Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is (B) CMIE

Correct option is (B) Active learning

Correct option is (D) All of these

Correct option is (D) Both (A) and (C)

Perfect competitive market:

1. Perfect market or perfect competitive market is defined by several characteristics.

Some are:

• Perfectly competitive market is such a market where firms accepts market price and sell their commodities
• Commodities are homogeneous (i.e. the qualities and characteristics of goods or services available in market do not vary between different suppliers)
• There are large number of buyers and sellers
• Buyers and sellers have complete knowledge of market situation
• Price is determined by demand and supply of a commodity. Firms sells commodity on this price only and no firm can influence this price. Hence, price is fixed and constant.

2. In perfect competition, market price (P) = Average Revenue (AR) = Marginal Revenue (MR) i.e. (P – AR = MR).
3. Under such market condition, if price of commodity is ₹ 50 then Average
Revenue and Marginal Revenue of firm will also be ₹ 50.
4. As a result, the curves of Average Revenue and Marginal Revenue of the firm are same and also parallel to X-axis. This curve is represented as DD in the diagram.

• In perfectly competitive market, Marginal Revenue (MR) and Average Revenue (AR) are constant and equal. As a result, both of them can be shown through one line or say curve DD.
• Moreover, in perfect competition, market price (P) = Average Revenue (AR) = Marginal Revenue (MR) i.e. (P = AR = MR). Hence, irrespective of any price MR and AR will remain parallel to the X-axis.
• All points on DD curve will show Average Revenue = Marginal Revenue.
• Also, the Average Revenue and Marginal Revenue curve merge into one another and since value of both the revenues are same and constant the slope of curve is zero.
• Hence, in perfect competition, average revenue curve and marginal revenue curve are same and parallel to the X-axis.

Correct option is (D) All of these

Correct option is (C) SAS

 \(A=\begin{bmatrix}2&3&4\\5&6&-10\\8&9&4\end{bmatrix}\)

∵ A = IA

⇒ \(\begin{bmatrix}2&3&4\\5&6&-10\\8&9&4\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}A\)

R₂ → 2R₂ - 5R₁

Applying R₃ → R₃ - 4R₁

\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&-3&-12\end{bmatrix}\)\(=\begin{bmatrix}1&0&0\\-5&2&0\\-4&0&1\end{bmatrix}A\)

Applying R₃ → R₃ - R₂

\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&0&28\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\-5&2&0\\1&-2&1\end{bmatrix}A\)

Applying R₃ → R₃/28

\(\begin{bmatrix}2&3&4\\0&-3&-40\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}1&0&0\\-5&2&0\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)

Applying R₁ → R₁ - 4R₃

R₂ → R₂ + 40R₃

\(\begin{bmatrix}2&3&0\\0&-3&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{6}{7}&\frac{2}{7}&\frac{-1}{7}\\\frac{-25}{7}&\frac{-6}{7}&\frac{10}{7}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)

Applying R₁ → R₁ + R₂

\(\begin{bmatrix}2&0&0\\0&-3&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{-19}{7}&\frac{-4}{7}&\frac{9}{7}\\\frac{-25}{7}&\frac{-6}{7}&\frac{10}{7}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)

Applying R₁ → R₁/2, R₂ → R₂/-3

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\) \(=\begin{bmatrix}\frac{-19}{14}&\frac{-2}{7}&\frac{9}{14}\\\frac{25}{21}&\frac{2}{7}&\frac{-10}{21}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}A\)

We obtain A^-1A = I

Therefore, A^-1 \(=\begin{bmatrix}\frac{-19}{14}&\frac{-2}{7}&\frac{9}{14}\\\frac{25}{21}&\frac{2}{7}&\frac{-10}{21}\\\frac{1}{28}&\frac{-1}{14}&\frac{1}{28}\end{bmatrix}\)

Dc’s are cos α, cos β and cos γ 

cos 90°, cos 135°, cos 45°

\(\Big(0, \frac{-1}{\sqrt 2}, \frac{1}{\sqrt 2}\Big)\)

Let height of cone be h.

And radius of hemisphere (r) = height of the hemisphere (H)

Volume of cone = Volume of hemisphere

\(\frac { 1 }{ 3 }\)πr²h = \(\frac { 2 }{ 3 }\)πr² × H

⇒ h = 2H

⇒ \(\frac { h }{ H }\) = \(\frac { 2 }{ 1 }\)

⇒ h : H = 2 : 1

Given,

Height of cone (h) = 8 cm

base radius (r) = 6 cm

Volume of the cone (V) = \(\frac { 1 }{ 3 }\)πr²h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 6 × 6 × 8

= \(\frac { 2112 }{ 7 }\)

= 301.7 cm³

Hence, volume of the cone = 301.7 cm³

Given,

Slant height of cone (l) = 12 meter

Lateral surface area = 1884.4 m²

Let the radius of base be r.

∴ πrl = 1884.4

\(\frac { 2112 }{ 7 }\) × 12 × r = 1884.4

r = \(\frac { 1884.4\times 7 }{ 22\times 12 }\)

= 50 m(approx)

Diameter (d)

Radius r = 12/2 = 6 cm

Perimeter of half circle = πr + diameter

= 3.14 x 6 + 12 (∵ π = 3.14)

= 30.84 cm.

Let radius of cirlce r cm

Circumference of circle = 44

⇒ 2πr = 44

⇒ 2 x 22/7 x r = 44

r = (44 x 7)/(22 x 7) = 7 cm

∴ Area of circle = πr²

= 22/7 x 7 x 7 

= 22 x 7

= 154 sq. cm

Radius of circle (r) = 21 cm

Circumference of circle = 2πr

= 2 x 22/7 x 21

= 2 x 22 x 3 

= 2 x 66 

= 132 cm

It is given that BO = OD

From the figure we know that AO is the median of △ ABD

So we get

Area of △ AOD = Area of △ AOB ……. (1)

We know that OC is the median of △ CBD

So we get

Area of △ DOC = Area of △ BOC ……. (2)

By adding both the equations

Area of △ AOD + Area of △ DOC = Area of △ AOB + Area of △ BOC

So we get

Area of △ ADC = Area of △ ABC

Therefore, it is proved that ar (△ ABC) = ar (△ ADC).

Radius of wheel (r) = 35 m

Circumference of wheel = 2πr

= 2 x 22/7 x 35 

= 2 x 22 x 5 

= 44 x 5

= 220 m

Wheel covers 220 metre distance revolution.

So to cover a distance of 880 metre number of revolution = 880/220 = 4 revolution.

Correct option is A) 15

Correct option is A) 88 cm

Length of minute hand is r = 14 m

Angle made by minute hand in 1 hour is θ = 360° = 2 π
\(\therefore\) Distance covered by minute hand in 1 hour

 = rθ = 14 x 2π = 14 x 2 x 22/7 = 88 cm

Draw two perpendiculars CP and DQ on AB. 

Now, 

ar(ΔABC) = 1/2 x AB x CP ⋅⋅⋅⋅⋅⋅⋅(1) 

ar(ΔABD) = 1/2 x AB x DQ ⋅⋅⋅⋅⋅⋅⋅(2) 

To prove the result, ar(ΔABC) = ar(ΔABD), we have to show that CP = DQ. 

In right angled triangles, ΔCPO and ΔDQO 

∠CPO = ∠DQO = 90° 

CO = OD (Given) 

∠COP = ∠DOQ (Vertically opposite angles) 

By AAS condition: 

ΔCP0 ≅ ΔDQO 

So, CP = DQ …………..(3) (By CPCT)

From equations (1), (2) and (3), we have 

ar(ΔABC) = ar(ΔABD)

Hence proved.

We know that

Area of △ ABD = ½ × BD × AL

Area of △ ADC = ½ × DC × AL

It is given that BD: DC = m: n

It can be written as

BD = DC × m/n

We know that

Area of △ ABD = ½ × BD × AL

By substituting BD

Area of △ ABD = ½ × (DC × m/n) × AL

So we get

Area of △ ABD = m/n × (1/2 × DC × AL)

It can be written as

Area of △ ABD = m/n × (Area of △ ADC))

We know that

Area of △ ABD/ Area of △ ADC = m/n

We can write it as

Area of △ ABD: Area of △ ADC = m: n

Therefore, it is proved that ar (△ ABD): ar (△ ADC) = m: n.

Correct option is B) 20

Radius of wheel is r = 7m

\(\therefore\) circumference of wheel is P = 2 π r  = 2 π x 7 = 14π m

\(\because\) Distance travel by wheel in one rotation  = 14 π m

\(\therefore\) Number of rotation required to travel distance of 14π m = 1

\(\therefore\) Number of rotation required to travel distance of 880 m

 = 1/14π x 880 = \(\frac{7}{14\times22}\times880=\frac{7\times40}{14}=\frac{40}2=20\)

Given that, 

CD bisected AB at O

To prove : 

Area (ΔABC) = Area (ΔABD)

Construction : 

CP perpendicular to AB and DQ perpendicular to AB

Proof : 

Area (ΔABC) = \(\frac{1}{2}\)(AB x CP) ...(i)

Area (ΔABD) = \(\frac{1}{2}\)(AB x DQ) ...(ii)

In ΔCPO and ΔDQO, 

We have,

∠CPO = ∠DQO 

(Each 90°) 

Given that, 

CO = DO 

∠COP = ∠DOQ

(Vertically opposite angle) 

Then, 

by AAS congruence rule,

ΔCPO ≅ ΔDQO

Therefore, 

CP = DQ 

(By c.p.c.t) 

Thus,

Area (ΔABC) = Area (ΔABD)

Hence, proved

ABCD is a parallelogram. 

When we join the diagonal of parallelogram, it divides it into two quadrilaterals. 

Step 1: Let AC is the diagonal, then, Area (ΔABC) = Area (ΔACD) = 1/2(Area of ||gm ABCD) 

Step 2: Let BD be another diagonal 

Area (ΔABD) = Area (ΔBCD) = 1/2( Area of llgm ABCD) 

Now, 

From Step 1 and step 2, we have 

Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = 1/2(Area of llgm ABCD) 

Hence Proved.

Option : (A)

Given : 

ABCD is a parallelogram 

E and F are the centroids of triangle ABD and BCD 

Since, 

The diagonals of parallelogram bisect each other 

AO is the median of triangle ABD 

And, 

CO is the median of triangle CBD 

EO = \(\frac{1}{3}\)AO 

(Since, centroid divides the median in the ratio 2:1)

Similarly, 

FO = \(\frac{1}{3}\)CO 

EO + FO = \(\frac{1}{3}\)AO +\(\frac{1}{3}\)CO 

= \(\frac{1}{3}\)(AO + CO) 

EF = \(\frac{1}{3}\)AC 

AE = \(\frac{1}{3}\)AO 

= \(\frac{2}{3}\) x \(\frac{1}{2}\) AC 

= \(\frac{1}{3}\)AC 

Therefore, 

EF = AE

We know that, 

Diagonal of parallelogram divides it into two quadrilaterals. 

Since, 

AC is the diagonal 

Then, 

Area (ΔABC) = Area (ΔACD) 

= \(\frac{1}{2}\) Area of parallelogram ABCD ...(i) 

Since, 

BD is the diagonal 

Then, 

Area (ΔABD) = Area (ΔBCD) 

= \(\frac{1}{2}\)Area of parallelogram ABCD ...(ii) 

Compare (i) and (ii), we get 

Therefore, 

Area (ΔABC) = Area (ΔACD) 

= Area (ΔABD) = Area (ΔBCD) 

= \(\frac{1}{2}\)Area of parallelogram ABCD

Answer : (b) 11

For the 1st A.P., 2, 5, 8, 11, ...... , 

First term (a₁) = 2, common difference (d₁) = 3 

∴ The sum of this A.P. to 2n terms

= \( \frac{2n}{2}[2a_1+(2n-1)d_1]\)

= n[4+(2n - 1)3]

= 4n + 6n² – 3n = 6n² + n = n (6n + 1)

For the second A.P., 57, 59, 61, 63, ....... , 

First term (b₁) = 57, common difference (d₂) = 2 

∴ The sum of this A.P. to n term is

\( Sum_n = \frac{n}{2}(2b_1+(n-1)d_2]\)

= \(\frac{n}{2}\)[114 + (n - 1) 2]

= 57n + n² – n = n² + 56n = n (n + 56)

Given, S_2n = Sum_n

⇒ n(6n + 1) = n(n + 56) 

⇒ 6n + 1 = n + 56 

⇒ 5n = 55 

⇒ n = 11.

(3ⁿ – 2n – 1) is divisible by 4 if and only if (3ⁿ – 2n – 1) is a multiple of 4.

Let P(n) ≡ (3ⁿ – 2n – 1) = 4m, where m ∈ N.

Step I: Put n = 1

∴ (3ⁿ – 2n – 1) = 3⁽¹⁾ – 2(1) – 1 = 0 = 4(0)

∴ (3ⁿ – 2n – 1) is a multiple of 4.

∴ P(n) is term for n = 1.

Step II:

Let us assume that P(n) is true for n = k.

∴ 3^k  – 2k – 1 = 4a, where a ∈ N

∴ 3^k = 4a + 2k + 1 ….(i)

Step III:

We have to prove that P(n) is term for n = k + 1,

i.e., to prove that

3^(k+1) – 2(k + 1) – 1 = 4b, where b ∈ N

P(k + 1) = 3^k+1 – 2(k + 1) – 1

= 3^k . 3 – 2k – 2 – 1

= (4a + 2k + 1) . 3 – 2k – 3 …….[From (i)]

= 12a + 6k + 3 – 2k – 3

= 12a + 4k = 4(3a + k)

= 4b, where b = (3a + k) ∈ N

∴ P(n) is term for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is term for all n ∈ N.

∴ 3ⁿ – 2n – 1 is divisible by 4, for all n ∈ N.

(2³ⁿ – 1) is divisible by 7 if and only if (2³ⁿ – 1) is a multiple of 7.

Let P(n) ≡ (2³ⁿ – 1) = 7m, where m ∈ N.

Step I:

Put n = 1

∴ 2³ⁿ – 1 = 2³⁽¹⁾ – 1 = 2  – 1 = 8 – 1 = 7

∴ (2³ⁿ – 1) is a multiple of 7.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k.

i.e., 2^3k  – 1 is a multiple of 7.

∴ 2^3k – 1 = 7a, where a ∈ N

∴ 2^3k = 7a + 1 ……(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

2^3(k+1)  – 1 = 7b, where b ∈ N.

∴ P(k + 1) = 2^3(k+1)  – 1

= 2^3k+3 – 1

= 2^3k . (2³) – 1

= (7a + 1)8 – 1 …..[From (i)]

= 56a + 8 – 1

= 56a + 7

= 7(8a + 1)

7b, where b = (8a + 1) ∈ N

∴ P(n) is true for n = k + 1.

Step IV: From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ (2⁴ⁿ – 1) is divisible by 7, for all n ∈ N.

Let P(n) ≡ (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ,

for all n ∈ N.

Step I: Put n = 1

L.H.S. = (cos θ + i sin θ)¹ = cos θ + i sin θ

R.H.S. = cos[(1)θ] + i sin[(1)θ] = cos θ + i sin θ

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k

 ∴ (cos θ + i sin θ)^k = cos kθ + i sin kθ …….(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that (cos θ + i sin θ)^k+1 = cos (k + 1)θ + i sin (k + 1)θ

L.H.S. = (cos θ + i sin θ)^k+1

= (cos θ + i sin θ)^k . (cos θ + i sin θ)

= (cos kθ + i sin kθ) . (cos θ + i sin θ) …… [From (i)]

= cos kθ cos θ + i sin θ cos kθ + i sin kθ cosθ – sin kθ sin θ ……[∵ i² = -1]

= (cos kθ cos θ – sin k θ sin θ) + i(sin kθ cos θ + cos kθ sin θ)

= cos(kθ + θ) + i sin(kθ + θ)

= cos(k + 1) θ + i sin (k + 1) θ

= R.H.S.

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

∴ (cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ), for all n ∈ N.

Iron and Steel industry is the basic industry. The raw materials are obtained both from Metallic and Non Metallic minerals.

Iron and Steel industries are located in close proximity to the coal fields or Iron ore mines.

The following factors are responsible for the concentration of Jute industries in the Hoogly region in West Bengal.

1. Raw materials: West Bengal is the largest producer of Jute. Availability of raw Jute for production. 

2. Processing: Jute require fresh water for processing. Abundant water supply is available by the riverines and continuous supply of fresh water is ensured due to Perennial nature. 

3. Transport: Well connected by the network of water ways, road ways and railways. 

4. Cheap labour: West Bengal is one of the densely populated area . So cheap labour is available.

5. Market: Kolkatta being one of the textile centre great demand for the product as well as port facilities available for export.

The cotton textile industries contribute about 7% of industrial output, 2% of India’s GDP and 15% of the country’s export earnings. It is one of the largest sources of employment generation in the country. At present there are 1,719 textile mills in the country. Out of which 188 mills are in public sector, 147 in cooperative sector and 1,284 in private sector. Currently, India is the third largest producer of cotton and has the largest loom arc and ring spindles in the world. At present, cotton textile industry is the largest organized modem industry of India. About 16% of the industrial capital, 14% of industrial production and over 20% of the industrial labour of the country are engaged in this industry.

The higher concentration of textile mills in and around Mumbai, makes it as “Manchester of India”. Presence of black cotton soil in Maharashtra, humid climate, presence of Mumbai port, availability of hydro power, good market and well developed transport facility favour the cotton textile industries in Mumbai. The major cotton textile industries are concentrated in the states of Maharashtra, Gujarat, West Bengal, Uttar Pradesh and Tamil nadu. Coimbatore is the most important centre in Tamil nadu with 200 mills out of its 435 and called as “Manchester ‘ of South India”. Erode, Tirupur, Karur, Chennai, Thirunelveli, Madurai, Thoothukudi, Salem and Virudhunagar are the other major cotton textiles centres in the state.

(b) Mumbai high

Manchester of India is Mumbai.

(d) Madhya Pradesh

(a) Petroleum

(b) Renewable resource

Correct Answer is; (a) 5

Any matter or energy derived from the environment that is used by living things including humans is called a natural resource.

Types of Natural Resources are: 

1. Renewable and 

2. Non – renewable resources.

Lignite is extracted in Tamil Nadu Neyveli.

(b) cotton dust

A. (iv) 

B. (iii) 

C. (i) 

D. (v) 

E. (ii)

Bauxite is an aluminium ore.

Answer:C

Bauxite formula is:Al2O3.2H2O

(d) Kanyakumari

(a) Bangalore

(b) Uttar pradesh and Bihar

Renewable resources: 

1. These resources can be replenished after utilization. 

2. Abundantly available in nature. 

3. Eg: Sun energy, Wind etc.

Non-Renewable resources:

1. These resources cannot be regained after utilization. 

2. Limited stock take millions of years for formation. 

3. Eg: Minerals

West coast: Mumbai high, Gujarat coast, Basseim,Aliabet (South of Bhavanagar), Ankleshwar, Cambay – Luni region, Ahmedabad -Kaloi region and Punjab – Haryana.

East Coast: Brahmaputra valley, Digboi, Nahoratitya, Moran-Hugrijan, Rudrasagar-Lawa (Assam region), Surrma Valley. Andaman and Nicobar, Gulf of Mannar, Punjab. Haryana, Baleshwar coast.