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This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. math exams: SAT Practice Questions - 1

Thursday, June 23, 2011

SAT Practice Questions - 1

  1. $2(5x-5)+5(2x+2)=$
    1. $0$
    2. $20x$
    3. $20x-10$
    4. $20x+10$
    5. $10x^{2}+20+x+20$

  2. If $x=a+2$, and $y=-8-a$ then $x+y=$

    1. $6$
    2. $10$
    3. $2a-6$
    4. $-10$
    5. $-6$

  3. If $x \ne -5$, then $\frac{x^{2}+3x-10}{x+5}-(x-2)=$

  4. If $(a-\frac{1}{a})^{2}=8$, then $a^{2}+\frac{1}{a^{2}}=$

    1. $8$
    2. $6$
    3. $10$
    4. $12$
    5. $100$

  5. $(x+y)^{2}=16$, and $x^{2}+y^{2}=6$ then $xy=$

    1. $5$
    2. $10$
    3. $6$
    4. $4$
    5. $16$

  6. $(x+y)=6$, and $x^{2}-y^{2}=2$ then $x-y=$

    1. $12$
    2. $4$
    3. $6$
    4. $\frac{1}{3}$
    5. $3$

  7. $\frac{15y+3}{3}-5y=$

    1. $1$
    2. $0$
    3. $10y+1$
    4. $3$
    5. $3y+1$


  8. if $b^{2}-a^{2}=9$ then $5(a-b)(a+b)=$

    1. $45$
    2. $\frac{9}{5}$
    3. $4$
    4. $50$
    5. $-45$

  9. When $c \ne 3$, then $\frac{c^{2}-9}{c-3}=$

    1. $c-3$
    2. $1$
    3. $c+3$
    4. $3-c$
    5. $o$

  10. If $b>0$, and $b^{2}-1=10 \times 12$, then $b=$

    1. $9$
    2. $1$
    3. $12$
    4. $10$
    5. $11$


    Answer Key
    1 B
    2 E
    3 0
    4 C
    5 A
    6 D
    7 A
    8 E
    9 C
    10 E
Posted by Ssafini : at 6:51 PM

1 comment:

  1. O.Z.December 6, 2012 at 9:20 AM

    I believe you have two mistakes above:

    First: the answer to 6 is "E"
    y^2-x^2=(x+y)(x-y)
    since, (x+y)=2 and we have that y^2-x^2=6
    then, (x-y)=6/2
    the result, in turn, should be 3 NOT 1/3

    Second:
    In #8, the answer should be A

    Again, y^2-x^2=(x+y)(x-y)
    so, we have that y^2-x^2=9
    In turn, we know that (x+y)(x-y)=9 as well.
    So, 5(x+y)(x-y)= 5*9= 45 NOT -45

    Correct me if I'm wrong!XD

    ReplyDelete

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