IMO - Mock Test

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IMO - Mock Test - 9

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Weekly Quiz Competition

Question 1
1.2 / -0

The given equations are solved on the basis of a certain system. On the same basis, find out the correct answer from the given options.

If 7 × 9 = 306, 12 × 8 = 609 and 16 × 4 = 406, then 19 × 5 = ?

A
905

B
506

C
509

D
605

Solution

For 7 × 9, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 306.
For 12 × 8, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 609.
For 16 × 4, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 406.
For 19 × 5, the answer is 95, but we have to write the answer in reverse order and add a 0 in between the numbers. So, answer = 509
Question 2
1.2 / -0

If A is ÷, B is ×, C is + and D is -, then which of the following statements is correct?

1. 1 C 8 D 16 B 20 A 4 = - 71
2. 1 C 8 D 16 B 20 A 4 = - 35
3. 1 C 8 D 16 B 20 A 4 = - 110
4. 1 C 8 D 16 B 20 A 4 = - 91

A
1

B
2

C
3

D
4

Solution

1 C 8 D 16 B 20 A 4 = 1 + 8 – 16 × 20 ÷ 4 = 1 + 8 – 320 ÷ 4 = 1 + 8 – 80 = 9 – 80 = - 71
Question 3
1.2 / -0

From the given alternative words, select the word which cannot be formed using the letters of the given word.

THERMODYNAMICS

A
MATHEMATICS

B
MOTHER

C
MODERN

D
DYNAMO

Solution

There are only one A and T in the given word. Therefore, the word MATHEMATICS cannot be formed.

T H E R M O D Y N A M I C S
⇒ M O T H E R

T H E R M O D Y N A M I C S
⇒ M O D E R N

T H E R M O D Y N A M I C S
⇒ D Y N A M O
Question 4
1.2 / -0

If all the letters are dropped, then which among the following would be 7^th from right?

8 C # 4 & 7 7 L & # 7 K L ! T ? P 4 2 W %

A
#

B
&

C
7

D
!

Solution

8 C # 4 & 7 7 L & # 7 K L ! T ? P 4 2 W %
After removing all the letters,
8 # 4 & 7 7 & # 7 ! ? 4 2 %
So, the right answer is #, which is 7^th from right.
Question 5
1.2 / -0

In a certain code language, 415 means 'milk is hot', 18 means 'hot soup' and 895 means 'soup is tasty'. Which number will indicate the word 'tasty'?

A
9

B
8

C
1

D
5

Solution

Neither 9 nor the word 'tasty' is repeated. Hence, the code for 'tasty' is 9.
Question 6
1.2 / -0

How many triangles are there in the following figure?

A
29

B
23

C
25

D
21

Solution

ΔGFH, ΔQEF, ΔHIN, ΔCNM, ΔERJ, ΔRJB, ΔJBL, ΔJKL, ΔIJK, ΔEIJ, ΔAGE, ΔGEI, ΔGCI, ΔCIO, ΔEBJ, ΔIJL, ΔIEB, ΔILB, ΔIEL, ΔEBL, ΔCJP, ΔEGL, ΔBQN, ΔCBD, ΔABC
25 triangles can be seen in the figure above. Hence, option 3 is the correct one.
Question 7
1.2 / -0

Find the odd one out.

BC125A, HI216G, PQ343O, GH242F

A
GH242F

B
PQ343O

C
HI216G

D
BC125A

Solution

The number in every term of the series, except GH242F, is the cube of 5, 6 and 7, respectively. But 242 is not a cube; hence, this is the answer.
Question 8
1.2 / -0

Pointing towards Rita, Nikhil said, "I am the only son of her mother's son." How is Rita related to Nikhil?

A
Aunt

B
Niece

C
Mother

D
Cousin

Solution

Rita is the aunt of Nikhil. Hence, option (1) is correct.
Question 9
1.2 / -0

Choose the cube that will be formed on folding the given question figure.

A
(1)

B
(2)

C
(3)

D
(4)

Solution

In the given figure, block A will always be on the left of block F. This is visible in option 3 only, as follows.
Question 10
1.2 / -0

Which of the following would be the correct answer for the missing numbers in the pattern?

A
121, 2197, 15

B
131, 2197, 15

C
121, 2195, 25

D
121, 2197, 25

Solution

All the numbers are moving from ones, squares and last into the cubes as follows:

The second line (inside the box) is the square of the numbers, which are below the box.
Question 11
1.2 / -0

Out the given options, choose the correct one to replace the question mark (?).

ACEG : BDFH : : IKMO : ?

A
JLNP

B
KLNP

C
JLNO

D
JLMN

Solution

Each letter is skipped in between the pairs of each letter-group.
Like in group ACEG, B is skipped between A and C, and F is skipped between E and G.
Similarly, for IKMO, it would be JLNP.
Question 12
1.2 / -0

Study the following figure and answer the question given below.

How many uneducated people are married?

A
8

B
9

C
10

D
11

Solution

Number of uneducated people who are married = 3 + 6 = 9
Question 13
1.2 / -0

Complete the following number series:

5, 26, 65, 122, 197, __

A
290

B
280

C
272

D
265

Solution

1 × 2 + 3 = 2 + 3 = 5
4 × 5 + 6 = 26
7 × 8 + 9 = 65
10 × 11 + 12 = 122
13 × 14 + 15 = 197
16 × 17 + 18 = 290
Question 14
1.2 / -0

Find the missing number in the given pattern:

A
8

B
11

C
6

D
2

Solution

The circle is divided into four parts and numbers coming in that part are deducted, and the square root of the result is calculated.
= 10;
= 4;
= 5.
Missing number = = 8
Question 15
1.2 / -0

Arrange the given words in the sequence in which they occur in the dictionary and then choose the correct sequence.

(a) Departmentalisation
(b) Disparages
(c) Disparage
(d) Disparts

A
a, c, b, d

B
a, b, c, d

C
b, c, d, a

D
a, d, c, b

Solution

According to alphabetic order, they will appear as follow:
(a) Departmentalisation
(c) Disparage
(b) Disparages
(d) Disparts
Question 16
1.2 / -0

What is the value of x in the given equation?

A
- 5

B
1

C
5

D
7

Solution

As the fractions are the same, so we are going to use the given formula on LHS.

a^m + aⁿ = a^{(m + n)}Here, m = 5 and n = - 6

Or

Or

As the bases are the same, powers can be compared.

So, - 1 = 1 - 2x
Or 2x = 1 + 1
Or 2x = 2
Or x = = 1
Question 17
1.2 / -0

If ΔVWX ~ ΔHIG, what is the ratio of the area of ΔVWX to the area of ΔHIG?

A

B

C

D

Solution

The ratio of sides of ΔVWX to those of ΔHIG is .
= = =
Question 18
1.2 / -0

If cos(90° - x) = - 0.6, what is the value of sin(x)?

A
- 0.6

B
0.6

C
- 0.10

D
0.10

Solution

90° - x and x are complementary, since their angles sum to 90°.
So, the sine of x is equal to the cosine of 90° - x.
sin(x) = cos(90° - x) = - 0.6
Question 19
1.2 / -0

Given a circle with centre O, find ABT if ATB = 60°.

A
120°

B
90°

C
60°

D
30°

Solution

Join AB.
ATB = 60°
So, TAB = 90° [Angle in a semicircle]
Now, using angle sum property in triangle TAB,
OTA + TAB + TBA = 180°
60° + 90° + TBA = 180°
So, TBA = 30°
Question 20
1.2 / -0

Find the value of y from the given two equations,

x = 8
- 9x - 9y = 9

A
- 9

B
9

C
5

D
- 5

Solution

x = 8 ...(i)
- 9x - 9y = 9 ...(ii)
Putting the value of x from equation (i) in equation (ii) we get,
- 9x - 9y = 9
- 72 - 9y = 9
- 9y = 81
y = - 9

Question 21

1.2 / -0

Class	2	5	7	8	10	12	13
Marks	4	6	12	P	11	7	5

If the mean of the following distribution is 8.35, what would be the value of P?

12

15

17

20 Solution

We have,

Class	Marks	Frequency
2	4	8
5	6	30
7	12	84
8	P	8P
10	11	110
12	7	84
13	5	65
SUM	45 + P	= 381 + P

Mean distribution = 8.35
We know,
Mean =

Sum of elements in the set = 381 + 8P
Number of elements in the set = Sum of marks = 45 + P
Therefore,
8.35 =
8.35(45 + P) = 381 + 8P
375.75 + 8.35P = 381 + 8P
8.35P – 8P = 381 - 375.75
0.35P = 5.25
P = = 15

Question 22
1.2 / -0

Simplify and express your answer using a single exponent.

(12c⁶)²

A
144 c¹²

B
10 c¹²

C
12 c¹²

D
144 c¹²⁰

Solution

The expression (12c⁶)²is raised to the power of 2. First, raise each factor to the power of 2.
Then, multiply the exponents.
(12c⁶)²= 12² (c⁶)²
= 144 (c⁶)²
= 144 c⁶^× ²= 144 c¹²
Question 23
1.2 / -0

In an A.P, there are three numbers. If the sum of these numbers is 36 and the product is 756, then which of the following will be the common difference?

A
17

B
15

C
11

D
9

Solution

Let the three numbers be a – d, a, a + d.
a - d + a + a + d = 36
3a = 36
a = 12
(a – d)(a)(a + d) = 756
a(a² – d²) = 756
12(12² – d²) = 756
12² – d² = 63
d² = 144 – 63
d² = 81
d = 9
Question 24
1.2 / -0

In the given figure, AB is the diameter of the circle with centre O. The tangent at C meets BA produced at P. If ABC = 40°, find the measures of CAB and CPA, respectively.

A
50° and 10°

B
40° and 20°

C
50° and 30°

D
60° and 10°

Solution

Given: AB is a diameter and CBA = 40°.
ACB = 90° (Angle in a semi-circle)
CAB = 180° – 90° – 40°= 50°
ABC = ACP = 40° (Angles in alternate segments)
Thus, in ΔPCB,
CPA + PCB + PBC = 180° (Sum of angles of a triangle)
CPA + (40° + 90°) + 40° = 180°
CPA = 180° – 170°
CPA = 10°
Question 25
1.2 / -0

The sum of n terms of an AP is n² + n. Find the common difference.

A
n – 1

B
4

C
2

D
– 4

Solution

First term (Put n = 1) = (1)² + 1 = 2
Sum of the first two terms (Put n = 2) = 2² + 2 = 6
2^nd term = 6 – 2 = 4
d = 4 – 2 = 2
Question 26
1.2 / -0

The ratio of the radii of two spheres is 2 : 1. What is the ratio of the surface areas of the two spheres?

A
2 : 1

B
4 : 1

C
1 : 4

D
8 : 1

Solution

Given: Radii of the two spheres are in the ratio 2 : 1.
If r₁ is the radius of the 1^st sphere and r₂ is the radius of the 2^ndsphere, then .
Now,
=
=
= ∴ Ratio = 4 : 1
Question 27
1.2 / -0

What type of lines does the pair of given linear equations, 5x + 2y - 9 = 0 and x + 3y - 8 = 0, represent?

A
Intersecting lines

B
Parallel lines

C
Coincident lines

D
None of these

Solution

Given linear equation is 5x + 2y - 9 = 0, x + 3y - 8 = 0
So, a₁= 5, a₂= 1, b₁= 2, b₂= 3
Now, ,
And,
So, lines are intersecting.
Question 28
1.2 / -0

What is the value of cos 1° cos 2° cos 3° ... cos 179°?

A
A positive real number

B
A negative real number

C
0

D
None of these

Solution

Since cos 90°= 0, therefore the value of cos 1° cos 2° cos 3° ... cos 90°... cos 179° = 0.
Question 29
1.2 / -0

What is the value of Q for which the system of equations has no solution?

Equations are:
x – Qy = 2 and 3x + 6y = - 5

A
- 2

B
- 1

C
0

D
1

Solution

x - Qy = 2 and 3x + 6y = - 5
So, a₁ = 1, b₁ = - Q and a₂ = 3, b₂ = 6
Now, for no solution,
or,
Or,
Or, Q = - 2
Question 30
1.2 / -0

An A.P. consists of 60 terms. If the first and last terms are 7 and 125 respectively, then find the 32^nd term.

A
55

B
59

C
65

D
69

Solution

Given, n = 60, a₁ = 7 and a₆₀ = 125
a₁ + 59d = 125
7 + 59d = 125
59d = 118
d = 118 ÷ 59 = 2
Now, a₃₂ = a₁ + 31d
= 7 + 31(2) = 7 + 62
Therefore, a₃₂ = 69
Question 31
1.2 / -0

Which of the following statements is true?

A
is a non-terminating decimal.

B
is a terminating decimal.

C
is a non-terminating decimal.

D
is a terminating decimal.

Solution

Hence, this is a terminating decimal as the denominator is a multiple of 10, i.e., denominator has prime factors of the form 2ⁿ × 5^m.
Question 32
1.2 / -0

The area of a parallelogram ABCD is 50 cm², where base = 5 cm. If the length of the base and height are increased by 40%, then what will be the area of the new parallelogram?

A
58 cm²

B
88 cm²

C
98 cm²

D
2250 cm²

Solution

Area of the parallelogram = bh
Area = 5 × h
50 = 5 × h
h = 10 cm
Hence, b = 5 cm and h = 10 cm
After increasing the lengths by 40%,
20% of 5 = 2 cm
20% of 10 cm = 4 cm
Therefore, the area of the new parallelogram = 7 × 14 = 98 cm²
Question 33
1.2 / -0

If the radius of a circle is doubled, then what will be the ratio of the area of the new circle to the area of the old circle?

A
1 : 4

B
4 : 1

C
2 : 1

D
1 : 2

Solution

Let the radius of the circle be r units.Area of the circle = πr²The radius of the circle when it is doubled = 2r
Area of the circle when the radius is doubled = π(2r²) = 4πr²
The ratio of the area of the new circle to that of the old circle = 4πr²: πr²The ratio of the areas of the new circle to the old circle is 4 : 1.
Question 34
1.2 / -0

Which of the following statements is true according to the fundamental theorem of arithmetic?

A
4ⁿ ends with digit zero at n = 275.

B
Every composite number is expressed as a product of primes and this prime factorisation is not unique.

C
28 can be expressed as 2 × 2 × 7 or 7 × 2 × 2.

D
In prime factorisation of a composite number, the order of factors is always unique.

Solution

Fundamental theorem of arithmetic states that every integer greater than 1, either is prime itself or is the product of prime numbers. This product is unique and those factors can be written in any order. Thus, 28 can be expressed as 2 × 2 × 7 or 7 × 2 × 2.
Question 35
1.2 / -0

For what value of are the following equations inconsistent?

x + 3y = - 3
12x + y =

A
8

B
-6

C
12

D
-12

Solution

a₁x + b₁y = c₁ ... (1)
a₂x + b₂y = c₂ ... (2)
A pair of linear equations is said to be inconsistent when .
Now,
x + 3y = - 3 … (3)
12x + y = … (4)
Comparing (1) and (2) with (3) and (4),

For inconsistency:

Or, =
Or,
Or, ² = 36
= 6 ... (5)
Also,
3² - 3
Or, ² - 3 - 3 0
² - 60
Or, (- 6) 0
Either 0 or 6
Question 36
1.2 / -0

40 circular plates, each of radius 5 cm and thickness 0.5 cm, are placed one above another to form a solid right cylinder. Find the total surface area (in cm²) of the cylinder so formed.

A
785.1

B
785.3

C
785.5

D
785.7

Solution

The total surface area of the cylinder = 2πr (r + h)

= cm²= cm²= 785.7 cm²
Question 37
1.2 / -0

Jake made a toy at home which is in the form of a cone of radius 4.5 cm, mounted on a hemisphere with the similar radius. The total height of the toy is 17.5 cm. Find the total surface area of the toy.

A
321.03 cm²

B
320.03 cm²

C
320 cm²

D
319.03 cm²

Solution

Total height of the toy = Height of the cone + Height of the hemisphere
Height of the hemisphere = Radius of the hemisphere = 4.5 cm
Height of the cone = (17.5 – 4.5) cm = 13 cm
Radius of the cone is 4.5 cm.
Slant height = cm
= cm
= cm = 13.7 cm

Surface area of the cone = πrl
= × 4.5 × 13.7 cm²= 193.75 cm²
Radius of the hemisphere = 4.5 cm
Surface area of the hemisphere = 2πr² = 2 × × 4.5 × 4.5 cm² = 127.2 cm²
Total surface area of the toy = Surface area of the cone + Surface area of the hemisphere
= 193.75 cm²+ 127.28 cm²= 321.03 cm²
Question 38
1.2 / -0

Directions For Questions

Directions: Read the following information and answer the given question.
Rohan, a friendly person, has a large open square grass field of side 18 m. He has a horse. In one corner of the field, he has also made a shed under which the horse stays. Rohan has tied the horse to a peg at one corner of the field. The length of the rope is 7 m.

...view full instructions

What would be the approximate increase in grazing area if the rope were 14 m long, instead of 7 m?

A
108 sq. m

B
116 sq. m

C
130 sq. m

D
123 sq. m

Solution

New area that can be grazed = sq. m = 154 sq. m
Old grazing area = sq. m = 38.5 sq. m
So, increase in grazing area = 154 - 38.5 = 115.5 sq. m = 116 sq. m (approx.)
Question 39
1.2 / -0

A water can is in the form of a circular cylinder having volume 448π cm³,and height 7 cm. Find the diameter of its base.

A
12 cm

B
14 cm

C
16 cm

D
18 cm

Solution

Height of the the right circular cylinder is 7 cm.
Volume of the right circular cylinder is 448π cm³.
πr²h = 448π
r²h = 448
7r²= 448
r² = 64
r = 8 cm
Therefore, the diameter of its base = (2 × 8) cm = 16 cm
Question 40
1.2 / -0

Kanav marks numbers on the marbles in such a way that there are 19 marbles numbered 1, 2, 3, 4, ....., 17, 18, 19, and they are put in a box and mixed thoroughly. One of his friends picks the marble from the box. What is the probability of getting an odd number?

A

B

C

D

Solution

Number of possible outcomes = 19
Number of favourable outcomes = 10 (1, 3, 5, 7, 9, 11, 13, 15, 17, 19)
Therefore, probability =
Question 41
1.2 / -0

Three metal cubes of volumes 125 cubic cm, 64 cubic cm and 27 cubic cm are melted to form a new cube. Find the edge length of the new cube formed.

A
6 cm

B
7 cm

C
8 cm

D
12 cm

Solution

V₁ = 125 cu. cm
V₂ = 64 cu. cm
V₃ = 27 cu. cm
V = V₁ + V₂ + V₃ = 125 + 64 + 27 cu. cm
V = 216 cu. cm
= 216 cu. cm
= 6 cm
Question 42
1.2 / -0

Rahul is cycling such that the wheels of the cycle are making 150 revolutions per minute. If the radius of each wheel is 30 cm, then calculate the speed (in kilometres per hour) with which Rahul is cycling?

A
16.97 km/hr

B
16.79 km/hr

C
16.59 km/hr

D
16.09 km/hr

Solution

Circumference of the wheel = 2πr = cm = km
Hence, the distance covered by the wheel in 1 revolution = Circumference of the wheel = km
So, the distance covered by the wheel in 150 revolutions = × 150 km
Distance covered by the wheel in 1 minute = × 150 km

The distance covered in 1 hour = × 150 × 60
= = = 16.97 km
Speed = Distance covered per hour = 16.97 km/hr
Question 43
1.2 / -0

In a cricket ground, during the drinks break, the water tank is used to serve water. The square water tank has its side equal to 40 m. There are four semi-circular grassy plots all around it. Find the cost of turfing the plots at Rs. 2.25 per m²using π as 3.14.

A
Rs. 5650

B
Rs. 5651

C
Rs. 5652

D
Rs. 5653

Solution

ABCD is the square plot of side 40 m.
Radius of each circle = = 20 m
The area of the semi-circular plots = Area of two circles with radius 20 m each
Hence, area of the four semi-circular plots = 2 × π (20)²m²= 800π m²
The cost of turfing the plot at Rs. 2.25 per m²= 800π × 2.25 m²
= Rs. 800 × 3.14 × 2.25 = Rs. 5652
Question 44
1.2 / -0

The radius of a cycle's wheel is 70 cm. How many times will the wheel revolve in order to cover a distance of 110 m?

A
25

B
20

C
15

D
10

Solution

The circumference of the wheel = 2πR = = 440 cm
The wheel will cover a distance of 440 cm in one revolution.
The distance to be covered by the wheel = 110 m = 1100 cm
Number of times the wheel will revolve to cover 110 m = = 25
Number of times the wheel will revolve to cover 110 m is 25.
Question 45
1.2 / -0

For a right circular ice cone, the radius of the base and the height are 7 cm and 24 cm, respectively. What is the total surface area of that cone?

A
701 cm²

B
702 cm²

C
703 cm²

D
704 cm²

Solution

The slant height of the cone = cm
= cm
= cm
= 25 cm
Total surface area of the right circular cone = πr(r + l)
= cm²= 704 cm²
Question 46
1.2 / -0

The combined age of a man and his wife is six times the combined age of their children. Two years ago, their combined age was ten times the combined age of their children, and six years hence, their combined age will be three times the combined age of their children. How many children do they have?

A
2

B
3

C
4

D
5

Solution

Let X = combined age of parents
Let Y = combined age of children
Let Z = number of children
Then,
X = 6Y;
(X – 4) = 10(Y – 2Z);
(X + 12) = 3(Y + 6Z)
On solving, we get
Z = 3
Question 47
1.2 / -0

If the p^th, q^thand r^th terms of an A.P. are x, y and z respectively, then x(q - r) + y(r - p) + z(p - q) is

A
- 1

B
0

C
1

D
3

Solution

a + (p - 1)d = x … (i)
a + (q - 1)d = y …(ii)
a + (r - 1)d = z …(iii)
Subtracting (ii) from (i), (iii) from (ii) and (i) from (iii), we get
(p - q)d = x - y (p - q) =
(q - r)d = y - z (q - r) =
(r - p)d = z - x (r - p) =
Given, x(q - r) + y(r - p) + z(p - q)
= + + = = = 0
Question 48
1.2 / -0

In the given figure, two circles with M and N as their centres have radii 25 cm and 9 cm, respectively, with 'PQ' as a common tangent. Find the length of PQ.

A
26 cm

B
30 cm

C
40 cm

D
46 cm

Solution

Draw RN || PQ
Now,
RM = 25 – 9 = 16
MN = MS + NS
= 25 + 9 = 34 cm
Using Pythagoras theorem, we get
RN² = PQ² = 34² – 16² = 1156 – 256 = 900
PQ = 30 cm
Question 49
1.2 / -0

Fill in the blanks:

All the circles are _____. All the squares are ______. All _____ triangles are similar. Two polygons with same number of sides are similar, if their corresponding angles are ______ and their corresponding sides are _____.

A
Similar, similar, equilateral, equal, proportional

B
Equilateral, similar, similar, equal, equilateral

C
Equilateral, similar, similar, equal, proportional

D
Equal, proportional, equilateral, equilateral, similar

Solution

All the circles are similar. All the squares are similar. All equilateral triangles are similar. Two polygons with same number of sides are similar, if their corresponding angles are equal and their corresponding sides are proportional.
Question 50
1.2 / -0

A vertical building and a tower are on the same ground level. From the top of the building, the angle of elevation of the top of the tower is 45° and the angle of depression of the foot of the tower is 60°. Find the height of the tower if the height of the building is 30 m.

A
35.77 m

B
40 m

C
47.32 m

D
50 m

Solution

In the diagram,
BC = 30 m = DE
ABD = 45°EBD = 60°

Let AD = x m
So, AE = (x + 30) m
In right-angled ΔABD,
tan 45° =
1 = BD = x m … (1)
In right-angled ΔEBD,
tan 60° =

BD = m …. (2)
From equations (1) and (2), we get
x = x = = m
∴ Height of the tower = AE = (x + 30) m
= (10+ 30) m
= (10 × 1.732 + 30) m = 47.32 m