Explore topic-wise InterviewSolutions in .

(i) False, because ax + by + c = 0 is a linear equation in two variables if both a and b are non-zero. 

(ii) False, because a linear equation in two variables has infinitely many solutions. 

(iii) False, the points (2, 0), (–3, 0) lie on the x-axis. The point (4, 2) lies in the first quadrant. The point (0, 5) lies on the y-axis. 

(iv) True, since the line parallel to y-axis at a distance a units to the left of y-axis is given by the equation x = – a. 

(v) False, because x = 0, y = 0 does not satisfy the equation.

The graph of the linear equation 3x + 4y = 12 cuts the x-axis at the point where y = 0. On putting y = 0 in the linear equation, we have 3x = 12, which gives x = 4. Thus, the required point is (4, 0).

The graph of the linear equation 3x + 4y = 12 cuts the y-axis at the point where x = 0. On putting x = 0 in the given equation, we have 4y = 12, which gives y = 3. Thus, the required point is (0, 3).

1. I think migrants are trouble shooters.

2. In almost all examples from our lesson, migrants served the purpose. 

3. They were nowhere trouble makers. 

4. But for them, most works cannot be easily completed. 

5. Though they lived in inhumane conditions, they worked hard. 

6. Apart from these, they eke out a living.

The coordinates of the points lying on the line parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of the x-axis are of the form (2, a). Putting x = 2, y = a in the equation x + y = 5, we get a = 3. Thus, the required point is (2, 3).

Correct option is (C) 184

Let the numbers are x & y.

\(\therefore\) Their difference = 115

\(\therefore\) x - y = 115       __________(1)

The ratio of two numbers is 3 : 8.

\(\therefore\) \(\frac yx=\frac 38\)      \((\because\) Largest number is x)

\(\Rightarrow\) 8y = 3x

\(\Rightarrow\) 8 (x - 115) = 3x     (From (1))

\(\Rightarrow\) 8x - 920 = 3x

\(\Rightarrow\) 8x - 3x = 920

\(\Rightarrow\) 5x = 920

\(\Rightarrow\) x = \(\frac{920}5\) = 184

\(\therefore\) Largest number is 184.

Correct option is C) 184

As the x-coordinate of the point is 5/2 times its ordinate, therefore, x = 5/2 y. 

Now putting x = 5/2 y in 2x + 5y = 20, we get, y = 2. Therefore, x = 5. Thus, the required point is (5, 2).

Correct option is (B) x = 2, y = -3

Given system of equations is

2x + 3y + 5 = 0           __________(1)

and 3x – 2y – 12 = 0  __________(2)

Multiply equation (1) by 2 and equation (2) by 3, we get

4x + 6y + 10 = 0     __________(3)

9x - 6y - 36 = 0      __________(4)

By adding equations (3) & (4), we get

13x - 26 = 0

\(\Rightarrow x=\frac{26}{13}=2\)

Then from (1), we have

\(y=\frac{-(2x+5)}3=\frac{-(2\times2+5)}3\)

\(=\frac{-9}3=-3\)

Hence, the solution of given system of equation is x = 2, y = -3.

Correct option is B) x = 2, y = -3

According to the question,

A linear equation such that each point on its graph has an ordinate(y) which is 3 times its abscissa(x).

So we get,

⇒ y = 3x.

Hence, y = 3x is the required linear equation.

Correct option is A) x = 1, y = 2

From the question, we have,

2x + 5y = 19 …(i)

According to the question,

Ordinate is 1½ times its abscissa

⇒ y = 1½ x = (3/2) x

Substituting y = (3/2)x in eq. (i)

We get,

2x + 5 (3/2) x = 19

(19/2)x = 19

x = 2

Substituting x = 2 in eq. (i)

We get

2x + 5y = 19

2(2) + 5y = 19

y = (19 – 4)/5 = 3

Hence, we get x =2 and y = 3

Thus, point (2, 3) is the required solution.

Given equations are

2x + y = 23 …(i)

4x – y = 19 …(ii)

On adding both equations, we get

6x = 42

So, x = 7

Put the value of x in Eq. (i), we get

2(7) + y = 23

y = 23 – 14

so, y = 9

Hence 5y – 2x = 5(9) – 2(7) = 45 – 14 = 31

y/x – 2 = 9/7 -2 = -5/7

Hence, the values of (5y – 2x) and y/x – 2 are 31 and -5/7 respectively.

Using property of rectangle,

We know that,

Lengths are equal,

i.e., CD = AB

Hence, x + 3y = 13 …(i)

Breadth are equal,

i.e., AD = BC

Hence, 3x + y = 7 …(ii)

On multiplying Eq. (ii) by 3 and then subtracting Eq. (i),

We get,

8x = 8

So, x = 1

On substituting x = 1 in Eq. (i),

We get,

y = 4

Therefore, the required values of x and y are 1 and 4, respectively.

If O be the origin, then

R = {(P, Q) : OP = OQ}

Reflexivity : ∀ point P ∈ A

OP = OP 

⇒ (P, P) ∈ R

i.e., R is reflexive.

Symmetry : Let P, Q ∈ A, such that (P, Q) ∈ R

OP = OQ

⇒ OQ = OP

⇒ (Q, P) ∈ R

i.e., R is symmetric.

Transitivity : Let P, Q, S ∈ A, 

such that (P, Q) ∈ R and (Q, S) ∈ R

OP = OQ and OQ = OS

OP = OS 

⇒ (P, S) ∈ R

i.e., R is transitive.

Now,  

we have R is reflexive, symmetric and transitive.

Therefore, R is an equivalence relation.

Let P, Q, R... be points in the set A, such that

(P, Q), (P, R)... ∈ R

⇒ OP = OQ; OP = OR; ... [where O is origin]

⇒ OP = OQ = OR = ...

i.e., All points P, Q, R ... ∈ A, which are related to P are equidistant from origin ‘O’.

Hence, set of all points of A related to P is the circle passing through P, having origin as centre.

It is given in the question that, 

Diameter of the cylindrical pillars = 48 cm 

Hence, 

The radius of the cylindrical pillars = \(\frac{48}{2}\)

= 24 cm 

= 0.24 m 

Height of the cylindrical pillars = 7 m 

Now, 

Lateral surface area of one pillar = \(\pi dh\)

\(=\frac{22}{7}\times0.48\times7\)

= 10.56 m² 

Now, 

The surface area to be painted = total surface area of 15 pillars 

= 15 × 10.56 

= 158.4 m² 

Therefore, 

The total cost of painting = Rs(158.4 × 2.5) 

= Rs 396

In the above given question, 

In order to find the greater capacity pack 

At first, we’ll calculate the volume of the cubic pack, 

Length of the side of pack, a = 5 cm 

Height of the pack, h = 14cm 

Hence, 

Volume of the pack = a²h 

= (5)²(14) 

= 5 × 5 × 14 

= 350 cm³ 

Now, 

We’ll calculate the volume of the cylindrical pack, 

Radius of the base, r = 35 cm 

Height of the cylinder, h = 12 cm 

Hence, 

Volume of the pack = \(\pi r^2h\)

= \(\frac{22}{7}\times35\times35\times12\)

= 22× 5× 35× 12 

= 462 cm³ 

Hence, 

It’s clear that the pack with the circular has a greater capacity than the than the pack with square base. 

And, 

The deference between their volume= 462 – 350 

= 112 cm³

(b) 5 < l < 5.5.

Original volume = (6 × 5 × 2) cm³ = 60 cm³ 

New volume = (60 – 20% of 60) cm³ = 48 cm³ 

Since the thickness remains the same, h = 2 cm 

⇒ l × b × 2 = 48 ⇒ l × b = 24 

New length = (6 – 10% of 6) cm = 5.4 cm 

New breadth = (5 – 10% of 5) cm = 4.5 cm 

Here l × b = 5.4 × 4.5 = 24.3 cm² 

∴ l should clearly lie between 5 and 5.5, i.e., 5 < l < 5.5.

2 – y = 8

=> 2 - 8 = y

=> - 6 = y

y = - 6

8.4 – x = - 2

=> 8.4 + 2 = x

= x = 10.4

5 – x = 3

=> 5 - 3 = X

X = 2

A transversal intersects two or more than two lines at distinct points.

If the sum of measures of two angles is 180^o, then they are supplementary.

If sum of measures of two angles is 90^o, then the angles are complementary.

Total measure of angles of the wheel = 360°

6 spokes = 360°

1 spoke = 60°

So, measure of the angle between 2 adjacent spokes = 2 x 60° = 120°

Consider the two adjacent angles as 5x and 4x.

We know that the two adjacent angles form a linear pair

So it can be written as

5x + 4x = 180^o

On further calculation

9x = 180^o

By division

x = 180/90

x = 20^o

Substituting the value of x in two adjacent angles

5x = 5 (20)^o = 100^o

4x = 4 (20)^o = 80^o

Therefore, the measure of each one of these angles is 100^o and 80^o

From the figure we know that ∠COE and ∠EOD form a linear pair

∠COE + ∠EOD = 180^o

It can also be written as

∠COE + ∠EOA + ∠AOD = 180^o

By substituting values in the above equation we get

5x + ∠EOA + 2x = 180^o

From the figure we know that ∠EOA and ∠BOF are vertically opposite angles

∠EOA = ∠BOF

So we get

5x + ∠BOF + 2x = 180^o

5x + 3x + 2x = 180^o

On further calculation

10x = 180^o

By division

x = 180/10 = 18

By substituting the value of x

∠AOD = 2x^o

So we get

∠AOD = 2 (18)^o = 36^o

∠EOA = ∠BOF = 3x^o

So we get

∠EOA = ∠BOF = 3 (18)^o = 54^o

∠COE = 5x^o

So we get

∠COE = 5 (18)^o = 90^o

A. 1 : 4

Radius of hemisphere initially = 6 cm

Radius of hemisphere finally = 12 cm

Surface area of hemisphere = 2 x π x r²

Thus ratio of the surface areas = (2 x π x 6²)/(2 x π x 12²)

= 1/4

We know that, 

Volume of a cuboid = Length × Breadth × Height 

Therefore, 

Volume of the room = 10 × 8 × 3.3 

= 264 m³ 

Space required by 1 person = 3 m³ 

Therefore, 

Total number of people that can be accommodated = \(\frac{264}{3}\)

= 88 

Hence, option B is correct

 (i) Let A = \(\begin{pmatrix} 5 &6 \\[0.3em] 7 & 8 \end{pmatrix}\)

order of A is 2 x 2.

ρ(A) ≤ 2 

Consider the second order minor 

\(\begin{vmatrix} 5 &6 \\[0.3em] 7 & 8 \end{vmatrix}\)= 40 – 42 = -2 ≠ 0 

There is a minor of order 2, which is not zero. 

ρ(A) = 2

(ii) Let A = \(\begin{pmatrix} 1 &-1 \\[0.3em] 3 & -6 \end{pmatrix}\)

Order of A is 2 × 2 

ρ(A) ≤ 2 

Consider the second order minor 

\(\begin{vmatrix} 1 & -1 \\[0.3em] 3 & -6 \end{vmatrix}\)= -6 + 3 = -3 ≠ 0

ρ(A) = 2

 (iii) Let A = \(\begin{pmatrix} 1 &4 \\[0.3em] 2 & 8 \end{pmatrix}\)

Since A is of order 2 × 2, ρ(A) ≤ 2 

Now \(\begin{vmatrix} 1 & 4 \\[0.3em] 2 & 8 \end{vmatrix}\) = 8 – 8 = 0 

Since second order minor vanishes ρ(A) ≠ 2 But first order minors, |1|, |4|, |2|, |8| are non zero. ρ(A) = 1

(iv) Let A =\( \begin{pmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{pmatrix} \)

Oradar of A is 3 x 3

ρ(A) ≤ 3 

Consider the third order minor

\( \begin{pmatrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{pmatrix} \)= 2(1 + 5) + 1 (3 + 5) + 1(3 – 1) 

= 2(6) + 8 + 2 

= 22 ≠ 0 

There is a minor of order 3, which is non zero. ρ(A) = 3

Given A = \(\begin{pmatrix}-2 & 1 & 3 & 4 \\0 & 1 & 1 & 2\\1 & 3 & 4 & 7\end{pmatrix}\)

order of A is 3 x 4.So p(A) ≤ 3

Consider the third order minor

\(\begin{vmatrix}-2 & 1 & 3 \\0 & 1 & 1 \\1 & 3 & 4\end{vmatrix}\)

= -2(4 - 3) - 1(0 - 1) + 3(0 - 1)

= -2 + 1 -3

= -4 ≠ 0

There exists a minor of order 3 which is not zero. So ρ(A) = 3

Given A = \(\begin{pmatrix}4 & 5 & 2 & 2 \\3 & 2 & 1 & 6\\4 & 4 & 8 & 0\end{pmatrix}\)

order if A is 3 x 4.ρ(A) ≤ 3

Consider the third order minor,

\(\begin{vmatrix}4 & 5 & 2 \\3 & 2 & 1 \\4 & 4 & 8\end{vmatrix}\)

= 4(16 – 4) – 5(24 – 4) + 2 (12 – 8) 

= 48 – 100 + 8 

= -44 ≠ 0 

There is a minor of order 3 which is not zero. ρ(A) = 3

Lens forms an image through converging light rays or diverging light rays.

A light ray will deviate from its path in some cases and does not deviate in some other cases.

A lens is formed when a transparent material is bounded by two spherical surfaces.

The mid point of lens is called pole (or) optical centre.

The lens having two spherical surfaces curved inward is called a double concave lens.

The lens having two spherical surfaces bulging outwards is called double convex lens.

B) The ray does not deviate from its path

The distance between the centre of curvature and curved surface is called radius of curvature

The centre of sphere which contains the part of curved surface is called centre of curvature.

It is thin at the middle and thicker at the edges.

It is thick at the middle as compared to edges.

The process by which a network of cross links is introduced into an elastomer is called vulcanization.

Vulcanization enhances the properties of natural rubber like tensile strength, stiffness, elasticity, toughness etc. Sulphur forms cross links between polyisoprene chains which results in improved properties of rubber.

• For neoprene vulcanizing agent is MgO. 
• For Buna-N vulcanizing agent is sulphur.

We have,

(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0

Upon expansion we get,

x² + 5x + 6 + x² – 5x +6 – 2x² – 2x =0

-2x + 12 = 0

By dividing the equation using -2 we get,

x – 6 = 0

x = 6

∴ x = 6

We have,

x/2 – 4/5 + x/5 + 3x/10 = 1/5

upon solving we get,

x/2 + x/5 + 3x/10 = 1/5 + 4/5

by taking LCM for 2, 5 and 10 which is 10

(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5

5x/10 + 2x/10 + 3x/10 = 1

(5x+2x+3x)/10 = 1

10x/10 = 1

x = 1

∴ x = 1

We have,

7/x + 35 = 1/10

7/x = 1/10 – 35

= ((1×1) – (35×10))/10

= (1 – 350)/10

7/x = -349/10

By using cross-multiplication we get,

x = -70/349

∴ x = -70/349

Correct option is  B) 38

Correct option is (B) 38

0.3x. + 0.4 = 0.28x + 1.16

\(\Rightarrow\) 0.3x - 0.28x = 1.16 - 0.4

\(\Rightarrow\) 0.02x = 0.76

\(\Rightarrow\) \(x=\frac{0.76}{0.02}=\frac{76}{2}=38.\)

Correct option is  A) 21

Correct option is (A) 21

2x – 7 = 35

\(\Rightarrow\) 2x = 35+7 = 42

\(\Rightarrow\) x = \(\frac{42}2\) = 21.

A) 10 × ten’s digit + unit’s digit

Correct option is (A) 10 × ten’s digit + unit’s digit 

Let ab be two digit number where a is ten's digit and b is unit's digit.

ab = 10a + b

B) 4x + 5 = 1