_______________________
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
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
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
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:
$\frac{1}{2}.x.6+\frac{1}{3}.y.6=4.6$
$3x+2y=24$
_____________________________
No comments:
Post a Comment