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Given:

A can fill the tank in 3 hours and B can empty the tank in 6 hours

Concept Used:

If time taken to fill the tank = X days/hours

∴ Part of tank fill in one day/hours = 1/X

Efficiency ∝ 1/Time

Calculation:

A can fill the tank in 3 hours

∴ Part of tank fill in one hours = 1/3

B can empty the tank in 6 hours

∴ Part of tank emptied in one hour = 1/6

Now, if both the pipes are open together

∴ Time taken to fill the tank = 1/3 – 1/6 = 1/6

So, both the pipes will fill the tank in 6 hours

Hence, option (2) is correct

Given:

x² – 4x + 3, 2x² – 3x + 1, and 2x² + 7x + 3

Concept:

The LCM of two or more given polynomials is the least polynomial which is exactly divisible by each of them.

Calculation:

x² – 4x + 3 = x² – 3x – x + 3

⇒ x(x – 3) – 1(x – 3)

⇒ (x – 1)(x – 3)

2x² – 3x + 1 = 2x² – 2x – x + 1

⇒ 2x(x – 1) – 1(x – 1)

⇒ (2x – 1)(x – 1)

2x² + 7x + 3 = 2x² + 6x + x + 3

⇒ 2x(x + 3) + 1(x + 3)

⇒ (2x + 1)(x + 3)

LCM of x² – 4x + 3, 2x² – 3x + 1, and 2x² + 7x + 3

⇒ (x – 1)(x – 3) (2x – 1)(x – 1) (2x + 1)(x + 3)

⇒ (2x – 1)(2x + 1) (x – 3)(x + 3) (x – 1)

⇒ (4x² – 1)(x² – 9)(x – 1)

∴ The required LCM is (4x² – 1)(x² – 9)(x – 1).

Given:

A contractor was engaged to construct a road in 16 days.

Formula used:

\(\frac{M_{1}\ × \ D_{1}}{W_{1}}\ =\ \frac{M_{2}\ × \ D_{2}}{W_{2}}\)

where M1 = M2 = number of man

D1  = D2 = number of days

W1 = W2 = part of work

Calculations: 

Remaining work = 1 - 5/8 = 3/8

Remaining time = 4 days

Using above formula we get,

⇒ \(\frac{20\ \times\ 12}{\frac{5}{8}}\ =\ \frac{M_{2}\ \times\ 4}{\frac{3}{8}}\)

⇒ M₂ = 36 

Number of additional workers = 36 – 20 = 16

∴ the correct answer is 16.

Given:

The first pipe can fill the pond in 45 minutes,

The second pipe can empty the pond in 1 hour

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Part of the pond filled in one minute

⇒ (1/45) – (1/60)

⇒ (4 – 3)/180

⇒ 1/180

(1/180) parts is filled in 1 minute

⇒ 1 – (1/45) = 44/45 parts is filled in (2 × 180 × 44)/45

⇒ 352 minutes

⇒ 5 hours 52 minutes

Remaining 1/45 parts will be filled in 1 minute

Total time taken = 5 hours 53 minutes

∴ The required time is 5 hours 53 minutes.

Given:

Time taken by pipe A = 20 minutes

Time taken by pipe B = 30 minutes

Time taken by pipe C = 40 minutes

Calculation:

Let the total capacity of the tank be LCM (20, 30, 40) = 120 litres

Now,

Tank filled by A in 1 minute = 120/20

⇒ 6 litre/min

Tank filled by B in 1 minute = 120/30

⇒ 4 litre/min

Tank filled by C in 1 minute = 120/40

⇒ 3 litre/min

Tank filled by (A + B + C) in 1 minute = (6 + 4 + 3)

⇒ 13 litre/min

Now,

According to the question,

For first minute pipe, A is opened = 6 litre

For second-minute pipe, B is opened = 4 litre

For third-minute pipe, C is opened = 3 litre

For fourth minute pipe, A is opened = 6 litre

For fifth-minute pipe, B is opened = 4 litre

For sixth-minute pipe, C is opened = 3 litre

Similarly, So…. On

Now,

We conclude that –

3 minutes = 13 litre

× 9                × 9

27 min   =      117 litre

Now,

The remaining work = 120 – 117

⇒ 3 litre

Now,

Again 28th minute A will be opened and fill the remaining tank

⇒ 3/6

⇒ 1/2

Now,

The tank fills in = (27 minute + ½ minute)

⇒ 55/2 minutes

∴ The tank will be filled in 55/2 minutes.

Given:

A swimming pool can be filled by tap P and tap Q together in 36 minutes

P open for 40 minutes and Q open for 30 minutes

Concept:

If a tap can fill a tank in x hours, then the tank filled by the tap in 1 hour = 1/x of the total tank.

Calculation:

Let the tap Q fill the tank in ‘a’ minutes.

Part of the swimming pool filled by tap P in 1 minute = \(\frac{1}{{36}} - \frac{1}{{\rm{a}}}\)

According to the question,

⇒ \({\rm{\;}}30 \times \frac{1}{{\rm{a}}}{\rm{}} + {\rm{}}40{\rm{}}\left( {\frac{1}{{36}} - \frac{1}{{\rm{a}}}} \right){\rm{\;}} = {\rm{}}1\)

⇒ \(\frac{{30}}{{\rm{a}}}{\rm{}} + {\rm{}}\frac{{10}}{9} - \frac{{40}}{{\rm{a}}}{\rm{}} = {\rm{}}1\)

⇒ \(\frac{{40}}{{\rm{a}}} - \frac{{30}}{{\rm{a}}}{\rm{}} = {\rm{}}\frac{{10}}{9} - 1\)

⇒ \(\frac{{10}}{{\rm{a}}}{\rm{}} = {\rm{}}\frac{1}{9}\)

⇒ a = 90 minutes

∴ Tap Q can fill the tank in 90 minutes.

Given:

Vinayak and Nitin, working separately can do a piece of work in 8 and 12 hours respectively.

Calculations:

Time taken by Vinayak to complete the work = 8 hours

Part of the work done by Vinayak in 1 day = 1/8

similarly, Part of the work done by Nitin in 1 day = 1/12

Part of the work done by Vinayak and Nitin in the first 2 hours = (1/8) + (1/12) = 5/24

Part of the work done in first 8 hours = (5 × 4)/24 = 5/6

Remaining work = 1 - 5/6 = 1/6

Now, it is the turn of Vinayak, part of the work done by Vinayak in 1 hour = 1/8

Remaining work = (1/6) - (1/8) = 1/24

Now, time taken by Nitin in completing this part of the work = 12/24 = 1/2 hours

Total time = 9 and 1/2 hours

The work starts at 9 am. Hence, the work will be completed at 6.30 pm.

∴ the correct answer is 6:30 pm.

Given:

15 ∶ x ∶∶ 5 ∶ y

Concept Used:

Product of mean numbers = Product of extreme numbers

Calculation:

15 ∶ x ∶∶ 5 ∶ y

⇒ 15 × y = 5 × x

⇒ 15/5 = x/y

⇒ x/y = 3/1

Then, x : y = 3 : 1

∴ The value of x : y = 3 : 1.

((2a - 4 + 8)/3, (4 + 3b + 14)/3, (6 - 10 + 2c)/3) = (0, 0, 0)

⇒ (2a + 4)/3 = 0 ⇒  a = - 2

similarly b = - 6 and c = 2.

Inverse: If a parallelogram is not a square then it is not a rhombus 

Converse: If a parallelogram is a rhombus then it is a square.

Slope =-1 ⇒ tanθ = -1 

angle made = 135°.

n (S) = 24 

P(visiting A before B) = 12/24 = 1/2

Equation BC = x + y – 3 = 0 

Length of the alitude from A (2, 3) = |(2 + 3 - 3)/√2| = |2/√2| = √2.

Taking the pair as x, x + 2, 

we get x > 50, 2x + 2 < 120 

⇒ x < 59 and writing the required pairs are (51, 53), (53, 55), (55, 57), (57, 59).

r = √2 and θ = π/4 

Polar form is √2 (cosπ/4 + isinπ/4)

Given:

0.002 × 0.5 = ?

Calculation:

0.002 × 0.5 = ?

⇒ (2 × 5)/(1000 × 10) = ?

⇒ 10/10000 = ?

⇒ ? = 0.001

∴ ? is 0.001.

(a) 20 Hz, 440 Hz, 8 Hz, 330 kHz to 300 GHz

Human voice lower hearing limit → 20Hz 

High quality music → 440Hz

Earth wave → 8Hz

Radio wave → 300kHz to 300 GHz

Given:

34.95 ÷ 240.016 ÷ 23.98 = ?

Calculation:

34.95 ÷ 240.016 ÷ 23.98 = ?

⇒ 34.95/240.16 ÷ 23.98 = ?

⇒ 0.14552/23.98 = ?

⇒ ? = 0.006

∴ ? is 0.006.

P(A) = 0.42 P(8) = 0.48, P(A ∩B) = 0.16 

(i) P(A₁) = 1 - 0.42 = 0.58 

(ii) P(B₁) = 1 - 0.48 = 0.52 

(iii) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) 

= 0.42 + 0.48 - 0.16 = 0.50 - 0.16 = 0.74.

Let us assume that √2 is rational i.e.. the gives statement is false. 

∴ √2 = 1/2 ≠ 0 a and b have no conunon factor. 

Squaring ∴ 2 =a²/b²  ⇒ [a² = 2b² ] ⇒ 2 divides a 

Again put a = 2C(C ∈ Z) 

∴ (2c)² = 2b² 

∴ 4c² = 2b² or [b² = 2c²]² divide b 

i.e., 2 divides both a and b hence our assumption is 

i.e.. a and b does f lot have common contradict out statement √2 is rational is false. 

∴ √2 is irrational. 

Equation of parabola y² = 8x compare with the standard form 

y² = 4ax ∴ 4a = 8 ∴ a = 2 

∴ focus = (a, 0) = (2, 0) 

Equation of directrix : x = -a. 

x = -2 or [x + 2 = 0] 

Latus rectum = 4a = 8

• The universal gas law is PV = nRT. P is pressure, V is volume, n is the amount of gas (mass, number of molecules, etc) R is a constant, and T is the temperature in Kelvins. 
• So since the pressure and the amount of gas is constant Volume is proportional to temperature.

When a surface is parallel to equipotential surface around a charge, the flux through the surface is maximum

Net electric flux is zero.

a = 8, a_n, = 26, n = 7, d = ?

Let the 5 AM's are A₁, A₂, A₃, A₄, and A₅

∴ (8), A₁, A₂, A₃, A_4, A₅ (26) ⇒ Take a_n, = a + (n - 1)d

26 = 8 + (7 - 1)d ⇒ 18 = 6d ∴ d = 3

∴ A_r, = a + d = 8 + 3 = 11 ⇒ A₂ = a + 2 = 8 + 6 =14

⇒ A₃ = 17, A₄ = 20, A₅ = 23.

8A₁, A₂, A₃, A₄, A₅, 26 

a = 8, n = 7, a_n = 26, d = ? 

a_n = a + (n-1)d ⇒ 26 = 8 + (7 – 1)d = 26 – 8 = 6d 

18 = 6d ∴ d=3 

Common difference 

∴ A₁ = a + d = 8 + 3 =11, A₂ =11 + 3 = 14, A₃ = 17, A₄ = 20, A₅ = 23 are the 5 AM’s

(A²+B²+2ABCOSø)^1/2=(A²+B²-2ABCOSø)^1/2​

So COSø=0

Ø=90°

Option D is correct..

In an inductor, when transient state is considered, the current will increase gradually as the inductor opposes the change in the current. But for deep understanding, let's have a look at newton's first law of motion..

“Every object in rest or motion tends to remain in its state unless its disturbed by external forces”

Now, initially, the inductor was idle but we tend to pass current through it. The inductor hates it. It tries to produce current in opposing direction so that it can cancel out the entering current so that it can remain in idle state. This is what we call as self inductance. The ability of an inductor to oppose the flow of current through itself..

But wait.. Where does the inductor get the source of opposing current it had produced? Doesn't it oppose law of conservation of energy?

No. We know that changing magnetic field can produce a current. So the source of opposing current is just the increasing magnetic field produced when the current increases gradually through the inductor. Hence, once steady state is reached, no opposing current is produced as the magnetic field becomes constant and only changing magnetic field with respect to time can produce a current..

So, the self inductance plays a role only in transient state.

Now, energy stored in the inductor means that it can produce the current even after the source is switched off. The energy here means the one used to pull the electrons and now you could have understood why it is referred to as “energy is stored in the inductor in the form of magnetic field”.

Let us move to capacitors. These are simpler than inductors.

We all know that capacitors are simply two metal plates separated by a distance with dielectric material in between them to increase capacitance. When negative potential is given to one plate and positive to another plate, the electrons accumulate on the negative plate of the capacitor as much as it can(determined by the capacitance value) provided that it is connected for sufficient time. When the source is switched off, the electrons accumulated there finds some other way through circuit to reach the positive plate of the other end in order to neutralize the charges. Hence “energy stored” here is taking advantage of the fact that electrons are attracted towards positive potential. So now its clear that “energy is stored in a capacitor in the form of electric field”.

Charges through inductor produces magnetic field and in capacitor produces electric field.

In case of Resistors:

Resistors are like a block to the flow of charges, and hence charges should spend some energy to cross them. And that energy is dissipated as heat in the resistor.

Hell yeah, charges through resistors produce heat.

But the problem here is, change in heat won't produce any energy to move the charges.. It just goes away ad heat.. Nothing else.

There may be a solution. But till now, scientists haven't discovered any method to make energy from the heat which is dissipating on the resistor.

So now, the problem is moving charges through resistor produces heat. So it can't store any energy.

b) Charge at B will suffer greater force

Lines are denser at the right end so the force will be greater at the points closer to that region.

(a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion. 

Here, I am giving you a correct format of "letter to the editor".

Yash marathe
Andheri west,
Mumbai
400001

Feb 05, 2019

Editor,
Times of India
10 Janpath Road,
Delhi

Subject: The downfall of educational interest of students in their studies

Sir,

I am Yash marathe studying in class .. in XYZ school, Mumbai. Every day I come across many students who were much good in their studies in past classes but they are loosing their interest in studies and their marks are going degraded each exam. I thought to write you a letter discussing the downfall of interest in studies of the students due to modern day technology.

In modern days as our technology is becoming advanced day by day, it is helping the mass as well as affecting in stacks. Students are wasting a lot of time in video games, social media, as these things grab their interest more than books and copies. But if all these factors are taken in a good way then the potential of students will increase to a greater extent. 

Sir, I want you to publish an editorial on this topic because a mass is getting affected by it. Hope you understand.

Yours truly,

Yash marathe
Student

(a) The Chechen Republic is a federal subject (a republic) of Russia. Darfur is a region in western Sudan. Swat is a valley and an administrative district in Pakistan.

The total hydel power potential of India is estimated at 150000 MW.

Wheat crops need low temperature during its sowing.

(c) A. Tropical evergreen forests- Above 200 cm

B. Tropical deciduous forests - 100-200 cm

C. Tropical Dry forest- 50-100 cm

D. Arid forest- Less than 50 cm

(a)

All these are the names of highest mountain peaks located in the countries respectively.

(a)

(c) The Steppes are areas of comparatively lower temperature and slightly more precipitation. Here annual average temperature is approximately 21°C. Temperate grasses are main vegetation of these regions. Norhern America’s Prairies and the steppes of Eurasia are the major areas of thesee grassland.

 750 parties are registered with the Election Commission.

Moran, Naharkatiya, Kalol and Mehsana are famous for Oil Field. 

(d) A – 4 ; B – 1 ; C – 2 ; D – 3

(d) Major natural regions of the world have not same type of climates. The different climates make different topographies.

Party based on ideological reasons; coercion and suppression of other parties is an ideological one- party system.

(c) A-iv,B-iii,C-ii,D-i

(a) Since United Kingdom, Norway, Netherland and Denmark are located on the coast, their temperature is higher than the places located on similar latitudes during the winter.

(c) It's not certain that political parties are only present in big countries.

One of China countries has single – party system.