Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is (c) It focused only on profitability

Correct option is (b) Likely entry of new competitors.

Correct option is (d) A mission statement typically identifies what the company's products or services are (what we do) and the customers and markets it serves (why we are here), whereas the focus of a strategic vision is on "where we are going and why."

Correct option is (d) Services and ideas

Correct option is (c) Habit.

Correct option is (a) Reduced work hours

Correct option is (d) Structural inertia

During the medieval India, there was tremendous, progress in agricultural activities. Irrigation facilities improved with the construction of canals and digging of wells. Advent of new machines and tools improved the possibility of crop production. Many varieties of paddy and other crops were being cultivated in India. The Chola kings who ruled south India at the beginning of the medieval period adopted several measures to ensure agricultural progress. The steps taken by the Chola kings include the following

• Construction of canals
• Steps to ensure water prosperity in river Kaveri
• Measures taken by the summit that worked. under the Subhas, the body for village administration.

• Sumatra and Java (Indonesia)
• Persia (Iran)
• Holland
• England
• China
• Portugal
• France

पन-विद्युत् (जल-विद्युत्) बनाने के लिए बनाये गए डैम या प्रोजेक्ट बहुउद्देश्यीय प्रोजेक्ट कहलाते हैं। ये एक से अधिक उद्देश्यों की पूर्ति करते हैं।

Correct option is (a) Cooperation

Several technological advancements and inventions took place in medieval period. Charkha (a domestic spinning wheel) was the key factor that facilitated the development of textile industry. It was in the fourteenth century that charkha began to be used in India. Indians adopted this technology from the Chinese. With the advent of charkha, the production of yarn in-creased by six times.

Weaving also advanced along with spinning. The invention of loom brought tremendous changes in weaving. The technology for the production of silk from silkworms also spread widely during that period. It was Bengal that pioneered the production of silk in the fourteenth century. 

The production of carpet and paper were other major handicrafts prevalent then. There was great demand for Indian carpets in the foreign market. In India paper began to be used in the thirteenth century.

Metallurgy and mining were other industries that flourished in this period. Those who were engaged in metallurgy mostly produced agriculture tools and war equipment. The production of horseshoe, and iron stirrup made drastic changes in war technology. Copper and gems were mined from Rajasthan and Golkonda respectively. Salt production was another important occupation that thrived much in this period. Seashore, salt rocks, and salt lakes were the major sources of salt. West Punjab and Sambhar in Rajasthan were the chief salt-producing centers.

Several technological advancement and inventions took place in medieval period. Charkha (a domestic spinning wheel) was the key factor that facilitated the development of textile industry. It was in the fourteenth century that charkha began to be used in India. Indians adopted this technology from the Chinese. With the advent of charkha, the production of yarn increased by six times.

Weaving also advanced along with spinning. The invention of loom brought tremendous changes in weaving. The technology for the production of silk from silkworms also spread widely during that period. It was Bengal that pioneered the production of silk in the fourteenth century. 

The production of carpet and paper were other major handicrafts prevalent then. There was great demand for Indian carpets in the foreign market. In India paper began to be used in the thirteenth century. Metallurgy and mining were other industries that flourished in this period.

Those who were engaged in metallurgy mostly produced agriculture tools and war equipments. The production of horseshoe, and iron stirrup made drastic changes in war technology. Copper and gems were mined from Rajasthan and Golkonda respectively. Salt production was another important occupation that thrived much in this period. Seashore, salt rocks, and salt lakes were the major sources of salt. West Punjab and Sambhar in Rajasthan were the chief salt-producing centers.

The hierarchy based on caste system prevailed in medieval period too. The Brahmins enjoyed a higher status. Increase in the number of castes was a major feature of this period. The newly formed occupational groups gradually evolved as castes.

Similar to that in North India, the castes in South India were Idankai and valenki. Even though there was economic progress in medieval period, very less could get benefit from it. Majority were affected by poverty. Caste system and slavery intensified inequalities in the social set up.

The social-economic status of women in Medieval India was not satisfactory. The practice of Sati and child marriage were existed in India. There were restrictions for remarriage. Since girls were married at a very tender age, they did not get the opportunity for education. The role of women in agricultural and non-agricultural sectors was significant. They involved themselves in all the stages of farming from sowing to harvesting. They were engaged in weaving, pottery, and embroidery too.

Even then, there were several women who had adorned higher political and social position. Noorjahan, the wife of the Mughal Emperor Jahangir and Sultana Rasiya were excellent administrators Gulbadan Begum (sister of Humayun), Jahanara (daughter of Shah Jahan), and Jeeja Bai (the mother of Shivaji) were women who held higher positions.

Correct option is D. All of the above

Correct option is D. It eliminates completely the work associated with recruitment on the part of the recruiter to his/her total convenience

a = 200, r = 1 + 10/100 = 11/10

Mosquitoes at the end of 1st year = 200 x 11/10

(i) Number of mosquitoes after 3 years

= 200 x 11/10 x (11/10)²

 = 200 (11/10)³

= 200 (1.1)³

(ii) Number of mosquitoes after 10 years = 200 (1.1)¹⁰

(iii) Number of mosquitoes after n years = 200

(1.1)ⁿ

(i) x – 6, 2x and x are in Geometric progression.

\(\therefore\) \(\frac{2x}{x-6}=\frac{x^2}{2x}\) 

4x = x² (x – 6)

4 = x – 6

x = 10

(ii) t₁ = x – 6 = 10 – 6 = 4

(iii) a = 4, r = 2x/(x - 6) = \(\frac{2(10)}4=5\)

t_n = ar^n-1

\(\therefore\) t_n = 4(5^n-1)

(b) 20m

Total area = Area of Δ AFC + Area of Δ AGD + Area of trapezium FCEH + Area of Δ BGE + Area of Δ DGB

= \(\frac12\) x AF x FC + \(\frac12\) x AG x DG + \(\frac12\) x (CF + EH) x HF + \(\frac12\) x BH x HE + \(\frac12\)x BG x DG

⇒ \(\frac12\) x 25 x 20 + \(\frac12\) x 50 x 40 + \(\frac12\) x (20 + x) x 50 + \(\frac12\) x 25 × x + \(\frac12\) x 50 x 40 = 3500

⇒ 250 + 1000 + (20 + x)25 + 12.5x + 1000 = 3500 

⇒ 2250 + 500 + 25x + 12.5x = 3500 

⇒ 37.5x = 3500 – 2750 = 750

⇒ x = \(\frac{750}{37.5}\) = 20m

C. 2

Condition for coincident lines is -

a₁/a₂ = b₁/b₂ = c₁/c₂ …(i)

Given lines, 3x - y + 8 = 0

and 6x - ky + 16 = 0;

Comparing with ax + by + c = 0;

Here, a₁ = 3, b₁ = - 1, c₁ = 8; a₂ = 6, b₂ = - k, c₂ = 16;

and

From Eq. (i), 3/6 = 1/k = 8/16

1/k = 1/2

So, k = 2

First, let us write the conditions for the applicability of Rolle’s theorem:

For a Real valued function ‘f’:

a) The function ‘f’ needs to be continuous in the closed interval [a, b].

b) The function ‘f’ needs differentiable on the open interval (a, b).

c) f(a) = f(b)

Then there exists at least one c in the open interval (a, b) such that f’(c) = 0.

Given function is:

⇒ y = x² on [– 2, 2]

We know that polynomials are continuous and differentiable over R.

Let’s check the values of y at the extremums

⇒ y(– 2) = (– 2)²

⇒ y(– 2) = 4

⇒ y(2) = (2)²

⇒ y(2) = 4

We got y(– 2) = y(2).

So, there exists a c such that f’(c) = 0.

For a curve g to have a tangent parallel to x – axis at point r, the criteria to be satisfied is g’(r) = 0.

⇒ y’(x) = 0

⇒ \(\frac{d(x^2)}{dx}=0\)

⇒ 2x = 0

⇒ x = 0

The value of y is

⇒ y = (0)²

⇒ y = 0

The point at which the curve has tangent parallel to x – axis is (0, 0).

(a) 20π

The curved surface area of a sphere of radius r intercepted between two parallel planes at a distance a and b from the centre of the sphere is 2πr (b – a) 

Given radius, r = 5; a = 2; b = 4 

Required surface area = 2πr (b – a) 

= 2π × 5 × (4 – 2) 

= 20π sq. units

Correct option is (D) All the above

Brick, Book and Match box all are in cuboidal shape.

D) All the above

Given,

Edge of cube = 2.1 m

We know that,

Surface area of cube = 6 × side²

= 6 × 2.1²

= 6 × 4.41

= 26.46 m²

Given details are,

Length of a cuboid = 4 m

Breadth of a cuboid = 2.5 m

Height of a cuboid = 50 cm = 0.50 m

By using the formula

Volume of a cuboid = l × b × h

= 4 × 2.5 × 0.50

= 5 m³

Given details are,

Dimensions of cuboidal box = 5cm × 5cm × 4cm

We know that,

Surface area of cuboid = 2 (lb + bh + hl) cm²

= 2 (5×5 + 5×4 + 4×5)

= 2 (25 + 20 + 20)

= 2 (65)

= 130 cm²

Total surface area (T.S.A) = 616 cm²

Let r be the radius of cylinder and h be the radius of cylinder. 

As per given statement: 

(curved surface area / (total surface area) = \(\frac{1}{2}\) 

or CSA =  \(\frac{1}{2}\)TSA 

CSA =  \(\frac{1}{2}\) x 616 = 308 

⇒ CSA = 308 cm² 

Now, 

TSA = 2πrh + 2πr² 

⇒ 616 = CSA + 2πr² 

⇒ 616 = 308 + 2πr² 

⇒ 2πr² = 616 – 308 

⇒ 2πr² = 308

⇒ r² = 308/2π

⇒ r² = 49 

or r = 7 cm …(1) 

As, CSA = 308 cm² 

2πrh = 308 

⇒ 2 x \(\frac{22}{7}\) x 7 x h = 308 (using (1)) 

⇒ h = 7 cm 

Now, 

Volume of cylinder = πr²h 

= \(\frac{22}{7}\) x 7 x 7 x 7 

= 1078 

Therefore, Volume of cylinder is 1078 cm².

Height of cylindrical tank (h) = 1 m

Base radius of cylindrical tank (r) = diameter/2 

= 140/2 cm 

= 70 cm 

= 0.7 m 

[1 m = 100 cm] 

Now, 

Area of sheet required = Total surface area of tank (TSA) = 2πr(h + r) 

=2 x 3.14 x 0.7(1 + 0.7) 

= 7.48 

Therefore, 7.48 m² metal sheet is required to make required closed cylindrical tank.

Given details are,

Surface area of cube = 150 m²

6 × side² = 150cm²

Side² = 150/6

= 25

Side = √25 = 5cm

∴ Volume of a cube = 5³ = 125m³

Given details are,

Volume of cube = 343 m³

Side of cube, a = ³√(343) = 7m

We know that,

Surface area of cube = 6 × side²

= 6 × 7²

= 6 × 49

= 294 m²

Given details are,

Edge of a cubic wooden box = 12 cm

We know that,

Surface area of cubic wooden box = 6 × side²

= 6 × 12²

= 6 × 144

= 864 cm²

Given details are,

Surface area of cube = 96 cm²

6 × side² = 96cm²

Side² = 96/6

= 16

Side = √16 = 4cm

∴ Volume of a cube = 4³ = 64cm³

Given details are,

Side of cube = 8 cm

Volume of cube = (side) ³

= 8³ 

= 512 cm³

Given details are,

Volume of cube = 216 dm³

Side of cube a = ³√(216) = 6dm

We know that,

Surface area of cube = 6 × side²

= 6 × 6²

= 6 × 36

= 216 dm²

Correct option is (B) collinear

If three and more points are lying on the same line then they are called collinear points.

B) collinear

Given details are,

Edge of one cube a₁ = 2 cm

Edge of second cube a₂ = 4 cm

So, volume v₁ = 2³ = 8cm³

Volume v₂ = 4³ = 64cm³

v₂ = 8v₁

Correct option is C. 64 cm³

Given,

Surface area of cube = 96 cm²

= 6a² = 96

= a² = \(\cfrac{96}6\) = 16

= a = \(\sqrt{16}\) = 4 cm

Volume of cube = a³ = 4 × 4 × 4 = 64 cm³

Given details are,

Dimensions of metal block = 5cm × 4cm × 3cm

Weight of block = 1 kg

Volume of box = 5×4×3 = 60 cm³

Dimension of new block = 15cm × 8cm × 3cm

Volume of new box = 15 × 8 × 3 = 360 cm³

We know that,

60cm³ = 1kg

360 cm³ = 6 × 60 cm³

= 6 × 1

= 6 kg

Correct option is a 3.5 cm

Given,

Volume of 1^st cube = \(\cfrac{1}8cm^3\)

Volume of second cube = 1 cm³

Volume of third cube = 8 cm³

∴edge of first cube = a₁ =\(\sqrt[3]{\cfrac18}\) = \(\cfrac12\) cm

∴ edge of second cube = a₂ = \(\sqrt[3]1\) = 1 cm

∴ edge of third cube = a₃ = \(\sqrt[3]8\) = 2 cm

Hence height of cubes together = a₁ + a₂ + a₃ = \(\cfrac12\) + 1 + 2

= \(\cfrac72\) = 3.5 cm

Correct answer is (A) 1^st quadrant

1105° = 3 x 360° + 25°

Since, 0° < 25° < 90°

∴ it lies in 1st quadrant.

Given details are,

Side of cube = 1.5 dm

Volume of cube = (side) ³

= 1.5³ 

= 3.375 dm³ 

= 3375 cm³

Given details are,

Edge length of cube A = 18 cm

Edge length of cube B = 24 cm

Edge length of cube C = 30 cm

Then,

Volume of cube A = v₁ = 18³ = 5832cm³

Volume of cube B = v₂ = 24³ = 13824cm³

Volume of cube C = v₃ = 30³ = 27000cm³

Total volume of cube A,B,C = 5832 + 13824 + 27000 = 46656 cm³

Let ‘a’ be the length of edge of newly formed cube.

a³ = 46656

a = ³√(46656)

= 36

∴ Edge of bigger cube is 36 cm.

Given details are,

Side of cube = 1.2 m

Volume of cube = (side) ³

= 1.2³ 

= 1.728 m³

Correct answer is (C) co-terminal angles.

45° - (-315°) = 45° + 315°

= 360°

which is integral multiple of 360°

∴ the angles are co-terminal.

Correct option is (A) 1000

1 litre = \(10^3cm^3=1000\,cm^3\)

Correct option is  A) 1000

Given,

Breadth of room is twice of its height, b = 2h or h = b/2 … (i)

Breadth is one half of length, b = l/2 or l = 2b … (ii)

Volume of the room = lbh = 512 dm³ … (iii)

By substituting (i) and (ii) in (iii)

2b × b × b/2 = 512

b³ = 512

b = ³√(512)

= 8

∴ Breadth of cube = b = 8 dm

Length of cube = 2b = 2×8 = 16 dm

Height of cube = b/2 = 8/2 = 4 dm

(D) 360°

The measure of co-terminal angles always differ by an integral multiple of 360°

Correct option is (D) a³

Volume of cube = \(a^3\)

Correct option is  D) a³

Given details are,

Dimensions of ice cream brick = 20 cm × 10cm × 7cm

Dimensions of fridge is = 100 cm × 50cm × 42 cm

So,

Number of bricks that can be put in fridge = volume of fridge / volume of one ice brick

= (100 × 50 × 42) / (20 × 10 × 7)

= 150 ice cream bricks