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Scholastic Aptitude Test (SAT). PREP. Practice problems Version 6. SAT. This is a good example of “thinking outside the box.” The obvious choice is 6, which is wrong. . The NOT so obvious but correct choice is D (7). SAT. 50. a. 50. a. x. x. 14. 16. 10. 14. 10.
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SAT This is a good example of “thinking outside the box.” The obvious choice is 6, which is wrong. The NOT so obvious but correct choice is D (7).
SAT 50 a 50 a x x 14 16 10 14 10 (a – x) 2 + 302 = 502 (48 – x) 2 + 900 = 2500 (48 – x) 2 = 1600 48 – x = 40 8 = x a2 + 142 = 502 a2 + 196 = 2500 a2 = 2304 a = 48
SAT What do the given values really represent? This question really says that $100 is 80% of the original selling price. So, $ 100 = 80 % of x 100 = .8 x 125 = x
SAT 1 2 3 4 5 6 7 8 9 10 12 11 The trick here is to make sure each non-parallel line intersects all 3 parallel lines and each other where there are NO COMMON intersections.
SAT x + y + 80 = y + z + 80 + 6 3 3 x + y + 80 = y + z + 80 + 18 x = z + 18 x – z = 18
SAT Choice option I. x + x ½ has no exponent rule for adding exponent expressions with like bases, which means choices B,C and E are NOT viable. Choice II works because x½ is the same as square root of x, which means A is not valid, leaving Choice D as the only possible choice.
SAT If 5 = x – 2 , then ( 5 )2 = (x – 2)2 5 = (x – 2)2 5 5
SAT x2 + kx + 12 = (x – 1 ) ( x – 12 ) = x2 – 13x + 12
SAT Since 2x – 3y = 5 and 4x2 – 12 xy + 9y2 = (2x – 3y)2 = 52 = 25
SAT Since 3x – 3 = x (x + 7) Then, 0 = x2 + 4 x + 3 and 0 = (x + 3) (x + 1) Thus, x = – 3 or – 1