Explore topic-wise InterviewSolutions in Current Affairs.

Amount receivd from sale of grass by a club should be tracted on Revenue Receipt.

75% is percentabe of non-current assets to total asset.

Income and expenditure A/C is Nominal A/C

A sinking fund is a part of Current liabilities

d) Assertion (A) is false but Reason (R)is true.

(d) All of above

Cash sales is operating activity

Work men’s compensation fund is a Provision 

In not-for-profit organization, excess of expenditure over income is called Deficit.

Correct option is: (B) Capital payment

Correct option is: (B) Revenue nature only

Issue of Debentures in cash is a Financing activities

Payment of honorarium of secretary is treated on Revenue Expenditure.

Income and Expenditure Account is prepared by Not for – profit Organisation

Correct option is: (A) Capital Receipt

Loss or sale of car  is operating expenses:

For non-trading organisation, honorarium is A Revinue Expenditure

Correct option is: (B) Revenue expenditure

For a non trading concern, honorarium is Expenses

Specific donation is  Capital Receipts 

Life membership fees received by club in shown in Balance Sheet.

A life membership fee received by a Club is shown in Balance sheet

Life membership fees received by a club is shown in Balance sheet

Excel workbook is collection of Worksheet

(b) Monera is an obsolete biological kingdom of the former five-kingdom system of scientific classification. It comprised most organisms with a prokaryotic cell organization. For this reason the kingdom was sometimes called Prokaryota or Prokaryotae.

Any four out of the following: 

(1) Column Charts 

(2) Bar Charts 

(3) Area Charts 

(4) Line Charts 

(5) Pie Charts

Aero models are classified according to the role and utility of the particular type. These are static models, gliders, control line models, and RC models. 

The four different types of Aero-Models are: - 

(a) Static models 

(b) Control line models 

(c) Radio Control (RC) models 

(d) Gliders 

OR

(a) Screwdriver 

(b) Hand drill

(c) Sandpaper and pins 

(d) Pliers 

(e) Knives with different blades 

(f) Different kind of saw 

(g) Files 

(h) Soldering irons 

(i) RC set (Transmitter, Receiver, Servos) etc

Charts allows to present data entered in the worksheet in a visual format using a variety of graph types. 

Before making a chart, first enter data into a worksheet.

Column Charts - They compare distinct items. In this chart category axis is horizontal (Xaxis) and value axis is vertical (Y-axis). It is important to keep the number of series in a column chart to a minimum. Many series cause the column to become narrow and difficult to analyze. Column charts are used for making comparisons.

Bar Charts - Compare different items or show single items at different intervals. It is similar to column chart except that the chart value axis is horizontal (X-axis) and category axis is vertical (Y-axis).

Area Charts – They show the relative contributions of each value to a total over time.

Line Charts - These charts are useful to compare the trends over time.

Pie Charts - The chart show the proportion of each part value to the total value in a data series. They are used to show proportional sales figures or representation of different categories of population like workers/non-workers, male/female, adults/children, etc.

Various Types of Charts. 

(a) Navigational Chart. 

(b) Ship`s boat charts. 

(c) Routing charts. 

(d) Magnetic charts. 

(e) Ocean sounding charts. 

(f) LD charts (lattice Decca). 

(g) Astronomical charts and diagrams. 

Instruments required for chart work 

a) Dividers 

b) Compass 

c) Parallel Ruler 

d) Protractor 

e) squares

Chart Scales. 

Charts are generally published in three different scales, they are:- 

(a) Small Scale Charts. These are charts covering a very vast area and the information such as sounding, lights etc. are not given in detail. These charts are generally used for passage planning and never should be used for navigation. 

(b) Medium Scale Charts. These charts are used for passage. The information for navigation including dangers is clearly shown on these charts. These charts cover a general area of about 50 – 70NM.

(c) Large Scale Charts. These charts are generally of harbours and their approaches. These charts contain all information’s required for precise navigation. These charts cover an area of 5 – 7NM. 

OR 

Radar. Radio aided Direction and ranging i.e. with the help of radio waves, the direction and range of objects are obtained. This Instrument is Used for Navigation.

Correct option is B) teeth

(b) 2 only 

As white colour reflect the solar radiation and absorb the infrared radiation. In winter season as body temperature is more than atmospheric temperature. Body radiate heat in infrared region which is not visible un is absorbed by sweater of white colour but it emit the same inless amount to the surrounding. Due which human body remain hot. Here, white colour absorb the infrared radiation from outside and keep the body hot.

(c) 5 

NTU = \(\cfrac{UA}{C_{min}}\)

U = 100 W/m²K

C = 1000 W/K = C_min 

A = 50 m²

NTU = \(\cfrac{UA}{C_{min}}\) = \(\cfrac{100\times50}{1000}\) = 5

(i) Wrap Text option is available under Text section of the Insert Tab. 

(ii) Any two text wrapping options from following 

1. In Line with text 

2. Square 

3. Through 

4. Top and Bottom 

5. Behind Text 

6. In Front of text

Correct option:

(A) 1 

(d) 1, 2 and 3

The choice of a refrigerant depends

1. If a refrigerant has large latent heat of evaporation the size of condenser evaporator and compressor reduces because less mass flow rate. 

2. If a specific volume a refrigerant is large at evaporator pressure, the size of compressor increases. Large specific volume refrigerant are compressed by centrifugal compressor and less specific by centrifugal compressor and less specific volume refrigeants by reciprocating compressor. 

3. Off course evaporator temperature (high for air conditioner as compared to cold storage) also influence the refrigerant choice.

(a) 1.27 kW 

Refrigeration plant on Carnot cycle takes load

i.e. Refrigeration effect, R = 4T=4 × 3.5 =14 kW 

The working temperature, 

T₁ = 27°C = 300K 

T₂ = 2°C = 275K 

\(\therefore\) COP is defined as,

\(\cfrac RW\) = \(\cfrac{T_2}{T_1-T_2}\)

\(\cfrac {14}W\) = \(\cfrac{275}{300-275}\) = \(\cfrac{275}{25}\) = 11

\(\therefore\) Power required to run the compressor,

p = \(\cfrac {14}{11}\) = 1.27 kW

W = sin⁴y cos⁴y

\(\frac{dW}{dy}\) = sin⁴y \(\frac{d}{dy}\) cos⁴y + (\(\frac{d}{dy}\) sin⁴y) cos⁴y

 = -4 sin⁴y cos³y sin y + 4 sin³y cos y cos⁴y

 = 4 sin³y cos³y (cos²y - sin²y)

 = 4 sin³y cos³y cos 2y

x = \(2cos^2(\frac t2)\) 

\(\frac{dx}{dt} = \) 2.2cos(t/2).(-sin(t/2)).1/2

 = -2 sin t/2 cos t/2

 = -sin t

E_A = \(\frac1{4\pi\varepsilon_0}\frac{q}{r^2}\)

E_A = 9 x 10⁹ x \(\frac{2}{5^2}\)

E_A = \(\frac{18\times10^9}{25}\) ⇒ 0.72 x 10⁹ N/C

E_B = \(\frac1{4\pi\varepsilon_0}\frac{q}{r^2}\) 

 = 9 x10⁹ x \(\frac2{5^2}\)

E_B = \(\frac{18}{25}\times10^9\)

E_B = 0.72 x 10⁹ N/C

E = E_A + E_B

E = 0.72 x 10⁹ + 0.72 x 10⁹

E = 1.44 x 10⁹ N/C

F = qE

F = 6 x 1.44 x 10⁹

F = 8.64 x 10⁹ N

Language is a system of communication that is used to express thoughts, ideas and feelings. Language is a vital part of human connection. Humans are the only ones that have mastered cognitive language connection.

A morpheme is the smallest unit of meaning in language providing the foundation of language and can not be divided into smaller parts carrying meaning. It has relatively the same stable meaning in different verbal environments. The study of morphemes is called Morphology.  There are two types of morphemes- Free Morphemes and Bound Morphemes.

• Free Morphemes can stand alone with specific meanings. It can be formed independently. For example; happy, get, when, etc. All of them have individual meaning and are free morphemes. 
• Free Morphemes can be categorized into two types: Lexical Morphemes and Grammatical / Functional Morphemes. 
  1. Lexical Morphemes are those that are large in number and independently meaningful. It includes nouns, adjectives and verbs. Example; boy, bad etc.
  2. Grammatical / Functional Morphemes consist of functional words in a language such as prepositions, pronouns, determiners, conjunctions. Example; but, on, and, above etc.  
• Bound Morphemes do not have any independent meaning and can be formed with help of free morphemes. Example; un, pre, en, etc. Bound Morphemes can be categorized into two sub-classes: Bound Roots and Affixes.
  1. Bound Roots have lexical meaning when included in other bound morphemes to form the content word. Example;- tain (retain, contain).
  2. Affixes naturally attached different types of words and used them to change the meaning of those words. Example;-ing in reading, writing, sleeping.

​Hence, the smallest unit of meaning in language is morphemes.

• Syntax:  Syntax is the arrangement of words and phrases to create well-formed sentences in a language.
• Phoneme: Phoneme is the smallest unit of speech distinguishing one word from another, as the element 't' in a cat which separates that word from "can", "cab".
• Pragmatics: Pragmatics is the branch of linguistics, which is the study of language.

(A) A-4, B-2, C-3, D-1

Let a=((1+2x+3x²)⁶+(1−4x²)⁶)

∴ Coefficient of x² in the expansion of the product (2−x)²((1+2x+3x²)6+(1−4x²)⁶)

    =2( Coefficient of x² in a)−1(Constant of expansion)

In the expansion of ((1+2x+3x²)⁶+(1−4x²)⁶), 

Constant =1+1=2

Coefficient of x²= Coefficient of x² in (⁶C₀​(1+2x)⁶(3x²)⁰)+ Coefficient of x² in (⁶C₁​(1+2x)⁵3)−⁶C₁​(4)

=⁶C₂​4+6×3−24

=6=+18−24=54

Then, coefficient of x² in (2−x)²((1+2x+3x²)6+(1−4x²)⁶)

=2×54−1(2)=108−2=106

Dual of (p ∨ q) ∧ T is (p ∧ q) ∨ F

f(x) = \(\frac{e^{2x-1}}{1+e^{2x-1}}\) 

f(1-x) = \(\frac{e^{2(1-x)-1}}{1+e^{2(1-x)-1}}\) = \(\frac{e^{1-2x}}{1+e^{1-2x}}\) = \(\frac{e^{-(2x-1)}}{1+e^{-(2x+1)}}\)

 = \(\frac1{e^{2x-1}+1}=\frac1{1+e^{2x-1}}\) 

\(\therefore\) f(x) + f(1 - x) = \(\frac{e^{2x-1}+1}{1+e^{2x-1}}=1\)

\(\therefore\) f(1/2009) + f(2008/2009) = 1

\(\therefore\) f(2/2009) + f(2007/2009) = 1

f(1004/2009) + f(1005/2009) = 1

\(\therefore\) f(1/2009) + f(2008/2009) + f(2/2009) + f(2007/2009) +....f(1004/2009) +f(1005/2009)

= 1 + 1 + ...1 (10004 times) = 1004

⇒ f(1/2009) + f(2/2009) + .... + f(1004/2009) + f(1005/2009) + ....+ f(2007/2009) + f(2008/2009) = 1004

For n = 1, xⁿ⁺¹ - xⁿ + 1 = x² - x + 1 which is exactly divided by x² - x +1.

\(\therefore\) options (c) & (d) are wrong.

For n = 4, xⁿ⁺¹ - xⁿ + 1 = x⁵ - x⁴ + 1 

And \(\frac{x^5-x^4+1}{x^2-x+1}\) = x³ - x - 1 + \(\frac2{x^2-x+1}\) does not divided by x² - x + 1.

For n = 7, xⁿ⁺¹ - xⁿ + 1 = x⁸ - x⁷ + 1

And \(\frac{x^8-x^7+1}{x^2-x+1}\) = x⁶ - x⁴ - x⁴ + x + 1

Hence, x² - x + 1 divides xⁿ⁺¹ - xⁿ + 1 for n = 7

which satisfies by option (b).

Hence, for dividing xⁿ⁺¹ - xⁿ + 1 by x² - x + 1

then n must be of the form 6k + 1.

\(\frac{2x+1}{(x-1)(x-2)(x-3)}=\frac A{x-1}+\frac B{x-2}+\frac c{x-3}\) 

⇒ 2x + 1 = A(x - 2)(x - 3) + B(x - 1)(x - 3) + c (x - 1)(x - 2)---(1)

Put  x = 1 in equation (1) we get

2A = 2 + 1 = 3 ⇒ A = 3/2

Put x = 2 in equation (1), we get

-B = 5 ⇒ B = -5

Put x = 3 in equation (1), we get

2c = 7 ⇒ c = 7/2

Hence, \(\frac{2x+1}{(x-1)(x-2)(x-3)}=\frac{3/2}{x-1}-\frac5{x-2}+\frac{7/2}{x-3}\)

complex rational statement is decomposed into two or more simpler fractions using partial fractions. To divide fractions into several sub-fractions, partial fractions ideas are employed.

Given:

$\frac{1}{(x-1)(x^2-9)}$

Find

Resolve the expression to a partial fraction.

Solution:

The given expression can be written as,

$\begin{array}{c}\frac{1}{(x-1)(x^2-9)}\end{array}$

\(\frac1{(x-1)(x^2-9)}\) = \(\frac1{(x-1)(x-3)(x+3)}\) = \(\frac A{(x-1)}+\frac {B}{x-3}+\frac C{x+3}\)

⇒ A (x - 3) (x + 3) + B (x - 1)(x + 3) + C(x - 1)(x - 3) = 1

Put x = 1, we get -8A = 1 ⇒ A = -1/8

Put x = 3, we get 12B  = 1 ⇒ B = 1/12

Put x = -3, we get 24C = 1 ⇒ C = 1/24

Hence, \(\frac1{(x-1)(x^2-9)}\)= \(\frac{1}{x-1}+\frac1{12}\frac1{x-3}+\frac{1}{24}\frac1{x+3}\)

f(x) = \(\frac{x^3}{(x^2+1)(x+1)}\) 

= \(\frac{x^3}{x^3+x^2+x+1}\) 

= 1 - \(\frac{x^2+x+1}{(x^2+1)(x+1)}\)

Now,

\(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = \(\frac{A}{x+1}\) + \(\frac{Bx+c}{x^2+1}\)

⇒ x² + x + 1 = A(x²+1)+(Bx+c)(x+1)

Put x = -1, we get

2A = 1-1+1 = 1

⇒ A = \(\frac{1}{2}\)

Put x = 0, we get

A + C = 1

⇒ C = 1 - A

= 1 - \(\frac{1}{2}\)

= \(\frac{1}{2}\)

Put x = 1, we get

2A + 2B + 2C = 3

⇒ 2C = 3 - 2A - 2B 

= 3 - 1 - 1 

= 1

⇒ C = \(\frac{1}{2}\) 

Hence,

\(\frac{x^2+x+1}{(x^2+1)(x+1)}\) = \(\frac{1}{2}\)(\(\frac{1}{x+1}\) + \(\frac{x+1}{x^2+1}\))

∴ f(x) = 1 - \(\frac{x^2+x+1}{(x^2+1)(x+1)}\)

= 1 - \(\frac{1}{2}\)(\(\frac{1}{x+1}\) + \(\frac{x+1}{x^2+1}\))