IMO - Mock Test

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IMO - Mock Test - 4

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Weekly Quiz Competition

Question 1
1 / -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 x 9 = 306, 12 x 8 = 609 and 16 x 4 = 406, then 19 x 5 = ?

A
905

B
506

C
509

D
605

Solution

For 7 x 9, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 306.

For 12 x 8, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 609.

For 16 x 4, the answer is written in reverse order by adding a 0 in between the numbers, i.e. 406.

For 19 x 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 / -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 / -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 4
1 / -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 5
1 / -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 6
1 / -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 7
1 / -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 8
1 / -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 9
1 / -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 10
1 / -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 c^6x2

= 144 c¹²
Question 11
1 / -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 12
1 / -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 13
1 / -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^o= 0, therefore the value of cos 1° cos 2° cos 3° ... cos 90^o... cos 179° = 0.
Question 14
1 / -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 15
1 / -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 x h

50 = 5 x 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 x 14 = 98 cm²
Question 16
1 / -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 = πr²
The 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: πr²
The ratio of the areas of the new circle to the old circle is 4 : 1.
Question 17
1 / -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 x 2 x 7 or 7 x 2 x 2.
Question 18
1 / -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 19
1 / -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
10/19

B
10/17

C
10/13

D
10/11

Solution

Number of possible outcomes = 19
Number of favourable outcomes = 10 (1, 3, 5, 7, 9, 11, 13, 15, 17, 19)
Therefore, probability = 10/19
Question 20
1 / -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 = 2 ×(22/7)×70 = 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 = 11000/440 = 25

Number of times the wheel will revolve to cover 110 m is 25.
Question 21
1 / -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 22
1 / -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.

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Re-Attempt Weekly Quiz Competition

IMO - Mock Test - 4

Result

IMO - Mock Test - 4

Score

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Rank

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SHARING IS CARING

Question 1

905

506

509

605

Solution

Question 2

1

2

3

4

Solution

Question 3

#

&

7

!

Solution

Question 4

GH242F

PQ343O

HI216G

BC125A

Solution

Question 5

JLNP

KLNP

JLNO

JLMN

Solution

Question 6

290

280

272

265

Solution

Question 7

a, c, b, d

a, b, c, d

b, c, d, a

a, d, c, b

Solution

Question 8

- 0.6

0.6

- 0.10

0.10

Solution

Question 9

- 9

9

5

- 5

Solution

Question 10

144 c12

10 c12

12 c12

144 c120

Solution

Question 11

17

15

11

9

Solution

Question 12

n – 1

4

2

– 4

Solution

144 c¹²

10 c¹²

12 c¹²

144 c¹²⁰

58 cm²

88 cm²

98 cm²

2250 cm²

4ⁿ ends with digit zero at n = 275.