IMO - Mock Test

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IMO - Mock Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Study the following arrangement carefully and answer the question given below.

B M % R 3 J @ K © D F 6 9 W 4 * N E P 2 $ A Y 5 I Q Z # 7 U G

How many such consonants are there in the above arrangement, each of which is immediately preceded by a symbol and immediately followed by a number?

A
None

B
One

C
Two

D
Three

Solution

B M % R 3 J @ K © D F 6 9 W 4 N E P 2 $ A Y 5 I Q Z # 7 U G
There is only one such combination which satisfies the above condition, i.e. "% R 3".
There is only one consonant, i.e., R.
Question 2
1 / -0

Complete the below series:

89, 56, 98, 63, 116, 77, 143, ____

A
179

B
98

C
91

D
89

Solution

There are two series:
First series = 89 + 9 = 98 + 18 = 116 + 27 = 143
Another series = 56+7 = 63 + 14 = 77 + 21 = 98
Question 3
1 / -0

How many terms are there in the given series?

63, 61, 59, 57, ..........., 25.

A
12

B
18

C
16

D
20

Solution

63 + 61 + 59 + 57 + .....+25
a = 63, d = -2 , a_n = 25
Therefore, a + (n – 1)d = a_n

63 + (n – 1)(-2) = 25

63 - 2n +2 = 25
65 - 25 = 2n

n = 20
Question 4
1 / -0

The sum of the first four consecutive even numbers is 60. Find out the 2^nd consecutive even number.

A
14

B
12

C
10

D
8

Solution

2a + 2a + 2 + 2a + 4 + 2a + 6 = 60
8a + 12 = 60
8a = 60 – 12
8a = 48
a = 48/8= 6
2^nd consecutive even number = 2a + 2
= 2 × 6 + 2 = 14
Question 5
1 / -0

The sum of two equal sides of an isosceles triangle is 5 times the third side. If the perimeter is 60 cm, then find the biggest side of the triangle.

A
6 cm

B
10 cm

C
15 cm

D
20 cm

Solution

Let two equal sides be x.
Let the third bigger side be y.
Now,
2x = 5y
Perimeter = x + x + y = 60 cm
2x + y = 60
5y + y = 60
6y = 60
y = 10 cm
Question 6
1 / -0

If z = x, y = w and u = v, then which of the following relationships is true?

A
z + y + u = x + w + v

B
z + z = x + v + y

C
z - x - y = w + u + v

D
z + y = w + v

Solution

z = x (Given) --1
y = w (Given)--2
u = v (Given)--3

By adding 1, 2 and 3, we get
z + y + u = x + w + v
Question 7
1 / -0

In a pack of 52 cards, what is the probability of getting an ace and a jack without replacement?

A
4/663

B
16/663

C
1/4

D
24/663

Solution

Number of aces in a pack = 4
The probability of getting an ace = 4/52
Probability of getting a jack out of 51 = 4/51
Now, the probability of getting an ace and a jack = 4/52 × 4/51
= 4/663
Question 8
1 / -0

What will be the 45^th term of an A.P., if it consists of 73 terms and the first and the last terms are 1 and 145, respectively?

A
115

B
101

C
99

D
89

Solution

Given: n = 73, a₁ = 1
So, a₁ + 72d = 145
1 + 72d = 145
72d = 144
d = 144 / 72
d = 2
a₄₅ = a₁+ 44d
= 1 + 44(2)
= 1 + 88
= 89
So, a₄₅ = 89
Question 9
1 / -0

In a car-bazaar, the total number of cars is 79. The percentage of white cars is 29.65%. How many cars in the car-bazaar are white?

A
23

B
25

C
27

D
29

Solution

29.65% = 0.2965
Approximate number of white cars in the car-bazaar = 0.2965 × 79 = 23.4235
Therefore, there are approximately 23 white cars in the car-bazaar.
Question 10
1 / -0

If an A.P. consists of 55 terms and if the first and last terms are 13 and 175, respectively, then determine the 49^th term.

A
97

B
109

C
123

D
157

Solution

Given: n = 55, a₁ = 13
and a₅₅ = 175
So, a₁ + 54d = 175
13 + 54d = 175
54d = 162
d = 162 / 54
d = 3
a₄₉ = a₁ + 48d = 13 + 48(3) = 13 + 144 = 157
So, a₄₉ = 157
Question 11
1 / -0

In a shop, 20% of the fruits are apples, 50% of the remaining are bananas and 30% of the remaining are grapes. The remaining 6,300 fruits are of different types. What is the total number of fruits in the shop?

A
22,000

B
22,500

C
23,000

D
23,500

Solution

20% of these fruits are apples = 0.2x apples
Therefore the remaining fruits = x − 0.2x = 0.8x fruits
50% of these remaining fruits are bananas = 50% of 0.8x = 0.4x fruits
Therefore, the remaining fruits = 0.8x − 0.4x = 0.4x fruits
30% of these remaining fruits are grapes = 30% of 0.4x = 0.12x fruits
Now the remaining fruits = 0.4x − 0.12x = 0.28x fruits
According to the question, the number of remaining fruits = 6300
0.28x = 6300

x = 6300/0.28
x = 22,500 fruits
Question 12
1 / -0

In an Art and Craft class, Aman creates 2 triangles with a cardboard. The base of one of the triangles is 15 cm and its height is 12 cm. What will be the height of the other triangle if its area is double the area of the first triangle and its base is 20 cm?

A
12 cm

B
16 cm

C
18 cm

D
19 cm

Solution

Area of the first triangle = 1/2x base x height
= 1/2x 15 x 12 = 90 cm²
Area of the second triangle = 2 x area of the first triangle
180 = 1/2x 20 x Height
Height = 18 cm
Question 13
1 / -0

The point which lies on X-axis at a distance of 6 units in the negative direction is

A
(0, 6)

B
(-6, 0)

C
(0, -6)

D
(6, 0)

Solution

The point which lies on X-axis at a distance of 6 units in the negative direction is (-6), which is the x-coordinate.
Now, looking at the options, no other option has its x-coordinate as (-6), except option 2.
Question 14
1 / -0

John builds a hall whose length, breadth and height are 11 m, 14 m and 12 m, respectively. What will be the total cost of painting the walls and the ceiling of the room if the cost of painting is Rs. 8.25 per m² ?

A
Rs. 6,220

B
Rs. 7,641

C
Rs. 10,080

D
Rs. 15,246

Solution

Surface area = 2(lb + bh + lh)
= 2(11 × 14 + 14 × 12 + 11 × 12)
= 2(154 + 168 + 132)
= 2(454)
= 908 m²
Area of (walls + ceiling) = 908 - Area of the floor
= 908 - 11 x 14
= 754
Total cost of painting = 776 × 8.25
= Rs. 6,220.5 or Rs 6220
Question 15
1 / -0

Two dice are thrown simultaneously. Find the sum of the probability of "getting a prime number as the sum" and probability of "getting a doublet of prime numbers".

A
1/2

B
2/3

C
5/4

D
None of these

Solution

Total number of outcomes = 6 × 6 = 36
Outcomes for getting a prime numbers as the sum are:
(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1) and (6, 5) = 15 outcomes
So, the probability of getting a prime number as the sum = 15/36
Further, prime doublets are (2, 2), (3, 3), (5, 5)
So, probability of getting a doublet of prime numbers = 3/36
Now, required probability = 15/36 + 3/36 = 1/2
Question 16
1 / -0

A Hula-Hoop with a radius of 35 cm is rolled on ground. To cover a distance of 110 m, how many revolutions should it make?

A
36

B
40

C
50

D
55

Solution

Radius = 35 cm
Diameter (d) = 35 x 2 = 70 cm
The circumference of the Hula-Hoop = πd
= 22/7x 70= 220 cm
The Hula-Hoop will cover a distance of 220 cm by revolving once.
The distance to be covered = 110 m = 11,000 cm
Therefore, number of times the hoop has to revolve to cover 110 m = 11000/220 = 50
Question 17
1 / -0

In a parallelogram ABCD, the base is twice the height, and the area is 338 cm². Find out the height of the parallelogram.

A
13 cm

B
18 cm

C
22 cm

D
26 cm

Solution

Area of a parallelogram = Base × Height
Base = 2(Height)
2h × h = 338
2h² = 338
h² = 338/2
h² = 169
h = √169
h = 13 cm

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Re-Attempt Weekly Quiz Competition

IMO - Mock Test - 3

Result

IMO - Mock Test - 3

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SHARING IS CARING

Question 1

None

One

Two

Three

Solution

Question 2

179

98

91

89

Solution

Question 3

12

18

16

20

Solution

Question 4

14

12

10

8

Solution

Question 5

6 cm

10 cm

15 cm

20 cm

Solution

Question 6

z + y + u = x + w + v

z + z = x + v + y

z - x - y = w + u + v

z + y = w + v

Solution

Question 7

4/663

16/663

1/4

24/663

Solution

Question 8

115

101

99

89

Solution

Question 9

23

25

27

29

Solution

Question 10

97

109

123

157

Solution

Question 11

22,000

22,500

23,000

23,500

Solution

Question 12

12 cm

16 cm

18 cm

19 cm

Solution