Explore topic-wise InterviewSolutions in Current Affairs.

(i) (e) line y = -5/6 x + 7/6 slope m = -5/6

(ii) (d) Ratio of Area Similar triangles = Ratio of squares of corresponding sides

(iii) (a) A.M. = 16+8/2 = 24/2 = 12

(iv) (b) No. of space diagonals = 4

(v) (c) tan²60° = (√3)² = 3

Option: (a) b³-a³/3

(b) tan (90° – 45) = tan45° = 1

Correct option is B) row

Correct option is A) wait and see

Correct option is C) leaves

• For most tea, yes. Boiling has a marked effect on the taste and benefits tea provides. Boiling changes oxygen levels too which is also an important factor. 
• Most low-grade tea is fine to boil, but green, white, Oolong(Chinese tea) and others yield greater benefits at lower temperatures.
• The amount of tea leaves used during steeping affects the strength of the tea liquor. Adding more leaves to the same amount of water increases the overall surface area of the tea, literally making more tea compounds available for dissolution.

1) The correct option is B (assertion and reason is correct but reason is not correct explanation of assertion) 

• Turbidity refers to the light scattering properties of a sample. Turbidity can be described as “haziness” or “milkyness,” and is caused by fine particles scattering light at more or less 90 degrees to the direction from which the light enters the sample. Turbidity is not to be confused with color, nor color with turbidity.
• Turbidity is normally measured using an electronic device in which a beam of ordinary white light is directed through a certain path length of the sample.
• Turbidity is the capacity of water to disperse and absorb light, preventing its transmission in depth. It depends on suspended and dissolved solids including pelagic organisms (plankton). Turbidity is a parameter measured in situ, and is directly related to the visual appearance of water.
• This is due to the fact that the food coloring is less dense than the milk, so it floats on the surface. The colors do not mix because they were not stirred. When the soap is added, the soap reduces the surface tension of the milk by dissolving the fat molecules

2)  The correct option is A (assertion and reason is correct and reason is correct explanation of assertion) 

• Sols are big enough to scatter a beam of light passing through them and make its path visible. The particles of a suspension scatter a beam of light passing through it and make its path visible. The solute particles do not settle down when left undisturbed, i.e., a solution is stable.
• Light appears to travels in straight line because its wavelength is very small .The light hits the screen in a region where straight lines can be drawn from the light source just clipping the hole in the barrier and then continuing them on to the screen.

3) The correct option is A (assertion and reason is correct and reason is correct explanation of assertion)

• Cellulose is an unbranched polymer of glucose residues put together via beta-1,4 linkages, which allow the molecule to form long and straight chains. This straight chain conformation is ideal for the formation of strong fibers.
• The glucose units in cellulose are linked by β glycosidic bonds, different than the α glycosidic bonds found in glycogen and starch. Cellulose has more hydrogen bonds between adjacent glucose units, both within a chain and between adjacent chains, making it a tougher fiber than glycogen or starch.
• Fibre assists your digestive system – keeping food moving through the gut and pushing waste out of the body. Animals, such as cows, sheep and horses, can digest cellulose, which is why they can get the energy and nutrients they need from grass.
• Cellulose is a polysaccharide (a form of carbohydrate) that has a structural role in animals and plants.Most animals can't digest cellulose because it is so hard to break down. Animals that eat only plants (herbivores) have special sacs in their digestive system to help break down cellulose.

The answer is option (1) The initial position of 5th maxima

The condition for the formation of bright fringe is:
path difference = dsinθ = n.λ, where n =0, ±1, ±2,....
The central maxima is at n = 0.
When one of the slit is covered with a thin film which increases the optical path difference by 5λ, the above equation can
be expressed as,
dsinθ+5λ = nλ;

dsinθ = (n-5)λ = n'λ where n' = 0, ±1, ±2,......
Thus, the central maxima will now be at n' = n-5 = 0. 

Therefore, n = 5. Thus, the central maxima shifts to the position of the 5th maxima.

Velocity of B is = 10 ,

so, velocity of string connecting to A , u is given by

10 = (u+0)/2

=>u = 20 m/s

now this Velocity will always be constant along the string
so, velocity of A along the string is
u'cos(37) = 20
=> u = 20/(cos37) = 20*5/4 = 25m/s

The correct answer is Cell wall and Cell membrane

• The Cell wall, the cell membrane and the vacuole all work together to maintain homeostasis.
• The Cell wall keeps the cell at a constant size, the cell membrane only allows certain materials into the cell, and the vacuole needs to maintain a certain size to keep the cell balanced.

Given A = {1,2,3, ….,13, 14}

R = {(x, y) : 3x - y =0}

∴ R= {1,3), (2, 6), (3, 9), (4, 12)}

Here (1, 1) ∉ R, (2, 2) ∉ R

R is not reflexive

Here (1, 3) ∈ R but (3, 1) ∉ R

∴ R is not symmetric

Here (1,3) and (3, 9)6 ∈ R but (1,9) ∉ R

∴ R is not transitive

∴ R is not an equivalence relation.

Solution: 

Value of the machine after 2 years = Rs.[50000*(1- 10/100)^2]
= Rs 50000*9*9/10*10 = Rs. 40500

Let total marks =x

Rachael scored marks =0.47x

0.47x=94

x= 94/0.47
​    x=200

Thus total marks will be 200

Let the total marks of the examination be = m

Raju gets 98 marks which is 58% of total marks

Now, m x 58% = 98

m = 168.9

m ≈ 169

sin²θ = (e^iθ − e^−iθ)²/- 4 = 1 − cos2θ/2.

Thus

u(r, θ) = (1 − r²cos2θ/25)/2.

Coordinates of D are:- {(1 + 9) / 2}, {(5 + 3) / 2} = (5, 4) 

The vertices of ∆ABD are:- A(3,10), B(1, 5), D(5, 4) 

Ar(ABD) = \(\frac{ 1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|\)

= 1/2 ({3 × 1} + {1 × -6} + {5 × 5})

= 1/2 (3 - 6 + 25)

= 22/2

= 11 sq. units

Therefore, the area formed by the ∆ABD is 11 sq.units. 

Comparing this equation with the general equation of the conic 

Ax² + Bxy + cy² + Dx + Ey +F = 0 

We get A = C also B = 0 

So the given conic is a circle.

The equation for the characteristic curves are dt/1 = dx/x . 

Thus the characteristic curves are given by 

t = ln x + c', or 

xe^−t = c. 

Therefore, the general solution of the equation is 

u(x, t) = f(xe^−t). 

To determine f, we set t = 0 and find that 

f(x) = e^−x2. 

Thus u(x, t) = e−x²e^−2t

y = e^loge^{(x +\(\sqrt{\text x^2-a^2}\) )}= x + \(\sqrt{\text x^2-a^2}\)

dy/dx = 1 + \(\frac{1}{2\sqrt{\text x^2-a^2}}\times2\text x = 1 + \frac{\text x}{\sqrt{\text x^2-a^2}}\)

=  -18+11x-x^2

=  -{x^2-11x+18}

=  -{x^2-9x-2x+18}

=  -{x(x-9)-2(x-9)}

=  -(x-9)(x-2)

therefore, value of x is equal to 2 or 9.

Answer (a,c,d) Na,Al & Fe

Deduction for Contribution to Pension Fund (Section 80CCC) : An Individual can claim deduction from gross total income equivalent to investment made of Rs. 1, 50, 000 whichever is less.

Deduction in Respect of Contribution to Pension Scheme of Central Government [Section 80CCD] : An Individual employed by central government or self employed can claim upto following:

Contribution by Employer: Deductible amount is contribution made by the employer to the employee during the year subjected to maximum of 10% of the salary of the employee.

Contribution by Employee: Deductible amount is contribution made by the employee during the year subjected to maximum of 10% of the salary of the employee.

i) It pre-pones the collection of tax. 

ii) Ensures a regular source of revenue to government. 

iii) Provides for a greater reach and wider base for tax.

Answer (a,b,c) Bauxite, Cryolite &  Feldspar

Answer (d) Bacon cell

1. (c) 2. (a) 3. (d) 4. (b)

(a) Sections 80-IA to 80RRB other than section 80P; or 

(b) Section10AA

Answer is 

I. (c) (0,0,1)

II. (d) (x/a) + (y/b) + (z/c) = 1

III. (a) 13

IV. (b) x + 2y - 3z -14 = 0

The intial  v is same

Angle of projections  is same

Use this formula , u² sin2Ø/ g

 So the ratio will be same i.e 1:1 

Let \(I = \int \frac{x^3}{(x^4 + 7)^8}dx\)

Let x⁴ + 7 = t

4x³ dx = dt

\(\therefore I = \frac14\int\frac{dt}{t^8} = \frac14 \times \frac{-1}{7}\left[\frac1{t^7}\right] + C\)

\(= \frac{-1}{28}\frac{1}{(x^4 + 7)^7} +C\)      \((\because t = x^4 + 7)\)

what we have to do  I mean with respect to dx or what please send full question 

311.04 x 5 = 1555.2

\(\frac{7}{9} + \frac{3}{-4}\)

= \(\frac{7}{9} + \frac{-3}{4}\)

= \(\frac{7\times 4+(-3)\times 9}{36}\)

= \(\frac{28-27}{36}\)

= \(\frac{1}{36}\)

Suppose we have log_a (b), we want to change it on exponential (e) base, then it can be written as:

log_a(b)=log_a(e)⋅log_e(b)

Let's say we have, y = c ⋅ f ( x ) , where c is a constant then, y ' = c ⋅ f ' ( x )

Now, this is quite straightforward to differentiate, as log₁₀(e) is constant, so only remaining function is log_e(x)

Hence,

y' = log₁₀(e) . 1/x

This means that given any x, there is only one y that can be paired with that x. A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b. All elements in B are used. By definition, to determine if a function is ONTO, you need to know information about both set A and B.

Option : (d)

\(\text{let}\,f(x_1)=f(x_2)\forall x_1,x_2\in R\)

\(\Rightarrow x^3_1=x_2^3\)

\(\Rightarrow x^3_1-x_2^3=0\)

\(\Rightarrow (x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0\)

\(\Rightarrow x_1=x_2\)  \((\because x_1^2+x_1x_2+x_2^2 \neq0)\)

\(\Rightarrow\) f is one - one

\(\text{let}\,f(x)=x^3=y\;\forall y\in R\)

\(\Rightarrow x=y^{\frac{1}{3}}\)

Therefore, every image \(y\in R\) has a unique pre image \(y^{\frac{1}{3}}\) in \(R\).

\(\Rightarrow f\) is onto

\(\therefore\) f is one-one and onto.

Let the number be n.

\(\therefore \frac n4 + \frac n5 = \frac n3 + 28\)

⇒ \(\frac{9n}{20} = \frac{84 + n}3{}\)

⇒ \(20n + 1680 = 27 n\)

⇒ \(7n = 1680\)

⇒ \(n = \frac{1680}{7}= 240\)

Hence, the required number is 240.

Correct option is (b).

Let assume contrary that √7 is a rational number, so it can be written in \(\frac pq \) form.

Let √7 = \(\frac pq \), gcd(p, q) = 1, p, q \(\in \mathbb{Z}\), q \(\ne \) 0

⇒ p = √7q

⇒ p² = 7q²     ......(1)

⇒ 7 divides p²

⇒ 7 divides p

⇒ p = 7m, m \(\in \mathbb{Z}\)

⇒ p² = 49m²

⇒ 7q² = 49m²     (From(1))

⇒ q² = 7m²

⇒ 7 divides q²

⇒ 7 divides q

\(\because \) 7 divides both p & q.

\(\therefore\) 7 is a factor of both p & q.

\(\therefore\) gcd(p, q) = 7 or multiple 7.

⇒ gcd(p, q) = 1

which is contradiction the fact that 

gcd(p, q) = 1

Hence, our assumption is wrong.

Hence, √7 is an irrational number.

Let 2-√7 is rational number.

⇒ -2 - √7 = \(\frac pq \),  q \(\ne \) 0, p, q \(\in \mathbb{Z}\)

⇒ \(\sqrt 7 = 2 - \frac pq = \frac{2q - p}{q} \in \mathbb Q\)     (\(\because \) 2q - p \(\in \mathbb{Z}\) & q \(\in \mathbb{Z}\),  q \(\ne \) 0)

which is contradiction 

(\(\because \) √7 is irrational and \(\frac {2q-p}q \) is rational And rational \(\ne \) irrational)

Hence, our assumption is wrong.

Hence, 2 -√7 is an irrational number.

Option : (C)

Z is minimum -24 at (0, 8)

Option : (a)

x = a secθ

⇒ \(\frac{dx}{d\theta}\) = a tanθ secθ

y = b tanθ

⇒ \(\frac{dy}{d\theta}\) = b sec²θ

∴ \(\frac{dy}{dx}\) = \(\frac{b}{a}\) cosecθ

⇒ \(\frac{d^2y}{dx^2}\) = \(\frac{-b}{a}\)cosecθ.cotθ.\(\frac{d\theta}{dx}\)

= \(\frac{-b}{a^2}\) cot³θ

∴ \(\frac{d^2y}{dx^2}\)]_{θ = \(\frac{\pi}{6}\)} = \(\frac{-3\sqrt 3b}{a^2}\)