IMO - Mock Test

Result

IMO - Mock Test - 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In a particular code language, TINKER is written as WNQPHW. How can RELIANT be written in the same code language?

A
UJONSDW

B
UJNODSW

C
UJONDSW

D
UJODSNW

Solution

For TINKER, T + 3 ---> W
I + 5 ---> N
N + 3 ---> Q
K + 5 ---> P
E + 3 ---> H
R + 5 ---> W
Similarly, for RELIANT
R + 3 --->U
E + 5 --->J
L + 3 --->O
I + 5 --->N
A + 3 --->D
N + 5 --->S
T + 3 --->W
Thus, the code is UJONDSW.
Question 2
1 / -0

If × is ÷, + is × , ÷ is + and - is -, then what will be the value of the expression 95 × 5 ÷ 2 - 7 + 2?

A
7

B
13

C
11

D
9

Solution

The given expression is 95 × 5 ÷ 2 - 7 + 2.

After substituting the signs, the expression becomes 95 ÷ 5 + 2 – 7 × 2 = 19 + 2 – 14 = 7
Question 3
1 / -0

Find out the missing number in the given series.

1, 5, 9, (?), 25, 37, 49

A
17

B
16

C
15

D
14

Solution

The given series is 1², 2² + 1, 3², 4² + 1, 5², 6² + 1 and 7².
So, the missing number is 4² + 1 = 16 + 1 = 17.
Question 4
1 / -0

In a school of 100 students, 24 play Football, 36 play Hockey and 9 play both Football and Hockey. How many students in the school have not enrolled in either Football or Hockey?

A
31

B
32

C
49

D
50

Solution

As there are 9 students who play both Football and Hockey, hence the remainder of students who play only football = 24 - 9 = 15
Remainder of students who play only Hockey = 36 - 9 = 27
Number of students who play either Hockey or Football = 15 + 27 + 9 = 51
Therefore, 100 - 51 = 49 students do not take part in either Hockey or Football.
Question 5
1 / -0

What number will come at the place of question mark?

16, 72, 192, 400, ?

A
592

B
600

C
720

D
800

Solution

4 × 2²= 16
8 × 3²= 72
12 × 4²= 192
16 × 5²= 400
20 × 6²= 720
Thus, the missing number is 720.
Question 6
1 / -0

What will come at the place of question mark?

7 : Heptagon : : ? : Tetradecagon

A
11

B
12

C
14

D
17

Solution

A 7 sided polygon is known as a heptagon.
Similarly, a 14 sided polygon is known as a tetradecagon.
Question 7
1 / -0

In a certain code language "REDDISH" is coded as TGFFKUJ. How will you code "CARROT" in that language?

A
ECTTQV

B
ECTTQN

C
ASDFGB

D
QWDFIH

Solution

We get each new letter by taking 2 jumps according to the alphabetical series.
For example, R + 2 = T, E + 2 = G and so on.
Going by the same logic, the code for CARROT is ECTTQV.
Question 8
1 / -0

Find the next term in the series.

12, 1, 15, 3, 18, 5, 21, ?

A
6

B
7

C
22

D
24

Solution

In the given series, two alternate series are there.

The first series follows a pattern:
12 + 3 = 15
15 + 3 = 18
18 + 3 = 21

The second series follows a pattern:
1 + 2 = 3
3 + 2 = 5
5 + 2 = 7

So, the next number will be 7.
Question 9
1 / -0

The sum of two positive numbers is 29 and the sum of their squares is one more than 420. Find the value of the bigger number.

A
14

B
15

C
16

D
25

Solution

Let one of the numbers be a and the other be b.
So, a + b = 29 ...(i)
a² + b² = 421 ...(ii)
a = 29 - b ...(iii)
Putting this value of a from equation (iii) into (ii), we get the following equation:
b² + (29 - b)² = 421 ...(iv)
Or, b = 14 or 15
If b = 14, we get a = 15 and vice versa.
Thus, the bigger number is 15.
Question 10
1 / -0

What will be the quadratic polynomial, if sum of zeroes is -2/5and product of zeros be 3?

A
5x² - 2x - 3

B
5x² - 2x + 3

C
5x² - 2x + 15

D
5x² + 2x + 15

Solution

Sum of zeros (S) = -2/5

Product of zeros (P) = 3

Quadratic polynomial will be k(x² - Sx + P), where k is LCM of denominator of S and P.

The quadratic equation is 5(x² - (-2/5) x + 3), where k = 5.
Hence, the quadratic equation is 5x² + 2x + 15.
Question 11
1 / -0

If 36y - 15xy + 32/16 x - 51y²= 0 and y = 3, then find the value of x.

A
- 8.16

B
- 6.24

C
8.16

D
51.2

Solution

If y = 3,

36y - 15xy + 32/16 x - 51y²= 0

108 - 45x + 32/16 x - 459 = 0

- 45x + 2x = 459 - 108

x = - 8.16
Question 12
1 / -0

The angles of the quadrilateral are in the ratio 15 : 9 : 11 : 5. Find out the measure of all four angles of quadrilateral.

A
139°, 80°, 92°, 45°

B
135°, 91°, 99°, 65°

C
135°, 81°, 99°, 45°

D
139°, 81°, 99°, 45°

Solution

Let the angles be 15x, 9x, 11x and 5x.
Applying angle sum property of a quadrilateral, we have
15x + 9x + 11x + 5x = 360^o
40x = 360^o
x = 360^o ÷ 40 = 9^o
Thus, the angles are
9^o× 15 = 135^o
9^o× 9 = 81^o
9^o× 11 = 99^o
9^o× 5 = 45^o
Question 13
1 / -0

Find the value of the expression (2x + 7y)² - (2x - 7y)².

A
56xy

B
28xy

C
98y²

D
8x²

Solution

(2x + 7y)² - (2x - 7y)²
= (2x + 7y + 2x - 7y)(2x + 7y - 2x + 7y)
= (4x)(14y)
= 56xy
Question 14
1 / -0

Two bus tickets from city X to Y and three tickets from city X to Z cost Rs. 77 but three tickets from city X to Y and two tickets from city X to Z cost Rs. 73. What are the fares for cities Y and Z from X, respectively?

A
Rs. 13 and Rs. 17

B
Rs. 17 and Rs. 13

C
Rs. 17 and Rs. 26

D
Rs. 26 and Rs. 17

Solution

Let Rs. x be the fare of city Y from city X and Rs. y be the fare of city Z from city X.
So, 2x + 3y = 77 as equation (i)

3x + 2y = 73 as equation (ii)

So, multiplying equation (i) by 3 and equation (ii) by 2 and then subtracting equation (ii) from equation (i), we get
5y = 85

So, y = 17 and now putting y = 17, we get x = 13.
Question 15
1 / -0

The point of the form (-a, a) always lies on

A
x + y = 0

B
x - y = 0

C
x - axis

D
y - axis

Solution

As -a = -(a), we have x = - y
Or x + y = 0
Question 16
1 / -0

How many tiles of length 30 cm and width 25 cm will be needed to cover the floor of length 20 m and width 15 m?

A
4

B
40

C
400

D
4000

Solution

Dimensions of the floor are 1500 cm x 2000 cm.
So, if the dimension 30 cm of the tile is kept parallel to the dimension 1500 cm of the floor, we shall have 1500/30 = 50 tiles in a row.
If the dimension 25 cm of the tile is kept parallel to the dimension 2000 cm of the floor, we shall have 2000/25 = 80 tiles in a row.
Thus, there shall be 50 rows of 80 tiles each.
Hence, the total number of tiles = 50 x 80 = 4000
Question 17
1 / -0

Which of the following can be the value of a + b, if a² - b² = 64 and a × b = 60?

A
10

B
16

C
20

D
60

Solution

a² - b² = 64 = 100 - 36 = 10² - 6²
Thus, a = 10 and b = 6 or a = -10 and b = -6, so that ab = 60.
Thus, a + b = 10 + 6 = 16 or -10 - 6 = -16
Question 18
1 / -0

Find the area of the new parallelogram whose length of base and its corresponding height are increased by 25%, if the area of a parallelogram PQRS is 80 cm²and its base is 4 cm.

A
100 cm²

B
120 cm²

C
125 cm²

D
150 cm²

Solution

Area of the parallelogram = base(b) × corresponding height(h)
Area = 4 × h
80 = 4 × h
h = 20 cm
Hence, b = 4 cm and h = 20 cm
After increasing the lengths by 25%,
25% of 4 = 1 cm
25% of 20 cm = 5 cm
Base of the new parallelogram = 5 cm
Height of the new parallelogram = 25 cm
Therefore, the area of the new parallelogram = 5 × 25 = 125 cm²
Question 19
1 / -0

Probability of an event can be

A
10/2

B
0

C
10/7

D
2

Solution

Probability of an event lies between 0 and 1.
Question 20
1 / -0

The enrollment of new students in xyz public school from the past 5 years are 1150, 1250, 1340, 1050, 2300. Find the range of the enrollment from the last five years.

A
3350

B
2550

C
1750

D
1250

Solution

Range = Highest value – Lowest value
Highest value = 2300
Lowest value = 1050
So, range = 2300 - 1050 = 1250
Question 21
1 / -0

Vicky is making necklaces. He has 18 red beads and 15 blue beads. If Vicky wants to make all the necklaces exactly the same with no beads left over, what is the greatest number of necklaces that he can make?

A
2

B
3

C
10

D
11

Solution

Number of red beads = 18 = 2 × 3 × 3
Number of blue beads = 15 = 3 × 5
Highest common factor is 3.
Number of necklaces made by 3 red beads each = 18/3 = 6
Number of necklaces made by 3 blue beads each = 15/3 = 5
So, the total number of necklaces = 6 + 5 = 11
Question 22
1 / -0

The perimeter of Sunil's rectangular plot is 62 m. If the length of the plot is increased by 2 m and the breadth is decreased by 1 m, then the area of the plot remains the same. Find the length of the plot.

A
20 m

B
11 m

C
22 m

D
14 m

Solution

Let length of the plot be x m and breadth of the plot be y m.
Perimeter = 62 m
2(x + y) = 62
(x + y) = x + y = 31
x = 31 - y ......(1)
Now, area = xy
Now, breadth is decreased by 1 m and length is increased by 2 m, and the area remains the same.
So, (x + 2)(y - 1) = xy
xy - x + 2y - 2 = xy
2y - x = 2 ..... (2)
Now put x = 31 - y in equation (2),
2y - (31 - y) = 2
2y - 31 + y = 2
3y - 33 = 0
y = 11
So, x = 31 - 11 = 20
Therefore, the length of the plot is 20 m.
Question 23
1 / -0

A container 2.5 m long, 3.5 m wide and 75 cm deep is to be made. It is opened at the top. Find:

(i) The area of the sheet required for making the container.
(ii) The cost of sheet for it, if a sheet measuring 1 m² costs Rs. 20.

A
(i) 17.75 m² (ii) Rs. 355

B
(i) 18.75 m² (ii) Rs. 375

C
(i) 17.75 m² (ii) Rs. 345

D
(i) 17.75 m² (ii) Rs. 455

Solution

Length of container (l) = 2.5 m
Width of container (b) = 3.5 m
Depth of container (h) = 0.75 m
(i) The area of sheet required to make the container is equal to the surface area of the container excluding the top.
Surface area of the container = Lateral surface area + Area of the base
= 2(l + b) × h + (l × b) = 2[(2.5 + 3.5) × 0.75] + (2.5 × 3.5) m²= (9 + 8.75) m²
= 17.75 m²
The sheet required required to make the container is 17.75 m².
(ii) Cost of 1 m² of sheet = Rs. 20
∴ Cost of 17.75 m² of sheet = Rs. (20 × 17.75) = Rs. 355
Question 24
1 / -0

A box contains 25 red marbles, 25 white marbles and 50 blue marbles. If one ball is drawn at random, find the probability of:

1. Getting a red marble
2. Getting a marble which is not red
3. Getting a white or red marble

A
1. 1/4,
2. 1/4,
3. 1/4

B
1. 1/4,
2. 3/4,
3. 1/2

C
1. 1/4,
2. 3/4,
3. 1/6

D
1. 1/4,
2. 1/2,
3. 1/2

Solution

Total outcomes = 100
Favourable outcomes for getting a red ball = 25

1. Probabilty of getting a red marble = 25/100 = 1/4

2. Probability for not getting a red marble = 1 - 1/4 = 3/4

3. Probability for getting a white or red marble = 1/4 + 1/4 = 1/2

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

IMO - Mock Test - 4

Result

IMO - Mock Test - 4

Score

-

Rank

-

SHARING IS CARING

Question 1

UJONSDW

UJNODSW

UJONDSW

UJODSNW

Solution

Question 2

7

13

11

9

Solution

Question 3

17

16

15

14

Solution

Question 4

31

32

49

50

Solution

Question 5

592

600

720

800

Solution

Question 6

11

12

14

17

Solution

Question 7

ECTTQV

ECTTQN

ASDFGB

QWDFIH

Solution

Question 8

6

7

22

24

Solution

Question 9

14

15

16

25

Solution

Question 10

5x2 - 2x - 3

5x2 - 2x + 3

5x2 - 2x + 15

5x2 + 2x + 15

Solution

Question 11

- 8.16

- 6.24

8.16

51.2

Solution

Question 12

139°, 80°, 92°, 45°

135°, 91°, 99°, 65°

135°, 81°, 99°, 45°

139°, 81°, 99°, 45°

Solution

5x² - 2x - 3

5x² - 2x + 3

5x² - 2x + 15

5x² + 2x + 15

98y²

8x²

100 cm²

120 cm²

125 cm²

150 cm²

(i) 17.75 m² (ii) Rs. 355

(i) 18.75 m² (ii) Rs. 375

(i) 17.75 m² (ii) Rs. 345

(i) 17.75 m² (ii) Rs. 455

1. 1/4,
2. 1/4,
3. 1/4

1. 1/4,
2. 3/4,
3. 1/2

1. 1/4,
2. 3/4,
3. 1/6