IMO - Mock Test

Result

IMO - Mock Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

In the following question, there is a certain relationship between the groups of letters to the left of : : and the same relationship holds between the groups of letters to its right, of which one group is missing. One of the four given options contains the missing group. Identify the correct option.

F H J : V X Z : : D F H : _____

A
I K M

B
T U X

C
T V W

D
T V X

Solution

F H J : V X Z : : D F H : T V X

This logic can be explained as follows:

F : (F + 16) = V
H : (H + 16) = X
J : (J + 16) = Z

Similarly,

D : (D + 16) = T
F : (F + 16) = V
H : (H + 16) = X
Question 2
1 / -0

Choose the missing term out of the given alternatives

BD, EI, HN, ?, NX

A
JR

B
KR

C
KS

D
LS

Solution

In each pair of letters, the first letter moves 3 steps forward and the second letter moves 5 steps forward.
BD, (B + 3)(D + 5) = EI, (E + 3)(I + 5) = HN, (H + 3)(N + 5) = KS, (K + 3)(N + 5) = NX
Question 3
1 / -0

Find the next number in the series:

8, 4.5, 13, 9.5, 18, 14.5, ?

A
28

B
20.5

C
19.5

D
23

Solution

There are 2 series hidden.
8 + 5 = 13
13 + 5 = 18
18 + 5 = 23; which is the missing number.
Also, 4.5 + 5 = 9.5
9.5 + 5 = 14.5, and so on
Question 4
1 / -0

Find the missing term in the series:

6857, 6863, _____, 6871, 6883

A
6869

B
6867

C
6865

D
6870

Solution

The numbers given in the series above are prime numbers. So, the missing prime number which comes in between 6863 and 6871 is 6869.
Question 5
1 / -0

Six friends P, Q, R, S, T and U went for a movie and got tickets in last row facing north. T and Q were sitting in the centre seats. R and P were sitting at the extreme ends of their row. U was sitting to the immediate left of T. Who was sitting to the immediate right of Q?

A
T

B
S

C
R

D
U

Solution

From the information given, the following arrangement can be made:

R/P U T Q S R/P

So, from the above arrangement it is clear that S is to the immediate right of Q.
Question 6
1 / -0

Rearrange the given letters below to make a word and then choose the group to which it belongs.

S R O E

A
Flower

B
City

C
Animal

D
Fruit

Solution

Rearranging the given letters, we can make a word called ROSE, which is a flower.
Question 7
1 / -0

If A is ÷, B is ×, C is +, and D is -, then what will be the solution for 35 C 15 D 10 A 5 B 2?

A
46

B
16

C
52

D
48

Solution

35 C 15 D 10 A 5 B 2
After substituting the letters with signs, 35 + 15 – 10 ÷ 5 × 2 = 35 + 15 – 2 × 2 = 35 + 15 – 4 = 50 - 4 = 4
Question 8
1 / -0

Sachin takes a number 4 and keeps it multiplying by 2 for nine times. What will be his answer?

A
0124

B
512

C
2048

D
1204

Solution

Here, the given problem can be written as:
2² x 2⁹ = 2¹¹ = 2048
Question 9
1 / -0

The abscissa of any point on y-axis is

A
0

B
1

C
-1

D
any number

Solution

All the x values will be zero on y-axis.
Question 10
1 / -0

How many natural pairs of x and y can satisfy the equation x - 3y = 0, where x and y are less than 15?

A
5

B
4

C
3

D
Infinite pairs can be there.

Solution

x - 3y = 0
Or x = 3y
When y = 0, then x = 0
One pair is (0, 0).
Similarly, putting the value of y as 1, 2, 3 and 4, we get the values of x as 3, 6, 9 and 12, respectively.
So, other pairs are (3, 1); (6, 2); (9, 3) and (12, 4).
Hence, the number of required natural pairs is 4.
Question 11
1 / -0

The sum of four consecutive integers is 5650. Which is the greatest of these integers?

A
1411

B
1414

C
1418

D
1420

Solution

Let us assume the 1^st integer to be n.

2^ndinteger: n+ 1

3^rdinteger: n+ 2

4^thinteger : n + 3

So, n + n + 1 + n + 2 + n + 3 = 4n + 6 = 5650

Or n = 1411

So, largest number = 1411 + 3 = 1414
Question 12
1 / -0

The marks of 21 students are given below:

Scores Frequency

91-100 9

81-90 6

71-80 3

61-70 0

51-60 2

41-50 1

Find the interval that contains the median.

A
41-50

B
91-100

C
71-80

D
81-90

Solution

The median marks are the marks of the 11^thperson from either end.
The 11^th person has scored marks between 81 and 90.
Thus, the desired interval is 81 - 90.
Question 13
1 / -0

Using the method of factorisation, find which of the following is/are the root(s) of the quadratic equation:

(2x + 3)²= 81

A
Only 3

B
3 and - 6

C
Only - 3

D
- 3 and 6

Solution

Step-1

(2x + 3)²= 81

Step-2

(2x + 3)²– (9)²= 0

Step-3

(2x + 3 - 9) (2x + 3 + 9) = 0

Step-4

(2x - 6) (2x + 12) = 0

Step-5

x = 3 , x = -6
Question 14
1 / -0

If x⁴⁷ + b⁴⁷ is divided by x + b, then what will be the remainder?

A
1

B
b⁴⁷

C
2b⁴⁷

D
0

Solution

p(x) = x⁴⁷ + b⁴⁷ is divided by q(x) = x + b.
So, remainder = p(–b)
Remainder = (–b)⁴⁷ + b⁴⁷ = –b⁴⁷ + b⁴⁷ = 0
Question 15
1 / -0

Heights of 10 students are given. Find the range.

119 cm, 118 cm, 115 cm, 111 cm, 114 cm, 118 cm, 120 cm, 117 cm, 110 cm, and 113 cm.

A
1 cm

B
3 cm

C
6 cm

D
10 cm

Solution

Range = Highest value - Lowest value= 120 cm - 110 cm = 10 cm
Question 16
1 / -0

Two numbers, when added, give 26 and when subtracted, give 10. Find the product of these two numbers?

A
26

B
214

C
124

D
144

Solution

Let one number be m.

Let the other number be n.

According to the question,

m + n = 26

m – n = 10

By solving these two equations,

m = 18

n = 8

Product = 18 x 8 = 144
Question 17
1 / -0

Convert the following decimal number into fractions in a simple form.

.244244244…

A
154/568

B
254/523

C
244/999

D
1000/225

Solution

Let p/q = 0.244244244.. ...(i)
Multiplying both sides by 1000, we get
1000 × p/q = 10 × 0.244244244..
1000 × p/q = 244.244244... ...(ii)
Subtracting (i) from (ii), we get
999 × p/q = 244
p/q = 244/999
Question 18
1 / -0

A car tyre makes 3,000 revolutions in a minute. How many degrees does it rotate through in one second?

A
18,000 degrees/second

B
20,000 degrees/second

C
20,433 degrees/second

D
30,787 degrees/second

Solution

3000 revolutions in 1 minute
Degrees rotated in one revolution = 360^o
Degrees rotated in 3000 revolutions = 360 x 3000 degrees = 10,80,000 degrees
Degrees rotated in one second = 1080000/60 degrees = 18,000 degrees
Question 19
1 / -0

In a Marketing class, 30 students like to do shopping on Amazon, and 17 students like to do shopping on Flip-kart. There are 10 students who like both the shopping websites. How many total students are there, if the data given above includes all the students?

A
40

B
37

C
31

D
23

Solution

Total number of students in class = Number of students who like Amazon + Number of students who like Flip-kart - Number of students who like both the shopping websites
= 30 + 17 – 10 = 37
Question 20
1 / -0

Chahat has a quadrilateral box whose angles are in the ratio 4 : 7 : 11 : 8. Which of the following are the angles of that box?

A
48º, 84º, 132º, 96º

B
44º, 77º, 111º, 88º

C
36º, 63º, 99º, 72º

D
24º, 42º, 66º, 48º

Solution

Let the common ratio between the angles be 'z'.

4z + 7z + 11z + 8z = 360º

30z = 360º

z = 12º

4z = 48º, 7z = 84º, 11z = 132º, 8z = 96º
Question 21
1 / -0

Anita is three years older than Bani while Bani is twice as old as Chandler. If the total of the ages of Anita, Bani and Chandler is 33, then how old is Bani?

A
14 years

B
12 years

C
11 years

D
10 years

Solution

Let Chandler's age be x years. Then, Bani's age = 2x years. Anita's age = (2x + 3) years

(2x + 3) + 2x + x = 33

5x = 30

x = 6

Hence, Bani's age = 2x = 12 years
Question 22
1 / -0

A and B have certain numbers of stamps. A says to B, ''If you give me one of your stamps, then we will have equal numbers of stamps." B replies, ''If you give me one of your stamps, then I will have twice as many as you will be left with''. Find the total number of the stamps that A and B have.

A
20

B
18

C
15

D
12

Solution

Let the number of the stamps with A be A and that with B be B.
So, A + 1 = B - 1 ...(i)
Also, B + 1 = 2(A - 1)...(ii)
Adding (i) and (ii), we get
A + B + 2 = B - 1 + A + A - 2
Or, A = 5
So, 6 = B - 1 or B = 7
Thus, A + B = 5 + 7 = 12
Question 23
1 / -0

Anuj bought 8 packets of rice. The weight marked was 1.5 kg but it actually contained the following weights (in kg):

1.56, 1.6, 1.31, 1.37, 1.49, 1.54, 1.45, 1.2

Find the probability that any of these bags, chosen at random, contained more than 1.5 kg of rice.

A
0.1875

B
0.375

C
0.5875

D
0.75

Solution

Total number of bags of rice = 8

Number of bags of rice that contained more than 1.5 kg of sugar = 3

(1.56, 1.6, 1.54)

Probability = Favourable outcomes/Total outcomes = 3/8 = 0.375
Question 24
1 / -0

5 years ago, the mean age of Cena's family of 4 was 20. If one member of the family dies today, then the average age of the family remains same today. At what age does the person expire?

A
40 years

B
20 years

C
60 years

D
15 years

Solution

The sum of their ages was 4 × 20 = 80 years

A member dies (meaning that there are now 3 family members)
We're told that the present average age of the family members remains the same.

This means that the sum of the ages is 3 × 20 = 60 years

We're asked about the present age of the dead member.

Since each of the original 4 family members has aged 5 years since the initial average was calculated, the sum of their present ages is 80 + 5 × 4 = 100

Since the current sum of the ages of 3 family members is 60, the age of the dead person = 100- 60 = 40 years old

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Re-Attempt Weekly Quiz Competition

Scores	Frequency
91-100	9
81-90	6
71-80	3
61-70	0
51-60	2
41-50	1

IMO - Mock Test - 3

Result

IMO - Mock Test - 3

Score

-

Rank

-

SHARING IS CARING

Question 1

I K M

T U X

T V W

T V X

Solution

Question 2

JR

KR

KS

LS

Solution

Question 3

28

20.5

19.5

23

Solution

Question 4

6869

6867

6865

6870

Solution

Question 5

T

S

R

U

Solution

Question 6

Flower

City

Animal

Fruit

Solution

Question 7

46

16

52

48

Solution

Question 8

0124

512

2048

1204

Solution

Question 9

0

1

-1

any number

Solution

Question 10

5

4

3

Infinite pairs can be there.

Solution

Question 11

1411

1414

1418

1420

Solution

Question 12

41-50

91-100

71-80

81-90

Solution

b⁴⁷

2b⁴⁷