UNIFIED COUNCIL
A n I S O 9 0 0 1 : 2008 C e r t i f i e d O r g a n i s a t i o n
UNIFIED CYBER OLYMPIAD - 2013 (LEVEL - 1) – (UPDATED)
Solutions for class : 10
Mental Ability
1.
2.
(C)
(A)
Of the given options, the triangles shown
in option (C) are not similar, as their
corresponding sides are not proportional.
⇒
22
216o 22
×
× r × 15 =
× (15)2
360o 7
7
⇒ r =9
Let the present ages of the father and the
son be ‘x’ years and ‘y’ years respectively.
∴ h=
l2 − r2 = 12
Five years ago, their ages
(x – 5) years and (y – 5) years.
∴V =
1 22
1 2
× 9 × 9 × 12
πr h = ×
3 7
3
were
According to the problem,
= 1018.3 cm3
(x – 5) = 7(y – 5)
(B)
⇒ x − 5 + 35 − 7y = 0
Since there are 200 plates,
n(S) = 200.
⇒ x − 7y + 30 = 0
Let B be the event of picking a defective
plate. Then, n(B) = 10.
⇒ x = 7y − 30
4.
...... (1)
The probability of picking a defective plate
Five years later, their ages will be (x + 5)
years and (y + 5) years.
According to the problem,
= P(B) =
5.
(D)
Area of first triangle
(x + 5) = 3(y + 5)
=
⇒ x + 5 = 3y + 15
⇒ x = 3y + 15 – 5
⇒ x = 3y + 10
=
7y – 30 = 3y + 10
1
× (2k − 2) × k = k ( k − 1)
2
Area of third triangle
⇒ 4y = 40
=
⇒ y = 10
⇒ x = 7y – 30
1
× ( 2k − 4 ) × k = k ( k − 2)
2
Common difference (d)
= 7(10) – 30
= k 2 − k − k 2 = −k
= 40
(or) = k 2 − 2k − k 2 + k = − k
∴ The present ages of the father and
the son are 40 years and 10 years
respectively.
(C)
1
× 2k × k = k 2
2
Area of second triangle
...... (2)
From (1) and (2), we have
3.
n(B)
10
=
= 0.05
n(S)
200
Sn =
C.S.A. = Area of the sector with angle 216o
S6 =
πR2 θ
⇒ πrl =
360
n
2a + (n − 1 ) d 
2
6
2k 2 + 5 ( − k )
2
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10th_UCO_I_SOL.
6.
(B)
OP = h, PQ = e and OQ =
= 3k (2k – 5)
∴ By Pythagoras’ theorem,
Let α, β and γ be the zeros of the required
polynomial.
 2 
h = e −
m
 2



1
α+β+γ =
4
αβ + βγ + γα =
⇒ h = m2 −
2 2
m
4
= m2 −
1 2
m
2
−3
9
and αβγ =
.
2
16
=
∴ The required polynomial
f (x) = x 3 − (α + β + γ ) x 2 + (αβ + βγ + γα ) x − αβγ
1 2 3
9
3
⇒ f (x) = x − x − x −
4
2
16
=
10. (A)
⇒ f (x ) = 16x 3 − 4x 2 − 24x − 9
∀ k ∈ R.]
8.
(D)
[Given e = m.]
1 2
m
2
m
2
Let OP be the tower of height h m and PQ
be the flag staff of height 6 m. Let the sun
make an angle θ with the ground. Let OA
= x m and AB = 2 3 m be the shadows of
the tower and the flag staff, respectively.
[Since, kf(x) has the same zeros as f(x)
(D)
2
2
Then according to the problem,
7.
2
m.
2
= 3  2k 2 − 5k 
Q
sin 2 θ − cos 2 θ = 1 is not true for all values
of θ . Hence it is not an identity.
6m
210 = 2 × 3 × 5 × 7
P
55 = 5 × 11
∴ L.C.M. = 2 × 3 × 5 × 7 × 11
h
= 2310
H.C.F. = 5
B
Difference of L.C.M and H.C.F
= 2310 – 5 = 2305
2305 − 210 × 6
= 19
55
Also,
∴ y 3 = 193 = 6859
9.
(C)
∴
Side of the square base of the given
pyramid = m
⇒ Its diagonal =
h
x
h+6
x+2 3
= tan θ
h
h+6
=
x x+2 3
⇒
e
o
O
⇒ 2 3h = 6x
p
m
xm
⇒ hx + 2 3h = hx + 6x
2m2
= 2m
h
A
Now, tan θ =
Given 2305 = 210 × 6 + 55y
⇒ y=
2 3m
h
6
=
= 3
x 2 3
⇒ tan θ = 3
Q
∴ θ = 60°
In the right triangle OPQ,
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10th_UCO_I_SOL.
11. (A)
Let the digits in the tens place and the ones
place be x and y respectively.
6x +
Then, according to the problem,
2
y+7 = 0
3
Here a1 = 2, b1 = k, c1 = −3 and
10x + y = 4(x + y) ⇒ y = 2x
a 2 = 6, b 2 =
and 10x + y = xy + 16
⇒ 10x + 2x = x(2x) + 16
⇒ 12x = 2x + 16
a1 b1
≠
a 2 b2
⇒ 2x – 12x + 16 = 0
2
⇒ x2 – 6x + 8 = 0
⇒ x2 – 4x – 2x + 8 = 0
⇒
2
k
≠
2
6
3
⇒
k≠
2 2
×
6 3
∴
k≠
2
9
⇒ x(x – 4) – 2(x – 4) = 0
⇒ (x – 4) (x – 2) = 0
⇒ x = 4 or 2
If x = 2, y = 2x = 4.
∴ The number is 24.
15. (C)
If x = 4, y = 8.
Radius of circle = 28 cm ÷ 2 = 14 cm
∴ The number is 48.
Perimeter of shaded part of figure
Hence, the required number among the
given options is 48.
 1 22

× 2 × 7 cm 
= 7 cm + 7 cm + 3 ×  ×
2
7


BC = 5 cm, (By Pythagoras' theorem)
= 14 cm + 66 cm
BP = x cm = RB
area
(∆ OBC )
1
× 12 × 5
2
⇒
D
F
E
K
U
LMNOPQRST
GHIJ
17. (C)
Difference between the digits of 37 is 7 – 3
= 4. In the others, this rule is not satisfied.
18. (D)
7 2 − 52 = 24 22 2 − 20 2 = 84
52 − 32 = 16 112 − 92 = 40
5
13
12
x+
x+
x = 30
2
2
2
62 − 42 = 20
92 − 72 = 32
2
 5 + 13 + 12 
⇒ 
 x = 30

2
⇒
Reasoning
16. (C)
1
1
1
⇒
× 5 × x + × 13 × x + × 12 × x
2
2
2
=
= 80 cm
+ area ( ∆ OCA ) + area
(∆ AOB ) = area (∆ ABC )
2
32 − 12 = 8 10 − 8 = 36
19. (C)
30
x = 30
2
20. (D)
⇒ x = 2 cm
13. (C)
2
, c2 = 7 .
3
For a unique solution,
2
12. (C)
..... (2)
GEYAAWT – GETAWAY means ‘a quick
departure’.
E8t4e9C
Let the mean of 50 observations be ‘x’. Then,
50 × x – 45 = (50 – 1) × x
21. (A)
⇒ x = 45
14. (D)
Given linear equations are
2x + ky – 3 = 0
..... (1)
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10th_UCO_I_SOL.
22. (D)
Vertex of the triangle in the middle square
is facing opposite to the other two.
Q
23. (C)
S
QPO
SRQ
R
24. (C)
U
P
W
UTS
T
R
V
YXW
WVU
X
T
28. (A)
V
4 13
7 6
5
9 8
12
10
11
The central line disappears and reappears
in alternate figures. In each step, half of the
outer figure disappears after a rotation of
90o.
(3 )(4 )
2
(2 )(6 )
2
2
2
(1 )(5 ) = 125
3
4
2
6
1
2
30. (C)
2 + 3, 4 + 7, 7 + 6, 5 + 12, 10 + 8, 9 + 8
→ 6
2 + 3 + 4 + 7, 4 + 7 + 8 + 9, 6 + 7 + 8 + 10,
7 + 6 + 5 + 12 → 4
2 + 3 + 4 + 7 + 13 + 6 + 5 + 12 → 1
∴ Total number of triangles
Jagat is the brother of Priya and Priya is
the daughter of Shyam. Therefore Shyam
is the father of Jagat. Ram is the brother of
Shyam. Therefore, Ram is the uncle of
Jagat.
Mahesh
Father
Shyam
Ram
Son
Brother
Priya Brother
Uncle
Shapes in the right diagonal interchange and
the two shapes in the left top corner and
right bottom corner interchange places and
top corner gets a new shape.
The opposite faces with the shaded parts of
the square faces have only one side painted.
31. (C)
IDE stands for Integrated Development
Environment
32. (B)
‘&’ is used to concatenate text.
33. (A)
!((1||0)&&!(1||1))
!(1)&&!(1) = !(1&&0) = !(0) = 1 = True
34. (D)
The first microprocessor was developed by
Intel.
35. (A)
Wi-Fi stands for Wireless Fidelity.
36. (A)
The output for the given HTML code is
1. Area
2. Perimeter
37. (C)
XML stands for eXtension Markup
Language.
38. (A)
The faded image behind the text in slides is
called ‘Watermark’.
39. (C)
.mdb is the extension of database files in
MS-Access.
Daughter
Jagat
5
Computers
= 12 + 6 + 4 + 4 + 1 = 27
Hence, there are 27 triangles.
2
Of the 27 smaller cubes, there are only 2
such cubes with only 1 face painted red as
only 2 faces of the larger cube are painted
red.
1 + 2 + 3, 8 + 10 + 11, 5 + 11 + 12, 1 + 8 + 9
→ 4
26. (B)
is RG.
29. (D)
Small triangles → 12
25. (C)
Code for white fill is G, and code for hexagon
is R.
Hence, the code for
2 3
1
27. (D)
40. (B)
1
2
b
d
3
e
4
a
5
c
41. (A)
Downloading refers to the transfer of files
from internet to the computer in use.
42. (B)
An operating system is the most essential
for the effective and efficient running of a
computer.
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10th_UCO_I_SOL.
43. (D)
> = 1000 AND < = 1200 represents values
in the range of 1000 and 1200.
44. (B)
The output of the given c++ program is 5.
47. (B)
The boys in the first row have no books.
45. (A)
Information technology is the application
that deals with computers and
telecommunication equipment to store,
retrieve and transmit data.
48. (C)
Rile = To irritate
49. (D)
‘Assistence’ is spelt incorrectly. The correct
spelling is ‘Assistance’.
50. (A)
He is zealous for freedom is the correct
statement.
English
46. (A)
Controversy, in which the second is
contemporary.
The alphabetical order of the given words
is Conceive – Contemporary – Contrive –
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10th_UCO_I_SOL.