UPCAT Math Reviewer Solution

8/6/2019 UPCAT Math Reviewer Solution

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UPCA T Page 1

REVIEWER at http://www.facebook.com/home.php?#!/pages/UPCAT-REVIEWER/112910555407543?ref=sgm

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UPC A T REVI EWE

ANSWER KEY AND SOLUTION TO

UPCAT MATHEMATICS REVIEWER(60 ITEMS)


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ANSWER KEY

1 D 13 C 25 D 37 D 49 B2 C 14 C 26 C 38 A 50 C

3 A 15 D 27 BONUS 39 D 51 D

4 B 16 C 28 B 40 B 52 B

5 C 17 D 29 B 41 C 53 C

6 B 18 D 30 A 42 D 54 C

7 B 19 C 31 A 43 C 55 D

8 D 20 A 32 A 44 C 56 A

9 B 21 D 33 D 45 C 57 C

10 C 22 C 34 A 46 D 58 B

11 A 23 B 35 A 47 A 59 A

12 D 24 C 36 C 48 A 60 D

SOLUTION

1. Amanda wants to buy a graphing calculator that is already on sale for P4200.The sale price is %30 below the original price. If she is able to get an additional

%15 off for being a student, how much money was saved from the original price?

(A) P1,890.00 (B) P1,170.00 (C) P2,000.00 (D) P2,430.00

Solution:

To get the original price divide P4200 by 0.7 the result is P6,000.00.

Amanda paid P4200 x 0.85 which is equal to P3,570.00

Amanda saved P6,000 P3,570.00 = P2,430.00.

Answer: D

100%-30%

100%-15%


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2. Let r and s be the roots of the quadratic equation .0211332 x x x x x x

Evaluate: 331

221

111

sr sr sr .

(A) 41 (B) 17

3 (C) 118 (D) 15

4

Solution :If we simplify the quadratic equation, the result is

.023265 222 x x x x x x .0183 2 x x

.3

1and3

8 rssr

Simplify 33

122

111

1sr sr sr

931

421

11

sr rssr rssr rs

938

331

1

438

231

1

138

31

1

938

331

1

438

231

1

138

31

1

4

1113

41

118

Answer : C

Recall:

If r and s are the roots of

02 cbxax then

sum of roots =ab

sr

product of roots =ac

rs


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3. Solve for x: 53 563025 x(A) 3 (B) 4 (C) 6 (D) 8

Solution:

53 563025 x 532 56655 x x divide both sides by

25

33 5665 x x3

3030x

3x

Answer: A

4. The x-intercepts of a parabola are 2 and 4, and the y-intercept is 8. If theparabola passes through the point (a,8), what is the value of a?

A) 5 (B) 6 (C) 7 (D) 8

8,a

0,4 0,2

8,0


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Solution: 42 x xk y , when x = 0, 4020k y 88k 1k

The equation of the parabola is 42 x x y 842 aa

8862

aa062 aa

06aa6or0 aa ,

a = 6

Answer: B

5. Sa isang lungsod, 2/5 ng mga lalaking may sapat na gulang ay kasal sa 5/7 ngbabaeng may sapat na gulang. Ang bilang ng mga kasal na kalalakihan at kasal nakababaihan ay pantay-pantay, at ang mga may sapat na gulang ng populasyon ayhigit sa 3400. Ano ang pinakamaliit na posibleng bilang ng mga may sapat nagulang na residente sa lungsod?

(A) 3442 (B) 3452 (C) 3432 (D) 3412

Solution:

Let x = number of meny = number of women

y x75

52

So y x1425


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Also x y2514

x and y are integersso n y 14 for a certain integer nthis implies that n x 25

3400y x34001425 nn

340039 n

18.8739

3400n

The smallest possible integer n is 88.

Therefore 3432883939 n y x

Answer: C

6. An acute angle is formed by two lines of slope 1 and 7. What is the slope of the line which bisects this angle?

(A) 23 (B) 2 (C) 3 (D) 4


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x y

x y 7

Solution:

The angle of inclination of the line x y 7 is O452

therefore 7452tan O

7tan452tan-1

45tan2tanO

O

7tan2-1 12tan

2tan7712tan

432tan

Using half-angle tangent formula

2

2tantan

2sin2cos1 [

432tan , using Pythagorean triple 3-4-5, 5

42cos and 532sin ]

53

541

45

45

1tan

If the angle of inclination of the line

bmx y is then

tanm

Recall:

B A

B A B A

tantan1tantan

tan

half-angle tangent formulas

B B

B B B

cos1sin

sincos1

2tan


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31

The angle of inclination of the angle bisector is O45 Therefore the slope of the angle bisector is equal to O45tan

OO

45tantan145tantan

45tan

tan11tan

311

131

2

Answer: B

7. The altitude to the hypotenuse of a triangle with angles of 30 and 60 is 3 units. What is thearea of the triangle in square units?

(A) 29 (B) 36 (C) 39 (D) 26

Solution:

30O

60 O

3 units

units 6

units 32


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base = 6 units

height = units 32

Area = hb2

1

32621

unitssquare36

Answer: B

8. What was the Biblical approximation to ?

(A) 3.14 (B)7

22(C) 3.1416 (D) 3

Answer: D

9. In the freshman class at Uno High School, there are 18 boys and 12 girls. The average height

of the boys is 170 cm, and that of the girls is 160 cm. What is the average height of all thestudents in the class?

(A) 165 cm (B) 166 cm (C) 167 cm (D) 168 cm

Average height = cm1218

1601217018

= cm166

Answer: B


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2 16 30

10. Having found that 2 x is a factor of 3023 23 x x x xP , which of the following isanother factor of xP ?(A) 3 x (B) 6.10 x(C) 1583 2 x x(D) none of the above

Solution:

We divide xP by 2 xWe use synthetic division

1583 2 x x is not factorable

Answer: C

11. Gawing simple: 3262 .(A) 2 (B) 3 (C) 4 (D) 6

Solution:

32

32

32

62

34

32622

32348

6

3 2 1 30

3 8 15 0


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32324

344

2

Answer: A

12. Nicole and Chris have the same grade average before the final exam. The final examcounted as 20% of the semester grade. Nicole got 7 percentage points higher than Chris on hersemester grade, and the final exam was worth 100 points. What is the number of points in thepositive difference between their final exam grades?

(A) 30 (B) 25 (C) 50 (D) 35

Solution:Let x be the number of points that Nicole got on her final examand y be the number of points that Chris got on his final exam.

Then

72.0 y x multiply both sides by 5

35y x

Answer: D


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13. Suppose there is a row of lockers numbered 1 through 50 with all the doors closed. Astudent is going to walk up and down the lockers opening and closing the locker doors. If a dooris closed the student will open it, and if it is open, it will be closed .If the student starts withmultiple of one, and continues through multiples of 50. How many lockers are open when thestudent is finished?

(A) 5 (B) 10 (C) 7 (D) 8

Solution:

If a number is a perfect square it has an odd number of factors if not it has an even number of factors.

Example is 9 the factors of 9 are 1,3,9 there are 3 factors and 3 is an odd number12 is not a perfect square. The factors of 12 are 1,2,3,4,6,12, there are 6 factors and 6 is an even number

Locker # 9 : multiple of 1 (open), multiple of 3 (close), multiple of 9 (open)

All lockers with a perfect square number are open when the student is finished the remaining lockers areclosed.

Locker # 12: multiple of 1(open), multiple of 2 (close), multiple of 3 (open), multiple of 4 (close),multiple of 6 (open), multiple of 12 (close)

The number of perfect squares less than 50 is 7 because 22 8507 .

7 lockers are open.

Answer: C

14. Ang isang mag-aalahas ay may 3 spheres ng ginto. Ang sukat ng kanilang radius ay 3mm, 4mm,at5mm .Kung ang lahat ng tatlong spheres ay tinunaw at bumuo ng panibagong sphere na ginto. Ano angsukat ng radius ng bagong sphere na ginto?

(A) 8mm (B) 9mm (C) 6 mm (D) 10 mm


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15. Congruent circles of centers A and B intersect such that AB is a radius of each circle. If AB=4 cm, what is the number of square centimeters in the area ACBD that is common to the twocircles?

(A) 343

16 (C) 38

316

(B) 343

32

(D) 383

32

Solution:

If 4 AB then 4 BC because radii of the same circle are equal

224

BX

ABCD Why? Because CD and AB are diagonals of rhombus ACBD . Diagonals of arhombus are perpendicular and bisect each other

O90CXBmO30 BCX because BC is twice BX

So O60CBX

Therefore O120CBD by symmetry

A B

C

D

Recall:

In a 30 O-60 O-90 O triangle.

The hypotenuse is twice the sideopposite of the 30 O angle.


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Area of shaded part = area of the sector with central angle CBD minus area of CBD

Area of sector:

Area of CBD:

Area of sector with central angle CBD

=

O2

360

r

=

O

O2

360120

4

=3

16

If 2 BX then 32CX

Therefore 34CD

Area of CBD = 2heightbase

2

BX CD

2

234

34

Recall:

In a 30 O-60 O-90 O triangle.

The side opposite of the 60 O

angle is 3 times the length of the side opposite of the 30 O

angle.


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Area of shaded part:

Area ACBD that is common to the two circles:

Answer: D

Area of shaded part = area of the sector with central angle CBD minus area of CB

343

16

area ACBD that is common to the two circles

=

343

162

= 383

32 because of symmetry


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16. The value of 11

3232 is(A) 5 (C) 1/5(B) -5 (D) -1/5

Solution:

113232

11132

1)1(32

132

15

51

Answer: C

17. What is the sum of the first 100 positive odd integers?(A) 25000 (C) 20000(B) 50000 (D) 10000

Solution:

nn aan

S 12[This is the formula for the first n terms of an arithmetic series]

The 100 th positive odd integer is 2(100) 1 = 199


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So 19912

100100S

20050

10000

Answer: D

18. Sa orasan ni Ginoong Abad ay 5:14 a.m. nang siya ay umalis ng bahay. Bumalik siyamakalipas ang 7 oras at 11 minuto, subalit nagtaka siya dahil 4:33 pa lang ng umaga sa orasan.Naalala niyang nagkaroon pala ng rotating brown out sa kanilang lugar kaya nang bumalik angkuryente ay kusang bumalik sa oras na 12:00 a.m. ang orasan. Anong oras ng umaga bumalik ang kuryente?

(A) 8:52 (C) 5:52(B) 6:52 (D) 7:52

Solution:

The time 7 hours and 11 minutes after 5:14 a.m. is 12:25p.m.

The time 4 hours and 33 minutes before 12:25 p.m. is 7:52 a.m.

Answer: D

19. Express in lowest terms:632

72

23

nnnn

(A)9n

n(B)

97

nn

C)9

2

nn

(D)

7

2

nn

Solution:

Factor the numerator and denominator and cancel the common factors

97

7632

7 22

23

nnnn

nnnn


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9

2

nn

Answer: C

20. How many square units are in the area defined by the set of points (x,y) in the first quadrantwhich satisfies 2010 y x ?

(A) 150 B) 152 (C) 154 (D) 144

Solution:

Answer: A

20 units

10 units

20 units

10 units

Area of shaded region = Area of big triangle

minus Area of small triangle

= unitssquare150502002

10102

2020


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21. Fibonacci gave a famous problem about reproducing rabbits. After n months, there is apopulation of nr rabbits, where 10r , 21r , 32r and so on. What was the rabbit population

at n=12 months?

(A) 55 (B) 89 (C) 144 (D) 233

Solution:

The 12 th term of the sequence 1,2,3,5,8,13,21,34,55,89,144,233 is 233

Answer: D

22. For what value of x does ? 17154321 x

(A) 17 (B) 16 (C) 18 (D) 21

Solution:

171

21 x x

18x , we disregard the negative root

Answer: C


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24. What is the measure of the obtuse angle formed by two intersecting angle bisectors of anequilateral triangle?

(A) 150 (B) 110 (C) 120 (D) 100

Solution:

Answer: C

25. We select 6 numbers at random, with replacement, from the set of integers from 1 to 300inclusive. What is the probability that the product of the 6 numbers is even?

(A) 3231 (B) 1 (C) 64

61 (D) 6463

Solution:Let A= event that the product of 6 numbers selected at random , with replacement, from the setof integers from 1 to 300 inclusive is even.

So A C = event that the product of 6 numbers selected at random, with replacement, from the setof integers from 1 to 300 inclusive is odd.

The product of 6 numbers is odd if each number is an odd number.6

300150

)(

C AP

6

2

1

641


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Therefore

)(1)( C AP AP

641

1

6463

Answer: D26. A deep well 5 feet in diameter is of unknown depth (to the water level). If a 5-foot post iserected at the edge of the well, the line of sight from the top of the post to the edge of the watersurface below will pass through a point 0.4 feet from the lip of the well below the post. Whatis the depth of the well (to the surface of the water)?

(A) 37.5 ft (B) 47.5 ft (C) 57.5 ft (D) 67.5 ft

Solution:

5.57 xAnswer: C

The figure is not drawn into scale.

ft AB 5

ft BP 5

ft BC 4.0

Let ft x BD

ADE ABC ~

Ratio of corresponding sides of similar trianglesare equal

DE AD

BC AB

55

4.05 x

2524.0 x


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27. Find integers p and q such that32

334is a root (zero) of the quadratic polynomial

q px x2 .

(A) p = -2 and q =11 (C) p = -12 and q = 5

(B) p = -6 and q = 8 (D) p = -4 and q=9

Solution:

First we simplify32

334

34)3(336348

32

32

32

334

1321

The other root of the quadratic polynomial is the conjugate of 3 21 which is 321 .

Sum of roots = 2)321()321(

Product of roots = 11121)32()1()321)(321( 22

Therefore the quadratic equation is

0roots)of (productroots)of (sum2 x x

01122 x x

2 p and 11q

Bonus: No answer in the given choices

321


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28. Barb has two watches, one of which loses 6 seconds every 24 hours and the other gains 1second per hour. He sets both of them to the correct time at 8:00 p.m. How many hours will passbefore the positive difference in the time shown is 4 hours?

(A) 11484 (B) 11520 (C) 11511 (D) 11518

Solution:

After 1 hour the difference is seconds45

246

1

hours11520hourshour1

minutes60minute1seconds60

secondshours

516

hoursseconds4

5hours4

Answer: B

29. Kung ang 203 x , ano ang halaga ng x9 ?(A) 30 (B) 400 (C) 180 (D) 729

Solution:203 x , we get the square of both sides

22 203 x

4009 x

Answer: B


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30. Kayang tapusin ng unang pintor ang trabaho sa loob ng 7 oras. Samantalang ang ikalawangpintor ay kayang tapusin ang parehong trabaho sa loob ng 13 oras. Ilang minuto matatapos angtrabaho kung magtutulungan ang dalawang pintor?

(A) 273 (B) 276 (C) 270 (D) 275

Solution:

1131

71

x x multiply both sides by LCD = 91

91713 x x

9120 x

hours2091

x

minutes273

Answer: A

31. Summa Cum Laundry Inc. now has 5000 customers. Its number of customers increases by20 % each year. At the end of how many full years will its number of customers first exceed

10,000?

(A) 4 (B) 3 (C) 5 (D) 6Solution:

Number of customers atthe end of 1st year 60002.150002nd year 72002.160003rd year 86402.17200

4th year 10000103682.18640

Answer: A


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32. Which of the following is equal to 63 454 .

(A) 3 128 (B) 3 56 (C) 9 216 (D) 6 112

Solution:

First term:33 22754

33 227

3 23

Second term:6 26 24

62

2

31

2

3 2

Therefore333333363 12826426424223454

Answer: A

33. On a number line, how many integers are no more than 8 units from 20 and also at least 10units from 35?

(A) 18 (B) 12 (C) 15 (D) 14

Solution:

A= Set of integers no more than 8 units from 20 = 28,27,26,25,24,23,22,21,20,19,18,17,16,15,14,13,12

B = Set of integers at least 10 units from 35 = ,...50,49,48,47,46,4525,24,23,22,21,20,19,18,17,16,15,14,13,12,11,10,9...,

25,24,23,22,21,20,19,18,17,16,15,14,13,12 B A

The number of elements in B A is 14.

Answer: D


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34. Ang limang magkakasunod na integer ay may average na x . Ano ang pinakamababanginteger?

(A) x-2 (B) x-3 (C) x-4 (D) x-5

Solution:

Since there are 5 consecutive integers and the average is x then there are 2 integers above x and2 integers below x . The integers are 2,1,,1,2 x x x x x . The smallest integer is 2 x .

Answer: A

35. The lines 55 xk y and 72x y are perpendicular if k=

(A) 11/2 (B) 5 (C) -2/9 (D) 9/2

Solution:

The product of the slopes should be equal to -1 because they are perpendicular.

152 k 1102k

112k

211k

Answer: A


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1

36. Find the length of the radius of the circle .0364 22 y y x x(A) 2 (B) 3 (C) 4 (D) 9

Solution:

0364 22 y y x x

364 22 y y x x 9439644 22 y y x x

1632 22 y x

4r

Answer: C

37. Given that1

0222 z y x

z y x, find the value of 444 z y x .

(A) 52 (B) 3

1 (C) 3 (D) 21

Solution:

First equation:

0z y x get the square of both sides

0222222 yz xz xy z y x

1222 yz xz xy divide both sides by 2

21

yz xz xy get the square of both sides

41222 222222222 xyz z xy yz x z y z x y x

Recall:

the equation of the circle is

222 r k yh x

center: k h,

length of radius: r


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0022 z y x xyz

41

So41222222 z y z x y x

Second equation:

1222 z y x get the square of both sides

1222 222222444 z y z x y x z y x

12 222222444 z y z x y x z y x

Therefore21444 z y x

Answer: D

38.Simplify: 22

11

cb

cb

(A) cbbc1

(C) bc1

(B)cb

bc (D) bc

cb

Solution:


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Multiply the numerator and denominator by bc the result is,

222211

cbbcbc

bcbc

cbcb

, factor the denominator and cancel the common factor

cbcbbccb

cbbc1

Answer: B

39. Let 6if 116if 11

x x

x x x f . Find 612 f f

(A) -4 (B) 28 (C) 40 (D) 18

Solution: 23111212 f 51166 f

18523612 f f

Answer: D


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We solve 0211 2x

0211 x

121 x

121 x

02 x

0 x

So if x=0, then the expression on the left side of the inequality is undefined

Therefore 0 x **

Multiply both sides of the inequality

92211

42

2

x x

xby 2211 x , the result is

22 211924 x x x We dont change the direction of the inequality because we multiplyboth sides by a positive number

x x x x 21222924 2

x x x x x 21184182244 22

182221184 x x x

Let x y 21

So x y 212

Also 162184 2y x


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And 7111822 2y x

The inequality 182221184 x x x in terms of y is

711162 22 y y y

0716112 23 y y y , we can use synthetic division to find the factors of the left side

0721 2 y y divide both sides by 21 y a positive number so dont change the direction

of the inequality [ So 1 y ]

072 y

2

7 y express this inequality in terms of the variable x

27

21 x square both sides [the resulting inequality is true because both sides are positive]

449

21 x

445

2 x

845 x ***

21

x *

0 x **

845

x ***

The intersection of the above inequalities is


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0845

21

xexcept x

Answer: C

42. All the sides of a given triangle are whole numbers. The perimeter of the triangle is 110inches and one side has measure 29 inches. What is the fewest number of inches that can be thelength of one of the remaining sides?

(A) 30 (C) 26

(B) 24 (D) 27

Solution:

Let a and b be the length of the other two sides where ba

The perimeter is 110 so 11029ba or 81ba

29ba

2981 bb

522b

26b

The smallest possible length of one of the remaining sides is inches27 .Answer: D

43. If a regular polygon has 27 diagonals, how many sides does it have?

(A) 12 (C) 9

(B) 6 (D) 7

Solution:

2723nn

543nn


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05432 nn

069 nn

6or9 nndisregard 6n

Answer: C44. If a and b are each chosen from the set 10,5,3,2,1 , what is the largest possible value of

ab

ba

?

(A) 20 (C)101

10

(B)2112 (D)

212

Solution:b=1 b=2 b=3 b=5 b=10

a=1 2 1/2 3 1/3 5 1/5 10 1/10a=2 2 1/2 2 1/6 2 9/10 5 1/5a=3 3 1/3 2 1/6 2 4/15 3 19/30a=5 5 1/5 2 9/10 2 4/15 2 1/2a=10 10 1/10 5 1/5 3 19/30 2 1/2

the largest possible value of ab

ba

is101

10 .

Answer: C


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45. Which of the following relations are functions?

I. 422 y x IV. 12y x

II. x y x 2 V. 12x y

III. x y 4 VI.22 y x

(A) I, II, III(B) II, III, IV(C) II, III, V(D) I, IV, VI

Solution: I. 422 y x IV. 12y x

II. x y x 2 is a linear function

III. x y 4 is a function V. 12x y is a quadratic function

VI.22 y x

Answer: C


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There are 6 diameters that can be formed using 12 points that are spaced evenly around a circle.Therefore the number of triangles having all their vertices at three of these 12 points, that are of the type 906030 is 2446 .

Answer: A

48. Find all the values of x satisfying the given conditions:3

31

x y ,

43

2x

y , and

321 y y(A) 39 (B) 49 (C) 59 (D) 69

Solution:

321 y y


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34

33

3 x xmultiply both sides by 12

363334 x x

3693124 x x

363 x

39 x

Answer: A

49. The lines x y 2 and x y2 are

(A) parallel (B) perpendicular (C) horizontal (D) vertical

Solution:

The slopes of the lines x y 2 and x y2 are 2 and21

respectively.

The slopes are negative reciprocal of each other. Therefore the two lines are perpendicular.

Answer: B

50. Suppose the following two quadratic equations052 k x x and 0392 k x x

have a non-zero root in common. What is the value of k?(A) 4 (B) 5 (C) 6 (D) 7Solution:


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Let p and q be the roots of 052 k x x

q and r be the roots of 0392 k x x

q is the common root

5q p 9r qk pq * k qr 3 **

Using * and **k k

pqqr 3

pr 3So the equation 9r q becomes 93 pq

pq 39 , substitute this in the equation 5q p

It becomes 539 p p

42 p

2 p , therefore 3q

632pqk Answer: C

(A) 4 (B) 5 (C) 6 (D) 7

51. Let x, y, and z represent lengths of the sides of a right triangle with y


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22

222 156 z y

z y x

2

22 156 x

x x

2

22 156 x

x x

2

221 x

x

21

Answer: D

52. The arithmetic mean of 14 numbers is what percent of the sum of the same 14 numbers?.Express your answer as a decimal to the nearest hundredth.

(A) 7.69% (B) 7.14% (C) 8.33% (D) 5.88%

Solution:

Let x be the sum of the 14 numbers

Arithmetic mean =14 x

14114

x

x

%14.7Answer: B

53. Let x x x f ln and x denotes the number of primes less than or equal to x . When

Gauss was 15 years old, he conjectured that as x goes to infinity, the ratio of x to x f approaches one. Use Gausss estimate to estimate the percentage of positive integers smallerthan 10010 which are prime. (Remark: the natural logarithm of 10 is about 2.3 . )

(A) 10% (B) 3% (C) 0.4% (D) 0.05%


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Solution:

percentage of positive integers smaller than 10010 which are prime

= 100100

10primearewhich10ansmaller thintegerspositiveof number

100

100

1010 f because as x goes to infinity x f x

100

100

100

1010ln

10

10010ln1

10ln1001

3.21001

2301

%4.0Answer: C


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54. Given that 381m and 64nm . Find the value of mn.(A) -1/2 (C) -3/4(B) -1 (D) -1/4

Solution:

381m

14 33 m

14m

41

m

43

mn

Answer: C

55. What is the sum of all the elements of this finite arithmetic sequence3, 6, 9, 12, 15, 18, ... , 42 ?

(A) 314 (B) 313 (C) 318 (D) 315

Solution:

14...654321342...181512963

We can use the formula for the sum of the first n natural numbers. 2

1...321

nnn

64nm

6441

n

344 n

3n


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21514

314...6543213

315

Answer: D

56. The mean, median and mode for this set of data are all equal:{ x, 68,30,58,52} . Find x .

(A) 52 (B) 68 (C) 58 (D) 30

Solution:

x x

552583068

x x 5208

2084 x

52 x

mean = median = mode = 52

Answer: A


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58. Ang produkto ng dalawang buong numero ay 100,000. Kung ang alinman sa mga numer o ayhindi multiple ng 10, ano ang kanilang suma(sum) ?

(A) 3158 (B) 3157 (C) 3159 (D) 3160

Solution:510100000

552

55 52

312532

3157312532

Answer: B

59. Solve for x: x

22

3019

3011

3019

(A) 11/30 (C) 2/5(B) 1/3 (D) 3/10

Solution:22

301930113019

x

3019

3011

3019

3011

3019

x

1

308

3019

30

11

Answer: A

Recall:

bababa 22


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60. A triangle with vertices A(0,0), B(8,0), and C(8,6) is graphed in a coordinate plane. The

triangular region determined by these points is rotated 360 about the x-axis forming ageometric solid. How many square units are in the total outside area of this geometric solid?

(A) 90 (C) 120

(B) 84 (D) 96

Solution:

Base area is:

2r A

The surface area of a cone is therefore given by:

rsr SA 2

sr r SA where r=radius of the base

h = height

s = length of the side

s can be calculated by using the Pythagorean theorem

22 hr s

r=6, h=8, 1086 22s

961066SA Answer: D

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