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SAT I Math Test #05SAT I Math Test #05SAT I Math Test #05SAT I Math Test #05SAT I Math Test #05SAT I Math Test #05SolutionSolutionSolutionSolutionSolutionSolutionSolutionSolution

SAT I Math Test No. 05SAT I Math Test No. 05SAT I Math Test No. 05SAT I Math Test No. 05SECTION 1SECTION 1SECTION 1SECTION 1

1. A perfect square is an integer that is a square of an1. A perfect square is an integer that is a square of aninteger. The integers h, k, and m are 16, 21 and 25,integer. The integers h, k, and m are 16, 21 and 25,not necessarily in that order, and each integer satisfiesnot necessarily in that order, and each integer satisfiesthe following condition.the following condition.•• k is a perfect square.• k is a perfect square.• m is an odd integer.• m is an odd integer.• h < k• h < kWhich of the following statements is TRUE?• h < kWhich of the following statements is TRUE?(A) h > m (B) m > k (C) m is a perfect square.(A) h > m (B) m > k (C) m is a perfect square.

(D) m is a multiple of 3. (E) h is a multiple of 3.(D) m is a multiple of 3. (E) h is a multiple of 3.

Since k is a perfect square, it is either 16 or 25.Since k is a perfect square, it is either 16 or 25.

But h < k, thus h = 16, k = 25.But h < k, thus h = 16, k = 25.

Also, m is an odd integer, thus m = 21.Also, m is an odd integer, thus m = 21.

So, h = 16, m = 21, k = 25.So, h = 16, m = 21, k = 25.So, h = 16, m = 21, k = 25.

The answer is (D)The answer is (D)


2. The length of one piece of rope is 5 inches less than2. The length of one piece of rope is 5 inches less than2. The length of one piece of rope is 5 inches less than3 times the length of a shorter piece. If the length of the3 times the length of a shorter piece. If the length of thelonger piece is 28 inches, what is the length, in inches,longer piece is 28 inches, what is the length, in inches,of the shorter piece?longer piece is 28 inches, what is the length, in inches,of the shorter piece?of the shorter piece?(A) 9 (B) 11 (C) 13 (D) 15 (E) 17(A) 9 (B) 11 (C) 13 (D) 15 (E) 17

L = 3S – 5, where L is the length of longer piece, and S is the length of a shorter one.

∴

L = 3S – 5, where L is the length of longer piece, and S is the length of a shorter one.

Now, L = 28.

∴

Now, L = 28.

∴ 28 = 3S – 5, or 3S = 33.∴ 28 = 3S – 5, or 3S = 33.∴

S = 11

∴

S = 11

Ans. (B)Ans. (B)


SUGAR CONTENT OF DEFFERENT JUICESSUGAR CONTENT OF DEFFERENT JUICES

Juice Size Amount of SugarJuice Size Amount of Sugar

Apple 10 oz 60 mgApple 10 oz 60 mg

Grape 8 oz 70 mgGrape 8 oz 70 mg

Orange 12 oz 45 mgOrange 12 oz 45 mg

3. Based on the table above, which of the following lists the3. Based on the table above, which of the following lists thebeverages from least to greatest sugar content per ounce?beverages from least to greatest sugar content per ounce?beverages from least to greatest sugar content per ounce?(A) A, G, O (B) A, O, G (C) G, A, O (D) G, O, A (E) O, A, G(A) A, G, O (B) A, O, G (C) G, A, O (D) G, O, A (E) O, A, G

The sugar content of each juice is as follows:The sugar content of each juice is as follows:

Apple = (60 mg)/(10 oz) = 6 mg/ozApple = (60 mg)/(10 oz) = 6 mg/oz

Grape = (70 mg)/(8 oz) = 8.75 mg/oz

∴

Grape = (70 mg)/(8 oz) = 8.75 mg/oz

Orange = (45 mg)/(12 oz) = 3.75 mg/oz

∴

Orange = (45 mg)/(12 oz) = 3.75 mg/oz

∴ 0 < A < G∴ 0 < A < G∴

Ans. (E)

∴

Ans. (E)


4. If k > 0, which of the following is closest in value to 0.73k?4. If k > 0, which of the following is closest in value to 0.73k?(A) k/7 (B) k/3 (C) (3k)/4 (D) k/2 (E) (3k)/7(A) k/7 (B) k/3 (C) (3k)/4 (D) k/2 (E) (3k)/7

(3/4)k = 0.75k ≈ 0.73k(3/4)k = 0.75k ≈ 0.73k

Ans. (C)Ans. (C)


5. If Ben collected n coins every day for 3 weeks,5. If Ben collected n coins every day for 3 weeks,how many coins did he collect?how many coins did he collect?(A) n + 3 (B) n - 3 (C) 3n (D) 7n (E) 21n(A) n + 3 (B) n - 3 (C) 3n (D) 7n (E) 21n

n × 7 days × 3 weeks = 21nn × 7 days × 3 weeks = 21n

Ans. (E)Ans. (E)


3, 4, 5, 6, 7, 8, 93, 4, 5, 6, 7, 8, 9

6. How many pairs of different numbers can be chosen6. How many pairs of different numbers can be chosen6. How many pairs of different numbers can be chosenfrom the list above so that the product of the two numbersfrom the list above so that the product of the two numbersis even? (Note: The pair 3, 4 is the same as the pair 4, 3.)is even? (Note: The pair 3, 4 is the same as the pair 4, 3.)is even? (Note: The pair 3, 4 is the same as the pair 4, 3.)(A) 6 (B) 9 (C) 10 (D) 15 (E) 21(A) 6 (B) 9 (C) 10 (D) 15 (E) 21

3 × 4, 3 × 6, 3 × 8,3 × 4, 3 × 6, 3 × 8,

4 × 5, 4 × 6, 4 × 7, 4 × 8, 4 ×9,4 × 5, 4 × 6, 4 × 7, 4 × 8, 4 ×9,

5 × 6, 5 × 8,5 × 6, 5 × 8,

6 × 7, 6 × 8, 6 × 9,6 × 7, 6 × 8, 6 × 9,

7 × 8,7 × 8,

8 × 98 × 9

Total = 15Total = 15

Ans. (D)Ans. (D)


7. Of 8 balls in a box, 3 are red and 5 are blue.7. Of 8 balls in a box, 3 are red and 5 are blue.Mike chooses 2 of these balls at random.Mike chooses 2 of these balls at random.If the first ball chosen is red, what is the probabilityIf the first ball chosen is red, what is the probabilitythat the second ball chosen is also red?that the second ball chosen is also red?(A) 3/8 (B) 3/28 (C) 1/4 (D) 5/8 (E) 5/7(A) 3/8 (B) 3/28 (C) 1/4 (D) 5/8 (E) 5/7

R = 3R = 3B = 5B = 5B = 5

T = 8T = 8

P(R – R) = 3/8 × 2/7 = 6/56 = 3/28P(R – R) = 3/8 × 2/7 = 6/56 = 3/28

Ans. (B)Ans. (B)


8. If a/k = b + 1, which of the following must equal k?8. If a/k = b + 1, which of the following must equal k?(A) a (B) a + 1 (C) b + 1 (D) a/(b + 1) (E) (a + 1)/b(A) a (B) a + 1 (C) b + 1 (D) a/(b + 1) (E) (a + 1)/b

a/k = b + 1, or k/a = 1/(b + 1)

∴

a/k = b + 1, or k/a = 1/(b + 1)

∴∴ k = a/(b + 1)∴ k = a/(b + 1)

Ans. (D)

∴

Ans. (D)


9. Ann, Barbara and Cindy can wear a hat colored yellow,9. Ann, Barbara and Cindy can wear a hat colored yellow,pink and blue. How many different combinations arepink and blue. How many different combinations arepossible?possible?(A) 4 (B) 6 (C) 9 (D) 12 (E) 18(A) 4 (B) 6 (C) 9 (D) 12 (E) 18

3 people × 3 hats = 93 people × 3 hats = 9

Ans. (C)Ans. (C)


10.If eight bags of flour with an average (arithmetic mean)10.If eight bags of flour with an average (arithmetic mean)weight of 10 pounds are put onto a shelf withweight of 10 pounds are put onto a shelf withten 8-pound bags of flour, what is the average weight,ten 8-pound bags of flour, what is the average weight,in pounds, of all 18 bags on the shelf?in pounds, of all 18 bags on the shelf?(A) 8 (B) 9 (C) 10 (D) 11 (E) 12(A) 8 (B) 9 (C) 10 (D) 11 (E) 12

Avg. = [(8 × 10) + (10 × 8)]/(18 bags) = 160/18 ≈ 9Avg. = [(8 × 10) + (10 × 8)]/(18 bags) = 160/18 ≈ 9

Ans. (B)Ans. (B)


11.Which one of the following cannot be a factor of 48?11.Which one of the following cannot be a factor of 48?(A) 2 (B) 4 (C) 6 (D) 7 (E) 24(A) 2 (B) 4 (C) 6 (D) 7 (E) 24

48 = 2 × 2 × 2 × 2 × 3

∴

48 = 2 × 2 × 2 × 2 × 3

∴ 7 cannot be a factor.∴ 7 cannot be a factor.∴

Ans. (C)

∴

Ans. (C)


12.If n > 0, for what value of n will n2 + 4 = 40?12.If n > 0, for what value of n will n + 4 = 40?(A) 3 (B) 4 (C) 6 (D) 8 (E) 16(A) 3 (B) 4 (C) 6 (D) 8 (E) 16

n2 = 36

∴

n2 = 36

∴∴ n = 6∴ n = 6

Ans. (C)

∴

Ans. (C)


13.If x is a positive even integer and 40 < 4x – 9 < 50,13.If x is a positive even integer and 40 < 4x – 9 < 50,what is the value of 4x – 9?what is the value of 4x – 9?(A) 45 (B) 46 (C) 47 (D) 48 (E) 49(A) 45 (B) 46 (C) 47 (D) 48 (E) 49

∴

40 < 4x – 9 < 50

∴

40 < 4x – 9 < 50

∴ 49 < 4x < 59, or 49/4 < x < 59/4 → 12.25 < x < 14.75∴ 49 < 4x < 59, or 49/4 < x < 59/4 → 12.25 < x < 14.75

Since x is even integers, we get x = 14.

∴

∴


∴

∴


∴ 4x – 9 = 4(14) – 9 = 56 – 9 = 47∴ 4x – 9 = 4(14) – 9 = 56 – 9 = 47

Ans. (C)

∴

Ans. (C)

∴


14.In n can be any odd integer such that 5 < √n < 8,14.In n can be any odd integer such that 5 < √n < 8,what is the difference between the largest possiblewhat is the difference between the largest possiblevalue of n and the smallest possible value of n?value of n and the smallest possible value of n?(A) 2 (B) 4 (C) 36 (D) 62 (E) 64(A) 2 (B) 4 (C) 36 (D) 62 (E) 64

5 < √n < 8

∴

5 < √n < 8

∴ 25 < n < 64∴ 25 < n < 64

∴

∴

But n = odd integer.

∴

∴

∴

But n = odd integer.

∴ the smallest n = 27, and the largest n = 63

∴

∴

∴ the smallest n = 27, and the largest n = 63

∴ 63 – 27 = 36

∴

∴ 63 – 27 = 36

∴

∴

Ans. (C)

∴

Ans. (C)

∴


15.The letters w, x, y, and z stand for numbers on the number15.The letters w, x, y, and z stand for numbers on the number15.The letters w, x, y, and z stand for numbers on the numberline above. Which of the following products is smallest?line above. Which of the following products is smallest?(A) wx (B) wy (C) wz (D) xy (E) yz(A) wx (B) wy (C) wz (D) xy (E) yz

To get the product of two numbers to be the smallest, the sign of the numbersTo get the product of two numbers to be the smallest, the sign of the numbers

must be opposite to get a negative number, while the absolute value of the productmust be opposite to get a negative number, while the absolute value of the product

must be the largest!!must be the largest!!

Therefore, wz = the smallest.Therefore, wz = the smallest.

Ans. (C)Ans. (C)


16.If the sum of the perimeters of two congruent squares16.If the sum of the perimeters of two congruent squaresis 40, what is the area of each square?is 40, what is the area of each square?(A) 2 (B) 4 (C) 5 (D) 10 (E) 25(A) 2 (B) 4 (C) 5 (D) 10 (E) 25

∴ P + P = 8x = 40

∴

∴ P1 + P2 = 8x = 40

∴

∴

∴ P1 + P2 = 8x = 40

∴ x = 5

∴

∴

∴ x = 5

∴ area = 52 = 25

∴

∴

∴ area = 52 = 25

∴

∴

Ans. (E)

∴

Ans. (E)


17.In the figure above, point R lies on17.In the figure above, point R lies onsegment O͞P. The area of the circlesegment O͞P. The area of the circlewith center O is 144π, and the areawith center O is 144π, and the areaof the circle with center R is 9π.of the circle with center R is 9π.

O͞What is the length of segment O͞R?What is the length of segment O͞R?(A) 12 (B) 9 (C) 6 (D) 3 (E) 2(A) 12 (B) 9 (C) 6 (D) 3 (E) 2

O͞Let the radius of the bigger circle be, O͞P = x,Let the radius of the bigger circle be, O͞P = x,

and the radius of the smaller circle, R͞P = y.and the radius of the smaller circle, R͞P = y.

The, 144π = πx2

∴

The, 144π = πx2

∴

∴

∴ x = 12, and 9π = πy2

∴

∴

∴ x = 12, and 9π = πy

∴ y = 3

∴

∴

∴ y = 3

∴ 12 – 3 = 9

∴

∴

∴ 12 – 3 = 9

∴

∴

Ans. (B)

∴

Ans. (B)


18.In a survey, 100 people were asked about two television18.In a survey, 100 people were asked about two televisionprograms, A and B. Of the people surveyed, 65 watchprograms, A and B. Of the people surveyed, 65 watchprogram A, 54 watch program B, and 34 watch bothprogram A, 54 watch program B, and 34 watch bothprograms. How many of the people surveyed watchprograms. How many of the people surveyed watchneither of them?neither of them?(A) 15 (B) 20 (C) 27 (D) 29 (E) 36(A) 15 (B) 20 (C) 27 (D) 29 (E) 36

Given Venn Diagram above, we get neither N = 100 – (AUB)Given Venn Diagram above, we get neither N = 100 – (AUB)

= 100 – [A + B – (A∩B)] = 100 – (65 + 54 – 34) = 15= 100 – [A + B – (A∩B)] = 100 – (65 + 54 – 34) = 15

Ans. (A)Ans. (A)


19.The volume of a rectangular box is 1 cubic foot.19.The volume of a rectangular box is 1 cubic foot.If the length of the box is 1/4 foot and the width is 1/2 foot,If the length of the box is 1/4 foot and the width is 1/2 foot,what is the height?what is the height?(A) 1/4 ft (B) 1/2 ft (C) 2 ft (D) 4 ft (E) 8 ft(A) 1/4 ft (B) 1/2 ft (C) 2 ft (D) 4 ft (E) 8 ft

V = 1 = 1/2 × 1/4 × h

∴

V = 1 = 1/2 × 1/4 × h

∴∴ h = 8∴ h = 8

Ans. (E)

∴

Ans. (E)

∴


20.In right triangle ABC above, B20.In right triangle ABC above,what is the length of side B͞C?

B

what is the length of side B͞C?(A) 2X (B) 3X (C) √5X(A) 2X (B) 3X (C) √5X

(D) √3X (E) 6X(D) √3X (E) 6X

3X3X

XA CXA C

B͞B͞C = √(x2 + (√3x)2) = √(4x2) = 2xB͞C = √(x + (√3x) ) = √(4x ) = 2x

Ans. (A)Ans. (A)


1. If | a | and | b | are integers less than 10 and | a/b | is1. If | a | and | b | are integers less than 10 and | a/b | isequivalent to 3/4, how many values of a are possible?equivalent to 3/4, how many values of a are possible?(A) One (B) Two (C) Three (D) Four (E) Five(A) One (B) Two (C) Three (D) Four (E) Five

The only possible values for | a/b | = 3/4 or 6/8, when | a | = 3 or 6.

∴


∴


∴ a = ±3, ±6∴ a = ±3, ±6

Ans. (D)

∴

Ans. (D)

∴


2. During a sale, a shopper pays the regular price for two2. During a sale, a shopper pays the regular price for twobottles of milk and pays $0.01 for the third bottle. If thebottles of milk and pays $0.01 for the third bottle. If theregular price of the milk is $1.87, how much per bottleregular price of the milk is $1.87, how much per bottledoes the shopper save by buying three bottles at this sale?does the shopper save by buying three bottles at this sale?(A) $0.01 (B) $0.62 (C) $0.93 (D) $1.86 (E) $1.87(A) $0.01 (B) $0.62 (C) $0.93 (D) $1.86 (E) $1.87

The regular price for 3 bottles is, 3 × $1.87 = $5.61

∴

The regular price for 3 bottles is, 3 × $1.87 = $5.61

Also, the sale price is (2 × $1.87) + ($0.01) = $3.75

∴

Also, the sale price is (2 × $1.87) + ($0.01) = $3.75

∴ The saving for 3 bottles is ($5.61 - $3.75) = $1.86

∴

∴ The saving for 3 bottles is ($5.61 - $3.75) = $1.86

∴

∴

∴ each bottle saving = $1.86 ÷ 3 = $0.62

∴

∴ each bottle saving = $1.86 ÷ 3 = $0.62

Ans. (B)

∴

Ans. (B)

SAT I Math Test No. SAT I Math Test No. 0505SAT I Math Test No. SAT I Math Test No. 0505SECTION SECTION 22SECTION SECTION 22

3. In the figure above,y

3. In the figure above,x > 90 and y = 2z.

yx > 90 and y = 2z.

y

If z is an integer,If z is an integer,what is the greatestwhat is the greatest

x zpossible value of y? x zpossible value of y?

(A) 57 (B) 58 (C) 59 (D) 60 (E) 61

x z

(A) 57 (B) 58 (C) 59 (D) 60 (E) 61

To get the max of angle y, we must have x to be the min,

∴

To get the max of angle y, we must have x to be the min,

∴ Given x > 90, x = 91∴ Given x > 90, x = 91

∴

∴ Given x > 90, x = 91

But z is an integer, and y + z = 2z + z = 3z must also be satisfied with (180 – x).

∴

∴

But z is an integer, and y + z = 2z + z = 3z must also be satisfied with (180 – x).

∴ x ≠ 91, since z cannot be integer.

∴

∴ x ≠ 91, since z cannot be integer.

∴

∴

The possible minimum value for x should be 93, so that 3z = 87.

∴

∴

∴

The possible minimum value for x should be 93, so that 3z = 87.

∴ z = 29

∴

∴ z = 29

∴ y = 58

∴

∴ y = 58

∴

∴

Ans. (B)

∴

Ans. (B)


4. The sales person said, "I will give you a discount of4. The sales person said, "I will give you a discount of25 percent from the listed price. Your final price is $299.“25 percent from the listed price. Your final price is $299.“What was the listed price?What was the listed price?(A) $224 (B) $349 (C) $399 (D) $449 (E) $499(A) $224 (B) $349 (C) $399 (D) $449 (E) $499

Let L = listed price, then $299 = (1 – 0.25)L

∴

Let L = listed price, then $299 = (1 – 0.25)L

∴ L = $299/0.75 = $399∴ L = $299/0.75 = $399∴

Ans. (C)

∴

Ans. (C)


5. In the figure above, line l is a5. In the figure above, line l is a

perpendicular bisector of lineperpendicular bisector of lineperpendicular bisector of line

segment PQ and PQ = 8.segment PQ and PQ = 8.

How many circles of radius 2√5How many circles of radius 2√5

having their centers on l can behaving their centers on l can behaving their centers on l can be

drawn that pass throughdrawn that pass through

both P and Q?both P and Q?both P and Q?(A) None (B) One (C) Two(A) None (B) One (C) Two

(D) Three (E) More than three(D) Three (E) More than three

( Next Slide ►► )( Next Slide ►► )


( ►► Continued from previouse slide)( ►► Continued from previouse slide)

circle with center O and r = 2√5circle with center O and r = 2√5

circle with center O´ and r = 2√5circle with center O´ and r = 2√5

( Next Slide ►► )( Next Slide ►► )


( ►► Continued from previouse slide)( ►► Continued from previouse slide)

P͞ P͞ R͞Given information that P͞Q = 8 and l is a P.B. line, we get P͞R = R͞Q = 4.Given information that P͞Q = 8 and l is a P.B. line, we get P͞R = R͞Q = 4.

Now, to get r = 2√5 = √20, we haveNow, to get r = 2√5 = √20, we have

Therefore, we get 2 circles with center on line lTherefore, we get 2 circles with center on line l

with point O = (0, 2) and O´ = (0, -2).with point O = (0, 2) and O´ = (0, -2).

Ans. (C)Ans. (C)


6. What is the surface area,6. What is the surface area,in square inches, of thein square inches, of therectangular solid above?rectangular solid above?(A) 52 (B) 104 (C) 156(A) 52 (B) 104 (C) 156

(D) 208 (E) 260(D) 208 (E) 260

S.A. = 2[(8 × 6) + (4 × 6) + (4 × 8)] = 208S.A. = 2[(8 × 6) + (4 × 6) + (4 × 8)] = 208

Ans. (D)Ans. (D)


7. What is the average (arithmetic mean) of 7 consecutive7. What is the average (arithmetic mean) of 7 consecutiveodd integers if the greatest of the integers is n?odd integers if the greatest of the integers is n?(A) n – 3 (B) n – 4.5 (C) n – 5 (D) n – 5.5 (E) n – 6(A) n – 3 (B) n – 4.5 (C) n – 5 (D) n – 5.5 (E) n – 6

Let’s just make n = 15, then 15, 13, 11, 9, 7, 5, 3 are the 7 consecutive odd integers.

∴


∴


∴Avg. = (sum of all)/(7 numbers) = 63/7 = 9∴Avg. = (sum of all)/(7 numbers) = 63/7 = 9

Now, try answer choice (A), (B), …

∴

Now, try answer choice (A), (B), …

∴

with n = 15 to get Avg. = 9.with n = 15 to get Avg. = 9.

The only choice is (E) n – 6 = 15 – 6 = 9.The only choice is (E) n – 6 = 15 – 6 = 9.


8. A camp cabin supply cabinet stocks numerals to use for8. A camp cabin supply cabinet stocks numerals to use fora cabin room numbers. If the stock consists of 10 each ofa cabin room numbers. If the stock consists of 10 each ofthe numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, how manythe numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9, how manyconsecutive room numbers, beginning with room numberconsecutive room numbers, beginning with room number1, can be formed?1, can be formed?(A) 17 (B) 18 (C) 19 (D) 22 (E) 23(A) 17 (B) 18 (C) 19 (D) 22 (E) 23

The cabin room numbers will start from 1 and it goes onThe cabin room numbers will start from 1 and it goes on

until it runs out of the numeral 1.until it runs out of the numeral 1.

That is,That is,

... 1 3 92 ... 1 3 92

, , … 1 0 1 1 1 2 1 7

, , …

∴

1 0 1 1 1 2 1 7

As we may count ‘s above, runs out on .

∴

1 1 71As we may count ‘s above, runs out on .

∴Ans. (A)

1 1 71

∴Ans. (A)∴∴


9. The sum of the positive even integers less than or equal to9. The sum of the positive even integers less than or equal to50 is subtracted from the sum of the positive odd integers50 is subtracted from the sum of the positive odd integersless than 50. What is the resulting difference?less than 50. What is the resulting difference?

– (2 + 4 + 6 + … + 48 + 50)– (2 + 4 + 6 + … + 48 + 50)– (2 + 4 + 6 + … + 48 + 50)

– (1 + 3 + 5 + … + 47 + 49)

∴

– (1 + 3 + 5 + … + 47 + 49)

– (1 + 1 + 1 + … + 1 + 1), which is 25 of 1’s.

∴

– (1 + 1 + 1 + … + 1 + 1), which is 25 of 1’s.

∴∴ 25∴ 25


△ ∠∠

10.In △ABC, the measure of ∠ ABC is 30° and AC < BC < AB.∠

10.In △ABC, the measure of ∠ ABC is 30° and AC < BC < AB.If the measure of ∠ BAC is x°,

△ ∠If the measure of ∠ BAC is x°,

△ ∠∠

what is one possible value of x?

△ ∠∠

what is one possible value of x?

∠ ∠ ∠

Since we have AC < BC < AB,

∠ ∠ ∠

∠

Since we have AC < BC < AB,

the angles become ∠ B < ∠ A < ∠ C.

∠

the angles become ∠ B < ∠ A < ∠ C.

Therefore, ∠ A = x° becomes the middle one.

∠ ∠

∠ ∠ ∠


∠ ∠

∴

∠ ∠ ∠


Now, ∠ A + ∠ C = 180° - 30° = 150°

∴

∠

Now, ∠ A + ∠ C = 180° - 30° = 150°

∴ the angle x must be greater than the smallest angle 30°,

∠

∠ ∠

∴ the angle x must be greater than the smallest angle 30°,

∴

∠ ∠

∴

but less than 1/2 × 150°, which is 75°.

∴

∴

but less than 1/2 × 150°, which is 75°.

∴ any angle between 30° and 75°.∴ any angle between 30° and 75°.∴∴


11.The shaded area shown in11.The shaded area shown inthe figure above is going tothe figure above is going tobe covered by square tilesbe covered by square tilesthat are 4 inches bythat are 4 inches by4 inches. How many tiles4 inches. How many tileswould be used to cover thewould be used to cover theshaded area?shaded area?

The total shaded are = 100 × 80 = 8000.

∴

The total shaded are = 100 × 80 = 8000.

∴∴ 8000 ÷ (4 × 4) = 500∴ 8000 ÷ (4 × 4) = 500


12.A bottle filled with juice weighs 10 pounds. When 1/3 of12.A bottle filled with juice weighs 10 pounds. When 1/3 ofthe juice remains in the bottle, it weighs 4 pounds.the juice remains in the bottle, it weighs 4 pounds.What is the weight of the empty bottle, in pounds?What is the weight of the empty bottle, in pounds?

Let B = the weight of the bottle, andLet B = the weight of the bottle, andLet B = the weight of the bottle, and

J = the weight of the juice.J = the weight of the juice.

Then, B + J = 10 - eq (1)Then, B + J = 10 - eq (1)

∴

Then, B + (1/3)J = 4 - eq (2)

∴

∴

Then, B + (1/3)J = 4 - eq (2)

∴ 3 × eq(2) – eq(1) = 2B = 2

∴

∴ 3 × eq(2) – eq(1) = 2B = 2

∴ B = 1

∴

∴ B = 1


C13.In the figure above, AC, CE, EB, BD, C13.In the figure above, AC, CE, EB, BD,and DA are line segments.

C

and DA are line segments. a

If a = 30 and b = 80,a

If a = 30 and b = 80,what is the value of x + y?what is the value of x + y?

b Dy

b Dc y

B

c

B

xxA EA E

∠ b = ∠ a + ∠ c, because ∠ b is the exterior angle, which is the sum of ∠ a and ∠ c.

∴ ∠

∠ b = ∠ a + ∠ c, because ∠ b is the exterior angle, which is the sum of ∠ a and ∠ c.

∴ ∠

∴ ∠

∠ ∠ ∠ ∠ ∠ ∠

∴ 80 = 30 + ∠ c

∴ ∠

∠ ∠ ∠

∠ ∠ ∠ ∠ ∠ ∠

∴ 80 = 30 + ∠ c

∴ ∠ c = 50°

∠ ∠ ∠

∴ ∠

∴ ∠ c = 50°

But angle ∠ c is also the exterior angle, which is the sum of ∠ x and ∠ y.

∴ ∠

∴ ∠

∴ ∠

But angle ∠ c is also the exterior angle, which is the sum of ∠ x and ∠ y.

∴ ∠

∴ ∠

∠ ∠ ∠

∴ ∠ c = 50° = x + y

∠ ∠ ∠

∴ ∠ c = 50° = x + y


14.If the product of five integers is even,14.If the product of five integers is even,at most how many of these five integers could be odd?at most how many of these five integers could be odd?

Consider odd × odd = odd and even × odd = even.

∴

Consider odd × odd = odd and even × odd = even.

∴

∴

∴ We get odd × odd × odd × odd × even = even.

∴

∴ We get odd × odd × odd × odd × even = even.

∴ 4 odds

∴

∴ 4 odds

∴

∴∴


15.When a number is multiplied by itself, the result is15.When a number is multiplied by itself, the result isequal to 13 less than 7 times twice the number.equal to 13 less than 7 times twice the number.What is one possible value of the number?What is one possible value of the number?

x2 = 7(2x) – 13

∴

x2 = 7(2x) – 13

∴

∴

x = 7(2x) – 13

∴ x2 – 14x + 13 = 0, (x – 13)(x – 1) = 0

∴

∴ x2 – 14x + 13 = 0, (x – 13)(x – 1) = 0

∴ x = 1 or 13

∴

∴ x = 1 or 13

∴

∴∴


16.A certain triangle has exactly two angles that have the16.A certain triangle has exactly two angles that have thesame measure. If the lengths of two of the sides of thesame measure. If the lengths of two of the sides of thetriangle are 20 and 40, what is the possible value for thetriangle are 20 and 40, what is the possible value for theperimeter of the triangle?perimeter of the triangle?

Since the triangle has exactly 2 angles with the same measure,Since the triangle has exactly 2 angles with the same measure,

it is an isosceles triangle as shown here.it is an isosceles triangle as shown here.

Therefore, P = 80 or 100.Therefore, P = 80 or 100.


A B C D EA B C D E

17.If the 5 cards shown above are placed in a row so that

A B C D E

17.If the 5 cards shown above are placed in a row so thatand must be next to each other, how manyB Cand must be next to each other, how manyB Cand must be next to each other, how many

different arrangements are possible?B Cdifferent arrangements are possible?

When we consider the two cards and as one group, always being together,B CWhen we consider the two cards and as one group, always being together,B C

then we have ( – ) – – – ,AB C D Ethen we have ( – ) – – – ,AB C D E

which we have 4! ways of arrangement.which we have 4! ways of arrangement.

Now, there are 2! ways of placing the two cards, ( – ).

∴

B CNow, there are 2! ways of placing the two cards, ( – ).

∴

B C

∴ we get 4! × 2! = 48∴ we get 4! × 2! = 48


18.For nonzero numbers a and b, let the operations18.For nonzero numbers a and b, let the operationsand be defined by a b = (a – b)/ab and a b =and be defined by a b = (a – b)/ab and a b =

(ab)/(a + b). What is the value of 5 (2 3)?(ab)/(a + b). What is the value of 5 (2 3)?

First, (2 3) = (2 × 3)/(2 + 3) = 6/5 = 1.2First, (2 3) = (2 × 3)/(2 + 3) = 6/5 = 1.2First, (2 3) = (2 × 3)/(2 + 3) = 6/5 = 1.2

Now, 5 (2 3) = 5 1.2 = (5 – 1.2)/(5 × 1.2) = 3.8/6 = 19/30 or 0.633Now, 5 (2 3) = 5 1.2 = (5 – 1.2)/(5 × 1.2) = 3.8/6 = 19/30 or 0.633


1. If x2 + 3x – 10 = 0 and x2 + 7x + 10 = 0,1. If x + 3x – 10 = 0 and x + 7x + 10 = 0,what is the value of x?what is the value of x?(A) -5 (B) 2 (C) -2 (D) 5(A) -5 (B) 2 (C) -2 (D) 5

(E) It cannot be determined from the information given.(E) It cannot be determined from the information given.

∴

∴

x2 + 3x – 10 = (x + 5)(x – 2) = 0 ∴ x = -5 or 2

∴

x + 3x – 10 = (x + 5)(x – 2) = 0 ∴ x = -5 or 2

x2 + 7x + 10 = (x + 5)(x + 2) = 0 ∴ x = -5 or -2

∴

x2 + 7x + 10 = (x + 5)(x + 2) = 0 ∴ x = -5 or -2

Therefore, x = -5 is the solution that satisfies both.

∴

∴

Therefore, x = -5 is the solution that satisfies both.

∴

Ans. (A)Ans. (A)


2. In the figure above, the area of square B CE2. In the figure above, the area of squareABCD is 4. What is the perimeter of

B CEABCD is 4. What is the perimeter ofisosceles triangle AED?isosceles triangle AED?(A) 2 (B) 2 + √3 (C) 2 + 2√3(A) 2 (B) 2 + √3 (C) 2 + 2√3

(D) 2 + √5 (E) 2 + 2√5(D) 2 + √5 (E) 2 + 2√5

A DA D

Using Pythagorean Th., we get A͞E = √(12 + 22) = √5

∴ △

A D

Using Pythagorean Th., we get A͞E = √(12 + 22) = √5

∴ △∴ Perimeter of △AED, P = 2 + 2√5∴ Perimeter of △AED, P = 2 + 2√5

Ans. (E)

∴ △

Ans. (E)


3. If the average (arithmetic mean) of x and y is 10 and the3. If the average (arithmetic mean) of x and y is 10 and theaverage of x, y, and z is 20, what is the value of z?average of x, y, and z is 20, what is the value of z?(A) 40 (B) 50 (C) 60 (D) 70 (E) 80(A) 40 (B) 50 (C) 60 (D) 70 (E) 80

∴

(x + y)/2 = 10

∴

(x + y)/2 = 10

∴ x + y = 20∴ x + y = 20

Also, (x + y + z)/3 = (20 + z)/3 = 20

∴

∴

Also, (x + y + z)/3 = (20 + z)/3 = 20

∴

∴

Also, (x + y + z)/3 = (20 + z)/3 = 20

∴ z = 40∴ z = 40

Ans. (A)

∴

Ans. (A)


4. If the lengths of the sides of a right triangle are4. If the lengths of the sides of a right triangle areconsecutive even integers, which of the followingconsecutive even integers, which of the followingcan be the perimeter of the triangle?can be the perimeter of the triangle?(A) 12 (B) 18 (C) 24 (D) 30 (E) 36(A) 12 (B) 18 (C) 24 (D) 30 (E) 36

x2 + (x + 2)2 = (x + 4)2

∴

x2 + (x + 2)2 = (x + 4)2

∴ x2 + x2 + 4x + 4 = x2 + 8x + 16

∴

∴ x2 + x2 + 4x + 4 = x2 + 8x + 16

∴

∴

∴ x2 – 4x – 12 = 0

∴

∴

∴ x – 4x – 12 = 0

or (x – 6)(x + 2) = 0

∴

∴

∴

or (x – 6)(x + 2) = 0

∴ x = -2 or 6

∴

∴ x = -2 or 6

∴

∴

Now, let x = 6, then 6, 8 and 10.

∴

∴

Now, let x = 6, then 6, 8 and 10.

∴ p = 6 + 8 + 10 = 24

∴

∴ p = 6 + 8 + 10 = 24

Ans. (C)

∴

Ans. (C)

∴


y5. In the semicircle above, y5. In the semicircle above,the center is at (3,0).the center is at (3,0).Which of the following areWhich of the following arex-coordinates of two pointsx-coordinates of two pointson this semicircle whoseon this semicircle whosey-coordinates are equal? O

xy-coordinates are equal? O

x

(A) -1 and 6 (B) -1 and 8(A) -1 and 6 (B) -1 and 8

(C) 0 and 6 (D) 2 and 6 (E) 2 and 8(C) 0 and 6 (D) 2 and 6 (E) 2 and 8

Since the semi-circle has center at (3, 0), and its radius r = 5,Since the semi-circle has center at (3, 0), and its radius r = 5,

we must look for the x-coordinates that are symmetric to x = 3 line.we must look for the x-coordinates that are symmetric to x = 3 line.

x = 0 and x = 6 are the points that are symmetric to x = 3 line.x = 0 and x = 6 are the points that are symmetric to x = 3 line.

Ans. (C)Ans. (C)


32,032 = 14 × 52 × 4432,032 = 14 × 52 × 44

6. Based on the information above,6. Based on the information above,6. Based on the information above,what is the greatest prime factor of 32,032?what is the greatest prime factor of 32,032?(A) 3 (B) 5 (C) 7 (D) 11 (E) 13(A) 3 (B) 5 (C) 7 (D) 11 (E) 13

32,032 = 14 × 52 × 44 = (7 × 2) × (2 × 2 × 13) × (2 × 2 × 11)

∴

32,032 = 14 × 52 × 44 = (7 × 2) × (2 × 2 × 13) × (2 × 2 × 11)

∴∴ The greatest prime factor is 13.∴ The greatest prime factor is 13.

Ans. (E)

∴

Ans. (E)


7. If n is an odd integer, x is the next odd integer greater7. If n is an odd integer, x is the next odd integer greaterthan n, and y is the next even integer greater than x,than n, and y is the next even integer greater than x,what is the value of xy in terms of n?what is the value of xy in terms of n?(A) 2n + 4 (B) 2n + 5 (C) n2 + 3n + 2 (D) n2 + 4n + 3 (E) n2 + 5n + 6(A) 2n + 4 (B) 2n + 5 (C) n2 + 3n + 2 (D) n2 + 4n + 3 (E) n2 + 5n + 6

x = n + 2, y = n + 3

∴

x = n + 2, y = n + 3

∴ xy = (n + 2)(n + 3) = n2 + 5n + 6∴ xy = (n + 2)(n + 3) = n2 + 5n + 6∴

Ans. (E)

∴

Ans. (E)


8. Three points not on a line are the vertices of a certain8. Three points not on a line are the vertices of a certaintriangle. If the lengths of the sides of the triangle aretriangle. If the lengths of the sides of the triangle areconsecutive odd integers, which of the following CANNOTconsecutive odd integers, which of the following CANNOTbe the length of the longest side of the triangle?be the length of the longest side of the triangle?(A) 5 (B) 7 (C) 9 (D) 11 (E) 13(A) 5 (B) 7 (C) 9 (D) 11 (E) 13

Remember, for any triangle, the sum of two sides is always greater than the other!!Remember, for any triangle, the sum of two sides is always greater than the other!!

Therefore, if we set the side length to be 1, 3 and 5, then, 1 + 3 < 5,Therefore, if we set the side length to be 1, 3 and 5, then, 1 + 3 < 5,

which cannot form a triangle.which cannot form a triangle.

Ans. (A)Ans. (A)


9. The probability of choosing a black ball from a box9. The probability of choosing a black ball from a box9. The probability of choosing a black ball from a boxcontaining only black balls and white balls is 4/7.containing only black balls and white balls is 4/7.If there are 12 white balls in the box,If there are 12 white balls in the box,how many black balls are in the box?If there are 12 white balls in the box,how many black balls are in the box?how many black balls are in the box?(A) 6 (B) 8 (C) 9 (D) 12 (E) 16(A) 6 (B) 8 (C) 9 (D) 12 (E) 16

∴

P (Black) = x/(x + 12) = 4/7

∴

∴

P (Black) = x/(x + 12) = 4/7

∴ 7x = 4x + 48

∴

∴ 7x = 4x + 48

∴ 3x = 48

∴

∴

∴ 3x = 48

∴

∴

∴

∴ x = 16

∴

∴ x = 16

Ans. (E)

∴

Ans. (E)


10.In the figure above,10.In the figure above,if x + y + z < 180 and DEif x + y + z < 180 and

z

DE

AC is parallel to ED, z

AC is parallel to ED,which of the followingwhich of the following

x y

must be true?x y

A B Cmust be true? A B C

(A) Line AE and BD will intersect when extended.(A) Line AE and BD will intersect when extended.

(B) Line BD is not perpendicular to line DC.(B) Line BD is not perpendicular to line DC.(C) x = z (D) y > x (E) y < x(C) x = z (D) y > x (E) y < x

Since △BDC has ∠ B + y + z = 180

∴ ∠

Since △BDC has ∠ B + y + z = 180

∴ ∠

△ ∠

∴ ∠ B = 180 – y – z

∴

△ ∠

∴ ∠ B = 180 – y – z

But, x + y + z < 180

∴

∴ ∠

But, x + y + z < 180

∴ x < 180 – y – z

∴ ∠

∴ ∠

∴ x < 180 – y – z

∴ ∠

∴

∴

∴ x < ∠ B

∴

∴

∴ x < ∠ B

∴ Lines AE and BD will intersect, when they are extended.

∴ ∠

∴ Lines AE and BD will intersect, when they are extended.

Ans. (A)

∴ ∠

∴

Ans. (A)


11.Raymond, working at a toy store, is paid k dollars each11.Raymond, working at a toy store, is paid k dollars eachday. From this amount he spends k/4 dollars per day.day. From this amount he spends k/4 dollars per day.In terms of k, how many days will it take RaymondIn terms of k, how many days will it take Raymondto save $1,000?to save $1,000?(A) k/4,000 (B) k/1,000 (C) 1,000/k (D) 4,000/k (E) 4,000/(3k)(A) k/4,000 (B) k/1,000 (C) 1,000/k (D) 4,000/k (E) 4,000/(3k)

∴

Raymond’s saving per day is, S = (k – k/4) = 3k/4 per day.

∴

Raymond’s saving per day is, S = (k – k/4) = 3k/4 per day.

∴ $1,000/[(3k)/4] = 4,000/(3k) days to save $1,000.∴ $1,000/[(3k)/4] = 4,000/(3k) days to save $1,000.

Ans. (E)

∴

Ans. (E)


7, -7, -10, ...7, -7, -10, ...

12.The first term in the sequence of numbers shown above12.The first term in the sequence of numbers shown above12.The first term in the sequence of numbers shown aboveis 7. Each even-numbered term is -1 times the previousis 7. Each even-numbered term is -1 times the previousterm and each odd-numbered term, after the first,term and each odd-numbered term, after the first,is 3 less than the previous term.is 3 less than the previous term.is 3 less than the previous term.What is the 47th term of the sequence?What is the 47th term of the sequence?(A) -10 (B) -7 (C) -1 (D) 7 (E) 10(A) -10 (B) -7 (C) -1 (D) 7 (E) 10

7, -7, -10, ...

∴ ∴

7, -7, -10, ...

∴ The fourth term is even term, which is (-1) times (-10). ∴ 10

∴

∴ The fourth term is even term, which is (-1) times (-10). ∴ 10

∴

∴ ∴

Also, the fifth term is odd term, which is 3 less than (10). ∴ 7

∴

∴ ∴

Also, the fifth term is odd term, which is 3 less than (10). ∴ 7

Therefore, it goes as, 7, (-)7, (-)10, 10, 7, (-)7, … , which repeats every four.

∴

∴ ∴

∴

Therefore, it goes as, 7, (-)7, (-)10, 10, 7, (-)7, … , which repeats every four.

∴ 47 ÷ 4 gives remainder of 3.

∴

∴


∴


∴ The third term, (-)10 is the answer.

∴

∴ The third term, (-)10 is the answer.

Ans. (A)

∴

∴

Ans. (A)

∴


13.When the radius of a given circle is lengthened by 1 inch,13.When the radius of a given circle is lengthened by 1 inch,the area is increased by 9π square inches. What is thethe area is increased by 9π square inches. What is thelength, in inches, of the radius of the original circle?length, in inches, of the radius of the original circle?(A) 4 (B) 6 (C) 8 (D) 9 (E) 10(A) 4 (B) 6 (C) 8 (D) 9 (E) 10

Let r = radius, and A = area of the original circle.Let r = radius, and A = area of the original circle.

Then, A + 9π = π(r + 1)2.Then, A + 9π = π(r + 1)2.

∴

But A = πr2.

∴

∴

But A = πr .

∴ πr2 + 9π = πr2 + 2πr + π

∴

∴ πr2 + 9π = πr2 + 2πr + π

∴ 2πr = 8π

∴

∴

∴ 2πr = 8π

∴

∴

∴

∴ r = 4

∴

∴ r = 4

Ans. (A)

∴

∴

Ans. (A)

∴


14. AB, BC, CD and AD are diameters14. AB, BC, CD and AD are diametersof the four circles shown above.of the four circles shown above.If AB = BC = CD = 2, what is theIf AB = BC = CD = 2, what is thearea of the shaded region?area of the shaded region?(A) 12π (B) 6π (C) (9/2)π (D) 3π (E) 2π(A) 12π (B) 6π (C) (9/2)π (D) 3π (E) 2π

Let R be the radius of the big circle, then R = 3.Let R be the radius of the big circle, then R = 3.

∴

Also, let r be the radius of the smaller circles, then r = 1.

∴

Also, let r be the radius of the smaller circles, then r = 1.

∴ the area of the shaded region,∴ the area of the shaded region,2 2

∴

A = (1/2)(π × 32) – 1/2 × 3 × (π × 12)

∴

A = (1/2)(π × 3 ) – 1/2 × 3 × (π × 1 )

= (9π)/2 – (3π)/2 = (6π)/2 = 3π= (9π)/2 – (3π)/2 = (6π)/2 = 3π

Ans. (D)Ans. (D)


15.If x = 2t2 and t3 = x, where t is a positive integer,15.If x = 2t and t = x, where t is a positive integer,what is the value of x?what is the value of x?(A) 2 (B) 8 (C) 6 (D) 5 (E) 4(A) 2 (B) 8 (C) 6 (D) 5 (E) 4

x = 2t2 = t3

∴

x = 2t2 = t3

∴

∴

x = 2t = t

∴ t3 – 2t2 = 0

∴

∴

∴ t3 – 2t2 = 0

∴ t2(t – 2) = 0

∴

∴

∴ t2(t – 2) = 0

∴

∴

∴

∴

∴ t = 0, 2

∴

∴

∴ t = 0, 2

∴ x = 2(2)2 = 8

∴

∴ x = 2(2)2 = 8

Ans. (B)

∴

∴

Ans. (B)


△16.In the figure above, △PQR is an isosceles Q16.In the figure above, △PQR is an isoscelestriangle with the vertex angle θ = 30°,

Q△triangle with the vertex angle θ = 30°,

30

△

and SR and TV are line segments. 30

△

and SR and TV are line segments.What is the value of x + y?What is the value of x + y?

x V(A) 60 (B) 75 (C) 20

xx

P

V(A) 60 (B) 75 (C) 20

(D) 45 (E) 15xxS

PR(D) 45 (E) 15 x

ySP

Ry

TT

Since △PQR is an isosceles △ with vertex angle of 30°, we get ∠ P = ∠ R = 75°.

∴

Since △PQR is an isosceles △ with vertex angle of 30°, we get ∠ P = ∠ R = 75°.

∴ x = 25°, and y = 2x = 50°

∴

△ △ ∠ ∠

∴ x = 25°, and y = 2x = 50°

∴

△ △ ∠ ∠

∴

∴ x + y = 75°

∴

∴ x + y = 75°

Ans. (B)

∴

∴

Ans. (B)

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