IMO - Mock Test

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IMO - Mock Test - 8

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Weekly Quiz Competition

Question 1
1.2 / -0

In the given question, two rows of numbers are given. The resultant number in each row is to be worked out separately based on the following rules and the question below the rows of numbers is to be answered. The operations on numbers progress from left to right.

Rules:
(i) If an odd number is followed by another composite odd number, they are to be multiplied.
(ii) If an even number is followed by an odd number which is not a perfect square, they are to be added.
(iii) If an even number is followed by a number which is a perfect square, the even number is to be subtracted from the perfect square.
(iv) If an odd number is followed by a prime odd number, the first number is to be divided by the second number.
(v) If an odd number is followed by an even number, the even number is to be subtracted from the odd number.

27 12 5

28 64 k

If the resultant of the first row is k, then what will be the resultant of the second row?

A
42

B
33

C
108

D
39

Solution

Here we have two rows: 27, 12, 5 and 28, 64, k
First we solve for 27, 12, 5.
So, from the given condition (v) we get 27 - 12 = 15
Now, we have 15, 5,
So, from the given condition (iv) we get = 15/5 = 3
Given result of first row is equal to k, then k = 3
From the second row 28, 64
So, from the given condition (iii) we get 64-28 = 36
From the given condition (ii), the result of second row is 36 + k = 36 + 3 = 39
Question 2
1.2 / -0

P is the mother of Q and daughter of R. S is the daughter of Q and sister of T. How is P related to T?

A
Paternal grandmother

B
Maternal grandmother

C
Data inadequate

D
None of these

Solution

We get the following diagram:

P is the maternal grandmother of T.
Question 3
1.2 / -0

P, Q and R are three points on the ground. Point P is north of point Q and PQR is 135° in anticlockwise direction. In which direction is point R from point Q?

A
North-East

B
North-West

C
South-East

D
South-West

Solution
Question 4
1.2 / -0

Find the water image of Fig. (X).

A

B

C

D

Solution

The reflected image will be upside down.
Question 5
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Select a figure from the options which satisfies the same conditions of placement of the dots as in Fig. (X).

A

B

C

D

Solution

First dot lies on square and circle.
Second dot lies on square, circle, rectangle and triangle.
Third dot lies on triangle only.
All these dot appear in the same manner in figure 4 only.
Question 6
1.2 / -0

Select a figure from the option figures which will continue the same series as established by the five Problem Figures.

A

B

C

D

Solution

Since the series completes with OSWAEI, option 2 is the correct answer.
Question 7
1.2 / -0

Directions: Study the following information carefully and answer the question.

Eight friends Q, R, S, T, V, W, Y and Z sit around a circular table, facing the centre. There are three males and five females in the group of friends. No two males are immediate neighbours of each other.
(i) V sits second to the right of his wife.
(ii) S sits third to the right of V.
(iii) W sits second to the right of her husband Z. Z is not an immediate neighbour of V's wife.
(iv) T is a male and Y is not an immediate neighbour of V.
(v) R sits second to the right of Q.

Which of the following statements is/are true regarding T?

A
T is an immediate neighbour of Z's wife.

B
No male is an immediate neighbour of T.

C
Q sits second to right of T.

D
All are true.

Solution

We get the following arrangement:

Hence, all the statements are true.
Question 8
1.2 / -0

Vijay is fifteenth from the front in a column of boys. There are thrice as many behind him as there are in front. How many boys are there between Vijay and the seventh boy from the end of the column?

A
33

B
34

C
35

D
Data inadequate

Solution

Vijay being 15^th in position, there are 14 boys in front of him and 42 (3 ×14) boys behind him. Hence there are 14 + 1 + 42 = 57 boys altogether standing. Now, 7^th from the end of the line is the 51^st boy.
Vijay is the 15^th boy from the beginning. Therefore, there are 50 - 15 = 35 boys or 51 - 16 = 35 boys in between Vijay and the seventh boy from the end of the column.
Question 9
1.2 / -0

Select a figure from the options, which when placed in the blank space of Fig. (X) would complete the pattern.

A

B

C

D

Solution

Hence, correct image is:
Question 10
1.2 / -0

Find out which of the options completes the figure matrix.

A

B

C

D

Solution

Observing the main body parts, we see that there are 3 figures/categories of different body parts (chest and stomach). Also the shape of the head is also changing in each. If we look at the foot of each category, we find that two figures in each have shoes.
Question 11
1.2 / -0

If 12% people in a village are suffering from cancer and 13% from blood pressure, which of the following diagram best represents the sick population of the village?

A

B

C

D

Solution

Clearly, 12% + 13% = 25%, i.e. ^th (or a quarter) of the total area of the circle.

Question 12

1.2 / -0

If all the consonants starting from B are given sequentially the value of even numbers such as B = 2, C = 4 and so on, and all the vowels are given double the value of the preceding vowel and the value of A is 5, then what is the value of REASONING?

162

177

185

187 Solution

A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P	Q	R	S
5	2	4	6	10	8	10	12	20	14	16	18	20	22	40	24	26	28	30

Now, REASONNG = 28 + 10 + 5 + 30 + 40 + 22 + 20 + 22 + 10 = 187

Question 13
1.2 / -0

Two positions of a dice are shown below. What number will be opposite to the number 4?

A
5

B
6

C
3

D
1

Solution

The numbers 1. 2, 3, 6 are adjacent to 4. So, 5 will be opposite to the number 4.
Question 14
1.2 / -0

If 2 is subtracted from each odd digit and 3 is added to each even digit in the number 3675249, then how many digits will appear twice in the new number thus formed?

A
1

B
2

C
3

D
4

Solution

The digits 7 and 5 appear twice.
Question 15
1.2 / -0

Select the figure in which figure (X) is exactly embedded as one of its part.

A

B

C

D

Solution

The figure X can be embedded exactly on (1) from the top as it is.
Question 16
1.2 / -0

Find the value of I, so that y - 2p is a factor of - 2y + lp.

A
0

B
1

C
2

D
3

Solution

Given,
(y- 2p) is factor of
At y = 2p we get given polynomial equation equal to zero.
2p-2(2p) + lp = 0
2p- 4p + lp = 0
2p - 4p + lp = 0
-2p + lp = 0
lp = 2p
l = 2
Question 17
1.2 / -0

If x⁴ + = 47, then find the value of x³ + .

A
7

B
18

C
6

D
12

Solution

x⁴ + (1/x⁴) = 47
x⁴ + (1/x^{4 )}+ 2 = 49
(x² + (1/x²))² = 49
x² + (1/x² ) = 7
x² + (1/x²) + 2 = 9
x + (1/x) = 3
x³ + (1/x³) + 3 × 3 = 27
x³ + (1/x³) = 18
Question 18
1.2 / -0

How many planes can be made to pass through three distinct points?

A
One if they are collinear

B
Infinite if they are collinear

C
Only one if they are non collinear

D
Both (2) and (3)

Solution

If the points are collinear then an infinite number of planes can be made to pass through them. If three distinct points are non-collinear then exactly one plane passes through them.
Question 19
1.2 / -0

In the given figure, B < A and C < D.

Which of the following statements is true regarding the relationship between AD and BC?

A
AD = BC

B
AD < BC

C
AD > BC

D
Nothing can be said.

Solution

B < A and C < D
A greater angle of a triangle is opposite a greater side.
Hence, we can conclude that side OB > OA.
Also, OC > OD
Thus, AD < BC
Question 20
1.2 / -0

In the given figure, AB II CD II EF. CE is joined and produced to G. If BAC = 130°, ACE = 140°, then find DCE and FEG respectively.

A
50°, 130°

B
90°, 90°

C
140°, 40°

D
45°, 135°

Solution

Angle ACD = 180° - 130° = 50°
Angle ECD = 140° - 50° = 90°
Angle FEG = 90° (Corresponding angle)
Question 21
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The graph of line y = 6 is

A
parallel to x-axis at a distance of 6 units from the origin.

B
parallel to y-axis at a distance of 6 units from the origin.

C
making an intercept of 6 units on the x-axis.

D
making an intercept of 6 units on both the axis.

Solution

Parallel to x-axis at a distance of 6 units from the origin.
Question 22
1.2 / -0

Directions: The given question is followed by three statements. You have to study the question and all the three statements to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.

What is the marked price of the suitcase?

I. When a discount of 15% is offered, the profit earned is 10.5%.
II. The cost price of the suitcase is 1500.
III. The marked price is 30% above the cost price.

A
I only

B
II and either I or III

C
Any one of the three

D
All I, II and III are required.

Solution

Using the cost price and discount, we can calculate the marked price. (Using I and II)
Also, using the cost price and the information in statement III that marked price is 30% above the cost price, we can calculate the marked price. (Using II and III)
Question 23
1.2 / -0

The line BE is a diameter of the given circle. If BAC = 33° and EBC = 57°. Then CAE =

A
57°

B
33°

C
48°

D
90°

Solution

Since BAE is a right angle in the semicircle,
BAE = EAC + CAB {CAB = 33°}
Therefore, CAE = 90° - 33° = 57°
Question 24
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ABCD is a field in the form of a quadrilateral whose sides are AB = 40 m, BC = 15 m, CD = 28 m and AD = 9 m. If DAB = 90°, than the area of the field is

A
306 m²

B
312 m²

C
126 m²

D
218 m²

Solution

Area of triangular part BAD = (1/2) x 9 x 40 = 180 m²
In triangular part BCD,
s = (41 + 15 + 28)/2 = 42
Area of triangular part BCD =
Arae of the field = 180 + 126 = 306 m²
Question 25
1.2 / -0

If then the value of k is

A
3

B
3²

C
3³

D

Solution

= k, 3^{2r + 0.5 + 0.5 - 0.5r - 1 + 0.5r} = k^r

(3²)^r = k^r

k = 3²
Question 26
1.2 / -0

The number of zeros of the zero polynomial is

A
0

B
1

C
2

D
infinite

Solution

A polynomial function of degree has n zeros, provided multiple zeros are counted more than once and provided complex zeros are counted.
Question 27
1.2 / -0

The sum of a number and its reciprocal is thrice the difference between the number and its reciprocal. The number is ________.

A

B

C

D
Both (1) and (2)

Solution

y² + 1 = 3y² - 3
2y² = 4
y =
Question 28
1.2 / -0

The probability of selecting a boy in a class is 0.6 and there are 45 students in a class. Find the number of girls in the class.

A
18

B
19

C
20

D
15

Solution

Total number of students = 45
Number of boys = 0.6 of 45 = 27
Number of girls = 45 - 27 = 18
Question 29
1.2 / -0

What percentage of a day is six hours and 45 minutes?

A
7.218%

B
8.3%

C
28.125%

D
None of these

Solution

Given time is 6 hours 45 minutes
This can also be represented as hours
As percentage of 'a day', i.e. 24 hours, it could be represented as, = 28.125%
Question 30
1.2 / -0

Find the value of a and b respectively, if

A
2, 1

B
1, 27

C
11, 28

D
2, 38

Solution

By rationalising, we get

=
=
Comparing with RHS of the given statement, we get
a = 1 and b = 27
Question 31
1.2 / -0

In the given figure (not drawn to scale), LMNO is a parallelogram and OPQR is a rhombus. Find NMH given that LMH is a straight line.

A
80°

B
60°

C
70°

D
50°

Solution

OPQR is a rhombus, PQ = QR = RO = OP
ROP = 180° - 70°= 110°POQ = 110°/2 = 55°
LMN = LON = 45° + 55° = 100°
∴ NMH = 180° - 100° = 80°
Question 32
1.2 / -0

In the given figure, 'O' is the center of circle, CAO = 25° and CBO = 35°. What is the value of AOB?

A
55°

B
110°

C
120°

D
Data insufficient

Solution

Since OA = OC = radius and OAC = OCA = 25°
AOC = 180° - (25° + 25°) = 130°
and OC = OB = radius, OBC = OCB = 35°
BOC = 180° - (35° + 35°) = 110°
Hence, AOC + BOC = 130°+ 110° = 240°,
AOB = 360° - 240° = 120°.
Question 33
1.2 / -0

John is of the same age as Mohan. Ram is also of the same age as Mohan. State the Euclid's axiom that illustrates the relative ages of John and Ram.

A
First Axiom

B
Second Axiom

C
Third Axiom

D
Fourth Axiom

Solution

John is of the same age as Mohan and Ram is also of the same age as of Mohan. Therefore, by First Euclid Axiom, "Things which are equal to the same thing are equal to one another". By this axiom, we can state that John is also of the same age as Ram.
Question 34
1.2 / -0

Directions: The pie chart below shows the number of fruits sold on a particular day at a fruit stall.

The ratio of the number of mangoes sold to the number of apples sold is 6 : 5. What percentage of the total sales came from the sale of mangoes?

A
20%

B
30%

C
45%

D
60%

Solution

Ratio of the number of mangoes sold to the number of apples sold = 6 : 5 = 6x : 5x
5x = 25%
x = 5%
6x = 30%
The correct answer is 30%.
Question 35
1.2 / -0

A cuboidal metal block of dimensions 20 cm × 16 cm × 12 cm weighs 6 kg. Find the weight of a block of the same metal of size 10 cm × 8 cm × 8 cm.

A
3 kg

B
8 kg

C
7 kg

D
1 kg

Solution

Given dimensions of the 6 kg block are 20 cm × 16 cm × 12 cm
So, volume of block = 20 cm × 16 cm × 12 cm = 3840 cm³
Weight per cubic cm = 6/3840 kg
Given dimensions of the 2^nd block are 10 cm × 8 cm × 8 cm
So, volume of the 2^nd block = 10 cm × 8 cm × 8 cm = 640 cm³
Therefore, weight of the 2^nd block = × 640 = 1 kg
Question 36
1.2 / -0

Ajay has a certain amount in his account. He gives half of it to his eldest son and one third of the remaining to his youngest son. What fraction of the original amount is left with him now?

A
1/3

B
2/3

C
3/4

D
1/6

Solution

Let the certain amount in his account be x.
Eldest son gets = x/2
Youngest son gets = 1/3 × (x/2) = x/6
Amount left with him now = (x/2) - (x/6) = x/3
The fraction of original amount left with him now = 1/3
Question 37
1.2 / -0

The average age of 6 sons of a family is 8 years. The average age of the sons together with their parents is 22 years. If the father is older than the mother by 8 years, then the age of the mother is

A
44 years

B
52 years

C
60 years

D
68 years

Solution

Total Age of Sons = 6 × 8 = 48
Total age of the sons together with their parents = 22 × 8 = 176
Total Age of parents = 176 - 48 = 128
Since father is older than the mother by 8 years,
Mother's age = (128 - 8)/2 = 120/2 = 60
Father's age is 68 years.
Mother's age is 60 years.
Question 38
1.2 / -0

A sum of Rs. 1550 is lent in two parts, one at 8% and another one at 6%. If the total annual income is Rs. 106, then find the money lent at each rate.

A
Rs. 750, Rs. 800

B
Rs. 600, Rs. 950

C
Rs. 650, Rs. 900

D
Rs. 850, Rs. 750

Solution

According to the question,
8x/100 + (1550 - x) × 6/100 = 106
8x + 9300 - 6x = 10600
2x = 10600 - 9300
x = 1300/2 = Rs. 650
(1550 - x) = Rs. 900
Question 39
1.2 / -0

A train travelling at 48 km/hr. completely crosses another train having half its length and travelling in opposite direction at 42 km/hr in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is ________

A
400 m

B
450 m

C
560 m

D
600 m

Solution

Let the length of the train traveling at 48 kmph be 2x meters.
And length of the platform is y meters.
Relative speed of train = (48 + 42) kmph
= (90 × 5/18) = 25 m/sec; and 48 kmph = 48 × 5/18 = 40/3 m/sec.
According to the question,
(2x + x)/25 = 12; or, 3x = 12 × 25 = 300; or, x = 300/3 = 100 m
Then, length of the train = 2x = 100 × 2 = 200 m.
(200 + y )/(40/3) = 45;
600 + 3y = 40 × 45; or, 3y = 1800 - 600 = 1200
y = 1200/3 = 400 m.
Length of the platform = 400 m
Question 40
1.2 / -0

Savita has Rs. 27 in the form of fifty-paisa and twenty-five paisa coins. She has twice as many twenty-five paisa coins as she has fifty-paisa coins. How many coins of each kind does she have?

A
27, 54

B
30, 60

C
25, 50

D
40, 80

Solution

Let the number of 50p coins be 'x'.
Then, the number of 25p coins = 2x
Hence, the total valuation is = 27
(x/2) + (x/2) = x = 27
OR
x = 27; (50p coins)
2x = 54; (25p coins)
Question 41
1.2 / -0

Four runners started running the race in the same direction around a circular path of 7 km. Their speeds are 4 km/hr, 3 km/hr, and 9 km/hr and 3.5 km/hr individually. If they have started their race at 6 o'clock in the morning, then at what time they will be at the starting point?

A
14 hours

B
13 hours

C
10 hours

D
15 hours

Solution

First we will find the time to be taken to complete one lap for each.
For 1^st,
The time taken to complete one lap =( 60 × 7 × 60 )/4 = 6300 seconds
For 2^nd,
The time taken to complete one lap = (60 × 7 × 60)/3 = 8400 seconds
For 3^rd,
The time taken to complete one lap = (60 × 7 × 60)/9 = 2800 seconds
For 4^th,
The time taken to complete one lap = (60 × 7 × 60)/3.5 = 7200 seconds
Now, LCM of these (6300, 8400, 2800, 7200) = 50400 Seconds = 14 hours
Question 42
1.2 / -0

In a bag, there are coins of 25 paisa, 10 paisa and 5 paisa in the ratio of 1 : 2 : 3. If there are Rs. 30 in all, then how many 5 paisa coins are there?

A
50

B
100

C
150

D
200

Solution

Let us suppose the number of coins of 25 paisa = x
Number of coins of 10 paisa = 2x
And number of coins of 5 paisa = 3x
So, 25 x + 10(2x) + 5(3x) = 3000 or x = 50
So, 3x = 150.
Question 43
1.2 / -0

A shopkeeper earns a profit of 12% on selling a book at 10% discount on the printed price. The ratio of the cost price and the printed price of the book are

A
45 : 56

B
50 : 61

C
99 : 125

D
None of these

Solution

Let the CP be 100
Hence, SP = 100 + 12% of 100 = 112
If the marked price be X, then 90% of X = 112

The required ratio = 100 : X = 90 : 112 = 45 : 56
Question 44
1.2 / -0

Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?

A
3 : 4

B
3 : 5

C
5 : 3

D
None of these

Solution

20W x 16 = 16M x 15
M : W = 20 : 15 = 4 : 3
Question 45
1.2 / -0

The dimensions of a photograph are 4 cm and 1.8 cms. If the breadth of the enlarged photo is 4.5 cm and it was enlarged proportionally, then what is the new length of new photograph?

A
6 cm

B
5.4 cm

C
10 cm

D
9 cm

Solution

Length of the enlarge photograph = (4.5/1.8) × 4 = 10 cm
Question 46
1.2 / -0

The polynomial p(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7 when divided by x + 1 leaves the remainder 19. Find

(a) The value of a.
(b) The remainder when p(x) is divided by x + 2.

A
(a) 1, (b) 13

B
(a) -3, (b) 48

C
(a) 5, (b) 62

D
(a) 8, (b) 43

Solution

According to question,
If the polynomial is divided by (x +1), then it leaves the remainder 19.
So, (-1)⁴ - 2(-1)³ + 3(-1)² + a + 3a - 7 = 19
This gives a = 5
Now, putting the value of 'a' in the given polynomial, we get
p(x) = (x)⁴ - 2(x)³ + 3(x)² - 5x + 8
Again, p(x) is divided by (x + 2).
So, putting x + 2 = 0 or x = -2 in p(x),
we get
16 + 16 + 12 + 10 + 8 = 62
Question 47
1.2 / -0

Which of the following options hold(s)?

Statement-1: If any two angles and the non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the triangles are congruent (AAS congruence criterion).
Statement-2: If in two right triangles, hypotenuse and one side of a triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent (RHS congruence criterion).

A
Statement-1 is true, but statement-2 is false.

B
Statement-1 is false, but statement-2 is true.

C
Both the statements are true.

D
Both the statements are false.

Solution

AAS stands for "angle, angle, side" and means that we have two triangles where we know that two angles and the non-included side are equal. If two angles and the non-included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent. Therefore, Statement-I is true.
The RHS criterion means that two triangles are congruent if they are both right-angled triangles, their hypotenuses are equal, and one other side is equal. Therefore, Statement-II is true as well.
Hence, option (3) is the correct answer.
Question 48
1.2 / -0

In the figure, A and B are the centres of two intersecting circles. Which Euclid's axiom will prove that ABC is an equilateral triangle?

A
If equals are added to equals, the wholes are equal.

B
Things which are double of the same things are equal to one another.

C
Things which are equal to the same thing are equal to one another.

D
If equals are subtracted from equals, the remainders are equal.

Solution

Since AB is the common radius for both the circles, we can conclude that AC = BC = AB = Radius.
Therefore, using Euclid's axiom, which states that "Things which are equal to the same thing are also equal to one another", we can prove that △ABC is an equilateral triangle.

Question 49

1.2 / -0

A die is rolled. If the number on the die is even, then a coin is tossed once; and if the number on the die is odd, then the coin is tossed twice. Match the events in Column I with their probabilities in Column II.

Column I	Column II
(P) Probability that two heads appear	(1)
(Q) Probability that at least 1 heads appears	(2) 0
(R) Probability that the die shows an even number and the coin shows exactly two heads	(3)
(S) Probability that the die shows an odd number and the coin shows at least one tails	(4)

(P) - 1, (Q) - 2, (R) - 3, (S) - 4

(P) - 3, (Q) - 1, (R) - 2, (S) - 4

(P) - 3, (Q) - 2, (R) - 1, (S) - 4

(P) - 4, (Q) - 3, (R) - 2, (S) - 1

Solution

A die is rolled initially.
If even number occurs on the die,a coin is tossed once.
If odd number occurs on the die,a coin is tossed twice.
We get following events:

(P) Probability that two heads appear = 3/18 = 1/6
(Q) Probability that at least one heads appears = 12/18 = 2/3
(R) Probability that the die shows an even number and the coin shows exactly two heads = 0/18 = 0
(S) Probability that the die shows an odd number and the coin shows at least one tails = 9/18 = 1/2

Question 50
1.2 / -0

The volume of the space inside a right circular conical tent is 138 m² and its vertical height is 4 m. Find the canvas required to make the tent and the cost of the canvas at the rate of Rs. 120 per m².

A
126.3 m², Rs. 15,164

B
126.3 m², Rs. 15,154

C
136.2 m², Rs. 16,344

D
142.3 m², Rs. 17,076

Solution

Given, h = 4m
V = 138 m³ = 968/7 m³
V (cone) =

33 = r²
r = 5.74 m

We know, I = = = 7 m = I
Area of the canvas = CSA = πrI = × 5.74 × 7
CSA = 126.28 ≈ 126.3 m²
Cost = 120 × 126.28
S = Rs. 15,153.6 ≈ Rs. 15,154