Page |1
42x 1  1024
42x.4  1024
42x  256
4 x  16
5.
1.
8x  2y  2
Solve for x if
Answer: 2
and
If 3
x+1
=81, find 3
2x-1
Answer: 243
163x  y  4y
Solution:
3 x 1  81
Solution:
3 x  3  81
8 x  2y  2  
3 x  27
163  y  4y 
3 x  33
From equation  :
x3
23x  2y  2
Thus,
3x  y  2 
32x 1  36 1
From equation :
32x 1  243
42(3x  y )  4y
6x  2y  y
6x  3y
6.
2x  y
Between equations  & :
Solution:
3x  y  2
2.
2
Solve for x if In (x +x-2) = In 2x+In(x-1)
Answer: 2


3x  2x  2
In x 2  x  2  In 2x  In  x  1
x2
In  x  2  x  1  In  2x  x  1
x  2  2x
If 8  3, find the value of 26x .
Answer: 9
x
x2
Solution:
7.
If log xy = 6 and log
From:
8x  3
Solve for x and y.
Answer: x = 10000,y = 100
23x  3
Solution:
log xy  6
Squaring both sides:
 
23x
2
 32
Solve for x: 16
Answer: 1/4
 x 1
8
16 x 1  8
2 
 x 1
8.
  2
3
4  4x
2
 23
4  4x  3
1
x
4
4.
2x+1
xy  106  
y 2  104
log  x / y   2
y  100
x  100 100   10,000
Find the value of x:
Log 6 + x log 4 = log 4 + log (32+4 x)
Answer: 3
Solution:


log 6  log 4  log 4  log  32  4 
log  6   4   log  4   32  4 
log 6  x log 4  log 4  log 32  4 x
x
x
x
If 4
=1024, find the value of 4 .
Answer: 16
 6  4 x  4  32   4  4 x
x
2  4   4  32 
Solution:
4 x  64
x3
9.
Mathematics
JRN/03152011
xy
106
 2
x / y 10
x
 102  
y
Solution:
4
Divide  by 
log10 xy  6
26x  9
3.
x
 2.
y
Find the value of x if
x
x
Page |2
x = log ca  log ab  log bc
logx 2 
Answer: 1
1
log2 x
1
 log2 x  2
log2 x
Solution:
logc a 
log a
log c
1   log2 x   2log2 x
loga b 
log b
log a
logb c 
log c
log b
log2 x 2  2log2 x  1  0
2  4  4 1
x
2
log2 x 
2
log2 x  1
log a log b log c
.
.
log c log a log b
x2
x 1
10. Solve for x if log x2 = (log x)2.
Answer: 1,100
14. Determine the k so that the equation 4x2+kx+1= 0 will have
just one real solution.
Answer: 4
Solution:
4x 2  kx  1  0
Solution:
log x 2  log x 
2
2 log x  log x 
A  4,
2
C 1
B  4AC  0  so that it willhave only one real value 
log x log x  2   0
k 2  4  4 1  0
Equate each factor to zero :
log x  0 ;
B  k,
2
k4
x  100  1
log x  2  0 ;
log x  2
2
x  102  100
15. Find the value of k in the quadratic equation 3x -kx+x-7k=0
if 3 is one of the roots.
Answer: 3
11. If log (9!) = 5.5598. What is the log of 10!
Answer: 6.5598
Solution:
3x 2  kx  x  7k  0
3  3   k  3   3  7k  0
Solution:
2
k3
log10!  log 10.9!
log10!  log10  log9!
log10!  1  5.5598
16. Find the absolute value of x if 4+4x  12
log10!  6.5598
Answer: 2,-4
12. If In x2 = (In x)2 solve for x:
Answer: e2
Solution:
In x 2  In x 
2
Solution:
4  4x  12
4  4x  12
4x  8
4x  16
x2
2,-4
x  4
2 In x  In x In x
2  In x
2  loge x
xe
17. Find the value of c from the given quadratic equation
x2 -4x+c=0 if the product of the roots is -5.
2
Answer: -5
Solution:
x 2  4x  C  0
A 1
B  4
C
x1  x 2 
A
C
 5  
1
C  5
13. If log x2 + log2x = 2 solve for x:
Answer: 2
Solution:
18. Find the value of k from the given quadratic equation
2x2  kx  6  0 if the sum of the roots is equal to 4.
Answer: 8
Mathematics
JRN/03152011
Page |3
n  n  1n  2  ... n  r  2  xn r 1yr 1
Solution:
x1  x 2  
4
r  1 !
12 1110  9  8  7  6  3 5 7
 x  1  25952256
7  6  5  4  3  2 1
B
A
k
2
792 x15  25952256
k8
x2
19. Find the value of h in the equation 2x  hx  4x  5h  0
so that the sum of the roots is 2.
Answer: 4
2
2
24. Find the coefficient of (x+y)10 containing the term x7y3.
Answer: 120
Solution:
Solution:
kxn m ym  kx 7 y 3 ;n  10 ;m  3
n!
k
n  m  !m!
B
A
 2  h x 2  4x  5h  0
r1  r2 
B
A
4
2
2h
2  h  2
2
k
k
10!
7!3!
10  9  8  7!
7!3  2 1
k  120
h4
20. Find the value of the constant h in the quadratic equation
3x2  hx  x  7h  0 if 3 is one of the roots.
x
Answer: 3
Solution:
r  5 ;m  r  1  5  1  4
4
4
n!

ex
e x
n

m
!m!


Solution:
  
3  3   h  3    3   7h  0
2
h3

21. Find the value of the constant “h” in the quadratic equation
2h x2  3x2  4x  5h  0 if the product of the roots is -4.
Answer: 4
8! o
e  70
4! 4!
26. Find the middle term of the expansion of (x2-5)8.
Answer:
43750 x8
Solution:
Solution:
 
8!
x2
4! 4!
middle term 
C
A
5h
4 
2h  3
8 h  12  5 h
r1 r2 
4
 5 4
middle term  70 x 8  5 
4
middle term  43750 x 8
h  4
27. Find the middle term of the expansion of

3 3
Answer: 924 x y
22. Find the sum of the coefficients of the expansion of (x+2)4.
Answer: 65
Solution:
Middle term 
Solution:
substitute x  1 but subtract the value of  2  .
4

1
1/ 2
6
28. Find the 6th term of the expansion of (2-x)14.
Answer:
-1025024 x5
12
Solution:
r6 ;
is equal to 25952256.
Answer: 2
m  r 1  5
n!

xn m ym
n  m  !m!
Solution:

14!
 2 9   x 5
9!5!
 1025024 x 5
Mathematics
JRN/03152011
6
4
23. Find the value of x if the 8th term of the expansion of
3
  y 
12! 1/ 2
x
6!6!
middle term  924 x 3 y 3
Sum of coefficient  1  2    2   65
4
x
-x 8
25. What is the 5th term of the expansion of (e +e ) ?
Answer: 70
x y

12
.
Page |4
S
29. Find the value of “x” in the following series of number
12,13,17,26, x.
Answer: 42
Solution:
12 13
1
17
 2 3 
2
12  1  13
2
13   2   17
2
17   3   36
30.
2
2
26
2
6
 no. of spheres in the long side of 
m  6

 rec tan gular base

 no. of spheres in the short side of 
n4

 rectangular base

x
 4 2
x  26   4 
n  n  1 3m  n  1
S
2
x  42
S
4  4  1 3  6   4  1
4  5 15 
6
6
 50 spheres
34. Spheres of the same radius are piled in the form of a pyramid
with a square base until there is just one sphere at the top
layer. If there are 4 spheres on each side of the square, find
the total number of spheres I the pile.
Answer: 30 spheres
Find the value of x of the given equation shown.
x + 2x +4x+8x+16x+…….256x=1022.
Answer: 2
Solution:
Solution:
Complete the sequence using common ratio = 2,
x  2x  4x  8x  16x  32x  64x  128x  256x  1022
S
Solve for x:
511x  1022
S
n  n  1 2n  1
6
4  4  1 8  1
6
S  30 spheres
x2
31. Find the value of x from the sequence number shown.
X+2x+3x+4x+….….8x=72.
Answer: 2
Solution:
By inspection the sequence forms an arithmetic progression.
Complete the sequence and solve for x:
35. Spheres of equal size are piled in the form of a complete
pyramid with a rectangular base, find the total number of
spheres in the pile if the number of spheres in the long side
is 5 and that of the short side is 4 until the top layer consists
of a single row of 2 spheres.
Answer: 40
Solution:
x  2x  3x  4x  5x  6x  7x  8x  72
S
Solving for x:
n  n  1 3m  n  1
6
 no. of spheres in the long side of 
m  5

 rec tan gular base

 no. of spheres in the short side of 
n4

 rectangular base

36 x  72
x2
32. Find the value of x from the series of numbers shown.
x + 3x + 9x + 27x…..729x=3279.
Answer: 3
S
Solution:
S
4  4  1 3  5   4  1
4  5 12 
6
6
S  40 Spheres
By inspection, the sequence forms a geometric progression.
Complete the sequence and solve for x:
x  3x  9 x  27 x  81x  243x  729x  3279
1093x  3279
36. Find the reminder if we divide
4y3  18y2  8y  4 by  2y  3  .
x 3
Answer: 11
33. If equal spheres are piled in a form of a complete pyramid
with a rectangular base, find the total number of spheres in
the pile if the number of spheres in the long side is 6 and
that of the short side is 4 until the top layer consist of a
single row of 3 spheres.
Answer: 50
Solution:
2y 2  6y  5
2y  3 4y  18y  8y  4
3
2
4y 3  6y 2
12y 2  8y
Solution:
12y 2  8y
 10y  4
 10y  5
 11re min der 
37
Mathematics
JRN/03152011
If 8+11a+8a3+18a2 is divided by 4a2+1+3a, the reminder is:
Page |5
Answer: 5
C
Solution:
Kxy 2
z
100 
2a  3
4a2  3a  1 8a3  18a2  11a  8
K  2 1
2
4
K  200
8a3  6a2  2a
C
 12a2  9a  8
 12a2  9a  3
C
 5  remin der 
Kxy 2
z
200  3  2 
2
5
C  480
38. If
1and
-2
are
rational
roots
of
the
equation
x4  x3  4x2  6x  12  0. what is the second depressed
equation?
Answer: x2+6=0
41. Twelve men can finish the job in 16 days. Five men were
working at the start and after 8 days 3 men were added.
How many days will it take to finish the job?
Answer: 27 days
Solution:
Solution:
1  1  4  6  12
1
12(16)=5(8)+(5+3)x
X=19 days
Total number of days to finish the job
=19+8
= 27 days
 1  2  6  12
1  2  6  12  0
x 3  2x 2  6x  12  0 1st depressed equation 
1  2  6  12
-2
 2  0  12
1 0  6  0
x 2  6  0  2nd depressed equation 
39. The radius of a curvature of a given curve varies directly with
x and inversely with the square of y. When x =2, y=3, the
radius of curvature is 100. Find the radius of curvature when
x =4 and y=6.
Answer: 50
42. The vibration frequency of string varies as the square root of
the tension and inversely as the product of the length and
diam. of the testing. If the testing is 3 feet long and 0.03 inch
diameter vibrates at 720times per second under 90 pounds
tension, at what frequency will a 2 ft, 0.025 inch string
vibrate under 50 pounds tension.
Answer: 966
Solution:
kx
K T
LD
K 90
720 
3  0.03 
y2
K  6.8305
w
Solution:
ry
100 
k  2
kx
K T
LD
6.8305 50
w
2  0.025 
y2
w  966
w
 3 2
k  450
r
r
450  4 
 6 2
r  50
40. The value of C varies directly with x and the square of y and
conversely with z. When x=2, y=1 and z=4,C=100. Find the
value of C when x=3,y=2 and z=5.
Answer: 480
43. Eight men can excavate15 cu.m. of drainage open canal in 7
hours. Three men can backfill 10 cu.m. in 4 hours. How long
will it take 10 men to excavate and backfill 20 cu.m. in the
same project?
Answer: 9.87 hrs.
Solution:
Solution:
For excavation:
No. of man hours to excavate/cu.m. of drainage
=
8 7
15
 3.733
Total no. of man-hours required to excavate 20 cu.m.
=3.733(20)
=74.667
Total no. of hrs. to excavate 20 cu.m. with 10 men.
=
74.667
 7.47 hrs.
10
For backfill:
Mathematics
JRN/03152011
Page |6
No. of man hours needed to backfill per cu.m. of
drainage =
3  4
10
 1.2
51. A piece of wire of length 36π is cut into two unequal parts.
Each part is then bent to form a circle. It is found that the
total area of the two circle is 80π. Find the diff. in radius of
two circles.
Answer: 6
Total no. of man hours required to backfill 20 cu.m.
=1.2(20)=24
Total no. of hours to backfill 20 cu.m. with 10 men
=
No. of days ahead =20-18
=2 days
24
 2.4 hrs.
10
Solution:
Total time = 7.47 + 2.4
Total time = 9.87 hrs
2πr + 2πR=36 π
r+R=18
2
2
πr + πR =180 π
2
2
r +R =180
2
2
(18-R) +R =180
R=12
r=18-12
r=6
45. A statistical clerk submitted the following reports. The average
rate of production of radios is 1.5 units for every 1.5 hrs.
work by 1.5 workers. How many radios were produced in
one month by 30 men working 200 hours during the month?
Answer: 4000
Solution:
Diff. in radius = 12-6
Diff. in radius = 6
No. of man hours to produce x radios
 200  300   60000
52. 12 cubic yards of crushed stone for surfacing three private
roads of different length is to be distributed in three piles so
that the second pile has 20 cu.ft. less than the first and the
third pile has 8 cu.ft. more than twice as much as the first.
Determine the volume of the biggest pile in cu.ft.
Answer: 176
No. of man hours to produce 1.5 radios
 1.5 1.5   2.25
By proportion
6000 2.25

x
1.5
x  4000 radios
Solution:
st
X=1 pile
nd
x-20=2 pile
rd
2x+8=3 pile
3
12 cu. yd. = 12 (3) =324 cu. ft.
x+x-20+2x+8=324
x=84 cu. ft.
2x+8=84(2)+8
2x+8=176 cu. ft. (largest volume)
46. A and B working together can finish a piece of work in 20
days. After working for 4 days, A quits and B finishes the
work in 24 days more, Find the no. of days that B could
finish the work alone.
Answer: 30 days
Solution:
1 1
1
 

A B 20
 1 1
 1
 A  B  4   B  24  1  


 
Between  & :
 24 
 1 
 20   4   B  1
 
B  30 days
53. Three cities are connected by roads forming a triangle, all of
diff. length. It is 30 km. around the circuit. One of the road is
10 km. long and the longest is 10 km. longer than the
shortest. What are the lengths of the longest and the
shortest of the three roads?
Answer: 15 and 5
Solution:
10+x+x+10=30
2x=10
x=5 shortest
x +10=15 longest
49. 28 men can finished the job in 60 days. At the start of the 16 th
day 5 men were laid off and after the 45th day 10 more men
were hired. How many days were they delayed in finishing
the job?
Answer: 2.27 days
54. The sum of two numbers is 30. If the larger number is divided
by the smaller number the quotient is equal to the smaller
number. Find the numbers.
Answer: 5 and 25
Solution:
28(60)=28(15) + (23)(30)+(33)x
X=17.27 days
Total no. of days they finish the job
=15+30+17.27
=62.27 days
Therefore they were delayed by 2.27 days
Solution:
50. 16 men has a contract to finish the job in 20 days. 20 men
were hired at the start and four quit the job after 8 days.
Determine the number of days delayed or ahead of the
scheduled time when they were able to finish the job.
Answer: 2 days ahead
x 2  30  x
x 2  x  30  0
 x  5  x  6   0
Solution:
x5
30  x  25
16(20) =20(8)+16x
x=10
Total no. of days to finish the job
=8+10 =18
Mathematics
JRN/03152011
x  smaller no.
30-x=larger no.
30  x
x
x
Page |7
55. One proposal in the Agrarian Reform Program is to have a
retention limit of 10 hectares. If a landowner was left with
10
hectares fewer than 40% of his land, after selling 6
hectares more than 70% of his land, what size of land did
he initially owned?
Answer: 40 hectares
Arithmetic mean :
8  10  12  14  15  18
6
x  12.833
x
Variance :
(8  12.833)2  (10  12.833)2  (12  12.833)2  (14  12.833) 2
Solution:
 (15  12.833)2  (18  12.833)2
6 1
Variance  12.694
x= original area he owns
x-(0.70x+6)=0.40x-10
0.30x-6=0.40x-10
0.10x=4
X=40 hectares
56. Twice the sum of two numbers is 28. The sum of the
squares of the two numbers is 100. The product of
the
two numbers is:
Answer: A. 48
Solution:
60. A boat travels upstream in two rivers A and B. The velocity
of
the current in river A is 3 kph while that of B is 1 kph.
It takes two times longer to travel 18 km in river A
than
it took to travel 10 km in river B. Calculate
the
speed of the boat in still water.
Answer: 21 kph
x=one no.
y=other no.
2(x+y)=28
2
2
X +y =100
X+y=14
2
2
(14-y) +y =100
2
2
196-28y+y +y =100
2
2y -28y+96=0
2
y -14y+48=0
(y-8)(y-6)=0
y=8
y=6
x=6
x=8
xy=8(6)
xy=48
Solution:
t A  time to travel 18 km in river A
tB  time to travel 10 km in B
t A  2tB
(18) 2(10)

x3
x 1
18x  18  20x  60
2x  42
x  21 kph
59. An airplane could travel a distance of 1000 miles with the
wind in the same time it could travel a distance of 800
miles against the wind. If the wind velocity is 40
mph,
what is the speed of the plane?
Answer: 360 mph
61. It takes a boat 3 times to travel upstream against a river
current that it takes the same boat to travel downstream. If
the speed of the boat is 40 kph, what is the speed of the
current?
Answer: 20 mph
Solution:
Solution:
Tup  3Tdown
1000
800

x  40 x  40
10x  400  8x  320
D
3D

40  x 40  x
x  20mph
2x  720
x  360mph
60. Compute the standard deviation of the following sets of
numbers 2,4,6,8,10,12.
Answer: 3.742
62. With a wind velocity of 40 kph, it takes an airplane as long
to travel 1,200 km with the wind as 900 km against it. How
fast can the airplane travel in still air?
Answer: 280 kph
Solution:
Solution:
900
1200

V  40 V  40
900V  36000  1200V  48000
2  4  6  8  10  12
Arithmetic mean 
6
Arithmetic mean  7
V  280kph
(2  7)  (4  7)  (8  7)  (10  7)  (12  7)
6 1
Variance  14
Variance 
2
2
2
2
2
63. Two racing cars A and B compete for a stretched of 10000
m. long. It took 40 min. for A to reach the finish line and B
50
min. to reach the finish line. How far was the B
behind when A reach the finish line?
Answer: 2000 m.
S tan darddeviation  14
S tan darddeviation  3.742
60. Compute the value of the variance of the following sets of
numbers 8,10,12,14,15,18.
Answer: 12.694
Solution:
Mathematics
JRN/03152011
Solution:
Page |8
X=40%
D
T
V
10000
 40
V1
69. The second hand of a clock is 4 inches long. Find the
speed of the tip of the second hand.
Answer: 0.42 in/sec.
V1  250 m / min. (vel.of A)
Solution:
10000
 50
V2
S  r
  
S  4  360  
  8
 180 
Solving for V:
V2  200 m / min.  vel.of B 
200(40 )  x  10000
x  2000 m.
65
V
Two cars A and B run a 10 km straight stretched. It took A
40
min. to reach the finish line with B 2 km. behind A.
How long would it take B to reach the finish line?
Answer: 50 min.
Solution:
V xT  D
V140
 10
60
V1  15 kph
V2 40
8
60
V2  12kph
D
t
V
10
t
hours 
12
10(60)
t
12
t  50 min.
70. A is as old as the combined age of his two brothers B and
C. But C is two years older than B. The combined age of
the
three last year was 3/4 their combined ages at
present. How old is B now?
Answer: 2
Solution:
Combined ages at present :
 x  2  x  2  x  x  4x  4
Combined ages last year :
 x  1  2  x  1  2  2x  1  4x  1
3
4x  1  (4x  4)
4
16x  4  12x  12
4x  8
x  2 years old
66. Bianca is twice as old as Pio and Neggie is twice as old as
Bianca. In ten years, their combined ages will be 58. How
old is Bianca now?
Answer: 8
Solution:
x+y = 60
0.35x+0.50y = 0.40 (60)
0.35x+0.50y = 24
0.35x+0.50y = 24
0.50x+0.50y = 30
-0.15x = -6
x = 40 gallons
 age of Bianca 
67. The gasoline tank of a car contains 50 liters of gasoline and
alcohol, the alcohol comprising 25%. How much of
the
mixture must be drawn off and replaced by alcohol so that
the tank contain a mixture of which 50% is alcohol?
Answer: 16.67 liters
72. A chemical engineer mixed two chemical solutions of
different strengths 30% and 50% of the chemical solutions
respectively. How many millimeters of the 30% strength
must be used to produce a mixture of 50 millimeters that
contains 42% of the chemical
Answer: 20 ml
Solution:
x  y  50
0.3x  0.50y  0.42(50)
Solution:
0.3x  0.50y  21 
0.25(50) - 0.25x+x = 0.50(50)
0.75x = 12.5
X=16.67 liters
0.50x  0.50y  0.50(50)
0.50x  0.50y  25  
68. A chemical engineer mixed 40 millimeters of 35%
hydrochloric acid solution with 20 millimeters of 50%
hydrochloric acid solution. What is the percentage of the
hydrochloric acid of the new solution?
Answer: 40%
Solution:
Between  & :
0.50x  0.50y  25
0.30x  0.50y  21
0.20x  4
x  20 ml
35(40)+50(20)=x(60)
Mathematics
JRN/03152011
age of B 
71. A chemist of a distillery experimented on two alcohol
solutions of different strengths, 35% alcohol and 50%
alcohol respectively. How many gallons containing
35%
alcohol
must be used to produce a mixture of 60
gallons that
contains 40% alcohol?
Answer: 40
Solution:
x  Pio ' s age
2x  Bianca ' s age
4x  Meggie ' s age
x  10  2x  10  4x  10  58
x  4 age of Pio
2x  8 years old
S 8

 0.42 in / sec
t 60
Page |9
a 1
73. If the value of the piece of property decreases by 10%
while the tax rate on the property increases by 10%, what
is the effect on the taxes?
Answer: Taxes decrease by 1%
d 1
S  105
n
2a  (n  1)(d)
2
n
105   2(1)  (n  1)(1)
2
210  2n  n2  n
S
Solution:
Let x = value of property
x-0.10x = 0.90x
= decreased value of property
y= orig. tax rate
y+10y=1.10y (increased tax rate)
Orig. tax = xy
New tax = (0.90x) (1.1y) = 0.99 xy
Therefore, the tax is decreased by 1%
n2  n  210  0
(n  15)(n  14)  0
N  14 (layers)
74. How many terms of the progression 3,5,7… must be in
order that their sum will be 2600?
Answer: 50
Solution:
a3
d  53  2
77. Find the sum of all the odd integers between 100 and
1000.
Answer: 247500
Solution:
S  2600
n
2a  (n  1)d
2
n
2600   2(3)  (n  1)(2)
2
5200  6n  2n2  2n
S
101,103...999
a  101
999  101  (n  1)(2)
2n  999  101  2
2n  4n  5200  0
2
n  450
n2  2n  2600  0
n
2a  (n  1)d
2
450
S
2(101)  (449)(2)
2
S  247500
S
(n  52)(n  50)  0
n  50
75. An Electronics engineering student got a score of 30% on
Test I of the five number test in Mathematics. On
the
last number he
got
90%
in
which
constant
difference more on each number than he had on
the
immediately preceding one. What was his average
score in Mathematics?
Answer: 60
Solution:
a  30
an  90
an  a  (n  1)d
90  30  (5  1)d
d  15
n
S   2a  (n  1)d
2
5
S   2(30)  4(15)
2
S  300
300
Average score 
5
Average  60
78. Find a positive value of x so that 4x, 5x+4 and 3x2-1 will be
in
A.P.
Answer: 3
Solution:
2nd  1st  3rd  2nd
(5x  4)  (4x)  (3x 2  1)  (5x  4)
x  4  3x 2  5x  5
3x 2  6x  9  0
x 2  2x  3  0
(x  3)(x  1)  0
x3
th
rd
79. If the 5 term in A.P. is 17 and the 3 term is 10, what is
the
8th term?
Answer: 27.5
Solution:
76. In a pile of logs, each layer contains one more log than the
layer above and the top contains just one log. If there are
105 logs in the pile, how many layers are there?
Answer: 14
Solution:
Mathematics
JRN/03152011
d  103  101  2 an  999
an  a  (n  1)d
P a g e | 10
a  4d  17
on, until the last layer which has 10 bricks.
total number of
bricks used up.
Answer: 639
a  2d  10
2d  7
a  61
an  10
10  61  (n  1)d
10  61  (n  1)( 3)
n  18
(7)
a  7d  3  7
2
a  7d  27.5  8th term 
n
(a1  an )
2
18
S
(61  10)
2
S  639
S
80. Find the sum of the first 40 even numbers.
Answer: 1640
Solution:
84. Determine the sum of the progression if there are 7
arithmetic mean between 3 and 35.
Answer: 171
2, 4,6.....n  40
a1 a  d, a  2d...
d2
Solution:
n  40
n
(a1  an )
2
9
S  (3  35)
2
S  171
S
n
2a  (n  1)(d)
2
40
S
2(2)  (39)(2)
2
S  1640
S
81. Find the quotient of the sum of all odd integers between
100 and 1000 when it is divided by 9.
Answer: 27500
85. The sum of an A.P. is 196. If the 1st term is 52 and the last
term is 4, determine the number of arithmetic mean
between 52 and 4.
Answer: 5
Solution:
a1  101 d  2 an  999
an  a  (n  1)d
999  101  (n  1)(2)
n  450
n
S  a  an 
2
450
S
101  999  247500
2
247500
Quotient 
 27500
9
Solution:
a1  52 an  4
S  196
n
S  (a1  an )
2
n
196  (52  4)
2
n7
There are 5 arithmetic meanbetween 52 and 4.
82. Find the sum of the numbers divisible by 6 which lie
between 75 and 190.
Answer: 2508
86. Find the positive value of x so that x, x2-5, 2x will be in
harmonic progression.
Answer: 3
an  186
78
186
 13
6
6  31
a1  a1  (n  1)d
186  78  (n  1)(6)
n  19
Solution:
1
1
1
,
,
in A.P.
2
X
2X
X 5
2nd  1st  3rd  2nd
1
1
1
1
 

x 2  5 x 2x x 2  5
x  x 2  5 x 2  5  2x

x(x 2  5)
2x(x 2  5)
2x  22  10  x 2  5  2x
n
(a1  an )
2
19
S
(78  186)
2
S  2508
S
3x 2  4x  15  0
(3x  5)(x  3)  0
x3
83. A stack of bricks has 61 bricks at the bottom layer, 58
bricks in the 2nd layer, 55 bricks in the 3rd layer and
so
Mathematics
JRN/03152011
d  3
an  a1  (n  1)d
(7)
a2
 10
2
a3
Solution:
a1  78
the
Solution:
7
2
a  7d  8th term
d
a2
Determine
P a g e | 11
87. There are 4 geometric means between 3 and 729. Find the
fourth term of the geometric progression.
Answer: 81
Solution:
a1  3
an  a1r
an  729
15
5
r3
r
n6
S
n 1
729  3 r 
a4   3  3 
5
n
3  729
3
3
n
88. The first and the last term of a G.P. is equal to 6 and 486
respectively. If the sum of all terms is 726, find the number
of
terms.
Answer: 5
Solution:
a6
S  726
an r  a
S
r 1
486r  6
726 
r 1
726r  726  486r  6
240r  720
r 3

3 1
 3n  1
 3
6
n6
91. The arithmetic mean and geometric mean of two numbers
are 10 and 8, respectively. Find their harmonic mean.
Answer: 6.4
an  486
Solution:
If the numbers are and y,
Then,
Arithmetic Mean:
AM 
xy
2
Geometric Mean:
GM  xy
n 1
Harmonic Mean:
486  6  3 
n 1
HM 
3n 1  81
4
x  y   2xy 
  xy

 2  x  y 
n5
 AMHM   GM
2
10 HM   8 
The Number 28,x+2, 112 form a G.P. What is the 10th
term?
Answer: 14336
2
HM  6.4
92. Expand (3cis108) .
Answer: 9cis216°
Solution:
 3136
rcis  r 2cis2  formula
2
 3cis108   32 cis 2(108)
2
x  54
x2
28
a  28
r
 9cis216
an  ar n 1
an  28  2 
2
2
Solution:
x  2 112

28
x2
 x  2
2xy
xy
 AMHM  
n 1  4
93. Two numbers differs by 40 and their arithmetic mean
exceeds their geometric mean by 2. What are these
numbers?
Answer: 81 and 121
9
an  14336
90. The sum of the terms in G.P. is 1820. How many terms are
there if the first term is 5, the second term is 15 and
the
third term is 45?
Answer: 6
Solution:
Mathematics
JRN/03152011

5 3n  1
1820  2 
a4  81
9.
r 1
1820 
a4  a1r 4 1
3n 1   3 

5
r 3
an  ar

a rn  1
Solution:
P a g e | 12
let x  one no.
100. A rubber ball is made to fall from a height of 50 ft. and is
observed to rebound 2/3 of the distance it falls. How far will
the ball travel before coming to rest if the ball
continues to fall
in this manner?
Answer: 250
x  40  other no.
x  x  40
 arithmetic mean
2
2x  40
 x  20  A.M.
2
Solution:
x  x  40   geometric mean
2
 50  2 
3
a1  66.67
a1 
A.M.  G.M.  2
x  20  x  x  40   2
x  18  x  x  40 
r
x 2  36x  324  x 2  40x
S
a
1 r
66.67
S
2
1
3
S  200
4x  324
x  81
x  40  121
Total dis tan ce  200  50
Total dis tan ce  250 ft.
94. If the 3rd term of a G.P. is 20, and the 6th term is 160, what
is the first term?
Answer: 5
101. The motion of a particle through a certain medium is such
that it moves two thirds as far each second as in the
preceding second. If it moves 6m of the first second, how
far will it move before coming to rest.
Answer: 18
Solution:
ar 2  3rd term
ar 5  6th term
ar 5  160
Solution:
ar 2  20
r3  8
r2
a6
2
3
S
a  2   20
2
a  5 first term
94. The number of bacteria in a certain culture doubles every
3 hrs. If there are N bacteria to start with, find the number
in
24 hrs.
Answer: 256 N
a9  N  2 
r
a
1 r
6
S
2
1
3
S  18 cm.
ar 2  20
Solution:
a1  N
a2  2N
a3  4N
2N 4N
r

N
2N
n 1
an  ar
2
3
102. A man who is on diet losses 24 lb. in 3 months, 16lb. in the
next 3 months and so on for a long time. What is the
maximum total weight loss?
Answer: 72
Solution:
a  24
16 2
r

24 3
a
S
1 r
24
S
2
1
3
S  72lb.
9 1
a9   2  N
8
a9  256N
99. Rationalize: (4  3i) /(2  i) .
Answer: 1  2i
104 Express 3+4i in trigonometric form
Answer: 5(Cos 53.13° + I Sin 53.13°)
Solution:
2
 4  3i   2  i  8  10i  3i
 2 i  2  i  
2
4 i



5  10i

5
 1  2i
Solution:
Mathematics
JRN/03152011
P a g e | 13
r  x2  y2
r
3
2
F  32  42  122  13
  4
2
r 5
110. Given the three vectors:
A  i  3j  4k
4
tan  
3
  53.13
B  2i  7j  k
C  i  4j  2k
Find the magnitude of the resultant vector, R.
Answer: 15
3  4i  r  cos   iSin  
3  4i  5  Cos 53.13  iSin 53.13 
Solution:
105.Find the distance between two given complex numbers:
Add the three vectors:
z1  3  4i
A  i  3j  4k
z2  2  2i
Answer:
B  2i  7j  k
C  i  4j  2k
R  2i  14j  5k
5
Then:
Solution:
d
 3  2   4  2
2
2
R  22  142  52  15
 5
106. Evaluate ln  3  4i  .
111. What is the cross product A  B of the vectors:
A  i  4j  6k
Answer: 1.61 0.927i
B  2i  3j  5k
Answer: 2i+7j-5k
Solution:
 x  iy   x  3; y  4
y 4

x 3
  
  53.13 
  0.927 rad
 180 
tan  
Solution:
i j k
i j k i j
1 4 6  1 4 6 1 4
2 3 5
2 3 5 2 3
r  32  42  5  magnitude
 i(4)(5)  j(6)(2)  k(1)(3)  (2)(4)(k)  (3)(6)(i)  (5)(1)(j)
ln  x  iy   lnr  i
 2i  7j  5k
ln  3  4i   ln 5  0.927i
 1.61  0.927i
107. Simplify: i29  i21  i .
Answer: 3i
Solution:
112.Which of the following is a vector quantity?
A.
kinetic energy
B.
electric field intensity
C.
entropy
D.
work
Answer: Electric Field Intensity
Note : i4n 1  i
i  i21  i  i4(7) 1  i4(5) i  i
29
 iii
 3i
108. Find the value of sin , if e^(0.92730i).
Answer: 0.80
Solution:
e
0.92730i

e
i
0.92730 180 

  53.13
113. Which of the following is a scalar quantity?
A.
temperature
B.
gravitational potential
C.
charge
D.
All of the above
Answer: All of the Above
114. Compute the number of 8 letter combination of all letters in
the alphabet.
Answer: 1562275
Solution:
There are 26 letters in the alphabet:
nCr 
Thus,
sin   sin53.13
26!
 1562275
26
  8  !8!
sin   0.80
115. Compute the number of 12 letter combination of all letters in
the alphabet.
109. Find the magnitude of the resultant force:
F  3i  4j  12k
Answer: 13
 Answer: 9657700
Solution:
Solution:
Mathematics
JRN/03152011
P a g e | 14
nCr 
26!
 26  12 !12!
P1  0.80.80.2  0.128
 9657700
The probability of passing the three subjects is:
P2  0.8 (0.8)(0.8)  (0.8)3  0.512
Thus, the probability of passing at least two subjects:
116. How many different signals each consisting of 6 flags hung
in a vertical line can be formed from 4 identical red flags and
2 identical blue flags?
Answer: 15
Pat least 2  0.128  0.512  0.64
122. With 50 items on a test each of which has 4 given answers,
how many possible answer patterns are there?
Solution:
n6
red  4
blue  2
Answer: 1.27  10
30
Solution:
No. of patterns  450  1.27  1030
P
n!
6!

 15
p!q! 4!2!
123.
117. Four couples are to eat at a round table with the men and
women alternating. If the hostess reserves a place for
herself, in how many ways can she assign seats to the
others?
Answer: 144
Solution:
5
1 +3 -5 -6 -5
-5 +10 -25
1 -2 +5 -31
Solution:
P   4  1 ! 4!  144
118. From the digits 0,1,2,3,4,5,6,7,8,9 , find the number of six
digit combination.
Answer: 210
Note: When the numbers in the third line are alternately
positive and negative, the trial number is a lower bound of
the real roots, therefore -5 is the lower bound.
124.
Solution:
nCr  10C6  210
The constant remainder when x30-2x+5 is divided by x+1.
Answer: 8
Solution:
f(x)=x30-2x+5
x=-1
f(x)=(-1)30-2(-1)+5
f(x)=8 (remainder)
119. The probability of getting at exactly 2 heads when a coin is
tossed four times.
Answer: 3/8 or 0.375
Solution:
125. In a certain party each one of the group drinks coke or beer
or whisky or all. Also 400 drink coke, 500 drink beer and 300
drink whisky. 100 drink coke and beer and 200 drink beer
and whisky. One who drinks whisky does not drink coke.
How many are in the group?
Answer: 900
P  nCr (p)r (q)n r
n4
Compute the lower bound of the real roots of the
polynomial equation x3+3x2-5x-6=0.
Answer: -5
r2
Thus,
2
2
 1  1
P  4C2      0.375
2  2
Beer
Solution:
200
120. A bag contains 3 yellow and 5 black balls. If 2 balls are
drawn in succession without replacement, find the probability
that the balls drawn are one yellow and one black.
Answer: 15/28 or 0.536
100
200
100
300
Whisky
Coke
Solution:
P  P1st yellow,sec ond red  P1st red,sec ond yellow
The number of persons in the group, N:
35 53
P     
87 87
N  100  200  200  100  300
N  900 persons
121. The probability that an examinee will pass in each subject in
the ECE Board exam is 0.80. What is the probability that an
examinee will pass at least 2 subjects?
Answer: 0.64
126.
Solution:
Solution:
There are three subjects in the ECE Board exam.
The probability of passing two subjects and failing in one is:
Mathematics
JRN/03152011
Eleven (11) men can finish the job in 15 days. Five (5) men
were working at the start and after 6 days four (4) men
were added. How many days will it take to finish the job?
Answer: 21 days
P a g e | 15
1115   5  6   9  x 
Solution:
x  wt. of gold
x  15 days
y  wt. of silver
Thus, total number of days is:
127.
 15  6
equation 1
 21 days
equation 2
A piece of wire of length 52 cm. is cut into two unequal
parts. Each part is then bent to form a square. It is found
that the total area of the two squares is 97 cm 2. Find the
difference between the sides of each square.
Answer: A. 5
x  y
x
y

 106  99
19
10
x
y

 7
19
10
10x  19y  1330
x 
10x19y  1330
Solution:
by subtracting 1 & 2
4x  4y  52
we obtain : 9x  684
x  y  13
2
x  76 kilos
2
x  y  97
132.
x2  13  x   97
2
x 9
y  13  9
The arithmetic mean of 80 numbers is 55. If the two
numbers namely 850 and 250 are removed, what is the
arithmetic mean of the remaining numbers?
Answer: 42.31
Solution:
y 4
Diff. in sides  9  4
S
 55
80
S  4400
Diff. in sides  5
128.
y  106
19x  19y  2014
One proposal in the Agrarian Reform Program is to have a
retention limit of 10 hectares. If a landowner was left with
10 hectares fewer than 40% of his land, after selling 6
hectares more than 70% of his land, what size of land did
he initially owned?
Answer: 40 hectares
4400  250  850
 A.M.
78
Arithmetic mean  A.M.  42.31
Solution:
133.
x  original area he owns
Find the determinant of an upper triangular matrix:
x  0.70x  6  0.40x  10
3
0.30x  6  0.40x  10
A 0
0.10x  4
0
2
1
1 2
0
4
x  40 hectares
Answer: -12
129.
A professor gives the following scores to her students.
Compute the arithmetic mean.
Frequency: 1 3 6 11 13 10 2
Score:
35 45 55 65 75 85 95
Answer: 70.2
Solution:
The determinant of an upper triangular matrix is equal to
the product of the diagonal.
Solution:
Determinant = 3 (-1) (4)
Determinant = -12
x  arithmetic mean
x 
1 35  3 45  6 55  11 65  13 75  10 85  2 95
1  3  6  11  13  10  2
134.
x  70.2
Find the determinant of the given matrix:
A
1
2
3
0
2
9 2
0
0
3 1
0 0
Answer: 6
130.
Round off 0.003086 to three significant figures.
Answer: 0.00309
131.
If 19 kilos of gold losses 1 kilo, and 10 kilos of silver losses
1kilo when weighed in water, find the weight of gold in a
bar of gold and silver weighing 106 kilos in air and 99 kilos
is water.
A. 76
B. 30
C. 82
D. 58
0
1
1
Solution:

A = (1) (2) (-3) (-1)
A=6
135.
Mathematics
JRN/03152011
The value of (1+i)6 is equal to:
Answer: -8i
Solution:
P a g e | 16
Answer: 32
Usin g Eulers Equation
Solution:
A  3i  4j  0k
x  iy  rei
1  i  12  12 ei
B  8i  0  0
45

180
  0.7854
1
tan  
1
  45
i
3
8
A xB=
j
4
0
k
0
0
i
3
8
j
4
0
1  i  2 e0.7854i
1  i6  
2

6
e
0.7854  6 i
A x B = (0+0+0) - (32k+0+0)
A x B = -32k
1  i6  84.7124i
1  i6  r  CosØ  SinØ 
4.7124 180 
Ø=
A x B = 32
140.
A

Ø=270
1  i6  8  Cos 270  Sin 270 
1  i6  8  0  i
1  i6  8i
137.
r2  3i  2j  k
AB  r2  r1
AB   3i  2j  k    2i  4j  5k 
AB  i  2j  4k
AB 
Solution:
k
12   2 2   4 2
AB  4.58
i
j
k
i
1 4 6 1 4 6 1
2 3 5
point
r1  2i  4j  5k
Answer: 2i+7j-5k
j
Find the magnitude of the vector having initial
(2,4,5) and terminal point B (3,2,1).
Answer: 4.58
Solution:
What is the cross product AB of the vectors?
A= i+4j+6k
B= 2i+3j+5k
i
 322
A xB=
141.
Compute the value of “b” if A and B are
perpendicular.
A=2i+bj+k
2 3 5 2
B=4i-2j-2k
Answer: 3
i  4  5   j  6  2   k 13   2  4  k  3 6 i  5 1 j 
 2i  7j  5k
138.
Solution:
So that A and B will be perpendicular
AB=0
A  B =2(4)+(b)(-2)+(1)(-2)=0
8-2b-2=0
2b=6
b=3
Assume the three force vectors intersect at a
single
point
F sub1= i+3j+4k
F sub2= 2i+7j-k
142.
Fsub3= -i+4j+2k
What is the magnitude of the resultant force vector,
Answer: 15
R?
How many 3 digit numbers can be formed from the
digits 2,4,6,8 and 9 if repetitions are allowed?
Answer: 125 ways
Solution:
Since repetitions are allowed each of the three
digits
can be filled in 5 ways. Since there are
three digit numbers.
Solution:
Add ,  and 
 F sub1= i+3j+4k
 F sub2= 2i+7j-k
P   5  5  5 
 Fsub3= -i+4j+2k
_____________
P  125 ways
143.
R=2i+14j+5k
Then, R=
 2 2  14 2   5 2
R  15
139.
Find the area of a parallelogram with two sides
identified by vectors from the origin to the points
(3,4)
and (8,0).
Mathematics
JRN/03152011
The permutation for two letters taken two at a
time
say P and Q are PQ and QP. Four letters
MNOP
taken four at a time gives four
permutations say
MNOP,
NOPM,
OPMN and
PMNO.
How many permutations are
there
for 9 letters taken 9
at a time?
Answer: 9
Solution:
P a g e | 17
P
n!
9!

n  1!  9  1!
9  8!
P
 8!

Solution :
9!
8!
Case11  H, 2  T, 3  T 
Case 2 1  T, 2  H, 3  T 
P  9 permutations
Case 3 1  T, 2  T, 3  H
144. How many different signals each consisting of
6
flags hung in a vertical line can be formed
from 4 identical
red flags and 2 identical blue flags?
Answer: 15
4 red
and
2 blue
1
8
1
8
1
8
1
3
8
3
P
8
P
Solution:
n6
1 1 1
. . 
2 2 2
1 1 1
P . . 
2 2 2
1 1 1
P . . 
2 2 2
P
149.
n!
P
q!r!
6!
P
4!2!
P  15
If a coin is tossed 5 times, find the probability
getting 3 heads.
Answer: 5/16
of
Solution:
Pr obability of getting a head 
1
2
1
2
Pr obability of getting 3 heads
Pr obability of getting a tail 
145.
A semiconductor company will hire 7 men and
women. In how many ways can the
company
choose
from 9 men and 6 women
qualified for
the
position?
Answer: 540
4
n5
who
r 3
3
Solution:
Number of ways of hiring men
n!
9!
nCr 
nCr 
nC  36 ways
n

r
!r!
9

 
 7 !7! r
P
5!
 1  1
 5  3  !3!  2   2 
P
5
16
2
Number of ways of hiring women :
nCr 
n!

150.
6!
n  r !r!  6  4 ! 4!
nCr  15 ways
Numbers
Probability
12
0.31
4
0.48
-6
0.21
Find the mean value of the given numbers
shown:
Answer: 4.38
n  15  36 
n  540 ways
Solution:
146.
In a licensure examination, an examinee may
select 7
problems from a set of 10 questions.
In how many
ways can he make his choice?
Answer: 120 ways
Mean value = 12(0.31)+4(0.48)+(-6)(0.21)
Mean value = 4.38
151.
Solution:
10!
7!3!
C 10,7   120 ways
C 10,7  
147. Find the probability of obtaining an even number
one roll of a dice.
Answer: 0.5
Compute the mean value of the numbers from
tabulated values shown.
(Numbers)
(Probability)
4.5
0.32
1.2
0.51
-2.3
0.17
Answer: 1.661
the
in
Solution:
Mean value= 4.5(0.32)+(1.2)(0.51)+(- 2.3)(0.17)
Mean value = 1.661
Solution:
There are three possible even numbers
 2,4 and 6 
Pr obability 
148.
152.
3
 0.5
6
A coin is tossed 3 times. What is the
getting 1 head and 2 tails?
Answer: 3/8
probability of
The mean value of the given set of numbers is
Find the value of x:
4.38.
Numbers
Probability
Answer: 4
-6
0.31
12
0.48
Solution:
Mean value=12(0.31)+x(0.48)+(-6)(0.21)
Mathematics
JRN/03152011
x
0.21
P a g e | 18
4.38=12(0.31)+0.48x-6(0.21)
x=4
158.
The two sides of a triangle are 40 m. and 50 m. r
espectively which is inscribed in a circle having a
radius of 12 m. If the area of the triangle is 2500
m 2,
compute the third side.
Answer: 60
A two sides of a parallelogram are 22m. and
12m.
respectively and one diagonal is 28m
long. Find the
length of the other diagonal.
Answer: 21.73
154.
Solution:
28  25.06
Solution:
Therefore 28 is a long diagonal
abc
4r
40  50  x 
2500 
4 12 
A
Usin g cosine law :
 22 2  12 2
x  60 m.
 28 
2
 25.06 long diagonal
  22   12   2  22 12  Cos
2
  180  10711'
  7249 '
x 2   22   12   2  22 12  Cos7249 '
2
x  21.73 m.
Solution:
abc
A
4r
8 10  c
2
  10711'
155. The area of a triangle inscribe in a circle is 39.19 cm2 and
the radius of the circumscribe circle is 7.14 cm. If the two sides
of the inscribe triangle are 8 cm.
and 10 cm respectively, find
the third side.
Answer: 14
39.19 
12 2   22 2
a2  b2 
159.
4  7.14 
c  14m.
2
(shorter diaginal)
Find the length of the side of a regular
inscribed in a circle of radius 10 cm.
Answer: 11.76
pentagon
Solution:
n  5 no. of sides of pentagon
156.
2  n   360
The area of triangle circumscribing a circle is 216 sq.cm.
if the radius of the circle is 6 cm. and the two sides of the
triangle are 12 cm. and 16 cm., compute the third side of
the triangle.
Answer: .44
2  5   360
2  72
  36
Solution:
216  6S
x
2r
x  2rSin
S  36
x  2 10  Sin36
abc
S
2
12  16  c
S
2
c  44 cm.
x  11.76 cm.
Sin 
A  rS
157.
160.
The area and perimeter of a triangle is 2310
sq.cm.
and 220 cm. respectively. A circle is
escribed outside the triangle having a radius of 55 cm. is
tangent
to one of
the sides of the triangle. Compute the
length of this
side.
Answer: 68cm.
Given 1coin with 5 cm. diameter and a large
supply
of coins with 2 cm. dia. What is the max.
number of smaller coins that maybe arrange tangentially
around
the larger without
any overlap?
Answer: 10
Solution:
A  r S  a
abc
2
220
S
2
S  110
1
3.5
  16.6
sin  
2  33.2
S
360
33.2
No. of smaller coins  10.84
No. of smaller coins 
say 10 coins to avoid overlap
A  r S  a
2310  55 110  a 
a  68m.
Mathematics
JRN/03152011
Solution: