac{F}{N+F}$

$R(N+F)=F$
$RN+RF=F$
$RN=F-RF$
$F(1-R)=RN$

$F=\frac{RN}{1-R}$

Answer: B

_____________________________


SAT Test 2015.
$a=1,052+1.08t$
The speed of a sound wave in air depends on the air temperature. The formula above shows the relationship between a, the speed of a sound wave, in feet per second, and t, the air temperature, in degrees Fahrenheit (°F).
Which of the following expresses the air temperature in terms of the speed of a sound wave?

A) $t=\frac{a-1,052}{1.08}$

B) $t=\frac{a+1,052}{1.08}$

C) $t=\frac{1,052-a}{1.08}$

D) $t=\frac{1.08}{a+1,052}$

Answer:

$a=1,052+1.08t$
$a-1,052=1.08t$

$t=\frac{a-1,052}{1.08}$

Answer: A

_____________________________


SAT Test 2015. If x > 3, which of the following is equivalent to

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

A) $\frac{2x+5}{x^2+5x+6}$

B) $\frac{x^2+5x+6}{2x+5}$

C) $2x+5$

D) $x^2+5x+6$

Answer:

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

$\frac{(x+2)(x+3)}{(x+3)+(x+2)}=$

$\frac{x^2+5x+6}{2x+5}$

Answer: B

_____________________________


SAT Test 2015. If $\frac{a}{b}=2$, what is the value of $\frac{4b}{a}$?
A) 0
B) 1
C) 2
D) 4

Answer:

$\frac{a}{b}=2$

$\frac{b}{a}=\frac{1}{2}$

$\frac{4b}{a}=\frac{4.1}{2}$

$\frac{4b}{a}=2$

Answer: C

_____________________________


SAT Test 2015. $(x^2y-3y^2+5xy^2)-(-x^2y+3xy^2-3y^2)$
Which of the following is equivalent to the expression above?
A) $4x^2y^2$
B) $8xy^2-6y^2$
C) $2x^2y+2xy^2$
D) $2x^2y+8xy^2-6y^2$

Answer:

$(x^2y-3y^2+5xy^2)-(-x^2y+3xy^2-3y^2)=$
$x^2y-3y^2+5xy^2+x^2y-3xy^2+3y^2=$
$x^2y+x^2y+5xy^2-3xy^2-3y^2+3y^2=$
$2x^2y+2xy^2$

Answer: C

_____________________________

SAT Practice Test - Math - Polynomials

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Polynomials.

_______________________


SAT 2015 Test. For a polynomial p(x), the value of p(3) is -2. Which of the following must be true about p(x)?
A) x-5 is a factor of p(x).
B) x-2 is a factor of p(x).
C) x+2 is a factor of p(x).
D) The remainder when p(x) is divided by x-3 is −2.

Answer:

 If the polynomial p(x) is divided by x-3, the result can be written as

$\frac{p(x)}{x-3}=q(x)+\frac{r}{x-3}$

$p(x)=(x-3)q(x)+r$

where q(x) is a polynomial and r is the remainder (in this case, r is a real number, since x-3 is a degree 1 polynomial).

It is given that $p(3)=-2$:

$p(x)=(x-3)q(x)+r$
$p(3)=(3-3)q(3)+r$
$-2=(0)q(3)+r$
$-2=r$

Answer: D

_____________________________


SAT Practice Test. The function f is defined by $f(x)=2x^3+3x^2+cx+8$, where c is a constant. In the x y plane, the graph of f intersects the x axis at the three points (-4, 0), (1/2, 0), and (p, 0). What is the value of c?
A. -18
B. -2
C. 2
D. 10

Answer:

If (-4, 0) is one solution, lt's solve for c, substituting -4 for x and 0 for y:
$f(x)=2x^3+3x^2+cx+8$
$0=2(-4)^3+3(-4)^2+c(-4)+8$
$0=-2.64+3.16-4c+8$
$4c=-72$
$c=-18$

Answer: A

_____________________________

SAT Practice Test - Math - Bisectors and Similarities in Triangles

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Bisectors and Similarities in Triangles.

_______________________


SAT Practice Test 2015.

In the figure above, AE||CD & and segment AD intersects segment CE at B. What is the length of
segment CE?

Answer:

Angles ABE and DBC have the same measure because they are vertical angles.
Since segment AE is parallel to segment CD, angles A and D are of the same measure (alternate interior angle theorem).
Thus, by the angle-angle theorem, triangle ABE is similar to triangle DBC. Therefore:

$\frac{BC}{BD}=\frac{BE}{AB}$

$(BC)(AB)=(BD)(BE)$
$(BC)(10)=(5)(8)$
$BC=4$

And segment CE measures:
$CE=BC+BE=4+8=12$

_____________________________


SAT 2015 Test.

In the xy-plane above, line A is parallel to line k. What is the value of p ?
A) 4
B) 5
C) 8
D) 10

Answer:

Since lines "l" and "k" are parallel, they have the same slope:

$\frac{2-0}{0-(-5)}=\frac{0-(-4)}{p-0}$

$\frac{2}{5}=\frac{4}{p}$

$2p=4(5)$
$p=10$

Answer: D

_____________________________


SAT 2015 Test.

A summer camp counselor wants to find a length, x, in feet, across a lake as represented in the sketch above. The lengths represented by AB, EB, BD, and CD on the sketch were determined to be 1800 feet, 1400 feet, 700 feet, and 800 feet, respectively. Segments AC and DE intersect at B, and ∠AEB and ∠CDB have the same measure. What is the value of x ?

Answer:

∠ABE and ∠CBD have the same measure (vertical angles);
∠AEB and ∠CDB have the same measure (given);
Therefore, ∠EAB and ∠DCB also have the same measure, and triangles EAB and DCB are similar.

Since the triangles are similar, the corresponding sides are in the same proportion; thus:

$\frac{CD}{AE}=\frac{BD}{EB}$

$\frac{800}{x}=\frac{700}{1400}$

$\frac{800}{x}=\frac{1}{2}$

$x=1600$ feet

_____________________________


SAT Practice Test.

The figure presents line segments A E and B D that intersect at point C. Line segments A B and D E are drawn resulting in two triangles A B C and E D C.

In the preceding figure, triangle A B C is similar to triangle E D C. Which of the following must be true?

A. Line segment A E is parallel to line segment B D.
B. Line segment A E is perpendicular to line segment B D.
C. Line segment A B is parallel to line segment D E.
D. Line segment A B is perpendicular to line segment D E.

Answer:

Given that triangle ABC is similar to triangle EDC, the corresponding angles $A\hat{B}C$ and $E\hat{D}C$ are congruent. If these two angles are congruent, the transversal BD must be crossing two parallel line segments (AB and DE).

The Alternate Interior Angles theorem states that, if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

Answer: C

_____________________________

SAT Practice Test - Math - Percentages, and Rates of Change

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Percentages and Rates of Change.

_______________________


SAT Practice Test 2015. The atomic weight of an unknown element, in atomic mass units (amu), is approximately 20% less than that of calcium. The atomic weight of calcium is 40 amu. Which of the following best approximates the atomic weight, in amu, of the unknown element?
A) 8
B) 20
C) 32
D) 48

Answer:

If the atomic weight of the unknown element is 20% less than that of calcium, than its atomic weight is 80% of the atomic weight of calcium, that is: 80%(40)=0.8(40)=32 amu

Answer: C

_____________________________


SAT Practice Test 2015. Katarina is a botanist studying the production of pears by two types of pear trees. She noticed that Type A trees produced 20 percent more pears than Type B trees did. Based on Katarina’s observation, if the Type A trees produced 144 pears, how many pears did the Type B trees produce?
A) 115
B) 120
C) 124
D) 173

Answer:

If Type A trees produced 20 percent more pears than Type B trees did, then:
$TypeA=1.2(TypeB)$

If the Type A trees produced 144 pears:
$144=1.2(TypeB)$
$TypeB=120$

Answer: B

_____________________________


SAT Practice Test 2015.

Which of the following best approximates the average rate of change in the annual budget for agriculture/natural resources in Kansas from 2008 to 2010?
A) 50,000,000 per year
B) 65,000,000 per year
C) 75,000,000 per year
D) 130,000,000 per year

Answer:

Budget 2008: 358,708
Budget 2010: 488,106
Increase in 2 years: 129,398
Average increase per year: 129,398/2=64,699

Answer: B

_____________________________


SAT Practice Test 2015. This question refers to the budget presented in the previous question.

Of the following, which program’s ratio of its 2007 budget to its 2010 budget is closest to the human resources program’s ratio of its 2007 budget to its 2010 budget?
A) Agriculture/natural resources
B) Education
C) Highways and transportation
D) Public safety

Answer:

Human Resources 2007: 4,051,050
Human Resources 2010: 5,921,379
Human Resources Ratio: 1,46

Agriculture/natural resources 2007: 373,904
Agriculture/natural resources 2010: 488,106
Agriculture/natural resources Ratio: 1,31

Education 2007: 2,164,607
Education 2010: 3,008,036
Education Ratio: 1,39

Highways and transportation 2007: 1,468,482
Highways and transportation 2010: 1,773,893
Highways and transportation Ratio: 1,21

Public safety 2007: 263,463
Public safety 2010: 464,233
Public safety Ratio: 1,76

Answer: B

_____________________________


SAT Practice Test 2015.

The data in the table above were produced by a sleep researcher studying the number of dreams people recall when asked to record their dreams for one week. Group X consisted of 100 people who observed early bedtimes, and Group Y consisted of 100 people who observed later bedtimes. If a person is chosen at random from those who recalled at least 1 dream, what is the probability that the person belonged to Group Y?
A) 68/100
B) 79/100
C) 79/164
D) 164/200

Answer:

Number ot people who recalled at least 1 dream in Group Y: 11+68=79.
Total number ot people who recalled at least 1 dream: 39+125=164.
The probability that the person belonged to Group Y: 79/164.

Answer: C

_____________________________


SAT Practice Test 2015. A survey was conducted among a randomly chosen sample of U. S. citizens about U. S. voter participation in the November 2012 presidential election. The following table displays a summary of the survey results.

According to the table, for which age group did the greatest percentage of people report that they had voted?
A. 18- to 34-year-olds
B. 35- to 54-year-olds
C. 55- to 74-year-olds
D. People 75 years old and over

Answer:

Let´s calcule the percentage for each age group:

18- to 34-year-olds: 30329/63008=48%
35- to 54-year-olds: 47085/74282=63%
55- to 74-year-olds: 43075/59998=71%
People 75 years old and over: 12459/17794=70%

Answer: C

_____________________________


SAT Practice Test. The following question refers to the information and table provided on the previous question.

Of the 18- to 34 year olds who reported voting, 500 people were selected at random to do a follow up survey where they were asked which candidate they voted for. There were 287 people in this follow up survey sample who said they voted for Candidate A, and the other 213 people voted for someone else. Using the data from both the follow up survey and the initial survey, which of the following is most likely to be an accurate statement?

A. About 123 million people 18 to 34 years old would report voting for Candidate A in the November 2012 presidential election.
B. About 76 million people 18 to 34 years old would report voting for Candidate A in the November 2012 presidential election.
C. About 36 million people 18 to 34 years old would report voting for Candidate A in the November 2012 presidential election.
D. About 17 million people 18 to 34 years old would report voting for Candidate A in the November 2012 presidential election.

Answer:

287 people out of 500 people of the 18- to 34 year old age group said they voted for Candidate A. That is about 57% of this sample.

57% out of 30,329,000 people is about 17 million people.

Answer: D

_____________________________

SAT Practice Test - Math - Ratios, Proportions, and Rule of Three

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Ratios, Proportions, and Rule of Three.

_______________________


SAT Practice Test 2015. A coastal geologist estimates that a certain country’s beaches are eroding at a rate of 1.5 feet per year. According to the geologist’s estimate, how long will it take, in years, for the country’s beaches to erode by 21 feet?

Answer:

This is a simple rule of three question.

The beach erodes 1.5 feet in 1 year.
The beach erodes 21 feet in x years.

Therefore,

$\frac{1.5}{21}=\frac{1}{x}$

$1.5x=21(1)$

$x=14$ years

Answer: 14

_____________________________


SAT Practice Test 2015.

Tony is planning to read a novel. The table above shows information about the novel, Tony’s reading speed, and the amount of time he plans to spend reading the novel each day. If Tony reads at the rates given in the table, which of the following is closest to the number of days it would take Tony to read the entire novel?
A) 6
B) 8
C) 23
D) 324

Answer:

Tony plans to read 180 minutes per day (3 hours), and he reads 250 words per minute. Therefore he reads 180*250=45000 words per day.

The novel has 346168 words. Thus it will take Tony 346168 / 45000 = 7.7 days to read the entire novel.

Answer: B

_____________________________


SAT Practice Test 2015. The amount of money a performer earns is directly proportional to the number of people attending the performance. The performer earns 120 dollars at a performance where 8 people attend. How much money (in dollars) will the performer earn when 20 people attend a performance?
A) 960
B) 480
C) 300
D) 240

Answer:

The performer earns 120 dollars, when 8 people attend.
The performer earns x dollars, when 20 people attend.

Therefore,

$\frac{120}{x}=\frac{8}{20}$

$8x=120(20)$
$x=300$

Answer: C

_____________________________


SAT Practice Test 2015. A quality control manager at a factory selects 7 lightbulbs at random for inspection out of every 400 lightbulbs produced. At this rate, how many lightbulbs will be inspected if the factory produces 20,000 lightbulbs?
A) 300
B) 350
C) 400
D) 450

Answer:

The manager selects 7 lightbulbs out of 400.
The manager will select x lightbulbs out of 20,000.

Therefore,

$\frac{7}{x}=\frac{400}{20000}$

$400x=7(20000)$
$x=7(50)$
$x=350$

Answer: B

_____________________________


SAT Practice Test 2015.

1 decagram = 10 grams
1,000 milligrams = 1 gram

A hospital stores one type of medicine in 2-decagram containers. Based on the information above, how many 1-milligram doses are there in one 2-decagram container?
A) 0.002
B) 200
C) 2,000
D) 20,000

Answer:

2 decagram = 2(10) grams = 20 grams
20grams = 20(1,000) milligrams = 20,000 miligrams = 20,000 doses

Answer: D

_____________________________


SAT Practice Test. An international bank issues its Traveler credit cards worldwide. When a customer makes a purchase using a Traveler card in a currency different from the customer’s home currency, the bank converts the purchase price at the daily foreign exchange rate and then charges a 4% fee on the converted cost.

Sara lives in the United States, but is on vacation in India. She used her Traveler card for a purchase that cost 602 rupees (Indian currency). The bank posted a charge of $9.88 to her account that included the 4% fee.

What foreign exchange rate, in Indian rupees per one U. S. dollar, did the bank use for Sara’s charge? Round your answer to the nearest whole number.

Answer:

Sara spent 602 rupees. If we divide this number by the exchange rate (r) in Indian rupees per one U. S. dollar, we will find the value of the purchase in dollar:

$P_{dollar}=602/r$

Now we have to add the 4% fee in order to find the charge to Sara's account:

$P_{dollar}.(1.04)=602/r.(1.04)=9.88$

$602.(1.04)=(9.88).r$

$r=63.36$ Indian rupees per one U. S. dollar

Answer: 63

_____________________________


SAT Practice Test. This question refers to the information provided in the previous question.

A bank in India sells a prepaid credit card worth 7,500 rupees. Sara can buy the prepaid card using dollars at the daily exchange rate with no fee, but she will lose any money left unspent on the prepaid card. What is the least number of the 7,500 rupees on the prepaid card Sara must spend for the prepaid card to be cheaper than charging all her purchases on the Traveler card? Round your answer to the nearest whole number of rupees.

Answer:

Using the exchange rate of 63 Indian rupees per one U. S. dollar, Sara will have to pay for the prepaid credit card: $(7500/63)$ U.S. dollar.

If Sara spends x Rupees in the Traveler card, her account will be charged $x/63.(1.04)$ dollars. We want this amount to be greater than the price of the prepaid credit card:

$x/63.(1.04)>7500/63$
$x.(1.04)>7500$
$x>7,211.5$

Answer: 7,212

_____________________________


SAT Practice Test. A typical image taken of the surface of Mars by a camera is 11.2 gigabits in size. A tracking station on Earth can receive data from the spacecraft at a data rate of 3 megabits per second for a maximum of 11 hours each day. If 1 gigabit equals 1,024 megabits, what is the maximum number of typical images that the tracking station could receive from the camera each day?
A. 3
B. 10
C. 56
D. 144

Answer:

If the tracking station can receive 3 megabits in one second; 11.2 gigabits will take x seconds to download.

$\frac{3}{1 sec}=\frac{11.2(1,024)}{x sec}$

$3=\frac{11.2(1,024)}{x}$

$x=\frac{11.2(1,024)}{3}=3,822$ sec=1.06h

This is how much time it takes for the station to receive one image.

Therefore, in 11 hours the station will be able to receive 10 images and a fraction of an image.

Answer: B

_____________________________


SAT Practice Test. The following table classifies 103 elements as metal, metalloid, or nonmetal and as solid, liquid, or gas at standard temperature and pressure.

What fraction of all solids and liquids in the preceding table are metalloids?

Answer:

The total number of solids and liquids is 92 (90+2).

The total number of solids and liquids that are also metalloids is 7 (7+0).

Therefore, the fraction of all solids and liquids in the preceding table that are metalloids is 7/92.


_____________________________

SAT Practice Test - Math - Means, Medians, and Statistics

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Means, Medians and Statistics.

_______________________


SAT Practice Test 2015. A sociologist chose 300 students at random from each of two schools and asked each student how many siblings he or she has. The results are shown in the table below.

There are a total of 2,400 students at Lincoln School and 3,300 students at Washington School.

What is the median number of siblings for all the students surveyed?
A) 0
B) 1
C) 2
D) 3

Answer:

The "median" is the value positioned in the middle of a list of values.

There are a total of 600 students in the survey:
260 students with 0 siblings (position 1 to 260 in the list)
190 students with 1 siblings (position 261 to 450 in the list)
90 students with 2 siblings (position 451 to 540 in the list)
40 students with 3 siblings (position 541 to 580 in the list)
20 students with 4 siblings (position 581 to 600 in the list)

The median will be the value in the position 300 or 301 in the list, that is, a student with 1 sibling.

Answer: B

_____________________________


SAT Practice Test 2015. The following question refers to the data in the previous question:

Based on the survey data, which of the following most accurately compares the expected total number of students with 4 siblings at the two schools?
A) The total number of students with 4 siblings is expected to be equal at the two schools.
B) The total number of students with 4 siblings at Lincoln School is expected to be 30 more than at Washington School.
C) The total number of students with 4 siblings at Washington School is expected to be 30 more than at Lincoln School.
D) The total number of students with 4 siblings at Washington School is expected to be 900 more than at Lincoln School.

Answer:

There are 300 students surveyed in Lincoln School, 10 of which have 4 siblings. Since there are a total of 2,400 students in Lincoln School, its expected total number of students with 4 siblings is:
(10/300)(2,400)=80 students.

There are 300 students surveyed in Washington School, 10 of which have 4 siblings. Since there are a total of 3,300 students in Washington School, its expected total number of students with 4 siblings is:
(10/300)(3,300)=110 students.

Answer: C

_____________________________


SAT Practice Test 2015. A survey was taken of the value of homes in a county, and it was found that the mean home value was 165,000 dollars and the median home value was 125,000 dollars. Which of the following situations could explain the difference between the mean and median home values in the county?
A) The homes have values that are close to each other.
B) There are a few homes that are valued much less than the rest.
C) There are a few homes that are valued much more than the rest.
D) Many of the homes have values between 125,000 and 165,000 dollars.

Answer:

The "mean" is the average, calculated by adding up all the values and dividing by the number of values.
The "median" is the value positioned in the middle of a list of values.

Whenever there are outliers in the data, the mean will be pulled in their direction while the median remains the same. The example in the question has a mean that is larger than the median, and so one possible explanation is that there are a few homes that are valued much more than the rest..

Answer: C

_____________________________


SAT Practice Test 2015.

The table above summarizes the results of 200 law school graduates who took the bar exam. If one of the surveyed graduates who passed the bar exam is chosen at random for an interview, what is the probability that the person chosen did not take the review course?
A) 18/25
B) 7/25
C) 25/200
D) 7/200

Answer:

There are a total of 25 graduates who passed the bar exam (18 who took the review course, 7 who did not take the review course). Thus the probability that the person chosen did not take the review course is 7/25.

Answer: B

_____________________________


SAT Practice Test 2015. A researcher conducted a survey to determine whether people in a certain large town prefer watching sports on television to attending the sporting event. The researcher asked 117 people who visited a local restaurant on a Saturday, and 7 people refused to respond. Which of the following factors makes it least likely that a reliable conclusion can be drawn about the sports-watching preferences of all people in the town?
A) Sample size
B) Population size
C) The number of people who refused to respond
D) Where the survey was given

Answer:

The results of the survey will be reliable only if the participants have been randomly selected from ALL people in that population. The fact that the research was conducted only with people who visited a local restaurant makes it least likely that a reliable conclusion can be drawn about the sports-watching preferences of all people in the town.

Answer: D

_____________________________


SAT 2015 Test. A square field measures 10 meters by 10 meters. Ten students each mark off a randomly selected region of the field; each region is square and has side lengths of 1 meter, and no two regions overlap. The students count the earthworms contained in the soil to a depth of 5 centimeters beneath the ground’s surface in each region. The results are shown in the table below.

Which of the following is a reasonable approximation of the number of earthworms to a depth of 5 centimeters beneath the ground’s surface in the entire field?
A) 150
B) 1,500
C) 15,000
D) 150,000

Answer:

Since the field is 10 meters by 10 meters, its total area is 100m².

The lowest number of earthworms found in 1m² is 107. In 100m² there would be 10,700 earthworms.

The largest  number of earthworms found in 1m² is 176. In 100m² there would be 17,600 earthworms.

Therefore the answer should be between 10,700 and 17,600 earthworms.

Answer: C

_____________________________


SAT 2015 Test.

The table above lists the lengths, to the nearest inch, of a random sample of 21 brown bullhead fish. The outlier measurement of 24 inches is an error. Of the mean, median, and range of the values listed, which will change the most if the 24-inch measurement is removed from the data?
A) Mean
B) Median
C) Range
D) They will all change by the same amount

Answer:

There are a total of 21 values; without the error, 20 values.

Mean with 21 values: $m$
Mean without the error: $(21m-24)/20=m+(m/20-24/20)$.
Since $8<m<24$, then $-0.8<m/20-24/20<0$
Change: from -0.8 to 0

Median with 21 values: the 11th fish is 12.
Median without the error: average of 10h and 11th fish is $(12+12)/2=12$
Change: 0

Range with 21 values: $24-8=16$
Range without the error: $16-8=8$
Change: 8

Answer: C

_____________________________


SAT 2015 Test.

A group of tenth-grade students responded to a survey that asked which math course they were currently enrolled in.
The survey data were broken down as shown in the table above. Which of the following categories accounts for approximately 19 percent of all the survey respondents?
A) Females taking Geometry
B) Females taking Algebra II
C) Males taking Geometry
D) Males taking Algebra I

Answer:

There are 310 survey respondents. 19% of 310 is 58.9.
The category with the closest number of students to 58.9 is "males taking Geometry" (59).

Answer: C

_____________________________


SAT 2015 Test.

Based on the histogram above, of the following, which is closest to the average (arithmetic mean) number of seeds per apple?
A) 4
B) 5
C) 6
D) 7

Answer:

From the histogram we have that,
2 Apples have 3 seeds each: 6 seeds total;
4 Apples have 5 seeds each: 20 seeds total;
1 Apple have 6 seeds each: 6 seeds total;
2 Apples have 7 seeds each: 14 seeds total;
3 Apples have 9 seeds each: 27 seeds total.

The total number of seeds in 12 apples is $6+20+6+14+27=73$.
Therefore the arithmetic mean is $73/12=6.08$ seeds per apple.

Answer: C

_____________________________


SAT Practice Test. A researcher wanted to know if there is an association between exercise and sleep for the population of 16 year olds in the United States. She obtained survey responses from a random sample of 2000 United States 16 year olds and found convincing evidence of a positive association between exercise and sleep. Which of the following conclusions is well supported by the data?

A. There is a positive association between exercise and sleep for 16 year olds in the United States.
B. There is a positive association between exercise and sleep for 16 year olds in the world.
C. Using exercise and sleep as defined by the study, an increase in sleep is caused by an increase of exercise for 16 year olds in the United States.
D. Using exercise and sleep as defined by the study, an increase in sleep is caused by an increase of exercise for 16 year olds in the world.

Answer:

It is given that the survey found convincing evidence of a positive association between exercise and sleep. The population of the study included only 16 year olds in the United States. Therefore, A is correct, and B is wrong.

Alternatives C and D mention cause and effect relationships, which can only be established when participants are assigned to different treatments. Based solely on the results of the survey it is not possible to assert that there is such a cause and effect relationship.

Answer: A

_____________________________


SAT Practice Test. At a primate reserve, the mean age of all the male primates is 15 years, and the mean age of all female primates is 19 years. Which of the following must be true about the mean age m of the combined group of male and female primates at the primate reserve?
A. $m=17$
B. $m>17$
C. $m<17$
D. $15<m<19$

Answer:

The mean age m of the combined group of male and female primates at the primate reserve will be between 15 and 19 years. If there are a lot more males than females, the mean will be closer to 15; if there are a lot more females than males, the mean will be closer to 19.

Answer: D

_____________________________


SAT Practice Test. A research assistant randomly selected 75 undergraduate students from the list of all students enrolled in the psychology degree program at a large university. She asked each of the 75 students, “How many minutes per day do you typically spend reading?” The mean reading time in the sample was 89 minutes, and the margin of error for this estimate was 4.28 minutes. Another research assistant intends to replicate the survey and will attempt to get a smaller margin of error. Which of the following samples will most likely result in a smaller margin of error for the estimated mean time students in the psychology degree program read per day?

A. 40 randomly selected undergraduate psychology degree program students
B. 40 randomly selected undergraduate students from all degree programs at the college
C. 300 randomly selected undergraduate psychology degree program students
D. 300 randomly selected undergraduate students from all degree programs at the college

Answer:

The margin of error is related to the inverse of the square root of the sample size:

$Error=\frac{k}{\sqrt{n}}$

Therefore, option A is wrong. Smaller sample sizes provide larger margins of error.

The first sample included only students enrolled in the psychology degree program. If the second sample includes students from all degree programs, it will be a different population than the original survey and therefore the impact to the mean and margin of error cannot be predicted. So options B and D are wrong.

Answer: C

_____________________________

SAT Practice Test - Math - Circles

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Circles.

_______________________


SAT 2015 Practice Test.

In the figure above, point O is the center of the circle, line segments LM and MN are tangent to the circle at points L and N, respectively, and the segments intersect at point M as shown. If the circumference of the circle is 96, what is the length of minor arc LN?

Answer:

Segments LM and MN are tangent to the circle. Thus angles OLM and ONM are right angles.

The sum of the internal angles in a convex quadrilateral is 360º. Therefore:
$LON+90+60+90=360$
$LON+240=360$
$LON=120$ degrees.

The length of minor arc LN is proportional to angle LON. If the circumference of the whole circle (360º) is 96, than arc LN (120º) measures:

$\frac{360}{120}=\frac{96}{LN}$

$360LN=96(120)$
$3LN=96$
$LN=32$

Answer: 32

_____________________________


SAT 2015 Test.

$x^2+y^2+4x-2y=-1$

The equation of a circle in the xy-plane is shown above. What is the radius of the circle?
A) 2
B) 3
C) 4
D) 9

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius.

Back to the given equation:
$x^2+y^2+4x-2y=-1$
$x^2+4x+y^2-2y=-1$
$x^2+4x+4-4+y^2-2y+1-1=-1$
$(x+2)^2-4+(y-1)^2-1=-1$
$(x+2)^2+(y-1)^2=-1+4+1$
$(x+2)^2+(y-1)^2=4$
$(x+2)^2+(y-1)^2=2^2$

Answer: A

_____________________________


SAT 2015 Test. Which of the following is an equation of a circle in the xy-plane with center (0, 4) and a radius with endpoint (4/3, 5)?
A) $x^2+(y-4)^2=25/9$
B) $x^2+(y+4)^2=25/9$
C) $x^2+(y-4)^2=5/3$
D) $x^2+(y+4)^2=3/5$

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius.
Since the center of the given circle is (0, 4):
$(x-a)^2+(y-b)^2=R^2$
$(x-0)^2+(y-4)^2=R^2$
$x^2+(y-4)^2=R^2$

The radius is the distance between (0, 4) and (4/3, 5):
$R^2=(5-4)^2+(4/3-0)^2$
$R^2=(1)^2+(4/3)^2$
$R^2=1+16/9$
$R^2=25/9$

Therefore the equation of the circle is:
$x^2 + (y – 4)^2= R^2$
$x^2 + (y – 4)^2 = 25/9$

Answer: A

_____________________________


SAT Practice Test.

The figure presents a metal nut with two hexagonal faces and six sides. The thickness of one side, from one hexagonal face to the other hexagonal face of the nut, is labeled as 1 centimeter.

The preceding figure shows a metal hex nut with two regular hexagonal faces and a thickness of 1 centimeter. The length of each side of a hexagonal face is 2 centimeters. A hole with a diameter of 2 centimeters is drilled through the nut. The density of the metal is 7.9 grams per cubic centimeter. What is the mass of this nut, to the nearest gram? (Density is mass divided by volume.)

Answer:

The following image is an hexagon with side 2cm:

An hexagon is composed by 6 equilateral triangles. In the figure above h is the hight of one of these tiangles. Let's calculate h using the Pythagoras Theorem in the rectangle triangle AMO:

$1^2+h^2=2^2$
$1+h^2=4$
$h^2=3$
$h=\sqrt{3}$

So the area of the equilateral triangle is $(1+1).h/2=\sqrt{3}$

The area of the hexagon is 6 times the area of each equilateral triangle: $A_{hexa}=6\sqrt{3}$.

The volume of the nut (before subtracing the volume of the hole) is the area of the hexagon times the thickness of the nut:
$V_{hexa}=6\sqrt{3}.1=6\sqrt{3}$

The volume of the hole is the area of the circle times the thickness of the nut:
$V_{hole}=\pi.1^2.1=\pi=3.14$

Therefore, the volume of the nut is:
$V_{hexa}-V_{hole}=6\sqrt{3}-3.14$
$V_{nut}=10.4-3.14=7,25cm^3$

The mass is the volume times the density:
$M_{nut}=7.25.(7.9)$
$M_{nut}=57,2$

Answer: 57g

_____________________________


SAT Practice Test. $x^2+y^2-6x+8y=144$.

The equation of a circle in the x y plane is shown above. What is the diameter of the circle?

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius. Let's develop this equation:

$(x – a)^2 + (y – b)^2 = R^2$
$x^2-2ax+a^2+y^2-2by+b^2 = R^2$
$x^2-2ax+y^2-2by = R^2-a^2-b^2$
$x^2+y^2-2ax-2by = R^2-a^2-b^2$

Comparing this result to the given equation:
$-2ax=-6x$, so $a=3$
$-2by=8y$, so $b=-4$

And
$R^2-a^2-b^2=144$
$R^2-3^2-(-4)^2=144$
$R^2-9-16=144$
$R^2=144+9+16$
$R^2=169$
$R=13$

And the diameter (twice the radius) is 26.

_____________________________


SAT Practice Test.

The figure presents a semicircle with a horizontal diameter labeled A B and a horizontal line segment labeled C D intersecting the semicircle at points C and D.

The semicircle above has a radius of r inches, and chord CD is parallel to the diameter AB. If the length of CD is 2/3 of the length of AB, what is the distance between the chord and the diameter in terms of r?

A. $\frac{1}{3}\pi.r$.

B. $\frac{2}{3}\pi.r$.

C. $\frac{\sqrt2}{2}.r$.

D. $\frac{\sqrt5}{3}.r$.

Answer:

In the following figure, O is the center of the circle, and M is the midpoint between C and D. The triangle OMD is rectangle:

Using the Pythagoras Theorem in triangle OMD:

$x^2+(\frac{2}{3}.r)^2=r^2$

$x^2+\frac{4}{9}.r^2=r^2$

$x^2=\frac{5}{9}.r^2$

$x=\frac{\sqrt5}{3}.r$

Answer: D

_____________________________

SAT Practice Test - Math - Sine and Cosine Functions

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Sine and Cosine Functions.

_______________________


SAT Practice Test 2015.

In the xy-plane above, O is the center of the circle, and the measure of ∠AOB is $\pi/a$ radians. What is the value of a?

Answer:

Segment AO measures:

$(AO)^2=(\sqrt{3})^2+1^2$
$(AO)^2=3+1=4$
$AO=2$

Therefore sine of ∠AOB is 1/2. And the measure of ∠AOB is 30°, which is equal to $\pi/6$ radians.

Answer: a=6

_____________________________


SAT Practice Test. In a right triangle, one angle measures x°, where sin x°=4/5. What is cos(90°-x°)?

Answer:

In a right triangle, if one of the angles adjacent to the hypotenuse measures x, the other angle adjacent to the hypotenuse measures 90-x.
Sine of x can be calculated by the dividing the oposite side by the hypotenuse. It turns out that the oposite side to angle x is the adjacent side to angle 90-x. So
$sin(x)=cos(90-x)$.
$cos(90-x)=4/5$

_____________________________


SAT Practice Test. An architect drew the following sketch while designing a house roof. The dimensions shown are for the interior of the triangle.

The figure presents a triangle with a horizontal base. Labels are given to two sides and to two angles. The left side of the triangle is labeled 24 feet. The base of the triangle is labeled 32 feet. The two lower interior angles are labeled x degrees. A note under the figure says that the figure is not drawn to scale.

What is the value of  $cos(x)$?

Answer:

Because the triangle is isosceles, a perpendicular from the top vertex to the opposite side will bisect the base, creating two right triangles. The side of these triangles close to angle x will, therefore, measure 16ft. The hypotenuse of these triangles measure 24ft. Using these measures:

$cos(x)=16/24$
$cos(x)=2/3$

_____________________________


SAT Practice Test. It is given that $sin(x)=a$, where x is the radian measure of an angle and $\pi/2<x<\pi$.
If $sin(w)=-a$, which of the following could be the value of w?
A. $\pi-x$
B. $x-\pi$
C. $2\pi+x$
D. $x-2\pi$

Answer:

In the following figure, x is the angle $A\hat{O}X$, and w can be either $A\hat{O}W$ or $A\hat{O}W'$:

$A\hat{O}W=\pi+(\pi-x)=2\pi-x=-x$ (no match in the given answers).

$A\hat{O}W'=-(\pi-x)=x-\pi$

Answer: B

_____________________________


SAT Practice Test. Which of the following is equal to $sin(\frac{\pi}{5})$?

A.   $-cos(\frac{\pi}{5})$.

B.   $-sin(\frac{\pi}{5})$.

C.   $cos(\frac{3\pi}{10})$.

D.   $sin(\frac{7\pi}{10})$.

Answer:

Important relation between sine and cosine: $sin(x)=cos(\frac{\pi}{2}-x)$. Therefore:

$sin(\frac{\pi}{5})=cos(\frac{\pi}{2}-\frac{\pi}{5})$

$sin(\frac{\pi}{5})=cos(\frac{5\pi}{10}-\frac{2\pi}{10})$

$sin(\frac{\pi}{5})=cos(\frac{3\pi}{10})$

Answer: C

_____________________________

SAT Practice Test - Math - Linear (and non-Linear) Systems of Equations

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Linear (and non-Linear) Systems of Equations.

_______________________


SAT Practice Test 2015. In one semester, Doug and Laura spent a combined 250 hours in the tutoring lab. If Doug spent 40 more hours in the lab than Laura did, how many hours did Laura spend in the lab?

Answer:

Let D and L be the number of hours Doug and Laura, respectively, spent in the lab.

Doug and Laura spent a combined 250 hours in the tutoring lab. Thus, the first equation is $D+L=250$.
Doug spent 40 more hours in the lab than Laura did. Thus, the second equation is $D=L+40$.

Substituting the second equation in the first one gives:
$D+L=250$
$(L+40)+L=250$
$2L=250-40$
$2L=210$
$L=105$ hours

Answer: 105

_____________________________


SAT Practice Test 2015.

$y=3$
$y=ax^2+b$

In the system of equations above, a and b are constants. For which of the following values of a and b does the system of equations have exactly two real solutions?
A) a = -2, b = 2
B) a = -2, b = 4
C) a = 2, b = 4
D) a = 4, b = 3

Answer:

$y=ax^2+b$, and $y=3$. Therefore,

$3=ax^2+b$
$ax^2=3-b$
$x^2=(3-b)/a$

The system will have two real solutions if $(3-b)/a>0$.

If $a>0$, then we should have $3-b>0$, or $b<3$. No choice satisfies this condition.

If $a<0$, then we should have $3-b<0$, or $b>3$. Choice B satisfies this condition.

Answer: B

_____________________________


SAT Practice Test 2015. A worker uses a forklift to move boxes that weigh either 40 pounds or 65 pounds each. Let x be the number of 40-pound boxes and y be the number of 65-pound boxes. The forklift can carry up to either 45 boxes or a weight of 2,400 pounds. Which of the following systems of inequalities represents this relationship?

A)
$40x+65y\leq2400$
$x+y\leq45$

B)
$x/40+y/65\leq2400$
$x+y\leq45$

C)
$40x+65y\leq45$
$x+y\leq2400$

D)
$x+y\leq2400$
$40x+65y\leq2400$

Answer:

The total number of boxes is $x+y$.
The forklift can carry up to 45 boxes. Therefore, one of the equations is $x+y\leq45$

The total weight of the 40-pound boxes is 40x.
The total weight of the 65-pound boxes is 65y.
The forklift can carry up to a weight of 2,400 pounds. Therefore, the other equation is $40x+65y\leq2400$

Answer: A

_____________________________


SAT Practice Test 2015.
ax by
$ax+by=12$
$2x+8y=60$
In the system of equations above, a and b are constants. If the system has infinitely many solutions, what is the value of a/b?

Answer:

In order for the system to have infinitely many solutions, the two equations must be equivalent.

Multiplying the first equation by 5:
$5ax+5by=5*12$
$5ax+5by=60$

Therefore, $5a=2$, and $5b=8$. Deviding these two equations:

$\frac{5a}{5b}=\frac{2}{8}$

$\frac{a}{b}=\frac{1}{4}=0.25$

_____________________________


SAT Practice Test 2015. The graph of a line in the xy-plane has slope 2 and contains the point (1, 8). The graph of a second line passes through the points (1, 2) and (2, 1). If the two lines intersect at the point (a, b), what is the value of a+b?
A) 4
B) 3
C) −1
D) −4

Answer:

The equation of the first line is $y=2x+b$, and it contains point (1, 8). Therefore:
$y=2x+b$
$8=2(1)+b$
$b=6$
And the first equation of the system is $y=2x+6$

The equation of the second line is $y=cx+d$, and it passes through the points (1, 2) and (2, 1). Therefore:
$2=c(1)+d$
$1=c(2)+d$
So $c=-1$ and $d=3$, and the equation of the second line is $y=-x+3$

Point (a, b) is the solution of the system with the equations of the two lines:
$y=2x+6$
$y=-x+3$

Subtracting the second equation from the first one:
$y-y=2x+6-(-x+3)$
$0=2x+6+x-3$
$0=3x+3$
$x=-1$

Substituing this value for x in the first equation:
$y=2x+6$
$y=2(-1)+6$
$y=-2+6$
$y=4$

So point (a, b) is (-1, 4), and $a+b=-1+4=3$.

Answer: B

_____________________________


SAT Test 2015.

$x+y=0$
$3x-2y=10$

Which of the following ordered pairs (x, y) satisfies the system of equations above?
A) (3, −2)
B) (2, −2)
C) (−2, 2)
D) (−2, −2)

Answer:

From the first equation: $x=-y$.

Substituting this relation in the second equation:
$3x-2y=10$
$3(-y)-2y=10$
$-3y-2y=10$
$-5y=10$
$y=-2$

And $x=2$.

Answer: B

_____________________________


SAT Test 2015. A food truck sells salads for 6.50 each and drinks for 2.00 each. The food truck’s revenue from selling a total of 209 salads and drinks in one day was 836.50. How many salads were sold that day?
A) 77
B) 93
C) 99
D) 105

Answer:

The food truck sold "s" salads and "d" drinks that day. So $s+d=209$, or $d=209-s$.
Total revenue is given by the equation: $6.50s+2d=836.50$.

Substituting $d=209-s$ in the second equation:
$6.50s+2d=836.50$
$6.50s+2(209-s)=836.50$
$6.50s+418-2s=836.50$
$4.50s=836.50-418$
$4.50s=418.50$
$s=93$

Answer: B

_____________________________


SAT Test 2015.
$x+y=-9$
$x+2y=-25$
According to the system of equations above, what is the value of x ?

Answer:

Subtracting the left and right sides of the first equation from the corresponding sides of the second equation gives
$(x+2y)-(x+y)=-25-(-9)$
$x+2y-x-y=-25+9$
$y=-16$

Substituting -16 for y in the first equation gives
$x+y=-9$
$x-16=-9$
$x=16-9$
$x=7$

_____________________________


SAT Test 2015.

$b=2.35+0.25x$
$c=1.75+0.40x$

In the equations above, b and c represent the price per pound, in dollars, of beef and chicken, respectively, x weeks after July 1 during last summer. What was the price per pound of beef when it was equal to the price per pound of chicken?
A) 2.60
B) 2.85
C) 2.95
D) 3.35

Answer:

If the prices of beef and chicken were the same, then
$b=c$
$2.35+0.25x=1.75+0.40x$
$2.35-1.75=0.40x-0.25x$
$0.6=0.15x$
$x=4$ weeks after July 1. That's when the price per pound of beef was equal to the price per pound of chicken.

Using this value of x in the equation $c=1.75+0.40x$:
$c=1.75+0.40.(4)$
$c=1.75+1.6$
$c=3.35$

Answer: D

_____________________________


SAT Test 2015.

$3x+4y=-23$
$2y-x=-19$

What is the solution (x, y) to the system of equations above?
A) (−5, −2)
B) (3, −8)
C) (4, −6)
D) (9, −6)

Answer:

Multiplying the second equation by -2:
$3x+4y=-23$
$-4y+2x=38$

Now adding the two equations:
$3x+4y-4y+2x=-23+38$
$5x=15$
$x=3$

Substituting x in the equation $2y-x=-19$:
$2y-3=-19$
$2y=-16$
$y=-8$

Answer: B

_____________________________


SAT Practice Test.
$x^2+y^2=153$
$y=-4x$
If (x, y) is a solution to the preceding system of equations, what is the value of $x^2$?
A.   negative 51
B. 3
C. 9
D. 144

Answer:

If $y=-4x$, then $y^2=16x^2$

Using this result in the first equation:

$x^2+y^2=153$
$x^2+16x^2=153$
$17x^2=153$
$x^2=9$

Answer: C

_____________________________


SAT Practice Test.

The figure presents the graph of a circle, a parabola, and a line in the x y plane. The horizontal axis is labeled x, the vertical axis is labeled y, and the origin is labeled O. The integers negative 3 through 3 appear on both axes.

The circle has its center at the origin and radius of approximately 2.2.

The parabola has its vertex on the y axis at negative 3 and opens upward.

The circle and parabola intersect at four points, of which two are below the x axis and two are above the x axis. Of the two points of intersection below the x axis, one is to the left of the y axis and one is to the right of the y axis. Of the two points of intersection above the x axis, one is to the left of the y axis and one is to the right of the y axis.

The line slants upward and to the right, and passes through two of the four points of intersection where the circle and parabola meet, one below the x axis and to the left of the y axis, and one above the x axis to the right of the y axis. In other words, the three graphs intersect at two points.

The following system of three equations is given beneath the figure.

$x^2+y^2=5$

$y=x^2-3$

$x-y=1$

A system of three equations and their graphs in the x y plane are shown above. How many solutions does the system have?

A. One

B. Two

C. Three

D. Four

Answer:

The solutions to the system of equations are the points where the circle, parabola, and line all intersect. The graph shows that there are only two points in which this happens: (2, 1) and (-1, -2).

Answer: B

_____________________________


SAT Practice Test.
$\frac{1}{2}.x-\frac{1}{4}.y=5$.
$a.x-3.y=20$.
In the system of linear equations above, a is a constant. If the system has no solution, what is the value of a?
A. 1/2
B. 2
C. 6
D. 12

Answer:

If the system of equations has no solution, the graphs of the equations must be parallel lines (lines with the same slope). This will happen when the constants multiplying $x$ and $y$ in one equation are multiples of the corresponding constants in the other equation. Therefore,

$\frac{(\frac{1}{2})}{a}=\frac{(\frac{1}{4})}{3}$

$3.\frac{1}{2}=a.\frac{1}{4}$

$\frac{3}{2}=\frac{a}{4}$

$a=6$

So the second equation is $6x-3y=20$.

If we multiply the first equation by 12, the result is $6x-3y=60$.

Note that the right side of the two equations are different ($20$ in the first one; $60$ in the second). If the right side of the two equations were the same, the two lines would coincide, and the system would have infinite solutions (instead of none).

Answer: C

_____________________________


SAT Practice Test.
$4x-y=3y+7$
$x+8y=4$
Based on the system of equations above, what is the value of the product x y?
A. -3/2
B. 1/4
C. 1/2
D. 11/9

Answer:

Simplifying the first equation,
$4x-y=3y+7$
$4x-y-3y=7$
$4x-4y=7$

Now multiplying this equation by 2:
$8x-8y=14$

Now let's add this transformed equation and the second given equation ($x+8y=4$):
$8x-8y+x+8y=14+4$
$8x+x=18$
$9x=18$
$x=2$

Now let's substitute 2 for x in the second given equation ($x+8y=4$):
$x+8y=4$
$2+8y=4$
$8y=2$
$y=1/4$

Finally, the product x y is:
$x.y=2.(1/4)=1/2$

Answer: C

_____________________________

SAT Practice Test. If $\frac{1}{2}.x+\frac{1}{3}.y=4$, what is the value of $3x+2y$?

Answer:

Let's multiply both sides of the equation by 6:
$\frac{1}{2}.x.6+\frac{1}{3}.y.6=4.6$
$3x+2y=24$

_____________________________

SAT Practice Test - Math - Complex Numbers

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving questions that focus on Complex Numbers.

_______________________


SAT Practice Test 2015. Which of the following complex numbers is equivalent to $\frac{3-5i}{8+2i}$? (Note: $i=\sqrt{-1}$)

A) $\frac{3}{8}-\frac{5i}{2}$

B) $\frac{3}{8}+\frac{5i}{2}$

C) $\frac{7}{34}-\frac{23i}{34}$

D) $\frac{7}{34}+\frac{23i}{34}$

Answer:

$\frac{3-5i}{8+2i}=$

$\frac{3-5i}{8+2i}*\frac{8-2i}{8-2i}=$

$\frac{24-6i-40i-10}{64+4}=$

$\frac{14-46i}{68}=$

$\frac{7-23i}{34}$

Answer: C

_____________________________


SAT Test 2015. For $i=\sqrt{-1}$ , what is the sum $(7+3i)+(-8+9i)$?
A) -1+12i
B) -1-6i
C) 15+12i
D) 15-6i

Answer:

$(7+3i)+(-8+9i)=$
$7-8+3i+9i=$
$-1+12i$

Answer: A

_____________________________


SAT Practice Test. Which of the following is equal to $(14-2i)(7+12i)$? (Note: $i=\sqrt{-1}$.)
A. 74
B. 122
C. 78+154i
D. 122+154 i

Answer:

$(14-2i)(7+12i)=$
$14.7+14.12i-2i.7-2i.12i=$
$98+168i-14i-24.i^2=$
$98+154i-24.(-1)=$
$98+154i+24=$
$122+154i$

Answer: D

_____________________________