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

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All SSAT Upper Level Math Resources

6 Diagnostic Tests 311 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : X And Y Intercept

Ellipse 1

Give the equation of the above ellipse.

Possible Answers:

$\frac{(x+6)^{2}}{16} + \frac{(y+2)^{2}}{64} = 1$

$\frac{(x-6)^{2}}{4} + \frac{(y-2)^{2}}{8} = 1$

$\frac{(x+6)^{2}}{4} + \frac{(y+2)^{2}}{8} = 1$

$\frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1$

$\frac{(x+6)^{2}}{64} + \frac{(y+2)^{2}}{256} = 1$

Correct answer:

$\frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1$

Explanation:

The equation of the ellipse with center (h,k) , horizontal axis of length , and vertical axis of length is

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

The ellipse has center (6,2) , horizontal axis of length 8, and vertical axis of length 16. Therefore,

h = 6,k = 2 , $a = \frac{8}{2} = 4$ , and $b = \frac{16}{2} = 8$ .

The equation of the ellipse is

$\frac{(x-6)^{2}}{4^{2}} + \frac{(y-2)^{2}}{8^{2}} = 1$

$\frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1$

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Example Question #5 : How To Find The Equation Of A Curve

Ellipse 1

Give the equation of the above ellipse.

Possible Answers:

$\frac{(x-1)^{2}}{5} + \frac{(y-1)^{2}}{3} = 1$

$\frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1$

$\frac{(x-1)^{2}}{100} + \frac{(y-1)^{2}}{36} = 1$

$\frac{(x+1)^{2}}{25} + \frac{(y+1)^{2}}{9} = 1$

$\frac{(x+1)^{2}}{100} + \frac{(y+1)^{2}}{36} = 1$

Correct answer:

$\frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1$

Explanation:

The equation of the ellipse with center (h,k) , horizontal axis of length , and vertical axis of length is

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

The ellipse has center (1,1) , horizontal axis of length 10, and vertical axis of length 6. Therefore,

h = k = 1 , $a = \frac{10}{2} = 5$ , and $b = \frac{6}{2} = 3$ .

The equation of the ellipse is

$\frac{(x-1)^{2}}{5^{2}} + \frac{(y-1)^{2}}{3^{2}} = 1$

$\frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1$

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Example Question #22 : X And Y Intercept

Ellipse 1

Give the equation of the above ellipse.

Possible Answers:

$\frac{(x+3)^{2}}{4} + \frac{(y+5)^{2}}{3} = 1$

$\frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1$

$\frac{(x+3)^{2}}{16} + \frac{(y+5)^{2}}{9} = 1$

$\frac{(x-3)^{2}}{4} + \frac{(y-5)^{2}}{3} = 1$

$\frac{(x-3)^{2}}{8} + \frac{(y-5)^{2}}{6} = 1$

Correct answer:

$\frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1$

Explanation:

The equation of the ellipse with center (h,k) , horizontal axis of length , and vertical axis of length is

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

The ellipse has center (3,5) , horizontal axis of length 8, and vertical axis of length 6. Therefore,

h = 3,k = 5 , $a = \frac{8}{2} = 4$ , and $b = \frac{6}{2} = 3$ .

The equation of the ellipse is

$\frac{(x-3)^{2}}{4^{2}} + \frac{(y-5)^{2}}{3^{2}} = 1$

$\frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1$

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Example Question #21 : X And Y Intercept

The -intercept and the only -intercept of a vertical parabola on the coordinate plane coincide with the -intercept and the -intercept of the line of the equation 3x+ 5y = 30 . Give the equation of the parabola.

Possible Answers:

Insufficient information is given to determine the equation.

$y=\frac{5}{18}x^{2}- \frac{10}{3}x+10$

$y = \frac{3}{50} x^{2}-\frac{6}{5}x+6$

$y = \frac{18}{5} x^{2}-\frac{216}{5}x+ \frac{648}{5}$

$y = \frac{50}{3} x^{2}-\frac{1,000}{3}x+ \frac{5,000}{3}$

Correct answer:

$y = \frac{3}{50} x^{2}-\frac{6}{5}x+6$

Explanation:

To find the -intercept, that is, the point of intersection with the -axis, of the line of equation 3x+ 5y = 30 , set x=0 and solve for :

3 (0)+ 5y = 30

5y = 30

y = 6

The -intercept is (0,6) .

The -intercept can be found by doing the opposite:

3x+ 5 (0) = 30

3x = 30

x=10

The -intercept is (10,0) .

The parabola has these intercepts as well. Also, since the vertical parabola has only one -intercept, that point doubles as its vertex as well.

The equation of a vertical parabola, in vertex form, is

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

where (h,k) is the vertex. Set h = 10,k=0 :

$y = a(x-10)^{2}+0$

$y = a(x-10)^{2}$

for some real . To find it, use the -intercept, setting x=0, y = 6

$6 = a(0-10)^{2}$

$6 = a( -10)^{2}$

6 = 100a

$a = \frac{6}{100} = \frac{3}{50}$

The parabola has equation $y =\frac{3}{50} (x-10)^{2}$ , which is rewritten as

$y =\frac{3}{50} (x^{2}-20x+100)$

$y = \frac{3}{50} x^{2}-\frac{6}{5}x+6$

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Example Question #1 : How To Find The Equation Of A Curve

An ellipse on the coordinate plane has as its center the point (-4, 9) . It passes through the points (-4, 5) and (11, 9) . Give its equation.

Possible Answers:

$\frac{(x+4)^{2}}{900} + \frac{(y-9)^{2}}{64} = 1$

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$

Insufficient information is given to determine the equation.

$\frac{(x-4)^{2}}{900} + \frac{(y+9)^{2}}{64} = 1$

$\frac{(x-4)^{2}}{225} + \frac{(y+9)^{2}}{16} = 1$

Correct answer:

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$

Explanation:

The equation of the ellipse with center (h,k) , horizontal axis of length , and vertical axis of length is

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

The center is (-4, 9) , so h = -4 and k = 9 .

To find , note that one endpoint of the horizontal axis is given by the point with the same -coordinate through which it passes, namely, (11, 9) . Half the length of this axis, which is , is the difference of the -coordinates, so a = 11- (-4) = 15 . Similarly, to find , note that one endpoint of the vertical axis is given by the point with the same -coordinate through which it passes, namely, (-4, 5) . Half the length of this axis, which is , is the difference of the -coordinates, so b = 9 - 5 = 4 .

The equation is

$\frac{(x-(-4))^{2}}{15^{2}} + \frac{(y-9)^{2}}{4^{2}} = 1$

$\frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1$ .

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Example Question #11 : How To Find The Equation Of A Curve

An ellipse passes through points (4, 6) , (10,6), (7, 2), (7, 10) .

Give its equation.

Possible Answers:

$\frac{(x-6)^{2}}{9} + \frac{(y-7)^{2}}{16} = 1$

$\frac{(x +7)^{2}}{16} + \frac{(y+6)^{2}}{9}= 1$

$\frac{(x +7)^{2}}{9} + \frac{(y+6)^{2}}{16} = 1$

$\frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1$

$\frac{(x-6)^{2}}{16} + \frac{(y-7)^{2}} {9}= 1$

Correct answer:

$\frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1$

Explanation:

The equation of the ellipse with center (h,k) , horizontal axis of length , and vertical axis of length is

$\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

(4, 6) and (10,6) are the endpoints of a horizontal line segment with midpoint

$\left ( \frac{4+10}{2}, \frac{6+6}{2} \right )$ , or (7,6)

and length 10-4= 6 .

(7, 2) and (7, 10) are the endpoints of a vertical line segment with midpoint

$\left ( \frac{7+7}{2}, \frac{2+10}{2} \right )$ , or (7,6)

and length 10-2 = 8

Because their midpoints coincide, these are the endpoints of the horizontal axis and vertical axis, respectively, of the ellipse, and the common midpoint (7,6) is the center.

Therefore,

h = 7 and k = 6 ;

2a = 6 and 2b= 8 ; consequently a = 3 and b = 4 .

The equation of the ellipse is

$\frac{(x-7)^{2}}{3^{2}} + \frac{(y-6)^{2}}{4^{2}} = 1$ , or

$\frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1$

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Example Question #512 : Ssat Upper Level Quantitative (Math)

A horizontal parabola on the coordinate plane (0,4) as its only -intercept; its -intercept is (6,0) .

Give its equation.

Possible Answers:

Insufficient information is given to determine the equation.

$x = \frac{1}{9} y^{2}-\frac{4}{3}y+4$

$x =- \frac{1}{9} y^{2}+4$

$x = \frac{8}{3}y^{2}-\frac{64}{3}y+\frac{128}{3}$

$x = \frac{3}{8} y^{2}-3y+6$

Correct answer:

$x = \frac{3}{8} y^{2}-3y+6$

Explanation:

If a horizontal parabola has only one -intercept, which here is (0,4) , that point doubles as its vertex as well.

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

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

where (h,k) is the vertex. Set h = 0,k=4 :

$x = a(y-4)^{2}+0$

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

To find , use the -intercept, setting x= 6, y=0 :

$6 = a(0-4)^{2}$

$6 = a( -4)^{2}$

6 = 16a

$a = \frac{6}{16} = \frac{3}{8}$

The equation, in vertex form, is $x = \frac{3}{8}(y-4)^{2}$ . In standard form:

$x = \frac{3}{8}(y^{2}-8y+16)$

$x = \frac{3}{8} y^{2}-3y+6$

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Example Question #511 : Ssat Upper Level Quantitative (Math)

A horizontal parabola on the coordinate plane has vertex (-3, -6) ; one of its -intercepts is (0, 4) .

Give its equation.

Possible Answers:

$x = \frac{9}{10} y^{2}+ \frac{27}{5} y+ \frac{81}{10}$

$x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}$

$x = \frac{6}{49} y^{2}+\frac{36}{49} y-\frac{240}{49}$

$x = \frac{7}{36} y^{2}+ \frac{7}{3} y+7$

Insufficient information is given to determine the equation.

Correct answer:

$x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}$

Explanation:

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

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

where (h,k) is the vertex. Set h = -3,k=-6 :

$x = a[y-(-6)]^{2}+ (-3)$

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

To find , use the known -intercept, setting x =0, y=4 :

$0 = a(4+6)^{2}-3$

$0 = a\cdot 10^{2}-3$

0 =100 a-3

100a= 3

$a = \frac{3}{100}$

The equation, in vertex form, is $x = \frac{3}{100}(y+6)^{2}-3$ ; in standard form:

$x = \frac{3}{100}(y^{2}+12+36)-3$

$x = \frac{3}{100} y^{2}+\frac{9}{25} y+\frac{27}{25} - \frac{75}{25}$

$x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}$

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Example Question #512 : Ssat Upper Level Quantitative (Math)

A vertical parabola on the coordinate plane has -intercepts (0,2) and (0,6) , and passes through (3, -2) .

Give its equation.

Possible Answers:

$x = \frac{2}{3}y^{2}- \frac{16}{3}y+8$

$x =- \frac{2}{3}y^{2}+ \frac{16}{3}y-8$

$x = -\frac{3}{32} y^{2}+ \frac{3}{4} y- \frac{9}{8}$

Insufficient information is given to determine the equation.

$x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}$

Correct answer:

$x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}$

Explanation:

A horizontal parabola which passes through (0,2) and (0,6) has as its equation

x = a(y-2)(y-6) .

To find , substitute the coordinates of the third point, setting x=3,y=-2 :

3= a(-2-2)(-2-6)

3= a(-4)(-8)

3= 32a

$a = \frac{3}{32}$

The equation is therefore $x = \frac{3}{32}(y-2)(y-6)$ , which is, in standard form:

$x = \frac{3}{32}(y^{2}-8y+12)$

$x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}$

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Example Question #515 : Ssat Upper Level Quantitative (Math)

A vertical parabola on the coordinate plane has -intercepts (2,0) and (6,0) , and passes through (3, -2) .

Give its equation.

Possible Answers:

$y = \frac{2}{3}x^{2}-\frac{16}{3}x+8$

$y = - \frac{3}{2}x^{2}+12x-18$

$y = x^{2}-8x+12$

$y = \frac{3}{2}x^{2}-12x+18$

$y =- \frac{2}{3}x^{2}+\frac{16}{3}x-8$

Correct answer:

$y = \frac{2}{3}x^{2}-\frac{16}{3}x+8$

Explanation:

A vertical parabola which passes through (2,0) and (6,0) has as its equation

y = a(x-2)(x-6)

To find , substitute the coordinates of the third point, setting x=3,y=-2 :

-2 = a(3-2)(3-6)

$-2 = a \cdot 1 \cdot (-3)$

-2 = -3a

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

The equation is $y = \frac{2}{3}(x-2)(x-6)$ ; expand to put it in standard form:

$y = \frac{2}{3}(x^{2}-8x+12)$

$y = \frac{2}{3}x^{2}-\frac{16}{3}x+8$

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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 #21 : X And Y Intercept

Example Question #5 : How To Find The Equation Of A Curve

Example Question #22 : X And Y Intercept

Example Question #21 : X And Y Intercept

Example Question #1 : How To Find The Equation Of A Curve

Example Question #11 : How To Find The Equation Of A Curve

Example Question #512 : Ssat Upper Level Quantitative (Math)

Example Question #511 : Ssat Upper Level Quantitative (Math)

Example Question #512 : Ssat Upper Level Quantitative (Math)

Example Question #515 : Ssat Upper Level Quantitative (Math)

All SSAT Upper Level Math Resources

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