IMO - Mock Test

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IMO - Mock Test - 6

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Question 1
1.2 / -0

Two rows of numbers are given. The resultant of 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 of numbers in each row progress from left to right.

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

9 15 50

12 25 24

If the resultant of first row is x and that of second row is y, then find the value of x y.

A
18

B
8

C
5

D
6

Solution

9 15 50

12 25 24

According to the given condition,
Row 1: (9 × 15), 50 = 135, 50
135 + 50 = 185
Row 2: (25 – 12), 24 = 13, 24
13 + 24 = 37

Now, = 5
Question 2
1.2 / -0

Pointing to a woman in a photograph, a man says, ''She is the grandmother of the son of my daughter-in-law's mother-in-law.'' How is the woman related to the man?

A
Mother

B
Mother-in-law

C
Sister

D
Wife

Solution

Daughter-in-law's mother-in-law = wife
Son of wife = son
Grandmother of son = mother
Hence, the woman is the mother of the man.
Question 3
1.2 / -0

A man travelled 400 metres straight from his office. He then turned left and travelled 500 metres straight, after which he turned left again and travelled 400 metres straight. Finally, he turned right and walked another 600 metres straight. How far is he from his office?

A
1.0 km

B
1.1 km

C
1.4 km

D
1.8 km

Solution

From the given information, we get the following diagram:

BC = Initial displacement (400 m)
CD = Displacement after left turn (500 m)
DA = Displacement after another left turn (400 m)
AE = Displacement after right turn (600 m)
Therefore,
BE = Final displacement from the office = 600 + 500 m = 1100 m = 1.1 km
Question 4
1.2 / -0

Select the correct mirror image of the given Fig. (X) if the mirror is placed along MN.

A

B

C

D

Solution

Hence, option (2) is the correct answer.
Question 5
1.2 / -0

Select a figure from the options which satisfies the same conditions of placement of the dots as in Figure (X).

A

B

C

D

Solution

In fig. (X), one of the dots lies in the region common to the circle and the triangle only, another dot lies in the region common to the circle, the square, and the triangle only. The third dot lies in the region common to the rectangle only. In each of the options (1), (3) and (4), there is no region common to the circle, the square and the triangle only. Only option (2) consists of all the three types of regions.
Question 6
1.2 / -0

There is a definite relationship between figures (1) and (2). Establish a similar relationship between figures (3) and (4) by selecting a suitable figure from the options that would replace the question mark (?) in figure (4).

A

B

C

D

Solution

The triangle slides downwards and the square gets replaced by symbol S though 45° rotation. Hence, only option (3) satisfies the above condition.

Question 7

1.2 / -0

Read the following information carefully and answer the question given below it.

(i) There is a group of six persons A, B, C, D, E and F in a family. They are Psychologist, Manager, Lawyer, Jeweler, Doctor and Engineer.
(ii) The doctor is the grandfather of F who is a Psychologist.
(iii) The Manager D is married to A.
(iv) C, the Jeweler, is married to the Lawyer.
(v) B is the mother of F and E.
(vi) There are two married couples in the family.

How is A related to E?

Brother

Uncle

Father

Grandfather

Solution

A: Doctor	D: Manager	Generation 1
C: Jeweler	B: Lawyer	Generation 2
E: Engineer	F: Psychologist	Generation 3

Since, E and F are siblings; hence they are at the bottom of the generation tree. Also, as per the given information, Doctor is the grandfather, implies, he is at the top of the family tree. Since, C, the jeweler, is married to the Lawyer; hence the only possible place available is the Generation 2. Also, it satisfies the condition of two married couples in the family.

Question 8
1.2 / -0

Arun is fifth from the left end and Navin is twelfth from the right end in a row of children. If Navin shifts by three places towards Arun, he becomes tenth from the left end. How many children are there in the row?

A
21

B
22

C
23

D
24

Solution

Required no. of children in the row = (12 + 3 + 10) - 1 = 25 - 1 = 24.
Question 9
1.2 / -0

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

A

B

C

D

Solution

After using the figure in option (4), Fig. (X) will look like:
Question 10
1.2 / -0

If 'Book' is called 'Pen', 'Pen' is called 'Paper', 'Paper' is called 'Eraser', 'Eraser' is called 'Scale' and 'Scale' is called 'Calculator', then with which does a student write?

A
Book

B
Paper

C
Pen

D
Calculator

Solution

A student writes with a pen.
Here, 'Pen' is called 'Paper'. So, the student writes with Paper. Hence, option (2) is the correct answer.
Question 11
1.2 / -0

Select a figure from the options that illustrates the relationship amongst ''pigeons, birds, dogs''.

A

B

C

D

Solution

Relationship means there should be an intersection among these. Pigeon and birds are of same category. Hence, most suitable option would be (1).
Question 12
1.2 / -0

If the first and the second letters of the word MISJUDGEMENTS are interchanged with the last and the second last letters respectively, and similarly the third and the fourth letters are interchanged with the third last and the fourth last letters respectively, and so on, then what will be the fifth letter to the right of the third letter from the left end?

A
E

B
G

C
D

D
T

Solution

After changing the letters of the given word it becomes,
^{_{MISJUDGEMENTS}_{STNEMEGDUJSIM}}
Now the fifth letter to the right of third letter from the left end is eighth (3 + 5) letter from the left end, i.e. D.
Question 13
1.2 / -0

Two positions of a dice are given below. When 1 dot is at the top, which number will be at the bottom?

A
2

B
3

C
4

D
6

Solution

In figure 1: Going in clockwise direction, 2 1 5
In figure 2: Going in clockwise direction, 2 6 3
i.e. 1 is opposite to 6, 5 and 3 are opposite to each other.
When 1 is at the top, then 6 number will be at the bottom.
Question 14
1.2 / -0

Directions: In the following question, the symbols @, ©, $ and % are used with the different meanings as given below:

A $ B means 'A is not smaller than B'.
A @ B means 'A is neither smaller than nor equal to B'.
A © B means 'A is neither smaller than nor greater than B'.
A % B means 'A is neither greater than nor equal to B'.

Now, assuming the following statements to be true, find out which of the three conclusions, I, II and III given below them is/are definitely true and give your answer accordingly.

Statements:
K © P, P @ Q, Q $ R
Conclusions:
I. K @ R
II. R % P
III. Q % K

A
Only I and II are true.

B
Only II and III are true.

C
Only III is true.

D
All I, II and III are true.

Solution

A $ B: A ≥ B
A @ B: A > B
A © B: A = B
A % B: A < B
Now,
Given K = P, P > Q, Q ≥ R
Or K = P > Q ≥ R
Therefore, K > R; R < P; Q < K.
Hence, all I, II and III are true.
Question 15
1.2 / -0

Select a number which will replace the question mark (?) in the number pattern given below.

A
16

B
15

C
28

D
21

Solution

First puzzle: 4 × 2 = 8, 8 + 6 = 14
Second puzzle: 3 × 5 = 15, 15 + 6 = 21
Third puzzle: 2 × 11 = 22, 22 + 6 = 28
Hence, option (3) is the correct answer.
Question 16
1.2 / -0

Find the remainder when 2x³- 9x² + x + 12 is divided by 2 + 3x.

A
0

B

C
-2

D

Solution

Let P(x) = 2x³ - 9x² + x + 12
By remainder theorem, when P(x) is divided by 2 + 3x, the remainder will be P(-2/3).

So, P(-2/3) = 2(-2/3)³ - 9(-2/3)² + (-2/3) + 12 =
Question 17
1.2 / -0

In which quadrant does the point P(x, y) lie if xy < 0?

A
I or II

B
II or IV

C
I or III

D
III or IV

Solution

For xy < 0, either x < 0 or y < 0.
If x < 0, P(-x, y) would lie in the second quadrant.
If y < 0, P(x, -y) would lie in the fourth quadrant.
Hence, option (2) is the correct answer.
Question 18
1.2 / -0

If x² + = 98, find the value of x³ + .

A
890

B
970

C
990

D
1110

Solution

a³ + b³ = (a + b)(a² - ab + b²)
x³ + (1/x³) = (x + 1/x){x² - (x)(1/x) + (1/x²)} = (x + 1/x) (x² + (1/x²) - 1) ... (1)
Also,
(x + 1/x)² = (x² + 2(x)(1/x) + 1/x²) = x² + (1/x²) + 2 = 98 + 2 = 100
So, x + (1/x) = 10
Now, x³ + (1/x³) = (10)(98 - 1) = 970 (using 1)
Question 19
1.2 / -0

The factors of 8a³ + b³ - 6ab + 1 are

A
(2a + b - 1)(4a² + b² + 1 - 3ab - 2a)

B
(2ab - b + 1)(4a² + b² - 4ab + 1 - 2a + b)

C
(2a + b + 1)(4a² + b² + 1 - 2ab - b - 2a)

D
(2a - 1 + b)(4a² + 1 - 4a - b - 2ab)

Solution

8a³+ b³ +1 - 6ab
= (2a)³ +b³+ 1 × 1 × 1 - 3 × 2a × b × 1
= (2a + b + 1){(2b)² + (b)²+ 1 × 1 - 2a × b - b × 1 - 1 × 2a}
= (2a + b + 1)(4b² + b² + 1 -2ab - b - 2a)
Question 20
1.2 / -0

Evaluate: (2x - y + 3z)(4x² + y² + 9z² + 2xy + 3yz - 6xz)

A
8x³ - y³ + 27z³ - 18xyz

B
8x³ - y³ + 27z³ + 18xyz

C
8x³ + y³ + 27z³ + 18xyz

D
8x³ + y³ - 27z³ + 18xyz

Solution

We know, (a - b + c)(a² + b²+ c² + ab + bc - ca) = a³ - b³ + c³ + 3abc
Here, a = 2x, b = -y, c = 3z
So, the given expression can be written as:
(2x - y + 3z)(4x² + y² + 9z² + 2xy + 3yz - 6xz) = (2x)³ - y³ + (3z)³ + 3(2x)(-y)(3z) = 8x³ - y³ + 27z³ - 18xyz
Question 21
1.2 / -0

A box of chocolates contains 5 chocolates with hard centre and 4 with soft centre. Amit takes a chocolate, selected at random, from the box and eats it. Ajay then takes a chocolate, selected at random, from the box. Find the probability that Amit and Ajay both choose a chocolate with hard centre.

A

B

C

D

Solution

There are 5 chocolates with hard centre and 4 with soft centre. So, total chocolates = 9

If Amit takes a chocolate, then probability of choosing a chocolate with hard centre =

Now, remaining chocolates = 8, in which 4 chocolates are with hard centre and 4 with soft centre.

If Ajay takes a chocolate, then probability of choosing a chocolate with hard centre =

So, required probability =
Question 22
1.2 / -0

In the given figure, ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On the base BC, a parallelogram DBEC of area same as that of ABC is constructed. Find the height DF of the parallelogram.

A
3 cm

B
5 cm

C
6 cm

D
7 cm

Solution

Semiperimeter of triangle ABC = s = = 10.5

So, area of the triangle = = 21 sq. cm
Now, area of the parallelogram = 21 cm²
So, 7 × DF = 21
i.e. DF = 3 cm
Question 23
1.2 / -0

The number of dimensions a point has is

A
0

B
1

C
2

D
3

Solution

A point has zero dimensions, as it can be marked anywhere.
Question 24
1.2 / -0

A solid iron rectangular block of dimensions (2.2 m × 1.2 m × 1 m) is cast into a hollow cylindrical pipe of internal radius 35 cm and thickness 5 cm. Find the length of the pipe.

A
20.5 m

B
24.5 m

C
22.4 m

D
18.4 m

Solution

Dimensions of the cuboidal block = 2.2 m × 1.2 m × 1 m = 220 cm × 120 cm × 100 cm
Volume of the cuboidal block = 220 × 120 × 100 cm³
Internal radius of the hollow cylinder (r) = 35 m
Thickness of the hollow cylinder = 5 cm
External radius of the hollow cylinder pipe = 35 + 5 = 40 cm
Let the length of the hollow cylinder pipe be h cm.
Volume of the hollow pipe = (R² - r²) h

= (40² - 35²) × h cm³

= × (40 - 35) × (40 + 35) × h cm³

= × 5 × 75 × h cm³

Here, volume of the cuboidal block = Volume of the hollow cylinder
220 × 120 × 100 = × 5 × 75 × h

h = = 2240 cm = 22.40 m
Length of the pipe = 22.4 m
Question 25
1.2 / -0

In the given figure, ABCD and ABEF are two cyclic quadrilaterals. If BCD = 110°, then BEF is equal to

A
55°

B
70°

C
90°

D
110°

Solution

We know that opposite angles of a cyclic quadrilateral are supplementary.
So, angle BAD = 70°
Hence, angle BEF = 110°
Question 26
1.2 / -0

Simplify:

A
3

B
2

C
4

D
0

Solution

Rationalising the denominator, we get

= --- (1)

Similarly, we can find the values as:

= --- (2)

And = --- (3)

Adding (1), (2) and (3), we get
Question 27
1.2 / -0

The figure below is the net of a prism made up of identical triangles. What is the total area of the faces of the prism if the side of the square is 6 cm?

A
75 cm²

B
84 cm²

C
95 cm²

D
56 cm²

Solution

Base of the triangle = side of the square = 6 cm

Height of the triangle =

Area of 4 triangles = 4 × ( × base × height) = 2 × 4 × 6 = 48 cm²

Area of the square = 6 × 6 = 36 cm²Total surface area = (48 + 36) cm² = 84 cm²
Hence, option (2) is the correct answer.
Question 28
1.2 / -0

The mean of 25 numbers is 8. If 2 is added to each number, what will be the new mean?

A
10

B
6

C
8

D
4

Solution

Total sum after adding 2 in each term = 200 + 2 × 25 = 250

So, mean = = 10

Alternative Method: We know that adding a constant value, c, to each term increases the mean by the constant.
Given old mean = 8
So, after adding 2 in each term, new mean = 8 + 2 = 10
Question 29
1.2 / -0

Euclid stated that all right angles are equal to each other in the form of a/an

A
axiom

B
definition

C
postulate

D
proof

Solution

Euclid stated that all right angles are equal to each other in the form of fourth postulate.
The given statement is a postulate.
Question 30
1.2 / -0

In the given figure,

(i) AB BF and EF BF
(ii) AC = BC
(iii) KD is perpendicular to BC and DE.

Find the measure of x.

A
75°

B
30°

C
60°

D
45°

Solution

In CBZ, 90° + 35° + = 180°
= 55°

In △KDC, 55° + 90° + = 180°
= 35°
Now, we know that KD is perpendicular to BC and DE.

Hence, 35° + = 90°
= 55°

Now, using the property that exterior angle is the sum of two interior angles in DZF, we get
90° + 25° = x + 55°
x = 60°
Question 31
1.2 / -0

If 2^x = 4^y = 8^z and , then the value of z is

A

B

C

D

Solution

Given: 2^x = 4^y = 8^z2^x = 2^2y = 2^3z
This gives x = 2y = 3z.
Putting these values in the other given statement: , we get

So, z =
Question 32
1.2 / -0

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.

A
125

B
50

C
100

D
150

Solution

Surface area of the cube = 6a²Each edge of the cube is increased by 50%.
New surface area of the cube = 6(1.5a)² = 2.25(6a²)
Percentage increase in the surface area of the cube = = 125%
Question 33
1.2 / -0

What is the area of the shaded part in the given figure? (Take )

A
72 cm²

B
108 cm²

C
324 cm²

D
648 cm²

Solution

Area of the shaded region = (where r = 18 cm)
So, area = 36 × 18 = 648 cm²
Question 34
1.2 / -0

In the given figure, AB II CD and EF II DQ. Determine PDQ, AED and DEF, respectively.

A
34°, 68°, 68°

B
68°, 34°, 68°

C
68°, 68°, 68°

D
34°, 34°, 68°

Solution

AED = 34° (Corresponding angles are equal)
DEF = 68° (Since AEB is a straight line, 180° - 34° - 78° = 68°)
PDQ = DEF = 68°(Corresponding angles are equal)
Question 35
1.2 / -0

The pie chart shows the grades attained by a group of students in a common test.

If 240 students sat for the common test and those who obtained grades D and E failed the test, then how many students passed the test?

A
42

B
182

C
215

D
204

Solution

Total number of students = 240
Number of students who obtained grade C = 25% of 240 = 60
Number of students who obtained grade D = 10% of 240 = 24
Number of students who obtained grade E = 5% of 240 = 12
Number of students who obtained grade A = 20% of 240 = 48
Percentage of students who obtained grade B = 100 - (25 + 10 + 5 + 20) = 40%
So, number of students appear in B = 40% of 240 = 96
Number of students passed = (C + A + B) = 60 + 48 + 96 = 204
Question 36
1.2 / -0

The taxi fare in a city is Rs. 25 for first kilometre and Rs. 10.5 for next each subsequent kilometre. A traveller is charged Rs. 109 as the fare. How many kilometres did he travel?

A
8 km

B
9 km

C
10 km

D
None of these

Solution

For 1 km, amount charged by the taxi = Rs. 25
Total fare paid by the traveller = Rs. 109
Remaining subsequent charges of distances covered by the taxi = 109 - 25 = Rs. 84
Distance covered = (84/10.5) + 1(Distance covered in first one km) = 8 + 1 = 9 km
Question 37
1.2 / -0

A bookseller procures 40 books for 3200 and sells them at a profit equal to the selling price of 8 books. What is the selling price of one dozen books, if the price of each book is same?

A
720

B
960

C
1200

D
1440

Solution

CP of 40 books = Rs. 3200

CP of one book = = Rs. 80

SP of 40 books = CP of 40 books + SP of 8 books
SP of 32 books = CP of 40 books
SP of 32 books = 3200
SP of one book = Rs. 100
SP of one dozen books = 12 × 100 = Rs. 1200
Question 38
1.2 / -0

Four years ago, the average age of a family of 6 members was 25 years. Meanwhile, a child was born in the family; still the average age of the whole family is same today. The present age of the child is

A
2 years

B
years

C
1 year

D
Data insufficient

Solution

Total present age of 6 persons excluding the child = 25 × 6 + 4 × 6 = 150 + 24 = 174 years
Let the age of the child be x years.
According to the question,

= 25

174 + x = 175
So, x = 1 year
Question 39
1.2 / -0

A moneylender borrows money at 4% p.a. and pays interest at the end of the year. He lends it at 6% p.a. compounded half-yearly and receives the interest at the end of the year. Thus, he gains Rs. 104.50 a year. The amount of money he borrows is

A
Rs. 4500

B
Rs. 5000

C
Rs. 5500

D
Rs. 6000

Solution

Let the borrowed money be Rs. x.
Using the formula of Cl and SI, we get

Amount he has to return =

Amount he will get = x

His gain = ^{= 104.50}

Solving the above equation, we get

x =

x = Rs. 5000
Question 40
1.2 / -0

A certain factory employed 600 men and 400 women and the average wage was Rs. 25.50 per day. If a woman got Rs. 5 less than a man, then what were the daily wages of a man and a woman, respectively?

A
Rs. 25; Rs. 20

B
Rs. 27.50; Rs. 22.50

C
Rs. 30; Rs. 25

D
Rs. 32.50; Rs. 27.50

Solution

Let the daily wage of a woman be Rs. x.
So, daily wage of a man = Rs. (x + 5)
According to the question,

= 25.50

600x + 3000 + 400x = 25,500
1000x = 25,500 - 3000 = 22,500
This gives x = Rs. 22.50.
So, daily wage of a woman = Rs. 22.5 and daily wage of a man = Rs. 27.5
Question 41
1.2 / -0

Three lightships flash simultaneously at 6:00 a.m. The first lightship flashes every 12 seconds, the second lightship every 30 seconds and the third lightship every 66 seconds. At what time will the three lightships next flash together?

A
6:09 a.m.

B
6:10 a.m.

C
6:11 a.m.

D
6:12 a.m.

Solution

LCM of 12, 30 and 66 is 660.
Hence, all the three lightships will flash together after 660 seconds.
Since, 60 seconds = 1 minute
Therefore, 660 seconds = 11 minutes
So, the time would be 6.11 a.m. when all the three lightships flash together next.
Question 42
1.2 / -0

A man earns Rs. 20 on the first day and spends Rs. 15 on the next day. He again earns Rs. 20 on the third day and spends Rs. 15 on the fourth day. If he continues to save like this, then how soon will he have Rs. 60 in hand?

A
On 17^th day

B
On 27^th day

C
On 30^th day

D
On 24^th day

Solution

The last day's earning = Rs. 20
So, the earning required before last day = 60 - 20 = Rs. 40
Now, 40/(20 - 15) = 8
So, total number of days = 8 × 2 + 1 = 17 days
Question 43
1.2 / -0

The price of a car depreciates in the first year by 25%, in the second year by 20%, in the third year by 15% and so on. The final price of the car after 3 years(in Rs.), if the present cost of the car is 10,00,000 is

A
7,80,000

B
5,10,000

C
6,90,000

D
1,70,000

Solution

The final price of the car after 3 years =

=
= 75 × 80 × 85
= Rs. 5,10,000
Question 44
1.2 / -0

A car travelling at of its usual speed covers 42 km in 1 hour 40 mins 48 secs. What is the usual speed of the car?

A
km/hr

B
25 km/hr

C
30 km/hr

D
37.5 km/hr

Solution

Let usual speed of car = x

Speed =

Time = 1 hr 40 min 48 sec = 1.68 hr

x =

x = 37.5 km/hr

Question 45

1.2 / -0

Cubical boxes of volume 15,625 cm³ each are put in a cubical store of side 2.5 m.

(i) How many such boxes can be put in the store?
(ii) What are the dimensions of the box?

(i)	(ii)
(A) 1250	15 cm
(B) 1000	15 cm
(C) 1250	25 cm
(D) 1000	25 cm

(A)

(B)

(C)

(D)

Solution

V = 15,625 cm³
Side = 2.5 m = 2.5 × 100 cm = 250 cm
Volume of cubical store = Side³ = (250)³ = 1,56,25,000 cm³
Number of boxes store contain = = 1000

(ii) Volume of the box = 15,625 cm³
Volume = Side³ = 15,625 cm³
Side³ = 25³
Upon solving we get,
Side = 25 cm
(i) 1000 and (ii) 25

Question 46
1.2 / -0

Consider the figure below.

The ratio of the area of I to that of II is 8 : 1. The area of III is as much as the area of II. If the difference between the area of I and area of II is 252 cm², then

(P) Area of I
(Q) Area of II + III

A
P - 96 cm², Q - 288 cm²

B
P - 288 cm², Q - 96 cm²

C
P - 188 cm², Q - 86 cm²

D
P - 86 cm², Q - 188 cm²

Solution

Let the area of II = x cm²
According to question,
Difference of area = (8 - 1) × sq unit = 252
7x = 252
So, x = 36
Thus, 8x = 8 × 36 = 288 cm²
(P) Area of I = 288 cm²

Area of III = of area of II = of 36 = 60 cm²

(Q) Area of II + III = 36 + 60 = 96 cm²
Question 47
1.2 / -0

Which of the following statements is INCORRECT?

A
If the altitudes of a triangle are equal, then it is equilateral.

B
If in a triangle, two sides are unequal, then the angle opposite to the longer side is greater than the angle opposite to the shorter side.

C
In a triangle, side opposite to the larger angle is longer than the side opposite to the smaller angle.

D
In a triangle, altitude from the vertex bisects the base.

Solution

1) If the altitudes of a triangle are equal, then it is equilateral. This is a true statement, as the area of triangle is proportional to base and it's height.
So, if the altitudes are equal then the base of the triangle is also equal and hence equilateral.
2) If in a triangle, two sides are unequal, then the angle opposite to the longer side is greater than the angle opposite to the shorter side. This is a true statement.
3) In a triangle, side opposite to the larger angle is longer than the side opposite to the smaller angle. This is triangle inequality.
4) In a triangle, altitude from the vertex bisects the base. This is not always true and is possible only when triangle is isosceles or equilateral.
Question 48
1.2 / -0

Fill in the blanks.

(P) Any point lying on x-axis is of the form ______.
(Q) The abscissa of a point on y-axis is _______.
(R) The point at which the two coordinate axes meet is called the ________.
(S) The perpendicular distance of the point (4, 5) from x-axis is ________.
(T) The perpendicular distance of the point (3, 7) from y-axis is ________.

A
(P) - (0, y), (Q) - 1, (R) Origin, (S) - 5, (T) - 3

B
(P) - (x, 0), (Q) - 0, (R) Origin, (S) - 5, (T) - 3

C
(P) - (x, 0), (Q) - 0, (R) Origin, (S) - 3, (T) - 5

D
(P) - (0, y), (Q) - 1, (R) Origin, (S) - 3, (T) - 5

Solution

(P) Any point lying on x-axis is of the form (x, 0) because any point that lie on X axis will have y coordinate 0.
(Q) The abscissa of a point on y-axis is 0 because abscissa of all the points on y axis is 0.
(R) The point at which the two coordinate axes meet is called the origin.
(S) The perpendicular distance of the point (4, 5) from x-axis is 5 because it is five units on Y axis.
(T) The perpendicular distance of the point (3, 7) from y-axis is 3 because it is 3 units on X axis.
Question 49
1.2 / -0

Following are the steps of construction of PQR, given that QR = 3 cm, PQR = 45° and QP - PR = 2 cm.
Arrange them and select the correct option.

(i) Make an angle XQR = 45° at point Q of base QR.
(ii) Join SR and draw the perpendicular bisector of SR say AB.
(iii) Draw the base QR of length 3 cm.
(iv) Let bisector AB intersects QX at P. Join PR.
(v) Cut the line segment QS = QP - PR = 2 cm from the ray QX.

A
^{_(iii)_(ii)_(i)_(v)_(iv)}

B
^{_(iii)_(i)_(ii)_(v)_(iv)}

C
^{_(iii)_(i)_(ii)_(iv)_(v)}

D
^{_(iii)_(i)_(v)_(ii)_(iv)}

Solution

Step I: Draw the line segment QR = 7 cm.
Step II: Make an angle XQR = 45° at point Q of base QR.
Step III: Cut the line segment QS = 2 cm (equal to PQ - PR) on ray QX.
Step IV: Join SR and draw the perpendicular bisector SR.
Step V: Let it intersect QX at point P. Join PR. PQR is the required triangle.
Question 50
1.2 / -0

Select the correct option.

Statement 1: (x + y)³ = x³ + y³ + 3xy(x + y) and (x² + y²) = (x + y)² + 2xy

Statement 2: If x² + = 7, then the value of x³ + = 18

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

Statement 1: (x + y)³ = x³ + y³ + 3xy(x + y) and (x² + y²) = (x + y)² + 2xy is true.

Statement 2: If x² + = 7, then the value of x³ + = 18
x² + = 7

Adding two on both sides, we get

x² + + 2 = 9

Cubing both sides, we get

Hence, both the statements are true.