, REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
MATHEMATICS
1. The equation y^2 = cx is the general equation of:
A. y’ = 2y/x B. y’ = 2x/y C. y’ = y/2x D. y’ = x/2y
SOLUTION:
y 2 = cx 0 = x(2yy ′ ) − y 2 )/x^2
y2 y 2 = 2xyy′
c= x
y2
Differentiate: y ′ = 2xy = y/2x
2. A line segment joining two points on a circle is called:
A. arc B. tangent C. sector D. chord
3. Sand is pouring to form a conical pile such that its altitude is always twice its radius. If the
volume of a conical pile is increasing at the rate of 25 pi cu.ft/min, how fast is the radius is
increasing when the radius is 5 feet?
A. 0.5 ft/min B. 0.5 pi ft/min C. 5 ft/min D. 5 pi ft/min
SOLUTION:
h = 2r, r = 5ft
1 1 2
Vcone = πr²h = 3 πr 2 (2r) = 3 πr³
3
2 dr
25ft³ = 3 π , 3πr² dt
dr
25π = 2π(5)² dt
dr 25π
= 2π(25) = 0.5 ft/min
dt
4. Evaluate ʃ ʃ 2r²sin Ө dr dӨ, 0 > r >sin Ө, > Ө > pi/2
A. pi/2 B. pi/8 C. pi/24 D. pi/48
SOLUTION:
π π
sin θ 2
∫02 ∫0 2r² sin θ cos ²θ drdθ =∫02 3 (sin θ)³ sin θ cos ²θ dθ
π π
sin θ 2
= ∫02 ∫0 2r² dr sin θ cos ²θ dθ = 3 ∫02 sin4 θ cos²θ dθ
π
2 sin θ 2 (3)(1)(1) π π
= ∫02 3 r² ∫0 sin θ cos ²θ dθ = 3 [(6)(4)(2)] 2 = 48
,5. A shopkeeper offers a 25% discount on the marked price on an item. In order to now cost $
48, what should the marked price be?
A. $ 12 C. $ 60
B. $ 36 D. $ 64
SOLUTION:
48 = (1 − 0.25)X
48
x = 0.75 = $ 64
6. An observer wishes to determine the height of a tower. He takes sights at the top of the
tower from A to B, which are 50 ft. apart, at the same elevation on a direct line with the
tower. The vertical angle at point A is 30 degrees and at point B is 40 degrees. What is the
height of the tower?
A. 85.60 ft B. 143.97 ft C. 110.29 ft D. 92.54 ft
SOLUTION:
β = 180 − 40 = 140°
α = 180 − 30 − 140 = 10°
50 x
= sin 30 ; x = 143.969621
sin θ
h = 143.969621 sin(40) = 92.54 ft
7. A tangent to a conic is a line
A. which is parallel to the normal
B. which touches the conic at only one point
C. which passed inside the conic
D. all of the above
8. Find the area of the triangle which the line 2x – 3y + 6 = 0 forms with the coordinate axes.
A. 3 B. 4 C. 5 D. 2
SOLUTION:
6
2x − 3(0) + 6 = 0 y=3=2
−6
x= = −3 1
2 A = 2 (3)(2) = 3 sq. units
2(0) − 3y + 6 = 0
, REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
MATHEMATICS
9. Find the general solution of (D² - D + 2)y = 0
A. y = e^x/2 (C1 sin sqrt. 7/2 x + C2 cos sqrt. 7/2 x)
B. y = e^x/2 (C1 sin sqrt. 7/2 x - C2 cos sqrt. 7/2 x)
C. y = e^x/2 (C1 cos sqrt. 7/2 x + C2 sin sqrt. 7/2 x)
D. y = e^x/2 (C1 cos sqrt. 7/2 x - C2 sin sqrt. 7/2 x)
SOLUTION:
(D2 − D + 2)y = 0 1 −7 7
m − 2 = √ 4 = √2 i
m² − m + 2 = 0
1 √7
1 2 7
m= + i
2 2
(m − 2) + 4 = 0
𝐲 = 𝐞𝐀𝐱 (𝐂𝟏 𝐜𝐨𝐬𝐁𝐱 + 𝐂𝟐 𝐬𝐢𝐧𝐁𝐱)
10. If 10 is subtracted from the opposite of a number, the difference is 5. What is the number?
A. 5 B.15 C.-5 D. -15
SOLUTION:
x - 10 = 5
Opposite of x – 10 = 5
15 – 10 = 5
∴ −5
11. If y = 5 – x, find x when y = 7
A. 12 B.-12 C. 2 D. -2
SOLUTION:
y = 5 – x, find x when y = 7
7=5–x
x = -7 + 5 = −2
12. A ranch has a cattle and horses in a ratio of 9:5. If there are 80 more head of cattle than
horses, how many animals are on the ranch?
A.140 B. 168 C. 238 D. 280
SOLUTION:
x 9
Cattle → x = 5 ; x = y + 80
y
Horses → y
AUGUST 2012
MATHEMATICS
1. The equation y^2 = cx is the general equation of:
A. y’ = 2y/x B. y’ = 2x/y C. y’ = y/2x D. y’ = x/2y
SOLUTION:
y 2 = cx 0 = x(2yy ′ ) − y 2 )/x^2
y2 y 2 = 2xyy′
c= x
y2
Differentiate: y ′ = 2xy = y/2x
2. A line segment joining two points on a circle is called:
A. arc B. tangent C. sector D. chord
3. Sand is pouring to form a conical pile such that its altitude is always twice its radius. If the
volume of a conical pile is increasing at the rate of 25 pi cu.ft/min, how fast is the radius is
increasing when the radius is 5 feet?
A. 0.5 ft/min B. 0.5 pi ft/min C. 5 ft/min D. 5 pi ft/min
SOLUTION:
h = 2r, r = 5ft
1 1 2
Vcone = πr²h = 3 πr 2 (2r) = 3 πr³
3
2 dr
25ft³ = 3 π , 3πr² dt
dr
25π = 2π(5)² dt
dr 25π
= 2π(25) = 0.5 ft/min
dt
4. Evaluate ʃ ʃ 2r²sin Ө dr dӨ, 0 > r >sin Ө, > Ө > pi/2
A. pi/2 B. pi/8 C. pi/24 D. pi/48
SOLUTION:
π π
sin θ 2
∫02 ∫0 2r² sin θ cos ²θ drdθ =∫02 3 (sin θ)³ sin θ cos ²θ dθ
π π
sin θ 2
= ∫02 ∫0 2r² dr sin θ cos ²θ dθ = 3 ∫02 sin4 θ cos²θ dθ
π
2 sin θ 2 (3)(1)(1) π π
= ∫02 3 r² ∫0 sin θ cos ²θ dθ = 3 [(6)(4)(2)] 2 = 48
,5. A shopkeeper offers a 25% discount on the marked price on an item. In order to now cost $
48, what should the marked price be?
A. $ 12 C. $ 60
B. $ 36 D. $ 64
SOLUTION:
48 = (1 − 0.25)X
48
x = 0.75 = $ 64
6. An observer wishes to determine the height of a tower. He takes sights at the top of the
tower from A to B, which are 50 ft. apart, at the same elevation on a direct line with the
tower. The vertical angle at point A is 30 degrees and at point B is 40 degrees. What is the
height of the tower?
A. 85.60 ft B. 143.97 ft C. 110.29 ft D. 92.54 ft
SOLUTION:
β = 180 − 40 = 140°
α = 180 − 30 − 140 = 10°
50 x
= sin 30 ; x = 143.969621
sin θ
h = 143.969621 sin(40) = 92.54 ft
7. A tangent to a conic is a line
A. which is parallel to the normal
B. which touches the conic at only one point
C. which passed inside the conic
D. all of the above
8. Find the area of the triangle which the line 2x – 3y + 6 = 0 forms with the coordinate axes.
A. 3 B. 4 C. 5 D. 2
SOLUTION:
6
2x − 3(0) + 6 = 0 y=3=2
−6
x= = −3 1
2 A = 2 (3)(2) = 3 sq. units
2(0) − 3y + 6 = 0
, REGISTERED ELECTRICAL ENGINEERS PRE-BOARD EXAMINATION
AUGUST 2012
MATHEMATICS
9. Find the general solution of (D² - D + 2)y = 0
A. y = e^x/2 (C1 sin sqrt. 7/2 x + C2 cos sqrt. 7/2 x)
B. y = e^x/2 (C1 sin sqrt. 7/2 x - C2 cos sqrt. 7/2 x)
C. y = e^x/2 (C1 cos sqrt. 7/2 x + C2 sin sqrt. 7/2 x)
D. y = e^x/2 (C1 cos sqrt. 7/2 x - C2 sin sqrt. 7/2 x)
SOLUTION:
(D2 − D + 2)y = 0 1 −7 7
m − 2 = √ 4 = √2 i
m² − m + 2 = 0
1 √7
1 2 7
m= + i
2 2
(m − 2) + 4 = 0
𝐲 = 𝐞𝐀𝐱 (𝐂𝟏 𝐜𝐨𝐬𝐁𝐱 + 𝐂𝟐 𝐬𝐢𝐧𝐁𝐱)
10. If 10 is subtracted from the opposite of a number, the difference is 5. What is the number?
A. 5 B.15 C.-5 D. -15
SOLUTION:
x - 10 = 5
Opposite of x – 10 = 5
15 – 10 = 5
∴ −5
11. If y = 5 – x, find x when y = 7
A. 12 B.-12 C. 2 D. -2
SOLUTION:
y = 5 – x, find x when y = 7
7=5–x
x = -7 + 5 = −2
12. A ranch has a cattle and horses in a ratio of 9:5. If there are 80 more head of cattle than
horses, how many animals are on the ranch?
A.140 B. 168 C. 238 D. 280
SOLUTION:
x 9
Cattle → x = 5 ; x = y + 80
y
Horses → y
