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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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Example Questions

Example Question #12 : How To Find The Equation Of A Curve

A horizontal parabola on the coordinate plane has -intercept (6,0) ; one of its -intercepts is (0, 2) .

Give its equation.

Possible Answers:

$x= y^{2}-5 y+ 6$

$x= -2 y^{2}+y + 6$

Insufficient information is given to determine the equation.

$x = - y^{2}- y+ 6$

$x =2y^{2}-7 y + 6$

Correct answer:

Insufficient information is given to determine the equation.

Explanation:

The equation of a horizontal parabola, in standard form, is

$x = ay^{2}+ by + c$

for some real a,b,c .

is the -coordinate of the -intercept, so c=6 , and the equation is

$x = ay^{2}+ by + 6$

Set x=0, y=2 :

$0 = a\cdot 2^{2}+ b \cdot 2 + 6$

0 = 4a+2b + 6

However, no other information is given, so the values of and cannot be determined for certain. The correct response is that insufficient information is given.

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Example Question #241 : Coordinate Geometry

A vertical parabola on the coordinate plane includes points (-2, 23), (1,5), and (3,13) .

Give its equation.

Possible Answers:

$y = 2x^{2}+10x-7$

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

$y = 2x^{2}-4x+7$

$y = x^{2}-4x+8$

$y = x^{2}+4x -8$

Correct answer:

$y = 2x^{2}-4x+7$

Explanation:

The standard form of the equation of a vertical parabola is

$y=ax^{2}+bx+c$

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a\cdot (-2)^{2}+b(-2)+c = 23$

4a-2b+c = 23

$a\cdot 1^{2}+b \cdot 1+c = 5$

a+b+c = 5

$a\cdot 3^{2}+b \cdot 3+c =13$

9a+3b+c = 13

The system

4a-2b+c = 23

a+b+c = 5

9a+3b+c = 13

can be solved through the elimination method.

First, multiply the second equation by and add to the third:

9a+3b+c =13

$\underline{ -a-b-c \; \; \; =- 5 }$

8a+2b

Next, multiply the second equation by and add to the first:

4a-2b+c = 23

$\underline{ -a-b-c \; =- 5 }$

3a-3b =18

Which can be divided by 3 on both sides to yield

a-b = 6

Now solve the two-by-two system

8a+2b = 8

a-b = 6

by substitution:

a = b+ 6

8 (b+6)+2b = 8

8b+48+2b = 8

10b+48 = 8

10b= -40

b= -4

Back-solve:

a-(-4) = 6

a+4= 6

a =2

Back-solve again:

2+(-4)+c = 5

-2+c=5

c=7

The equation of the parabola is therefore

$y = 2x^{2}-4x+7$ .

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Example Question #14 : How To Find The Equation Of A Curve

A vertical parabola on the coordinate plane shares one -intercept with the line of the equation 2x+4y = 8 , and the other with the line of the equation 4x-2y = 12 . It also passes through (6, -4) . Give the equation of the parabola.

Possible Answers:

$y = -\frac{2}{3}x^{2}+\frac{14}{3}x-8$

$y = \frac{2}{3}x^{2}-\frac{14}{3}x+8$

The correct answer is not among the other responses.

$y = -\frac{3}{2}x^{2}+\frac{21}{2}x-18$

$y = \frac{3}{2}x^{2}- \frac{21}{2}x+18$

Correct answer:

$y = -\frac{2}{3}x^{2}+\frac{14}{3}x-8$

Explanation:

First, find the -intercepts—the points of intersection with the -axis—of the lines by substituting 0 for in both equations.

2x+4y = 8

2x+4 (0) = 8

2x = 8

x=4

(4,0) is the -intercept of this line.

4x-2y = 12

4x-2 (0)= 12

4x= 12

x= 3

(3,0) is the -intercept of this line.

The parabola has -intercepts at (3,0) and (4,0) , so its equation can be expressed as

y = a(x-3)(x-4)

for some real . To find it, substitute using the coordinates of the third point, setting x= 6, y = -4 :

-4 = a(6-3)(6-4)

$-4 = a \cdot 3 \cdot 2$

-4 = 6 a

$a = \frac{-4}{6} = -\frac{2}{3}$ .

The equation is $y = -\frac{2}{3}(x-3)(x-4)$ , which, in standard form, can be rewritten as:

$y = -\frac{2}{3}(x^{2}-7x+12)$

$y = -\frac{2}{3}x^{2}+\frac{14}{3}x-8$

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Example Question #15 : How To Find The Equation Of A Curve

A horizontal parabola on the coordinate plane includes points (-11, -1) (-1,1) , and (10,2) .

Give its equation.

Possible Answers:

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

$x=2y^{2}-11y+8$

$x=2y^{2}+5y-8$

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

$x= y^{2}-6y+4$

Correct answer:

$x=2y^{2}+5y-8$

Explanation:

The standard form of the equation of a horizontal parabola is

$x=ay^{2}+by+ c$

If the values of and from each ordered pair are substituted in succession, three equations in three variables are formed:

$a \cdot (-1)^{2}+b \cdot (-1)+ c = -11$

a -b + c = -11

$a \cdot (1)^{2}+b \cdot 1+ c = -1$

a+b + c = -1

$a \cdot (2)^{2}+b \cdot 2+ c = 10$

4a+2b+ c = 10

The three-by-three linear system

a -b + c = -11

a+b + c = -1

4a+2b+ c = 10

can be solved by way of the elimination method.

can be found first, by multiplying the first equation by and add it to the second:

a+b + c = -1

$\underline{-a+b - c = 11}$

=10

$2b \div 2 = 10 \div 2$

b = 5

Substitute 5 for in the last two equations to form a two-by-two linear system:

a+5+ c = -1

a+5+ c - 5= -1 - 5

a + c = -6

$4a+2 \cdot 5+ c = 10$

4a+10+ c = 10

4a+10+ c- 10 = 10 - 10

4a +c = 0

The system

a + c = -6

4a +c = 0

can be solved by way of the substitution method;

4a +c = 0

4a +c -4a = 0 -4a

c = -4a

a + (-4a) = -6

-3a = -6

$-3a\div (-3) = -6 \div (-3)$

a = 2

Substitute 2 for in the top equation:

2 + c = -6

2 + c - 2= -6 - 2

c= -8

The equation is $x=2y^{2}+5y-8$ .

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Example Question #301 : Geometry

Circle

Give the equation of the above circle.

Possible Answers:

$(x+2)^{2}+ (y+4)^{2}= 25$

$(x-2)^{2}+ (y-4)^{2}= 5$

$(x+2)^{2}+ (y+4)^{2}= 5$

$(x-2)^{2}+ (y-4)^{2}= 25$

None of the other choices is correct.

Correct answer:

$(x-2)^{2}+ (y-4)^{2}= 25$

Explanation:

A circle with center (h, k) and radius has equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The circle has center (2,4) and radius 5, so substitute:

$(x-2)^{2}+ (y-4)^{2}= 5^{2}$

$(x-2)^{2}+ (y-4)^{2}= 25$

Report an Error

Example Question #302 : Geometry

A circle on the coordinate plane has a diameter whose endpoints are (-3, 7) and (9, 5) . Give its equation.

Possible Answers:

$(x-3)^{2}+ (y-6)^{2}=148$

$(x+3)^{2}+ (y+6)^{2}=148$

$(x-3)^{2}+ (y-6)^{2}=37$

$(x+3)^{2}+ (y+6)^{2}=37$

$(x-3)^{2}+ (y-6)^{2}=14$

Correct answer:

$(x-3)^{2}+ (y-6)^{2}=37$

Explanation:

A circle with center (h, k) and radius has equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{-3+9}{2}, \frac{7+5}{2} \right )$

(3, 6)

Therefore, h = 3 and k = 6

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using (3, 6) and (9, 5) . We will concern ourcelves with finding the square of the radius $r^{2}$ :

$r^{2}= \left ( x_{2}-x_{1}\right )^{2} + \left ( y_{2}-y_{1}\right )^{2}$

$r^{2}= \left (9-3\right )^{2} + \left ( 5-6\right )^{2}$

$r^{2}=6^{2} + \left (-1\right )^{2}$

$r^{2}=36+ 1= 37$

Substitute:

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

$(x-3)^{2}+ (y-6)^{2}=37$

Report an Error

Example Question #3 : How To Find The Equation Of A Circle

Circle

Give the equation of the above circle.

Possible Answers:

$x^{2}+4x + y ^{2}-6y+9=0$

$x^{2}-4x + y ^{2}+6y+9=0$

$x^{2}-4x + y ^{2}+6y-3=0$

$x^{2}-6x + y ^{2}+4y-3=0$

$x^{2}+6x + y ^{2}-4y-3=0$

Correct answer:

$x^{2}+6x + y ^{2}-4y-3=0$

Explanation:

A circle with center (h, k) and radius has equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The circle has center (-3, 2) and radius 4, so substitute:

$(x-(-3))^{2}+ (y-2)^{2}= 4^{2}$

$(x+3)^{2}+ (y-2)^{2}= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4= 16$

$x^{2}+6x + 9+ y ^{2}-4y+4- 16= 16-16$

$x^{2}+6x + y ^{2}-4y-3=0$

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Example Question #3 : Circles

A circle on the coordinate plane has a diameter whose endpoints are (2, 7) and (-8, 3) . Give its equation.

Possible Answers:

$x^{2}+6x + y^{2}-10y-82= 0$

$x^{2}+6x + y^{2}-10y+5 = 0$

$x^{2}+6x + y^{2}-10y+63= 0$

$x^{2}-6x + y^{2}+10y-82= 0$

$x^{2}-6x + y^{2}+10y+5 = 0$

Correct answer:

$x^{2}+6x + y^{2}-10y+5 = 0$

Explanation:

A circle with center (h, k) and radius has equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The midpoint of a diameter of the circle is its center, so use the midpoint formula to find this:

$\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2} \right )$

$\left ( \frac{2+(-8)}{2}, \frac{7+3}{2} \right )$

(-3, 5)

Therefore, h = -3 and k = 5 .

The radus is the distance between the center and one endpoint, so take advantage of the distance formula using (-3, 5) and (2, 7) . We will concern ourcelves with finding the square of the radius $r^{2}$ :

$r^{2}= \left ( x_{2}-x_{1}\right )^{2} + \left ( y_{2}-y_{1}\right )^{2}$

$r^{2}= \left ( 2- (-3)\right )^{2} + \left ( 7-5\right )^{2}$

$r^{2}= 5^{2} +2^{2}$

$r^{2}=25+ 4 = 29$

Substitute:

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

$(x-(-3))^{2}+ (y-5)^{2}= 29$

$(x+3)^{2}+ (y-5)^{2}= 29$

Expand:

$x^{2}+6x+9+ y^{2}-10y+25= 29$

$x^{2}+6x + y^{2}-10y+34= 29$

$x^{2}+6x + y^{2}-10y+34- 29 = 0$

$x^{2}+6x + y^{2}-10y+5 = 0$

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Example Question #2 : How To Find The Equation Of A Circle

A circle on the coordinate plane has center (4, -6) and circumference $20 \pi$ . Give its equation.

Possible Answers:

$(x+4)^{2}+ (y-6)^{2}=10$

$(x+4)^{2}+ (y-6)^{2}=400$

$(x+4)^{2}+ (y-6)^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=10$

Correct answer:

$(x-4)^{2}+ (y+6)^{2}=100$

Explanation:

A circle with center (h, k) and radius has equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The center is (4, -6) , so h = 4, k=-6 .

To find , use the circumference formula:

$2 \pi r = C$

$2 \pi r = 20 \pi$

r = 10

$r^{2} = 100$

Substitute:

$(x-4)^{2}+ (y-(-6))^{2}=100$

$(x-4)^{2}+ (y+6)^{2}=100$

Report an Error

Example Question #303 : Geometry

A circle on the coordinate plane has center (2, 5) and area $36 \pi$ . Give its equation.

Possible Answers:

$(x-2)^{2}+ (y-5)^{2}= 36$

$(x+2)^{2}+ (y+5)^{2}= 18$

$(x-2)^{2}+ (y-5)^{2}=18$

$(x-2)^{2}+ (y-5)^{2}= 6$

$(x+2)^{2}+ (y+5)^{2}= 6$

Correct answer:

$(x-2)^{2}+ (y-5)^{2}= 36$

Explanation:

A circle with center (h, k) and radius has the equation

$(x-h)^{2}+ (y-k)^{2}= r^{2}$

The center is (2, 5) , so h = 2, k = 5 .

The area is $36 \pi$ , so to find $r^{2}$ , use the area formula:

$\pi r^{2}= A$

$\pi r^{2}= 36 \pi$

$r^{2}= 36$

The equation of the line is therefore:

$(x-2)^{2}+ (y-5)^{2}= 36$

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #12 : How To Find The Equation Of A Curve

Example Question #241 : Coordinate Geometry

Example Question #14 : How To Find The Equation Of A Curve

Example Question #15 : How To Find The Equation Of A Curve

Example Question #301 : Geometry

Example Question #302 : Geometry

Example Question #3 : How To Find The Equation Of A Circle

Example Question #3 : Circles

Example Question #2 : How To Find The Equation Of A Circle

Example Question #303 : Geometry

All SSAT Upper Level Math Resources

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