Solution UPCAT Practice Test 1

UPCAT REVIEWER PRACTICE TEST 1

SOLUTION

1

If you have questions or homework in Math (from elementary to college level) send to us at [email protected]. Do let us know the deadline of your assignment. Well submit the solution on time through your e-mail address.


SOLUTION

2


1. Find the contrapositive of the following statement. If a figure has three sides, it is a triangle.

(A) If a figure does not have three sides, it is a triangle. (B) If a figure is a triangle, then it does not have three sides (C) If a figure is not a triangle, then it does not have three sides. (D) If a figure has three sides, it is not a triangle.

Solution:

Recall:

The contrapositive of the statement If p then q is If not q then not p

So the contrapositive of If a figure has three sides, it is a triangle is If a figure is not a triangle, then it does not have three sides

Answer: C

2. Solve for x: 46 xx

(A) no solution (B) 100 (C) 5 (D) 16

25

Solution:

46 xx transpose x to the right

xx 46 get the square of both sides


SOLUTION

3


xxx 8166

108 x divide both sides by 8

4

5x get the square of both sides

16

25x

Checking:

46 xx

4?

16

256

16

25

4?

16

25

16

9625

4?

16

25

16

121

4?

4

5

4

11

4?

4

16

44Answer: D


SOLUTION

4


3. Find the length of diagonal AC in the rectangular solid shown. Dimensions are in feet.

(A) ftd 229 (B) ftd7 (C) ftd229 (D) ftd7

Solution:

Using Pythagorean Theorem

222 5 ABd

252 dAB

Now, we apply the Pythagorean Theorem to ABC

222 ACBCAB

d

2

C

B

A 5

d

2

C

B

A 5

A

B

d

5


SOLUTION

5


222 225 ACd 2922 dAC

292 dAC

Answer: C

4. The area of a regular octagon is 230cm . What is the area of a regular octagon with sides four

times as large?

(A) 2545 cm (B) 2480 cm (C)

23600 cm (D) 2120 cm

Solution:

Given: 21 30 cmA

?2 A

12 4ss or 41

2 s

s

2

1

2

1

2

s

s

A

A

2

1

2 4A

A

12 16 AA

30162 A

Answer: B

22 480 cmA


SOLUTION

6


5. Simplify: 7373 (A) -4 (B) 58 (C) 10 (D) -40

Solution:

22 737373 73

4

Answer: A

6. If the sum of the roots of 0532 xx is added to the product of its roots, the result is

(A) -2 (B) -8 (C) -15 (D) 15

Recall:

22 bababa


SOLUTION

7


Solution:

In 0532 xx a=1, b=3, and c=-5So

Sum of roots = 1

3 = -3

Product of roots = 1

5= -5

Sum + Product of roots = (-3) + (-5) = -8

Answer: B

Recall:

In the quadratic equation

02 cbxax , where 0a

Sum of roots = a

b

Product of roots = a

cWhy?

The roots of 02 cbxax using quadratic formula are:

a

acbb

2

42 and

a

acbb

2

42

Sum of roots = a

acbb

2

42 +

a

acbb

2

42

= a

b

2

2

= a

b

product of roots =

a

acbb

a

acbb

2

4

2

4 22

2

222

2

4

a

acbb

2

22

2

4

a

acbb

24

4

a

ac

a

c


SOLUTION

8


7. The roots of the equation 42 2 xx are (A) real, rational, and unequal (B) real and irrational (C) real, rational, and equal (D) imaginary

Solution:

Express 42 2 xx in the form 02 cbxax , and compute the discriminant acb 42

The b2 4ac term is called the discriminant and it helps to determine how many and what kind of roots you see in the solution

Value of b2

4ac Is it a perfect square? Nature of the Roots

b2 4ac > 0 yes 2 real roots, rational

b2 4ac > 0 no 2 real roots, irrational

b2 4ac < 0 not possible 2 imaginary roots

b2 4ac = 0 not possible 1 real root

So 42 2 xx becomes 042 2 xx , a = 2, b = -1, c = -4

3342414 22 acbThe result is 33 which is greater than zero and not a perfect square.

therefore the roots are real and irrational.

Answer: B


SOLUTION

9


8. Which statement must be true if a parabola represented by the equation cbxaxy 2 does not intersect the x-axis?

(A) ,042 acb and acb 42 is not a perfect square

(B) ,042 acb and acb 42 is a perfect square

(C) 042 acb

(D) 042 acb

Solution:

To get the x-intercept/s of cbxaxy 2 we let y = 0, and solve for x.

cbxaxy 2 does not intersect the x-axis when the roots of 02 cbxax are not real or imaginary.

That happens when 042 acbAnswer: C


SOLUTION

10


9. The value of

1

3

2

0

27

3

is

(A) 9 (B)

9

1 (C) 9 (D) 9

1

Solution:1

3

2

1

3

2

0

27

1

27

3

3

2

27

23 27

23

9

Answer: C

10. What is the last term in the expansion of 52yx ? (A) 52 y (B) 532y (C) 5y (D) 510y

Solution:

The last term in the expansion of 52yx is 52y

which is equal to 532y

Answer: B


SOLUTION

11


11. The larger root of the equation 043 xx is (A) -3 (B) -4 (C) 4 (D) 3

Solution:

043 xx03 x or 04 x

3x or 4x

The larger of -3 and 4 is 4.

Answer: C

12. Express xx

1

1

1 as a single fraction.

(A) xx

x

2

32 (B)

xx

x

2

12 (C)

12

2

x (D) 23

x

Solution:

1

11

1

1

xxxx

xx

112

xx

x

xx

x

2

12

Answer: B


SOLUTION

12


13. Ano ang kabuuan ng walang katapusang geometric series na ?...6696.0116.186.11.3 (A) 8.75 (B) 9.75 (C) 4.75 (D) 7.75

Solution:

Let ...6696.0116.186.11.3 x

...116.186.11.36.01.3 x xx 6.01.3

1.34.0 x

4.0

1.3x

75.7or 4

31x

We can also use the formula r

aSn

1

1 , where 1a = first term and r = common ratio

6.01

1.3

nS

75.74

31

4.0

1.3 nS

Answer: D


SOLUTION

13


14. Which equation represents a hyperbola?

(A) 216xy (B) 216 xy (C) 22 16 xy (D)

xy

16

Solution:

(A) 216xy is a parabola

(B) 216 xy is also a parabola and it opens downward.

(C) 22 16 xy can be expressed as 1622 yx .

Therefore it is a circle.

(D) x

y16 Its graph is a hyperbola.

Answer: D

Recall:

The graph of the quadratic function

cbxaxy 2 , where a, b, and c are constants and 0a , is a parabola that opens upward if a>0, and a parabola that opens downward if a


SOLUTION

14


15. Which expression is equivalent to the complex fraction

21

2

x

xx

x

?

(A) 2

x (B)

2

2

xx

(C) 4

22 x

x (D)

x

2

Solution:

xxx

x

x

x

xx

x

22

2

21

2

2

x

Answer: A

16. What is the radian measure of the angle formed by the hands of the clock at 2:00 pm?

(A)

2

(B)

3

(C)

4

(D)

6

Solution: The degree measure formed by the hands of the clock at 2:00 is 60O. To convert 60O to radian

measure, multiply 60O by 180

.

rad3180

60O

O

Answer: B


SOLUTION

15


17. The expression 15 3[2 + 6(3)] simplifies to

(A) -45 (B) -33 (C) 63 (D) 192

Solution:

182315362315 16315 4815

63Answer: C

18. Ano ang halaga ng 13

1

12

m

m

m ?

(A) 15 (B) 55 (C) 57 (D) 245

Solution:

21013

1

75312

m

m

m

4951

55Answer: B


SOLUTION

16


19. Ang pagsusulit sa asignaturang HEKASI ay may 10 katanungan na nagkakahalaga ng 5 puntos bawat isa, 7 mga katanungan na nagkakahalaga ng 6 na puntos sa bawat isa, at 4 na mga katanungan na nagkakahalaga ng 2 puntos sa bawat isa. Wala sa mga tanong na ito ang bibigyan ng bahagyang kredito. Gaano karaming mga puntos sa pagitan ng 0 at 100 ang imposibleng iskor?

(A) 3 (B) 2 (C) 4 (D) 7

Solution:

There are 10 questions worth 5 points each so multiples of 5 including 0 up to 50 are possible scores. [5x0=0, 5x1=5, 5x2=10,,5x10=50].We encircle the numbers.

0 1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 5051 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 7071 72 73 74 75 76 77 78 79 8081 82 83 84 85 86 87 88 89 9091 92 93 94 95 96 97 98 99 100

We add multiples of 6 from 6 to 42 to the encircled numbers because there are 7 question worth 6 points each. We underline the resulting numbers.

0 1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 5051 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 7071 72 73 74 75 76 77 78 79 8081 82 83 84 85 86 87 88 89 9091 92 93 94 95 96 97 98 99 100


SOLUTION

17


We add multiples of 2 from 2 to 8 to the encircled numbers and also to the underlined numbers because there are 4 questions worth 2 points each. We highlight the resulting numbers.

0 1 2 3 4 5 6 7 8 9 1011 12 13 14 15 16 17 18 19 2021 22 23 24 25 26 27 28 29 3031 32 33 34 35 36 37 38 39 4041 42 43 44 45 46 47 48 49 5051 52 53 54 55 56 57 58 59 6061 62 63 64 65 66 67 68 69 7071 72 73 74 75 76 77 78 79 8081 82 83 84 85 86 87 88 89 9091 92 93 94 95 96 97 98 99 100

If the number is not encircled, highlighted or underlined then the number is an impossible score. The impossible scores between 0 and 100 are 1, 3, 97, and 99.

There are four impossible scores between 0 and 100.Answer: C


SOLUTION

18


20. Ang isang malaking istante ng libro ay maaaring naglalaman sa pagitan ng 57 at 564 na mga libro.

Eksaktong 61 ay librong matematika at eksaktong 9

1 ay librong physics. Ano ang positibong kaibahan

sa pagitan ng pinakamataas at ang pinakamaliit na posibleng bilang ng mga libro na maaaring naka-imbak sa istante?

(A) 468 (B) 486 (C) 504 (D) 522

Solution:

Eksaktong 61 ay librong matematika the number of books in Mathematics is divisible by 6

eksaktong 91 ay librong physics the number of books in Physics is divisible by 9

Number of Math and Physics books combined is divisible by both 6 and 9. Therefore, total number of books is divisible by 18.

The smallest number more than 57 that is divisible by 18 is 72.

The largest number less than 564 that is divisible by 18 is 558.

The difference between the largest and smallest possible number of books is 558 72 = 486

Answer: B


SOLUTION

19


21. Simplify: 5

3560

3

52

(A)

15

1529 (B)

3

157 (C)

15

157 (D) 3

1529

Solution:

First term: 3

52

3

52 , Multiply the numerator and denominator by to rationalize.

3

3

3

52

3

152

60:TermSecond

154

154

152

Third term: 5

35

5

5

5

35

3


SOLUTION

20


5

155

15

151523

152

5

3560

3

52

LCD is 3

3

157

Answer: B

22. Given: R is the midpoint of MS

MSTR If you outlined a proof that shows TSTM , which would NOT be used?

(A) TSRTMR by the SAS congruency postulate (B) TSTM by CPCTC (C) TSRTMR by the ASA congruency postulate (D)

T

SRM

T

SRM


SOLUTION

21


Solution:

Below is a proof that shows TSTM

Statement Reason

1. R is the midpoint of MS 1. Given

RSMR 2. 2. A midpoint cuts a segment into two congruent segments

MSTR 3. 3. Givenanglesright areSRTandMRT4. sright formlines4.

SRTMRT5. aresright All5.TRTR 6. 6. reflexive

7. TSRTMR 7. SASTSTM 8. 8. CPCTC

All of the choices were used to prove that

TSTM EXCEPT (C) TSRTMR by the ASA congruency postulate

ANSWER: C

S

A

S


SOLUTION

22


23. Refer to the figure shown. State the congruency postulate that can be used to prove that WXVTUV .

Given: WVTV and XVUV

(A) SSS (B) SAS (C) ASA (D) AAS

Solution:

Included angle and two included sides are congruent. Therefore by SAS WXVTUV

Answer: B not C as given in the answer key

X

WV

U

T

X

WV

U

T


SOLUTION

23


24. Find OM if LO bisects NLM , 20LM , 3NO , . and 5LN .

(A) 10.23 (B) 0.75 (C) 12 (D) 33.33

Solution:

LN

NO

LMOM

5

3

20

OM

12 OM

ANSWER: C

M

O

N

L


SOLUTION

24


25. What value of x will give the maximum value for 377 2 xx ?

(A) 0 (B) 1 (C) 21

(D) 2

3

Solution:

Let 3772 xxy

7a , 7b , 3c

Since a is negative, then

a

bh

2

14

7

2

1 , will give the maximum value

for 377 2 xx .

ANSWER: C

Recall:

In cbxaxy 2 , the vertex is kh,If a is positive then h will give the minimum value for y , and the minimum value is equal

to k .

If a is negative then h will give the maximum value for y , and the maximum value is equal

to k .

a

bh

2

, a

back

4

4 2


SOLUTION

25


26. Written in simplest form yx

yx2

2

4

4

is

(A) 1 (B) 0 (C) yx

yx2

2

4

4

(D) -1

Solution:

44

4

42

2

2

2

yx

yx

yx

yx

1

ANSWER: D

27. Which expression is equivalent to 27

27

?

(A) 5

9 (B) -1 (C)

5

1429 (D)

14

211

Solution:

Multiply the numerator and denominator by the conjugate of the denominator.

The conjugate of 27 is 27

27

21427

27

27

27

27

5

1429

ANSWER: C


SOLUTION

26


28. Given two lines whose equations are 083 yx and 092 kyx , determine the value of k such that the two lines are perpendicular.

(A) 3

2 (B) 6 (C) 8 (D) 9

Solution:

We express 083 yx in the form bmxy , the result is 83 xySo its slope is -3.

We express 092 kyx in the form bmxy , the result is k

xk

y92

So its slope is k

2.

We get the product of their slopes and equate to -1.

123

k, if we solve for k, the answer is

6k

Answer: B

Recall:

In bmxy , m is the slope of the line

Two lines are perpendicular if their slopes are negative reciprocals of each other.

or if the product of their slopes is equal to -1.


SOLUTION

27


29. Solve for x: 22 64256 xx

(A) 11

6 (B) 5

6 (C) 5

1 (D) 0

Solution:

Express both sides of the equation in the same base

22 64256 xx

2324 44 xx

638 44 xx

638 xx

65 x

5

6 x

Answer: B

Recall: law of exponent for powers

mnnm aa


SOLUTION

28


30. Find the square root of 4452 234 xxxx . (A) 22 xx (B) 22

2 xx (C) 232 xx (D) 2

2 x

Solution:

44522 23422 xxxxxxAnswer: A

31. The product of the square roots of two consecutive positive numbers is 142 , what is their sum? (A) 15 (B) 17 (C) 19 (D) 21

Solution:

Let x and 1x be the two consecutive positive numbers.

1421 xx

561441 xx 561 xx , so 7x because 7(7+1) = 56

The two consecutive positive numbers are 7 and 8. Their sum is 15.

Answer: A


SOLUTION

29


4

32. Given the formula 329

5 FC ; find F when C is 20.

(A) 15 (B) 17 (C) 68 (D) 21

Solution:

329

5 FC

329

520 F multiply both sides by

5

9

329

5

5

920

5

9 F

3236 F

68F

Answer: C


SOLUTION

30


33. What number added to 6% of itself equals 31.8? (A)29.892 (B) 31.74 (C)30 (D) 31

Solution:

Let x be the number

8.3106.0 xx

8.3106.1 x

06.1

8.31x

30xAnswer: C


SOLUTION

31


34. Perform the indicated operations: 22 32332 aaaa (A) 2 (B) 0 (C) -3 (D) a2

Solution:

9663912432332 22222 aaaaaaaaaa 9124 2 aa

Answer: B

35. What must be the value of m if 5x is a factor of 352 2 mxx ?

(A) 3 (B) 5 (C) 7 (D) 10

Solution:

Let 352)( 2 mxxxf

5x is a factor of 352)( 2 mxxxf when 0)5( f [We use the Remainder Theorem]

So 35552)5( 2 mf m515

30515 mm

Answer: A

aa 63 2

962 aa

0


SOLUTION

32


36. Reduce

ba

b

a

b

ab

ab2 to a single fraction in its lowest terms.

(A) b

ba 2 (B)

a

ba 2 (C)

a

ba 2 (D)

b

ba 2

Solution:

ba

b

a

b

ab

ab2

, we distribute -2

ba

b

a

b

ab

ab

22

ba

ba

ba

b

a

b

22

ba

bab

a

b

)(22

ba

ab

a

b

2

ab

ab

a

b

2

Similar fractions, [they have the same denominators] so we combine the numerators.

ab

ab

is also equal to ba

ba

1


SOLUTION

33


3x2

a

a

a

b 2

a

ba 2

Answer: C

37. Find the quotient if 6532 23 xxx is divided by 232 xx .

(A) 32 x (B) 32 x (C) 32 x (D) 32 x

Solution:

232 xx 6532 23 xxx

Answer: B

693 2 xx

6

a

a1

Step 1.Divide 32x by 2x , the result is x2

Step 2.Multiply x2 by 232 xx , the result is xxx 462 23

xxx 462 23

xx 93 2

Step 3.Subtract xxx 462 23 from

6532 23 xxx , the result is xx 93 2

Step 4.Bring down +6

Step 5.Divide 23x by 2x , the result is 3

Step 6.Multiply 3 by 232 xx , the result is 693 2 xx

Step 7.Subtract 693 2 xx from 693 2 xx , the result is 0, there is no

remainder.

0

Last Step8. The quotient is

32 x


SOLUTION

34


38. Ang mga bahay sa Tinio Street ay may sunud-sunod na bilang mula 1 hanggang 447. Ilang tanso na numero ang kailangan upang magawa ang lahat ng bilang ng mga bahay ?

(A) 1232 (B) 1231 (C) 1236 (D) 1233

Solution:

ANSWER: D

39. Solve for x: 34257

311

14

1

5

7 xxx

(A) 4 (B) 11 (C) 18 (D) 25

Solution:

34257

311

14

1

5

7 xxx multiply both sides by 70 = LCD

3425

7

37011

14

1

5

770 xx

x

distribute 70

2380253011598 xxx23807503055598 xxx

1630305593 xx157563 x

25xANSWER: D


SOLUTION

35


40. The length of a room is 8 feet greater than its width; if each dimension is increased by 2 feet, the area will be increased by 60 square feet. Find the area of the floor.

(A) 65 (B) 105 (C) 153 (D) 180

Solution:

Area = 8xx

New Area = 608102 xxxx 608102 xxxx 6082012 22 xxxx

404 x

10 x

Original area of the floor = 21801810 ft

ANSWER: D

2x

10x

x

8x


SOLUTION

36


41. Find the greatest common factor of 963 2 xx , 15216 2 xx , and 66 3 x .

(A) 13 x (B) 33 x (C) 13 x (D) 33 xSolution:

133323963 22 xxxxxx 1523572315216 22 xxxxxx

11321666 233 xxxxx

GCF = 13 xAnswer: A

42. Find the equation of the line that passes through the point (-2,-5) and is parallel to the line 5x-4y=2.

(A) 5x-4y=-17 (B) -2x-5y=2 (C) 5x+4y=2 (D) 5x-4y=10

Solution:

The lines 5x-4y=2 and 5x-4y=k [k is constant] are parallel.

To solve for k, we substitute (-2,-5) in 5x-4y=k

k 5425k 2010

10k

Therefore the equation of the line that passes through the point (-2,-5) and is parallel to the line 5x-4y=2

is 1045 yxANSWER: D


SOLUTION

37


43. Simplify:

3

2

532

2

234

x

py

y

px

(A) 14

46

16 p

yx (B) 1511

1216

py

x (C) 1910

6

4 py

x (D) 1913

12

16 py

x

Solution:

6

159

4

4623

2

532

2

23 44

x

py

y

px

x

py

y

px

6

159

4

4624

x

py

y

px

1913

12

16 py

x

ANSWER: D

44. Simplify:

27212721 7676 yxyx

(A) 7494236 yxyx (B) 74936 yx (C) 7494236 yxyx (D) 74936 yx

Solution:

22

722

12

72

12

72

1767676

yxyxyx

74936 yx

ANSWER: B

Recall:

22 bababa


SOLUTION

38


45. Simplify: 38243 36

xyyx

xyyx

(A) 34 yx (B) xy (C) 65 yx (D) 45 yx

Solution:

4

3342

382

43 36

xy

yxxyyx

xyyx

xyyx

45 yx

ANSWER: D


SOLUTION

39


46. Use similar triangles to find x.

(A) 8/9 ft (B) 5.4 ft (C) 15 ft (D) ft321

Solution:

9

5

3x

ft321or

3

5x

ANSWER: D

3 ft9 ft

5 ft

x


SOLUTION

40


47. Given: BCPQ // . Find the length of AC .

(A) 17 (B) 21 (C) 23 (D) 18

Solution:

AC

AQAB

AP

Let AQ = x

1214

6

xx

cross multiply

72614 xx

728 x

9x

AC = x+12 = 9 + 12 = 21

Answer: B

B

12

6

QP

A

C

8


SOLUTION

41


48. The numbers 27, 36, and 45 represents the length of the sides of a/an

(A) acute triangle (B) obtuse triangle (C) no triangle (D) right triangle

Solution:

3-4-5 is a Pythagorean Triple, multiply by 9

27-36-45

Therefore the numbers 27, 36, and 45 represents the length of the sides of a right triangle.

ANSWER: D

49. In the figure shown, square WXYZ is inscribed in circle O. Also, XYOM and .7OM Find the area of the shaded region.

(A) 4949 (B) 49249 (C) 19698 (D) 196147

YX

ZW


SOLUTION

42


Solution:

142 OMYZ

XZ is a diagonal of the square, therefore

2142 YZXZ

XZ is also the diameter of the circle, so the radius is one-half of its measure.

Radius of circle O = 27

Area of shaded region = Area of circle Area of square

= 22 1427 = 19698

Answer: C

YX

ZW


SOLUTION

43


50. Simplify:

3

1

8

72

3

1

6

5

(A) 8

3 (B)

5

4 (C)

3

2 (D)

6

1

Solution:

3

1

8

72448

820

24

24

3

1

8

72

3

1

6

5

82148

28

35

28

5

4

Answer: B


SOLUTION

44


51. Evaluate 055522 34242 yxyxyxyx if 3x and 4y . (A) -45 (B) -81 (C) -36 (D) -27

Solution:

055522 34242 yxyxyxyx

yxyx 42 22 substitute 3x and 4y .

434423 22 14329

27

ANSWER: D

1


SOLUTION

45


52. The length of AC is 6

15 meters. The length of BC is

2

12 meters. Find AB.

(A) 2 m (B) 4

17 m (C)

3

27 m (D)

3

22 m

Solution:

ACBCAB

6

15

2

12 AB

2

12

6

15 AB

6

32

6

15

6

15

6

31

6

16

m3

22or

3

8ANSWER: D

B CA


SOLUTION

46


53. What is the sum of 2 and 18 ?

(A) 52i

(B) 25i

(C) 24i

(D) i6

Solution:

232182 ii

24i

Answer: C

54. Ano ang ika-7-term sa isang geometric sequence kung ang unang term ay 81 at ang ika-11-term ay

?729

1

(A) 27

1

(B)

9

1

(C)

3

1

(D) 1

Solution:

Given: 811 a , 7291

11 a , ?7 a

In geometric sequence the nth term is 11 nn raa

111111

raa

1081729

1r


SOLUTION

47


10

81729

1r

1046 33

1r

10103

1r

1010

3

1r

3

1r

Therefore

1717

raa

6

3

181

6

4

3

3

23

1

9

1

Answer: B


SOLUTION

48


55. Kung ang 25% ng isang numero ay 75. Ano naman ang 30% ng numero?

(A) 80 (B) 90 (C) 100 (D) 851

Solution:

Let x be the number7525.0 x , multiply both sides by 4

300x

Therefore

Answer: B

903.0 x


SOLUTION

49


56. Based on the diagram below, which statement is true?

(A) a // b (B) a // c (C) b // c (D) d // e

Solution:

Therefore d//e

ANSWER: D

e

d

cba

120

11560

110


SOLUTION

50


57. Ayon sa isang sinaunang paniniwala, kapag ang isang kaibigan ay dumalaw sa isang may sakit na tao,

301 ng kanyang pagkakasakit ay nawawala . Ano ang pinakamababang bilang ng kaibigan ang

kailangang bumisita sa may sakit upang maalis ang 98% o higit pa ng kanyang pagkakasakit?

(A) 114 (B) 115 (C) 116 (D) 117

Solution:

02.030

29

x

get the log of both sides

02.0log30

29log

x

30

29log

02.0logx

3936.115xThe smallest integer greater than 115.3936 is 116.

Answer: C

100% - 98%

30

11

Recall:

ana bn

b loglog


SOLUTION

51


58. Si Sarah ay gagawa ng isang keyk at ilang mga cookies. Ang keyk ay nangangailangan ng 8

3 tasa ng

asukal at ang mga cookies ay nangangailangan ng 5

3 tasa ng asukal. Si Sarah ay may 16

15 tasa ng

asukal. Siya ba ay may sapat na asukal?

(A) Siya ay may sapat na asukal

(B) Kailangan pa niya ng 8

1 tasa ng asukal.

(C) Kailangan pa niya ng 80

3 tasa ng asukal.

(D) Kailangan pa niya ng 19

4 tasa ng asukal.

Solution:

Sarah needs 16

15

5

3

8

3

cups of sugar

16

15

40

39

16

15

40

39

80

7578

80

3

Answer: C


SOLUTION

52


59. Find the distance from the point (2,3) to the line .5 yx

(A) 1

(B)3

2

(C)

2

3

(D) 23

Solution:

22

11

BA

CByAxd

22 1153121

2

6

23ANSWER: D


SOLUTION

53


60. If 41 x

x , what is the value of 2

2 1

xx ?

(A) 16

(B) 15

(C) 14

(D) 12

Solution:

41 x

x get the square of both sides

1611

22

2 xx

xx

141

22

xx

Answer: C

.

Solution UPCAT Practice Test 1

Documents