Solution UPCAT Practice Test 1

UPCAT REVIEWER PRACTICE TEST 1

SOLUTION

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 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:

x6  x  4

(A) no solution

(B) 100

(C) 5

(D)

25 16

Solution:

x6  x  4

transpose

x6  4 x

get the square of both sides

x to the right

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION x  6  16  8 x  x 8 x  10 x x

divide both sides by 8

5 4

get the square of both sides

25 16

Checking:

x6  x  4 25 25 ? 6  4 16 16  25  96 25 ?  4 16 16  121 25 ?  4 16 16  11 5 ?  4 4 4 16 ? 4 4  44 Answer: D

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UPCAT REVIEWER PRACTICE TEST 1

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

B

2 A (A) 29  d

2

5

d (B) 7  d ft

ft

(C)

29  d 2 ft

(D)

7  d ft

Solution:

C

B

2 A

Using Pythagorean Theorem

d  5   AB  2

2

5

d

A

2

d

 AB  d 2  25 5

B

Now, we apply the Pythagorean Theorem to  ABC

 AB 2  BC 2   AC 2

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UPCAT REVIEWER PRACTICE TEST 1

d

2

SOLUTION



 25  2 2   AC 

 AC 2

2

 d 2  29

AC  d 2  29 Answer: C 2

4. The area of a regular octagon is 30cm . What is the area of a regular octagon with sides four times as large? 2

(A) 545 cm

(B) 480 cm

2

2

(C) 3600 cm

2

(D) 120 cm

Solution: Given: A1  30 cm 2

A2  ? s2 4 s1

s2  4s1 or A2  s2    A1  s1 

2

A2  42 A1

A2  16 A1 A2  16 30 A2  480 cm 2 Answer: B

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UPCAT REVIEWER PRACTICE TEST 1

5. Simplify: (A) -4



3 7



3 7



(B) 58

SOLUTION (C) 10

(D) -40

Solution:



3 7



  3   7 

3 7 

2

Recall:

2

a  b a  b   a 2  b 2

 37  4 Answer: A

6. If the sum of the roots of x 2  3 x  5  0 is added to the product of its roots, the result is (A) -2

(B) -8

(C) -15

(D) 15

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION Solution:

The roots of ax 2  bx  c  0 using quadratic formula are:

Recall:

 b  b 2  4ac  b  b 2  4ac and 2a 2a

In the quadratic equation

ax 2  bx  c  0 , where a  0

b Sum of roots =  a

Sum of roots = Why?

c Product of roots = a

In x 2  3 x  5  0 a=1, b=3, and c=-5 So Sum of roots = 

3 = -3 1

5 Product of roots = = -5 1 Sum + Product of roots = (-3) + (-5) = -8 Answer: B

 b  b 2  4ac  b  b 2  4ac + 2a 2a

=

 2b 2a

=

b a

product of roots =

  b  b 2  4ac    b  b 2  4ac        2 a 2 a    2   b    b 2  4ac   2a 2

2



b 2  b 2  4ac  2a 2 

4ac 4a 2



c a



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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 7. The roots of the equation 2 x 2  x  4 are (A) real, rational, and unequal (B) real and irrational (C) real, rational, and equal (D) imaginary Solution: Express 2 x 2  x  4 in the form ax 2  bx  c  0 , and compute the discriminant b 2  4 ac 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 2

Value of b – 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 2 x 2  x  4 becomes 2 x 2  x  4  0 , a = 2, b = -1, c = -4

b 2  4ac   1  42  4   33 2

The result is 33 which is greater than zero and not a perfect square. therefore the roots are real and irrational. Answer: B

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 8. Which statement must be true if a parabola represented by the equation y  ax  bx  c does not 2

intersect the x-axis? (A) b  4ac  0, and b 2  4ac is not a perfect square 2

(B) b  4ac  0, and b 2  4ac is a perfect square 2

(C) b 2  4ac  0 (D) b 2  4ac  0

Solution: To get the x-intercept/s of y  ax  bx  c we let y = 0, and solve for x. 2

y  ax2  bx  c does not intersect the x-axis when the roots of ax 2  bx  c  0 are not real or imaginary. That happens when b 2  4ac  0 Answer: C

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION  0  3 9. The value of  2  27 3

   

1

is

(A)  9

(B) 

1 9

(C) 9

(D)

1 9

Solution:

 0  3  2  27 3

1

    1   2   27 3

   

1

2

 27 3

 27 



3

2

 32

9 Answer: C 5

10. What is the last term in the expansion of  x  2 y  ? (A) 2 y 5

(B) 32y

5

(C) y

5

(D) 10y

5

Solution: 5

The last term in the expansion of  x  2 y  is

which is equal to

2 y 5

32 y 5

Answer: B

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 11. The larger root of the equation  x  3x  4  0 is (A) -3

(B) -4

(C) 4

(D) 3

Solution:

x  3x  4  0 x  3  0 or x  4  0 x  3 or x  4

The larger of -3 and 4 is 4. Answer: C

12. Express

(A)

1 1  as a single fraction. x 1 x

2x  3 x2  x

(B)

2x  1 x2  x

(C)

2 2x  1

(D)

3 x2

Solution:

1 1 x   x  1   x 1 x xx  1 

2x  1 xx  1

2x  1 x2  x

 Answer: B

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 13. Ano ang kabuuan ng walang katapusang geometric series na

3.1  1.86  1.116  0.6696  ... ? (A) 8.75

(B) 9.75

(C) 4.75

(D) 7.75

Solution: Let

x  3.1  1.86  1.116  0.6696  ...

x  3.1  0.6 3.1  1.86  1.116  ... x  3.1  0.6 x

0.4 x  3.1 x

3.1 0.4

x

31 or 7.75 4

We can also use the formula S n 

Sn 

3 .1 1  0 .6

Sn 

3.1 31   7.75 0 .4 4

a1 , where a1 = first term and r = common ratio 1 r

Answer: D

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UPCAT REVIEWER PRACTICE TEST 1

SOLUTION 14. Which equation represents a hyperbola? (A) y  16x 2

(B) y  16  x 2

(D) y 

(C) y 2  16  x 2

16 x

Solution: (A) y  16x 2 is a parabola

Recall: The graph of the quadratic function

y  ax 2  bx  c , where a, b, and c are constants and a  0 , is a parabola that opens upward if a>0, and a parabola that opens downward if a