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Let the breadth of the rectangle = x and length = y

sin 35° = \(\frac{x}{12}\)

x = 12 × sin 35 = 12 × 0.57 = 6.84 cm.

cos 35° = \(\frac{y}{12}\)

y = 12 × cos 35 = 12 × 0.82 = 9.84 cm.

Perimeter = 2(6.84 + 9.84) = 2 × 16.68 = 33.36 cm

tan 50° = \(\frac{opposite\,side}{adjacent\,side}\) = \(\frac{x}{24}\)

x = 2.4 × tan 50° = 2.4 × 1.1918 = 2.86

Triangle side ratio is 1 : : 2

Altitude = hypotenuse \(\times\frac{\sqrt{3}}{2}=4\times\frac{\sqrt{3}}{2}\)

Area = \(\frac{1}{2}\times23.46=3.46\) cm²

Angle of the right 45°, 45°, 90°

Ratio of sides = 1 : 1 : √2

hypotenuse = 20cm

∴ The other two sides are \(\frac{20}{\sqrt{2}}\) cm each

∴ Area of the right angled triangle = \(\frac{1}{2}\times\frac{20}{\sqrt{2}}\times\frac{20}{\sqrt{2}}\) = \(\frac{1}{2}\times\frac{20\times20}{2}=100\) cm²

In DABC, ∠B = 150°, AC = 3 cm Draw diameter AD and join CD 

∠ADC = 180 – 150 = 30°, ∠ACD = 90° 

Angles of DADC are 

30°, 60°, 90°

30° 60° 90° 

1 : √3 : 2 

↓    ↓     ↓ 

3 3√3 6

Diameter, AD = 6 cm

a. 4. 4, √32 

b. 2, 2, √8 

c. 3, 3, √18 

d. 10, 10, √200 

e. 1, 1, √2

Area of the parallelogram = \(\frac{1}{2}\) × AC × BD

In ΔABD, ∠A = 60°

ΔABD is an equilateral triangle.

BD = 10

Area of the parallelogram ABCD = \(\frac{1}{2}\) × 10 × 12 = 60m²

For a right-angled triangle one of the angles is 30° then other one is 60°. 

Side which is opposite to the angle of 90° is twice of the side which is opposite to the angle of 30°

Center of circumcircle is the midpoint of the side, which is opposite to the angle 90° that means half.

∴ Radius of the circumcircle is equal to the side opposite to 30°.

The sides of the isosceles right-angle triangle with angles 45: 45 :90 will have sides proportional to [opposite to corresponding angles] 1: 1: √2

∠C = 90° 

The sides of the right angle triangle with angles 30°, 60°, 90° are proportional 1 : √3: 2

Correct option is (D) ONPMQ

Concept:

\(\rm \frac{log a}{log b}=log_ba\)

Calculation:

Here, \(\rm (2^x-1)^2=4(2^x-2)\)

\(⇒ \rm 2^{2x}-(2)2^x+1=(4)2^x-8\)

\(⇒ \rm 2^{2x}-(6)2^x+9=0\)

Now, let 2^x = y

∴\(\rm y^2-6y+9=0\)

\(⇒ \rm y^2-3y-3y+9=0\)

\(⇒ \rm y(y-3)-3(y-3)=0\)

(y - 3) (y - 3) = 0

⇒ y = 3

⇒ 2x = 3

x log 2 = log(3)

⇒ x = \(\rm \frac{log 3}{log2}\)

= \(\rm log_23\)

Concepts:

If  A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2^pq 

Calculation:

Given: A = {x, y, z} and B = {1, 2}

⇒ n(A) = 3 and n(B) = 2

As we know that, if A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2pq 

Here, p = 3 and q = 2

So, the number of relations from A to B = 2⁶ 

Hence, option 2 is the correct answer.

Concept:

Domain:

Let f : A → B be a function, then the set A is called as the domain of the function f.

Co-domain:
Let f : A → B be a function, then the set B is called as the co-domain of the function f.

Range:
Let f : A → B, then the range of the function f consists of those elements in B which have at least one preimage in A. It is denoted as f (A) i.e f (A) = {b ∈ B | f (a) = b for some a A}

Note: Range is the subset of the codomain of f.

Calculation:

Given: A = {9, 10, 11, 12, 13} and f: A → N is a function where N is the set of natural numbers such that f(n) = the highest prime factor of n

We know that, the prime factorization of: 9 = 3², 10 = 2 × 5, 11 = 11 × 1, 12 = 2² × 3 and 13 = 13 × 1.

According to the definition of the function f we have,

⇒ f(9) = 3, f(10) = 5, f(11) = 11, f(12) = 3, f(13) = 13

As we know that, if f : A → B, then the range of the function f consists of those elements in B which have at least one preimage in A. It is denoted as f (A) i.e f (A) = {b ∈ B | f (a) = b for some a A}

⇒ Range of f = {3, 5, 11, 13}

Hence, option 3 is the correct answer.

2^m – 2ⁿ = 112 

m = 7, n = 4 

(2⁷ – 2⁴ = 112) 

m × n = 7 × 4 = 28

Answer is (3) 3

f(x) = ((2^{1 - x} + 2^{1 + x} + 3^x + 3^-x)/2)

Using AM ≥ GM

f(x) ≥ 3

In addition to that, Ce has 4 electrons in its outer orbital, thus leaving only one unpaired electron on the 6s orbital doesn't sound really stable

In Kjeldahl’s method, nitrogen present is estimated as NH₃

The chemical substances produced or derived from micro organisms which can inhibit the growth or even destroy the micro organisms are called antibiotics. e.g.Penicillin.

N₂O gas liberated when zinc reacts with dil. HNO₃.

Correct option  (1)  [RhCl(CO)(PPh₃)(NH₃)]

Explanation:

[RhCl(CO)(PPh₃)(NH₃)] is inactive as it is square planer complex.

In ammonia (NH₃), the molecules are involved in the intermolecular hydrogen bonding on account of the polarity of N-H bonds. But the same is not possible in molecules of Phosphine (PH₃), since the polarity of P-H bond is negligible as compared to N-H bond. This is due to lesser electronegativity of phosphorus as compared to nitrogen. Therefore, boiling point of ammonia is more than that of phosphine.

Due to Hydrogen bonding.

(4)  If both assertion and reason are false.

Exports are goods produced within the domestic territory so treated as a part of domestic income. 

(1) Iodine solution

 correct option (1)  SO^-2₃

Correct option (1) Xe

Explanation:

Because IP of Xe and oxygen are almost same and atomic size Xe is large.

In chloroacetic acid Cl^- ion has -I effect which decreases the electron density of O-H bond in carboxylic group while in case of acetic acid CH₃ group has +I effect which increases the electron density of O-H bond. Therefore chloro acetic acid is more acidic than acetic acid.

Xe forms compounds with more electronegative elements in which fulfill valence shell p-orbital in to empty d-orbital e.g. XeF₂, XeF₄, XeF₆ etc.

The bond length of HF is very small and HF is more electronegative element while the bond length of HI is very very big and HI less electronegative element. Thus HF is weaker than HI in acetic acid.

Solvay process  a catalyst is not used.

Bond length of H-F is shorter than H-Cl. Hence H-F is weaker acid than HCl.  

(d)  All of these 

The mole fraction is solute is 1/76.

Oxidation number of sulphur is +4 in SO_2. Which is intermediate of minimum O^' N O,N, of sulphur –2 and maximum O' N +6. Hence SO₂ acts as oxidising and reducing agent both. 

\(\sigma\)(x, y) = p

\(\because\sigma\)(ax + b) = |a| \(\sigma\) (x)

\(\therefore\) \(\sigma\) (3x + 5, 3y + 5) = \(\sigma\)(3(x, y) + (5, 51)) = 3 \(\sigma\)(x, y)

= 3p

Using Chain rule ,

\(d {tan\sqrt x \over \sqrt x} \times d {\sqrt x \over x}\)

Since derivative of \(\tan x = \sec^{2}x\) and \(\sqrt x = {1 \over 2\sqrt x}\)

therefore,  derivative of \(\tan \sqrt x = \sec^{2}\sqrt x\)

Hence , derivative of \(\tan \sqrt x\) is  \({\sec^{2} \sqrt x \over 2\sqrt x}\)

K_b is not the colligative property .

The temperature at which vapour pressure of liquid becomes equal to atmospheric pressure is called boiling point of the liquid. The vapour pressure of liquid is lowered when a non-volatile solute is added to it. Therefore, the temperature of solution is rise to increase the vapour pressure equal to atmospheric pressure.

To find ,  \(\int \cos^4x dx\)

Since we know that 

\(\cos^2x = {1 \over 2}(\cos2x+1)\)

so , \(\cos^4x = {1 \over 4}(\cos^22x+1+2\cos2x)\)

We can also expand \(\cos^22x \)

⇒ \(\cos^22x = {1 \over 2}(\cos4x + 1)\)

Which means \(\cos^4x = {1 \over 8}(\cos4x) + {1 \over 2}(\cos2x) + {3 \over 8}\)

Now we have to evaluate the integral of : 

\({1 \over 8}\int\cos4xdx + {1 \over 2}\int\cos2xdx + {3 \over 8}\int dx\)

Since , Integral of \(\cos x = \sin x\)

⇒\({1 \over 8}\times{1 \over 4}\int\cos4x (d(4x)) + {1 \over 2}\times{1 \over 2}\int\cos2x(d(2x)) + {3 \over 8}\int dx\)

⇒\({\sin4x \over 32} + {\sin2x \over 4} + {3x \over 8} + C\)

Hence , \(\int \cos^4x dx\) = \({\sin4x \over 32} + {\sin2x \over 4} + {3x \over 8} + C\)

Correct option is: (C) Base year

∫ydx = ∫4 cos 4x dx

 = 4[\(\frac{sin4x}4\)] = sin 4x

Correct option is: (D) All of the above

The challenges being faced in the implementation of GST are as follows:

(a) Establishing and up gradation of IT Framework 

(b) Meeting implementation challenges 

(c) Tax administration 

(d) Effective coordination between Centre and State Tax Administrations 

(e) Training of officials and trade and industry 

(f) Spreading accounting and IT Literacy 

(g) Reorganization of audit procedures

Following are the steps to Computation of total income and tax liability:

(i) Compute the income of an individual under 5 heads of income on the basis of his residential status. 

(ii) Income of any other person, if includible under section 60 to 64, will be included under respective heads. 

(iii) Set off of the losses if permissible, while aggregating the income under 5 heads of income. 

(iv) Carry forward and set off of the losses of the past years, if permissible, from such income. 

(v) The income computed as above will be Gross Total Income from which permissible deductions under section 80C to 80U are allowed. However, no deduction under these sections will be allowed from short term capital gain covered under section 111A, any long term capital gain and winning of lotteries etc., though these incomes are part of gross total income. 

(vi) The balance income after allowing the deductions is known as total income which will be rounded off to the nearest Rs. 10.

(vii) Computation of tax on such total income at the prescribed rates of tax and adding education cess and secondary and higher education cess thereon and allowing relief under section 89, if any. 

(viii) Deducting the TDS and advance tax paid for the relevant assessment year. The balance is the net tax payable is to be rounded off to the nearest Rs. 10. This tax has to be paid as self-assessment tax before submitting the income tax return.

The challenges that were faced during the implementation of GST are as follows:

1. Establishment and upgradation of IT framework: The number of taxpayers is likely to go up significantly. Also, the process of tracking inter(orintra) State transactions will be online. The type of clearinghouse mechanism through GSTN considered in the dual model GST will handle large volumes of data.

For this purpose, the Goods and Service Tax Network (GSTN) has been set up by the Government to create enabling environment for smooth introduction and implementation of GST.

2. Meeting implementation challenges: The implementation of GST systems and procedures would not be very lengthy and complex process in a long run. A GST Implementation Advisory Committee has been constituted within CBEC for overall supervision and monitoring of progress towards implementation of GST.

3. Tax administration: The Central Board of Excise and Customs(CBEC)and the State Tax Administrations will responsible for implementing CGST and SGST respectively. For implementing dual GST, arobust and integrated tax administration is required to efficiently track supply of goods an dservicesa cross the country as also to precisely account for the taxes. Putting in risk management system will give meaningful results only when there will be an efficient tax administration.

4. Effective coordination between Centre & State tax Administrations: The assignment of concurrent jurisdiction to the Centre and the States for the levy of GST would require a unique institutional mechanism that would ensure that decisions about the structure, design and operation of GST are taken jointly by them. For it to be effective, such a mechanism also needs to have Constitutional force. There is a need of harmonization of processes & procedures between CGST / SGST & IGST Law.

5. Training of Officials and Trade &Industry: Until the people from trade and industry are adequately educated about the laws and procedures related to GST, there is possibility of non compliance and tax evasion. Training / familiarization of trade industry on a large scale will also be required.

\((\frac{dy}{dx})^2 + \cfrac{1}{\frac{dy}{dx}} = 2\)

⇒ \((\frac{dy}{dx})^3\) + 1 = \(2\frac{dy}{dx}\) (Multiply both sides by \(\frac{dy}{dx}\))

⇒ \((\frac{dy}{dx})^3\) - \(2\frac{dy}{dx}\) +1 = 0

which is differential equation in which highest order derivative is \(\frac{dy}{dx}\) and power raised to 3.

\(\therefore\) order = 1 & degree = 3.

The prescribed rate is 30%. No surcharge, education cess or SHEC shall beadded.