Section 2= Math

1 . D

2 . 8

3 . 8

4 . 8

PSATSample Test

a a a a - a o a a a a

5

The hundredths' digit is the second digit to the right of the decimal point, which inthis case is 8.

Check each choice, one at a time. (A) equals 9x2. (B) however, doesn't: 4x + 5x =9X because when you're adding like variables you sum the coefficients and leavethe variable as it is. Choices (C), (D), and (E) all equal 9x2.

You must always do the same thing to both sides of an equation1 ' l

i?x) = t(10)2 x ' 1 0

X = 5

gSI"/" is one of the percents whose fractional equivalent should be familiar to you.31 1

SSIZ = t,

,o you're looking for the region on the pie chart that is approximatolt 1'- ' r

3

of the circle.

5. G The area of a square is equal to the length of its side squared. Eagh of the twosquares has an area of 25, so each square has a side length of \/25 = 5. From thesketch below you can see that the original rectangle has a width of 5 and a length of5 + 5 = 1 0 .

The perimeter of a rectangle is equal to2(l+ w), where /= length and w= width.The perimeter of the original rectangle is 2(10 + 5) = 2(15) = 30.

6. B This question talks about rates and times, so you'll need to use the formularate x time = distance. John traveled at 48 miles per hour for 2 hours and15 minutes , or 2.25 hours. He traveled 48 x 2.25 = 108 miles. The question askswhat rate he'd have to drive at to make this same trip in 2 hours, which is

108 mi les= 54 miles per hour.2 hours

7 . A This question can be solved very quickly if you recognize that (x + 1)(x - 1) is adifference of squares. In general, (a + b)(a- b) = a2 - E.Here, (x+ 1)(x- t; =x2-1. You're given that x2 =7, so x2- 1=7 - 1 = 6. l f you didl ' t see this, youcould have used your calculator to come up with the value of Y7, and plugged thatin for x to solve.

8. D This question is a little tricky, since it calls on you to deal with percents in theabstract. To make the question more concrete, Pick Numbers. Since it's a percentproblem, the best number to pick is 100. But because there are 2 variables, x and y,you might be confused as to which of them you should make 100. Since x is given

as 66 { R"r."n, ot y, il you let }z = 100, then x is simply uu 3

Now you need toJ

determine what percent of y x is. Percent = ::?;1L x 100%. Since you're looking forwnole

a percent of x, x is the whole and y is the part. Plugging in gives you percent =

# . rco%ot #& x 100%. Since dividing by a fraction is the same asood g

multiplying by its reciprocal, 100 x #

t rcO% = s9.

= 15Oo/o, (D). Even if you

couldn't entirely solve this question, you could have used common sense to

eliminate a few choices. Since x is only AAl"n of y, yis greater than x. Therefore, y' 3must be more than 100% of X and (A) and (B) could both be eliminated as too'small.

9. A Look at the prime factorizations of 4, 6, and 10:

4 = 2 x 26 = 2 x 3

1 0 = 2 x 5

A multiple of all three of these numbers will have to contain each of these primefactors raised to the highest exponent with which it is used. In other words, it mustcontain two factors of 2, one factor of 3, and one factor of 5. Now look at the answerchoices, and find the prime factorization of each:

(A) I = 2 x 2 x 2 or 23. Since you determined that the number you were looking foronly had to have 2 factors of 2, it doesn't have to be divisible by 8.

Even if you could not solve this question systematically, you could have tried comingup with a numberwhich is a mult ip le oI 4,6, and 10 and which is not a mult ip le ofthe correct answer choice. With a little experimenting you would find that 60 is amultiple ol 4, 6, and 10 but not a multiple of 8.

10. E The least common denominator of all the fractions is 64. Rewriting each fraction

with a denominator of 64 gives you #. #* # * #. &* fr, which sums to

I Og and 64 have no common factor other than 1, so this fraction cannot be64reduced. Therefore, the numerator of the sum is 63.

PSATSample Test

a a a a a a a a a a a

6

1 1 . E

12. B

1 3 . A

14. C

PSATSample Test

a a a a - a a a a a a

7

The vertices of the square are labeled D, E, F, and G in the figure above. Theperimeter of LABC is equal to the sum AE EB, BF, FG, GC, CD, and DA. SquareDEFG has an area of 3. The area of a square is equal to its side length squared, soFGz - 3, and FG =li. Since AABC is equilateral, all three of its angles are 60degrees. Since DE is parallel to BC, A^AED is also an equilateral triangle. In squareDEFG, DE = FG, so DE = V3. In equilateral triangle AED, AE = DA = DE, so AE =DA=Y3. Now look at right triangle EFB. ZEFB has a measure of 90'because thisangle is supplementary to zEFG of the square. You know that angle EBF has ameasure of 60" because it is an angle of equilateral triangle ABC. So the third angleof triangle EBF, IFEB, must be 180" - 90' - 60" = 30o. So right triangle EFB is a30-60-90 triangle whose sides are in the ratio x : xXE:2xJn LEFB, the longerleg, Ef is also a side of the square, so it has a length of V3. The length of theshorter leg, BF, is 1 and the hypotenuse, EB, is 2. Now AB = AE + EB = Y3 + 2.Since all three sides of equilateral triangle ABC ye equal, the perimeter of triangleABCis AB + BC + CA = 3AB =3(V3 + 2) =3V3 + 6.

To simplify a radical, factor out any perfect squares in it. Va8 = V16 x 3 = V16 "

f 3=+Vg. s imitar ly, f i =\ /4xg=\/4 "

Vd =2{3. So you have 4V3 +z\/g+ V g = 7 f 3 .

To work with the symbol, just plug the given values into the expression ,' * I.

The left side of the equation becomes t'* tand

the right side of the equation

becomes 5' * *.This gives you 9 + 2 = 25 * !, or 11 = 25* {. You're asked to2 2 2

solve for n, so isolate it on one side of the equation by subtracting 25 from both

sides. This gives you -1 4 = +.

Multiplying both sides by 2, -28 = n.

You're given that the average of three different numbers is 4, and you're asked to findthe greatest possible value for one of those integers. Since you know th'e actualaverage, and the number of terms, you can solve for the sum of the three numbers byplugging the values into the average formula: (average)(number of terms) = (sum ofterms). Therefore the sum of the terms is ( )(3) = 12.To find the greatest possible valuefor one of the terms, you need to take the smallest possible values for the other twoterms. The question states that allthree terms are positive integers, so the smallestpossibility for one of these terms is 1. lt can't be 0, since 0 is neither positive ornegative. Now you need to find the smallest possible value for the second term. All 3terms must be different, so this term can't also be 1. The smallest it can be is 2. Thiswould make the third term 12 - 1 - 2 = 9.

1 5 . E

1 6 . E

1 7 . B

1 8 . A

You're given the length of the hypotenuse and the length of one leg of a righttriangle, and are asked to solve for the the length of the other leg, so you shouldimmediately think of the Pythagorean Theorem. The theorem states that in a righttriangle, the square of the hypotenuse is equal to the sum of the squares of thelegs. Here, the hypotenuse is x + 3 and one of the legs is x + 2. Let the length ofthe leg you're looking for be called n. Plugging into the theorem gives you:

r f + ( x + 2 ) 2 = ( x + 3 ) 2r f = ( x + 3 ) 2 - ( x + 2 ) 2r f = ( x 2 + 6 x + 9 ) - ( x 2 + 4 x + 4 )rf = 2x;t_F_n = Y 2 x + 5

The best way to solve a remainders problem is to pick a number for the variable.Since n leaves a remainder of 7 when divided by 9, n is 7 more than a multiple of9, so let n equal 9 + 7 or 16. When n = 16, 5n = 80. 80 divided by 9 is 8 with aremainder of B.

The best way to solve this question is to try out the answer choices to see if oneworks. There is no way to move 1 marble from Box A to Box B so that the ratio of blackmarbles to white marbles will be the same in both boxes. However, if you move 2 blackmarbles from Box A to Box B, then Box A will contain 2 black marbles and1 white marble and Box B will contain 6 black marbles and 3 white marbles. Thereforethe ratio of black marbles to white marbles in both boxes will be 2 to 1.

You're given the equation V = !ovs and are asked what the value of V is multiplied3

by i ' t r is multiplied Oy !.ty

it and see what happens. Multiplying ,by +gives

you

{. elug this into the given equation:2 u = +4;)'=i4#)

=t4E)= (*)(*-'

Since the original value of V is tor",

you can see that multiplying ,by +causes

V

to be multiplied by 1.tt

19. E You're asked to find the length of a chord of a circle, which also happens to be theside of a triangle drawn within that circle. See what you can figure out from thegiven information. You're told that the circle has area 36a and since the area of acircle is equal to z times its radius squared, the radius of the circle is 6. This meansthat radii OA = OC = 6, and LOAC is isosceles. At this point you can eliminate acouple of answer choices. Remember, diagrams on the SAT are drawn to scale

psAT unless otherwise noted, so you can see tnat AC is greater than either OA or OC,

.93Tgl{"..:. and is therefore greater than 6. Eliminate (A) and (D) for being too small. You also- --6----- knowthatthe longestchord of acirc le is a diameter, onethat passesthrough i ts

center. Since chord AC doesn't pass through center O it must be less than thecircle's diameter, which is 12 (d = 2l).Therefore you can eliminate (C) as being toolarge. You've barely started to solve, and you've already worked your way down to 2choices, (B) and (E). lf you were pressed for time or stumped as to how to proceed,you'd have great guessing odds.

What else can you figure out from the given information? You're told that arc ABC

has length 4n. The circumference of the circle is 2trr, or 12n, so this arc is # = !' 1 2 n 3of the entire circumference. That means that the central angle of the arc, IAOC,

1represents

J of the circle's total angle measure, ot

ry = 120". Since LOAC is

isosceles with OA = OC, ZOAC = Z-OCA. Therefore each of these angles is equal+^ 180" - 120" _ 60',o ----r: = =; = 30o. lf you draw a perpendicular line from O to AC, you

divide LOAC into two right triangles.

Since each contains a 30" angle as shown, the third angle in each triangle must be60'. So each is a 30-60-90 triangle, with sides in the ratio of x: xV3: 2x. Sincethe hypotenuse of each right triangle is 6, the side opposite the 60'angle is 3V3.That means AC is 3V3 + 3V3 = 6V3.

20. D This is a good question to solve by Picking Numbers, and since it's a percentproblem it 's usually bestto pick 100. Letthe average population of the 9 otherdistricts be 100. The population of District 4 is 3 times that, or 3 x 100 = 300.The question asks what percent of the city's population lives in District A.

Percent = #!- x 100%, where the part is the population of District A and thewhole

whole is the population of the entire city, which is the sum of the population ofDistrict A and the 9 other districts. Plugging this in gives you:

300300 + 9(1 00)

PSATSample Test

a a a a - a a a a a a

. 9

x 1OO"/o, or #

x 100%.

This reduc". to ]

x 100% or 25"/o.

1 . DTo divide by a fraction, multiply by its reciprocal.

_ 2 x _ x4 2

4.

5 .

6.

7.

PISATSample Test

a a a a t a a a a o o

1 4

Section 4= Math

1( 4 \\E)

2 . ERemember the order of operations. PEMDAS can help you get through longproblems and prevent careless mistakes.x = l - 5 l r * l - + - s l r 7x = 5 2 + l - z l x zX = 2 5 + 7 x 7x = 2 5 + 4 9X = 7 4

ENote that the question is not asking for the value of n. 4n - 15 = 17, so 4n = 32 andfl = 8. Then 3n + 9 = 3(8) + I = 24+ 9 = 33.

cThis question can be rewritte n u" { = 9. Cro.. multiply to find 27 x = 135, then'15 xdivide both sides by 27 to find X = 5. This problem is also a good candidate for

Backsolving.

cThe first step is to translate the question into an equation. Call the original number

x. Then write the question as x - 4 = 9x. Subtract x from both sides of the equation

to find -4 = 8x. Divide both sides by 8 to find x = -+z

BLearn the vocabulary of the PSAT math section before Test Day, so that you are notconfused and have an edge on complicated problems. Since set J is the set of allnegative odd integers and set K is the set of all negative even integers, the union ofthese two sets is a set that combines all of the values of both sets into one set. So,the union of these sets would be all negative odd integers and all negative evenintegers, which simplifies to all negative integers.

ERemember that a-b = 1-

ab '

m - 2 1 1 1= B'= 64

8' c { s"l;#il"

(A) and (D) are both less than I yard, so they are too short to tie up the bag and ""li'5-'".iil 'z

can be eliminated. To find out which of the remaining lengths is shortest, and therefore

will have the least amount of remaining rope, give them all a common denominator.

fiffi = €3, +(3) = ffi, anaS(f) = ffi. rn" shortest option ronger than f vard. 2rs

5 yaro.

The explanations intentionally skip from 8 to 16, since questions 9-15 do notappear on the test.

1 6 . 2Note that the question asks for the value of x. Substitule 2y for x in the firstequation. 2y + 4y= 6, so 6y = 6 and y = 1. x = 2y = 2(1) = 2.

9 or .754

lf Alice completes her homework in 8 hours, she completes * of it every hour. Afterworking for 6 hours, she has completed U r *

= 3

= f; ot it.u

I 296The product of 8 and 9 is72, so half their product is 36. Since 36 is the square root ofthe number you're looking for, the number must be 362 = 36 x 36 = 1,296.

Average = =$!HL, so in this case the average is- numDer ol terms'

1 7 .

1 8 .

1e. fi ot .zoe

1 1 6 44 - 6 2 4 ' 2 4

(#)_ 1 0 _ 1 _ 5-

2 4 ^ Z - 2 4

You could also grid in the decimal equivalent of ], .eOe.z+

20. ILooking at the addition problem, you see that A plus A in the units column yields an8 in the units column of the sum. Therefore A could equal either 4 (since4 + 4 =8) or 9 (since 9 + 9 = 18). lf A were 4, then you wouldn't have to carry a 1 tothe tens column, and 4 plus B would equal 18. There is no value of B for which thiscould be true, so try A = 9. lt A = 9, then A plus A in the units column would sum to18, leaving an 8 in the units column of the sum, and requiring you to cafty a 1 to thetens column. Therefore you'd have 1 + 9 + B = 1B in the sum. This is true for anyv a l u e o f B ( 1 + 9 + 3 = 1 3 , 1 + 9 + 7 - 1 7 , e t c . ) S o t h e v a l u e o f 4 m u s t b e g .

21. 36LCAD = ICAE - ZDAE. You're told that ZCAE = 73, so if you can find ZDAE youcan solve for ICAD. AB is perpendicular Io AE, so ZBAE = 90. Also, ZBAE =ZBAD + ZDAE, and since IBAD = 53, 90 = 53 + IDAE, and 37 = ZDAE. Now thatyou have the measure of IDAE you can solve for ICAD: ZCAD = 73 - 37 = 36.

PSATSample Test

a a a a a a a a a a a

t 6

22. 45You're asked to find the perimeter of the rectangle. Perimeter = 2(l + w), where / andu/ represent length and width respectively. So you need to find these measurementsto solve. You're not given any information about the rectangle, but you are told thatthe two circles inscribed within it have circumference 9n. Therefore, there must besome way to use the information about the circles to determine the length and widthof the rectangle. lf you draw in some radii, as shown in the figure below, you'll noticethat the length of the rectangle is equal to 3 radii and that the width is equal to 2radii,

Circumference is equal lo 2nr, where r is the radius of the circle. So 2nr = 9n,2r = 9, and r = 4.5. The length of the rectangle is 3(a.5) = 13.5, and the width is2(4.5) = 9. Plugging into the formula, you find that the perimeter of the rectangleis 2(13.5 + 9) = 2(22.5) = 45.

23. 5Since 9 cookies were left uneaten, the number of cookies that were eaten is 64 - 9= 55. You're told that each guest ate n cookies, so n times the number of guests atthe party must equal 55. Since the number of guests must be a whole number, thenumber of cookies must also be a whole number. and both numbers must be factorsof 55. The factors of 55 are 1, 5, 11 and 55. You're told that the number of guests atthe party is greater than 9, so it must be either 11 or 55, which would make n either5 or 1 respectively. You're told that n must be greater than 1, so it must thereforebe 5.

24. 48You're given a lot of information here, and if you just take a little time to sift throughit, you'll be able to solve. You know how many boxes of candy were sold, how muchmoney was raised, and that some boxes cost $2 and the rest cost $5. You're askedto solve for the number of boxes sold at $5, so let that number equal x. Since a totalof 150 boxes were sold, 150 - x is the number sold at $2. lf you multiply the numberof boxes sold at each price by their respective prices and then add the two dollaramounts, you'll get the total amount of money raised. You know that this total is$444, so you can set up an equation to solve for x

5 x + 2 ( 1 5 0 - x ) = 4 4 45x+ 300 - 2x= 444

3x+ 300 = 4443x = 144x = 4 8

2s t" ,

So the number of boxes sold at $5 is 48. ,"li#L"

a a a o a o a o o o o

lf you weren't sure how to proceed on this question, your calculator would be a great 17ally. You could use it to try out various possibilities for the number of boxes sold ateach price. You know that 150 boxes were sold, so you might try 75 at each price:(75 x $5) + (75 x $Z; = 937U + $150 = $525. Only $444 was raised, so there musthave been fewer boxes sold at $S. try 50 and 100: (50 x $5) + (100 x $2) = $250 +$200 = $450. This is still too big, but just by a little, so you know you're close. A fewmore tries on your calculator should help you find that 48 boxes were sold at $5.

12.2 ot 12.3

The perimeter of a rectangle is 2(/ + w), where / b the length and w is the width.Since the perimeter of this rectangle is 1 4, 2(l + A = 14. Dividing both sides by 2,I + w = 7. Think about different rectangles which have a length and a width whichadd up to 7. l f /= 6 and w=1, then the area of the rectangle is 6 x 1, or 6. l f /= 5and w = 2, the area is 5x2, or 10. l f l=4and w = 3, the area is 4x3,or 12. Not ice,that the closer the sides of the rectangle come to being equal, the greater the areabecomes. The area will be a maximum when the rectangle is a square. In this caseI = w = 3.5. The area of such a square is 3.5 x 3.5 = 12.25.