Preparing For The Accuplacer Test - TestPrep-Online

TestPrep-Online 1 11 .9 19. 191 1 .1 0The result calculated is; therefore, the correctanswer is D.If you chose any other answer, review your calculation, specifically in addingdecimals.Level 2:3. Kate wanted to buy a red dress for the prom. Online, the dress cost 176whereas, at the store, the price was 25% higher. Kate decided to wait for thedress to go on sale at the store and buy it there. When the dress in the store wason sale, the price decreased by 20%. What was the store’s sale price of the dress?A.B.C.D. 275 220 140.8 176THE CORRECT ANSWER IS D.Increasing a number byis equivalent to multiplying it by, whiledecreasing a number byis equivalent to multiplying it by.The store’s original price of the dress was 25% more than the online price, thus itwas equivalent to multiplying the online price by 1.25. The store’s sale price ofthe dress decreased by 20% and was equivalent to multiplying the store’s originalprice by 0.8.Now, set up the equation:1 7 6 1 .2 5 0 .8 1 7 6 54 4 1765According to the calculations, the store’s sale price is equal to the online price of 176; therefore, the correct answer is D.If you chose answer B, you only calculated the store’s original price; if you choseanswer C, you calculated the store’s sale price from the online price and not fromthe store’s original price; if you chose answer A, you did not set up the equationproperly. Review how to build and solve percentage problems.Copyright: www.testprep-online.com16

TestPrep-Online4. The perimeter of the blue rectangle is 24 inches; one side is equal to 8 inches.How many red squares are needed to fill the blue square if the area of the redsquare is 0.25 square inches?A.B.C.D.63264128THE CORRECT ANSWER IS D.Rectangles have both length and width, and the area is calculated by multiplyingthem: Area Length Width whereas a rectangle's perimeter is calculated byadding all the edges: Perimeter 2 (Length Width).In order to solve this type of problem, summarize the information given in thequestion:P e r im e t e r B lu e 2 4S id e B l u e a 8A re a R e d 0 .2 5P e r im e t e r 2 a 2 b24 2 8 2b2b 24 16 8b 8 2 4In order to find out how many red squares are needed to fill the blue rectangle,calculate the blue rectangle's area:A re a a bA r e a B lu e 8 4 3 2In order to calculate how many red squares are needed to fill the blue rectangledivide their areas:A r e a B lu e A r e a R e d 3 2 0 . 2 5 ?Convert the equation into fractions in order to simplify the calculation:3 2 0 .2 5 321 2510 0 321 14 321 4 1In order to fill the blue rectangle you needcorrect answer is D.32 41 1128 128 1281red squares; therefore, theCopyright: www.testprep-online.com17

TestPrep-OnlineIf you chose answer A, you multiplied the areas, instead of dividing them. If youchose answers B or C, review your calculation of the blue rectangle's sides and itsarea.Level 3:5.?A.B.C.D.1.85314.1914.491386THE CORRECT ANSWER IS B.First, decide whether to convert the decimal to a fraction or the fraction to adecimal. In this exercise it is easier to convert the fraction to a decimal:.Rewrite the equation:?Begin solving the equation by breaking down each part:First, multiply:?In order to multiply the decimals, convert them to whole numbers by moving thedecimal point:0 . 5 6 2 . 8 8 5 6 2 8 8 Now multiply:4 44 1 4 2 8856 17 2 8144 0 0161 2 8Do not forget to move the decimal point back:1 6 1 2 8 . 1 . 6 1 2 8 Next, continue with the division of the calculated result and the next decimal:?In order to divide the two decimals, move the decimal point in the divisor all theway to the right to make it a whole number. Move the decimal point the samenumber of places in the dividend.1 . 6 1 2 8 0 . 1 2 1 1 6 . 1 2 8 1 2 Copyright: www.testprep-online.com18

TestPrep-OnlineNow divide:0 1 3 .4 41 2 1 6 1 .2 80 1 2 4136 5248 48480Last, add the cumulative result calculated to the last decimal number of theequation:?1 1 3 .4 40 .7 51 4 .1 9The final result is; therefore, the correctanswer is B.If you chose answers A or D – review the order of operations in your calculation.Remember that multiplication and division operations are carried out beforeaddition and subtraction operations. If you chose answer C, review yourcalculation.6. If, thenPis betweenA.B.C.D.THE CORRECT ANSWER IS C.According to the order of operations, first solve the additions in the parentheses,then multiply:()?The first parenthetical expression is solved by finding the lowest commondenominator of the fractions. In this case, it is 20:1 41 101 1 5 5 4 1 5 420 20220 4 5 Copyright: www.testprep-online.com19

TestPrep-OnlineTo solve the second parenthetical expression, use long addition:11 0 .9 5730 .3 8001 .3 373Now, rewrite the exercise: ( )?In order to multiply the numbers, convert the fraction to a decimal:( )()( )?To simplify, move the decimal point:0. 5 1. 3 3 7 3 5 13373 Multiply: 1 1 3 1 1 3 3 7 356 6 8 6 5Remember, you must return the decimal point the same number of spaces youmoved it in order to discover the correct value: 66 8 6 5 . 0 . 6 6 8 6 5 ()To find the range in which P is located, convert the answers to decimals:1 0 .1101 2 0 .5101 425 0 .2 51003 42575 0 .7 510034 275 2 .7 5100Check the answers:A.B.C.D.10 P 0 P 0 . 1101 P 1012 0 . 1 P 0 . 2 54 P 213 0 . 5 P 0 . 7 543 P 3 2 . 7 5 P 34Copyright: www.testprep-online.com20

TestPrep-OnlineP is located in the range of the third answer:is bigger than0.5 and smaller than 0.75. You may place the result on a number line to see theanswer clearly: 00 .10 .20 .30 .40 .50 .6 0 .70 .80 .910 .6 7If you chose any other answer please review the solution.Elementary AlgebraLevel 1:7. What is the value of the expressionand?A. 84whenB. 81C. 14D. 84THE CORRECT ANSWER IS A.Method A:Substitute the values for x and y. (2 x 5 y )(5 x 3 y ) ? x 3 y 2 2 3 5 ( 2 ) 5 3 3 ( 2 ) 6 ( 1 0 ) 1 5 ( 6 ) (6 1 0 )(1 5 6 ) ( 4 )( 2 1 ) 8 4Method B:Simplify the equation first, and then substitute the values for x and y:Copyright: www.testprep-online.com21

TestPrep-Online(2 x 5 y )(5 x 3 y ) 2 x 5 x 5 x 3 y (2 x 3 y ) 10x2 25xy 6 xy 15y 1 0 x 1 9 x y 1 5 y x 322 10x22 10 9 19 ( 6) 15 4 19xy 15y22 y 2 21 0 (3 ) 1 9 (3 )( 2 ) 1 5( 2 ) 90 114 60 84Therefore,and the correct answer is A.If you chose any other answer, review your calculations.8. A. B. C. D. THE CORRECT ANSWER IS C.To calculate this equation, use square roots rules:ab a bandaa b b 0 bSimplify the equation:5 105 10 25 10 22 5 5 251According to the calculation, the correct answer is C.If you chose any other answer, review the square roots rules.Level 2:9. Simplify the inequality:A.B.C.D. THE CORRECT ANSWER IS B.Copyright: www.testprep-online.com22

TestPrep-OnlineTo solve this problem, apply the inequality rules of operations. Multiply bothsides of the inequality in order to work with whole numbers: 1 2 .4 x 8 x 0 .4 x 5 .2 2 .7 x 10 1 2 .4 x 1 0 8 x 1 0 0 .4 x 1 0 5 .2 1 0 2 .7 x 1 0 124 x 80x 4 x 52 27 xSimplify the inequality by canceling down and moving like-terms: 124 x 80x 4 x 52 27 x 44 x 31x 52 31x 44 x 31x 31x 52 31x 13 13x 52x 52 13 x 4Remember, when you divide an inequality by a negative number, reverse theinequality sign.According to the calculations, the correct answer is B.If you chose answer A, you forgot to reverse the inequality sign when dividing bya negative number. If you chose answer D, review the definition of absolutevalue. If you chose answer C, review your calculations.10. Each house in New Town costs five times the equivalent of a house in Old Town.Linda bought five houses in New Town and two houses in Old Town.Which of the following could she have bought for the same price?A. One house in New Town plus twenty five houses in Old TownB. Seven houses in New TownC. Three houses in New Town plus fourteen houses in Old TownD. Two houses in New Town plus seventeen houses in Old TownTHE CORRECT ANSWER IS D.To solve this problem, express the data given in an algebraic equation:One house in New Town is equivalent in cost to 5 houses in Old Town.Represent the cost of 1 house in Old Town by x; accordingly, the cost of onehouse in New Town will be 5x.Linda spent the equivalent of 5 houses in New Town plus 2 houses in Old Town 5 5x 2x 25x 2x 27 xCheck the answers to find which of the options costs 27x:Copyright: www.testprep-online.com23

TestPrep-OnlineA. One house in New Town plus twenty five houses in Old Town 1 5x 25x 5x 25x 30 xB. Seven houses in New Town 7 5x 35xC. Three houses in New Town plus fourteen houses in Old Town 3 5x 14 x 15x 14 x 29 xD. Two houses in New Town plus seventeen houses in Old Town 2 5x 17 x 10 x 17 x 27 xTherefore, the only correct answer is D.If you chose any other answer, review your calculations and the way youexpressed the information given in the algebraic equations.Level 3:11. Which of the following statements is not possible given the following rules? A. Both are negative.B. They are equal.C. Both are positive.D. One is negative and the other is positive.THE CORRECT ANSWER IS C.Remember that the absolute value is positive:In order to solve this problem, check the options given in the answers:Option A: Both numbers are negative a 0 and bP o s it iv enum berN e g a t iv enum ber2b 0 P o s it iv enum berN e g a t iv enum ber2a b aThis inequality is possible.Option B: They are equala bSubstitute into the two inequalities:Copyright: www.testprep-online.com24

TestPrep-Online2a a2b bThere are three possible cases for a b : 1)positive, 3) a b and both are negative.a b 0 , 2) a band both areAssess each of the options:a)a b 0a b 02a a2b b 2 0 02 0 0 0 00 0This is an incorrect inequality.b)a band both are positive(a b ) a n d (a 0 ) a n d (b 0 )P o s it iv enum berP o s it iv enum ber2a P o s it iv enum berP o s it iv enum ber2ba b 2a a2b bConsequently, this is also an incorrect inequality.c)a band both are negative(a b ) a n d (a 0 ) a n d (b 0 )P o s it iv enum berN e g a t iv enum ber2a P o s it iv enum berN e g a t iv enum ber2ba bThis inequality is possible; hence, this answer is possible whenare negative.Copyright: www.testprep-online.coma band both25

TestPrep-OnlineOption C: Both numbers are positive a 0 and b 0 P o s it iv enum berP o s it iv enum ber 2abS im p lif yP o s it iv enum berP o s it iv enum ber 2bP o s it iv enum ber 2a P o s it iv enum ber2baP o s it iv enum berbP o s it iv enum ber P o s it iv enum berF ro mt h e 2 in e q u a lit ie s 2aP o s it iv enum ber bP o s it iv enum ber 2bP o s it iv enum ber aaThis inequality is incorrect as it is impossible to getnumber.2a a, when a is a positiveOption D: One number is positive and the other is negative a 0 and bP o s it iv enum berN e g a t iv enum ber2b 0 P o s it iv enum berP o s it iv enum ber2a b aThis is possible if the absolute value of a 0 and ais bigger than b . 0 bP o s it iv enum berP o s it iv enum ber2bis bigger than a .P o s it iv enum berN e g a t iv enum ber2a bb aThis is possible if the absolute value ofTherefore, option D is possible when the conditions are met.Consequently, when both numbers are positive, the given inequalities are notpossible; therefore, the correct answer is C.If you chose any other answer, review the explanation and the rules of absolutevalue and inequality.12. IfA. 1and, what is the value of?B. 2.5C. 4D. 6THE CORRECT ANSWER IS C.To solve this problem, find the x and y values from the two system equations.Method A:Copyright: www.testprep-online.com26

TestPrep-OnlineExpress x or y from one equation and substitute it in the other: 2y 5 x 5 4y x 212 y 5 x 5 5 x 5 2 y x 5 2y54 y x 2 1 4 y 5 2y 215Substitute x in order to find y:4y 5 2y 21 554y 5 5 2y 5 21 552 0 y (5 2 y ) 1 0 520y 5 2y 10522y 110y 5Substituteinto one of the given equations and find x:2y 5x 5y 5 2 5 5x 510 5x 55x 5 10 5x 5 15Method B:Subtract one equation from the other in order to cancel x or y:Copyright: www.testprep-online.com27

TestPrep-Online 2 y 5 x 5 2 4 y x 2 1 2y 2 5 x 2 5 2 4y x 21 4y 10x 10 4y x 21 4y 10x 4y x 10 21 4y 4y 10x x 1111x 11x 1Substituteinto one of the given equations, and find y:y 5x 5x 1 2y 5 1 52y 5 52y 5 5 10y 10 52Calculate:and; therefore,.The correct answer is C.If you chose any other answer, review your calculations.College-Level MathLevel 1:13. For all ,A.B.C. 1D.E.?Copyright: www.testprep-online.com28

TestPrep-OnlineTHE CORRECT ANSWER IS: DRemember basic trigonometric identities:Now, rewrite the equation:Accordingly, the explanation for the equation is:therefore, the correct answer is D.If you chose any other answer, review basic trigonometry identities.14. What is the value ofifandA. 2B. 4C. 14D. 16E. 18;?THE CORRECT ANSWER IS C.Place the valuein the given function:2 4g 0 .5 20 .5 1 1 2 0 .5 3 4 0 .5 ( 0 .5 0 .5 ) Next place the calculated value(in 4 2 2 16 2 14:)The final result is (); therefore, the correct answer is C.If you chose answer B, you probably used 0.5 instead of ( 0.5). If you chose anyother answer, review your calculations.Level 2:15. What are the first term and the common difference in a mathematical sequencein which the sum of 10 terms is equal to 120, and the sum of 15 terms is equalto 255?A.;B.;C.;D.;E. Cannot be determined from the given informationCopyright: www.testprep-online.com29

TestPrep-OnlineTHE CORRECT ANSWER IS B.In order to solve this problem, use a formula suited to mathematical sequences:nSn 2a1 2 n 1 d , in which d is the common difference,term, n is the number of terms in the sequence, andsequence.Method A:First, draw information from the question:Use the formula to express n 1 d 2S n n 2a1 n 1 d 2S n n2Sn 2 2a1 n2S nn n 1 d 2a1 n 1 dandS 15 2 5 5. n 1 d n 1 d a122nis the sum of the: 2a1 Sn S 10 1 2 0is the firsta1 Sn 2nThen, use the given data to solve the equations: S 10 a1 10 S 15 a1 15 1 0 1 d120 2 15 10 1 d255 29d2 1514d2From these two equations, find d:120 109d 2255 1415d2 4 .5 d 7 d 1 7 1 22 .5d 5d 2Now placein one of the equations and find:d 2a1 120 109d 12 29 2 12 9 32Method B:Use the formulatwo equationsSn n2S 10 1 2 0 2a1 and n 1 d S 15 2 5 5, substitute the given data, and form.Copyright: www.testprep-online.com30

TestPrep-Online10 120 2a1 1 0 1 d 2 2 5 5 1 5 2a 1 5 1 d 1 2 1 2 0 5 2 a 1 9 d 2 5 5 7 . 5 2 a 1 1 4 d 1 2 0 1 0a1 4 5d 2 5 5 1 5a1 1 0 5d 5 2 4 2a1 9d 5 1 3a1 2 1dMultiply the first equation by 3 and the second by 2, and then subtract one fromthe other to cancel : 2 4 2 a 1 9 d 5 1 3 a 1 2 1 d 3 2 2 4 3 2a1 3 9d 3 5 1 2 3a1 2 2 1d 2 7 2 6 a1 2 7 d 1 0 2 6 a1 4 2d 7 2 1 0 2 6 a1 2 7 d 6 a1 4 2d 3 0 6 a1 2 7 d 6 a1 4 2d 3 0 1 5d 30 15 d d 2Substituteinto one of the equations to find:2 4 2a1 9dCopyright: www.testprep-online.com:31

TestPrep-Online 2 4 2a1 9d d 2 2 4 2a1 9 22 4 2a1 1 82 4 1 8 2a1 1 8 1 86 2a1a1 3From the calculations;; therefore, the correct answer is B.If you chose answer A, you mixed and d. If you chose answers C or D, checkyour calculations. If you chose answer E, review the arithmetic sequence rules.16.;is a natural number.A.B. 9C.D.E.THE CORRECT ANSWER IS B.In order to solve this exercise use the following exponents’ rules:n(a )m an m, andaaan am an m,n amn mSimplify the equation by using the rules above:3b 29(3bbb 2 ) 6b2b b3 3 (3 3 2From the calculations:bb 2 ) 63b 29b 3 9 (33b(3bbb 2 ) 6bb 9 (3 33b2bbb 2 3 ) 6b 9( 332b2b 6 6b) 9bb 2 ) 62bbb 9; therefore, the correct answer is B.If you chose one of the other answers, review the exponents’ rules.Copyright: www.testprep-online.com32

TestPrep-OnlineLevel 3:17. Calculate the yellow area in the drawing below:A.B.C.D. E. ()THE CORRECT ANSWER IS C.To solve this problem, calculate the yellow area by computing the whole squarearea and subtracting the area of the green circle:Calculate the yellow square side using the Pythagorean Theorem ( a2 cy e llo w 2 1 2 a a 3 3 2c y e llo w 1 2 a a 3 3 2 b2 c2):22

preparation for your Accuplacer test. TestPrep-Online will soon be offering a complete preparation kit for the Accuplacer, which will include a study guide for each section of the test, along with sample questions