New SAT Math Workbook

B represent the integers as x x 2 and x 4 3x x

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Unformatted text preview: 2870 35 x + 35 x + 210 = 2870 70 x = 2660 x = 38 If he walked for 2 hours at 4 miles per hour, he walked for 8 miles. 5. (D) R 36 31 The rate of the faster car was x + 6 or 44 m.p.h. 2. (C) Before noon After noon R 50 40 · · T =D x 50x 8–x 40(8 – x ) T x x = D 36x 31x The 8 hours must be divided into 2 parts. 50 x + 40 (8 − x ) = 350 50 x + 320 − 40 x = 350 10 x = 30 x=3 They travel the same number of hours. 36 x − 31x = 30 5 x = 30 x=6 If he traveled 3 hours before noon, he left at 9 A.M. 3. (E) R 600 650 · T x x– 1 2 = D 600x 650(x – 1 ) 2 This problem may be reasoned without an equation. If the faster car gains 5 miles per hour on the slower car, it will gain 30 miles in 6 hours. The later plane traveled 1 600 x = 650 x − 2 600 x = 650 x − 325 325 = 50 x 61 = x 2 1 hour less. 2 The plane that left at 3 P.M. traveled for 6 2 hours. The time is then 9:30 P.M. 1 www.petersons.com 194 Chapter 12 Exercise 8 1. (E) In x days, he has painted of the barn. 5 To find what part is still unpainted, subtract the 5 part completed from 1. Think of 1 as . 5 x 5− x −= 55 5 5 x Retest 1. (B) Represent the integers as x, x + 2, and x + 4. 3x = ( x + 4 ) + 10 2 x = 14 x=7 x+2=9 2. (B) Mary x 6 + Ruth x 9 2. =1 (C) Represent the original fraction by x . 3x Multiply by 18. 3x + 2 x = 18 5 x = 18 3 x=3 5 x +8 8 = 3x − 6 9 Cross multiply. 9 x + 72 = 24 x − 48 120 = 15 x x=8 3x = 24 3. (E) Inlet x 3 − Drain x 6 =1 The original fraction is 3. (C) Original Added New 200 = 80 + 2 x 120 = 2 x x = 60 8 . 24 Multiply by 6. 2x − x = 6 x=6 No. of Quarts 40 x 40 + x Percent 5 0 2 Amount of 200 0 80 + 2x · Alcohol = Alcohol Notice the two fractions are subtracted, as the drainpipe does not help the inlet pipe but works against it. 4. (B) Tractor 2 4 Plow x 12 + =1 This can be done without algebra, as half the job was completed by the tractor; therefore, the second fraction must also be equal to 1 . x is 2 therefore 6. 5. (C) Michael 2 6 Barry 2 x 4. (D) Let x = Charles' age now x + 11 = Miriam's age now x + 3 = Charles' age in 3 years x + 14 = Miriam's age in 3 years g x + 14 = 2 ( x + 3) x + 14 = 2 x...
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