Colloids can be classified on various bases:

(i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids, or gases, we can have eight types of colloids.
(ii) On the basis of the dispersion medium, sols can be divided as:

(iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).

भारत में लाल मिट्टी कर्नाटक, आन्ध्र प्रदेश, मध्य प्रदेश, महाराष्ट्र के दक्षिण-पूर्वी भागों, तमिलनाडु, मेघालय, ओडिशा में पायी जाती है।

1.   स्वामी विवेकानंद : जीवनी :: गुलाब सिंह : कहानी

2.   स्वामी विवेकानंद : जगदीश चंद्र :: चेरापूँजी से आया हूँ : प्रदीप पंत ।

3.   मौसी : कहानी :: भीम और राक्षस : एकांकी

4.   मौसी : भीष्म साहनी :: भीम और राक्षस : विष्णु प्रभाकर

5.   खेलो कूदो स्वस्थ रहो : निबंध :: नफे के चक्कर में : लोककथा

6.   जय जय भारत माता : मैथिलीशरण गुप्त :: चलना हमारा काम है। : शिवमंगलसिंह सुमन

7.   पर्यावरण बचाओ : डॉ.परशुराम शुक्ला :: पूर्वाक्षर का पूर्वाग्रह : पी.एस. रामानुजम

8.   भीम और राक्षस : एकांकी :: सडक की रक्षा-सबकी सुरक्षा : कहानी

9.   भीम : कुंती :: स्वामी विवेकानंद : भुवनेश्वरी देवी ।

10.  तरूवर फल : नहीं खात है :: सरवर : पियहि न पान

11.  यह मेरा : बेटा :: यह मेरी : बेटी

12.  पाठ पढाना : सबक सिखाना :: भाग्य फूटना : बुरा होना

13.  शारीरिक खेल : शारीरिक श्रम :: मानसिक खेल : मानसिक श्रम ।

14.  मेघालय की राजधानी : शिलंग :: मणिपुर की राजधानी : चेरापूंजी

15.  क्रिकेट : मैदानी खेल :: शतरंज : घरेलू खेल

16.  दौडना : वैयक्तिक खेल :: फुटबाल : सामूहिक खेल

17.  तुम्हारा नाम क्या है ? : प्रश्नार्थक वाक्य :: पानी ले आओ। : आज्ञार्थक वाक्य

18.  साथी, हाथ बढाना : साहीर लुधियानवी :: अनुशासन ही शासन है : डॉ. रमेश मिलन

1. b. C-B 

2. c. Only B 

3. c. Only C 

4. b. It may increase food production to meet the rising demands. 

5. b .Decreased through the years 

6. d. Ground water

bx + c

Given quadratic polynomial is f(x)=ax^ 2 +

The zeroes of the polynomial are aß zeros = - b/a

Sum of the alpha + beta = - b/a Poduct of the alpha beta=> c a Squaring both sides, zeros = c/a

Now, alpha + beta = - b/a

=>( alpha+ beta)^ 2 =-( b a )^ 2

=> alpha^ 2 + beta^ 2 +2 alpha beta= b^ 2 a^ 2

=>a^ 2 + beta^ 2 +2( c a )= b^ 2 a^ 2

=> alpha^ 2 + beta^ 2 = b^ 2 a^ 2 - 2c a

=> alpha^ 2 + beta^ 2 = b^ 2 -2ca a^ 2

=>a^ 2 + beta^ 2 = b^ 2 -2ac a^ 2

<span class="vlist" styl

C = 10 \(\mu\)

d = 0.5 mm

v = 20 v

q = ?

Charge on each plate 

q = cv

q = 10 x 20

q = 200\(\mu\)c

It is in the 0/0 form 

Lt x → π/4    0 - f'(π/4) / 1 - 0

8³ – 8 = 504,

9² – 9 = 72,

10³ – 10 = 990,

11² – 11 = 110, 

12³ – 12 = 1716,

13² – 13 = 156

The wrong term is 1714

Given that,

The series follows the following pattern:

3, 14, 39, 84, 156, 258

1³ + 1² + 1 = 3

23 + 22 + 2 = 14

33 + 32 + 3 = 39

43 + 42 + 4 = 84

53 + 52 + 5 = 155

63 + 62 + 6 = 258

∴ 156 is the wrong in the given patterrn

Given that,

256, 260, 251, 267, 243, 278

256

⇒ 256 + 2² = 260

⇒ 260 - 32 = 251

⇒ 251 + 42 = 267

⇒ 267 - 52 = 242

⇒ 242 + 62 = 278

∴ 243 is the wrong number in the given pattern

Given:

18, 19, 27, 54, ?, 243

Calculation:

The series follows the following pattern

18 + 1³ = 19

19 + 2³ = 27

27 + 3³ = 54

54 + 4³ = 118

118 + 5³ = 243

∴ The Required term in the series will be 118

Given series is 8+12+18+...

Since, elements of the series are even numbers, therefore, there sum is always even.

It can't be 243/4.

Hence, no terms of the given series is 243/4.

\(\frac{dy}{dx}-\frac1{x}y = xy^2\) 

⇒ \(\frac1{y^2}\frac{dy}{dx}-\frac1{x}.\frac1{y}=x\)----(1)

Let \(\frac1y=z\)

Then \(-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}\) 

Then from (1), we obtain

\(-\frac{dz}{dx}-\frac1{x}z=x\) 

⇒ \(\frac{dz}{dx}+\frac1{x}z=-x\) 

\(\therefore\) I.F. = e^\(\int\)pdx = e^{\(\int\) (1/x)dx} = e^{log x} = x

\(\therefore\) Complete solution is

y x I.F. = \(\int\)(I.F.). Q dx

⇒ y \(\times\) x = \(\int\)x \(\times\) -x dx

⇒ yx = -\(\int\)x²dx

⇒ yx = \(-\frac{x^3}3+C\) 

⇒ y = \(-\frac{x^2}3+\frac{C}x\) 

Hence, solution of given differential equation is

y = \(-\frac{x^2}3+\frac{C}x\)

Given differential equation is

x² \(\frac{d^2y}{dx^2}+5x\frac{dy}{dx}-5y=2log x\)

Let x = e^z then x\(\frac{dy}{dx}=\frac{dy}{dz}\) = Dy, where \(\frac{d}{dx}=D\)

and \(x^2\frac{d^2y}{dx^2}\) = D(D - 1)y

\(\therefore\) Given differential equation converts into

D(D - 1)y + 5Dy - 5y = 2z (\(\because\) log x = z)

⇒ (D² - D + 5D - 5)y = 2z

⇒ (D² + 4D - 5)y = 2z----(1)

It's auxilarly equation is

m² + 4m - 5 = 0

⇒ (m + 5)(m - 1) = 0

⇒ m + 5 = 0 or m - 1 = 0

⇒ m = -5 or m = 1

\(\therefore\) C.F. = C₁e^-5z + C₂e^z

& P.I. = \(\frac{1}{D^2+4D-5}2z\) 

\(=\frac{-2}5\cfrac1{1-\frac{D^2+4D}5}z\) 

\(=\frac{-2}5(1-\frac{D^2+4D}5)^{-1}z\)

\(=\frac{-2}5(1+\frac{D^2+4D}5+\frac{(D^2+4D)^2}{5^2}+....)z\)

(\(\because\) (1 - x)^-1 = 1 + x + x²+....)

\(=\frac{-2}5(z+\frac15D^2z+\frac45 Dz)\) 

\(=\frac{-2}5(z+0+\frac45)\) (\(\because\) Dz = \(\frac{dz}{dz}=1\))

\(=\frac{-2}5z-\frac8{25}\) 

\(\therefore\) Complete solution of differential equation(1) is

y = C.F. + P. I.

 = C₁(e^z)^-5 + C₂e^z- \(\frac25z-\frac8{25}\)

 = C₁x^-5 + C₂ x - \(\frac25\) log x - \(\frac8{25}\)

(\(\because\) e^z = x ⇒ z = log x)

 = \(\frac{C_1}{x^5}+C_2x - \frac25logx-\frac8{25}\) 

which is solution of given differential equation.

\(\because\) A.M. \(\geq\) G.M.

\(\therefore\) \(\frac{cos A+cos B+cos C}3\geq(cos A. cos B. cos C)^{1/3}\)

But we know that Cos A + Cos B + Cos C \(\leq\) 3/2

\(\therefore\) (cos A. Cos B. Cos C)^1/3 \(\leq\) \(\frac{3/2}3=\frac12\)

\(\therefore\) Cos A. Cos B. Cos C \(\leq\) \((\frac1{2})^3\) 

⇒ Cos A. Cos B. Cos C \(\leq\) \(\frac18\) 

Hence, maximum value of Cos A. Cos B. Cos C is 1/8

Given:

134, 245, 356, 467

Calculation:

⇒ 134 + 111 = 245

⇒ 245 + 111 = 356

⇒ 356 + 111 = 467

⇒ 467 + 111 = 578

∴ The missing number is 578.

S.P of the item = Rs (Q + 45)

So, ((Q + 45) – Q)/Q) × 100% = 12.5%

⇒ 45/Q = 0.125

⇒ Q = 360

S.P of the item = Rs. 405

M.P of the item = Rs. 360(1 + R/100)

S.P of the item = Rs 360(1 + R/100)(1 – S/100)

So, 360 (1 + R/100)(1 – T/100) = 405

⇒ (1 + R/100)(1 – S/100) = 9/8 ----(i)

In new scenario :

S.P of the item = 360(1 + (4R/5)/100) (1 – R/100)

Since, there is a loss of Rs 36, S.P = 360 – 36 = 324

So, 360(1 + (4R/5)/100) (1 – R/100) = 324

Let R/100 = x

So, (1 + 4x/5) (1 – x) = 9/10

⇒ 1 – x + 4x/5 – 4x²/5 = 9/10

⇒ 10 – 10x + 8x – 8x² = 9

⇒ 8x² + 2x – 1 = 0

⇒ (2x + 1) (4x – 1) = 0

⇒ x = -1/2 or ¼

If x = -1/2, R = -50, but the percentage markup cannot be negative,

So, x = ¼, B = 25

Putting in (i)

(1 + ¼)(1 – S/100) = 9/8

⇒ 1 – S/100 = 9/10

⇒ C = 10

S.P of the item if it was sold without discount =

360 × (125/100) = Rs 90

So, all (Q), (R), (S) and (T) are determined.

Given:

Selling price of 10 booklets= Cost price of 12 booklets

Formula used:

Gain% = ((S.P. – C.P.)/C.P) × 100

Calculation:

S.P. of 10 = C.P. of 12

S.P/C.P = 12/10

Let C.P. be 10x

So, S.P. = 12x

Gain% = ((12x – 10x)/10x) × 100

⇒ Gain% = 20%

∴ Gain% is 20%

Given:

The marked price of TV = 40% higher than CP

Discount is given by the Trader = 20%

Formula used:

S.P = M.P × (100 – D%)/100

S.P = C.P × (100 + P%)/100

P = S.P – C.P

P% = (P/C.P) × 100

Where,

S.P → Selling price

M.P → Marked price

D% → Discount%

P% → Profit%

P → Profit

C.P → Cost price

Calculation:

Let the price of T.V be 100x

Marked price is 40% more = 100x + 100x × (40/100) = 140x

S.P of the T.V after 20% discount = 140x × 80/100 = 112x

Profit = 112x – 100x = 12x

Profit% = (12x/100x) × 100

⇒ Profit% = 12%

∴ Profit% is 12%

Given:

Total profit = 36,000

Suraj invested = Rs 20,000

Deepak invested = Rs 25,000

Calculation:

Quantity I:

Prabhakar invested 5,000 less than suraj

⇒ Prabhakar invested = Rs 15000

⇒ Prabhakar : Suraj : Deeoak = 15000 : 20000 : 25000

⇒ P : S : D = 3 : 4 : 5

⇒ Prabhakar’s share = (3 / 12) × 36000

⇒ Prabhakar’s share = Rs 9,000

Quantity II:

Total amount = Rs 39,000

Suraj = Prabhakar

⇒ P : S : D = 20000 : 20000 : 25000

⇒ P : S : D = 4 : 4 : 5

⇒ Prabhakar’s share = (4 / 13) × 39000

⇒ Prabhakar’s share = Rs 12,000

∴ Quanity I < Quantity II

Let the sum of Suraj be S,

Let the sum of Aditya be A, and

Let the sum of Ravi be R.

Hence, from the given conditions, we get:

S = 4R      ----(i)

R = 16A

⇒ A = R/16      ----(ii)

So, we can obtain the ratio of the sums of Aditya and Suraj as:

A/S = (R/16)/4R = 1/64

∴ The required ratio of the sums of Aditya and Suraj is 1 ∶ 64

Given:

Total profit made by Prabhakar and Suraj = Rs. 11000

1/3 of Suraj’s profit = 2/5 Prabhakar’s profit

Calculation:

1/3 of Suraj’s profit = 2/5 Prabhakar’s profit

Suraj’s profit ∶ Prabhakar’s profit = 6 ∶ 5

Let profit of Suraj be 6x

So, the Profit of Prabhakar = 5x

Total of Suraj and Prabhakar's profit = 6x + 5x = 11x

⇒ 11x = Rs. 11000

⇒ x = Rs. 11000/11

⇒ x = Rs. 1000

⇒ Prabhakar's profit = 1000 × 5 = Rs. 5000

∴ Profit of prabhakar is Rs. 5000

Given:

Ratio of the marked price(MP) of a trouser and a jacket = 5 : 7

Discount on trouser = 30%

Total discount on both of them = 50%

Formula used:

Discount% = (Discount/MP) × 100

Calculation:

Let the MP of trouser and jacket be 5x and 7x

Discount on trouser = 0.3 × 5x = 1.5x

Total discount on both = 0.5 × (5x + 7x) = 6x

⇒ Discount on jacket = 6x - 1.5x = 4.5x

Discount% on jacket = (4.5x/7x) × 100 = 64.28

∴ Discount on jacket = 64.28%

Given:

SP of phone = Rs. 11,620

Loss% = 17%

Formula used:

CP = SP × [100/(100 – Loss%)] 

Calculation:

CP = Rs. 11,620 × [100/(100 – 17)]

⇒ Rs. 11620 × (100/83)

⇒ Rs. 14,000

∴ The C.P of phone was Rs. 14,000

Given:

MRP = 20% more than C.P

Discount = 30%

Concept Used:

Cost price × (100 + Profit %)/100 = Selling price

Discount % = {(MRP - SP)/MRP} × 100

Calculation:

Cost price = CP,  Selling price = SP,  Marked price = MRP

Let CP be 100.

MRP = 20% more than CP

⇒ MRP = 120

Discount = 120 × 30% 

⇒ 36

SP = 120 - 36

⇒ 84 

⇒ Loss = {(100 - 84)/100} × 100

∴ Loss percent is 16%

Given:

Marked price of the table = ₹ 5000

Successive discount = 22 percent and 37 percent

Selling price of the table = ₹ 4200

Transportation charges = ₹ 543

Formula used:

Profit = Selling price – Cost price

Discount = Marked price – Selling price

Profit percentage = (Profit/Cost price) × 100

Discount percentage = (Discount/Marked price) × 100

Calculation:

Cost price of the table = 5000 × [(100 – 22)/100] × [(100 – 37)/100]

⇒ 5000 × (78/100) × (63/100) = 2457

Total cost price of the table = 2457 + 543 = 3000

Profit on table = 4200 – 3000 = 1200

Profit percentage = (1200/3000) × 100 = 40%

∴ The profit percentage after selling the table is 40%.

Given:

Ramesh bought 200 spoons at a total price of Rs. 2400.

He sold 150 spoons at Rs. 14 per spoon.

Formula Used:

Profit or Gain = Selling price – Cost Price

Loss = Cost Price – Selling Price

Profit percentage = (Profit /Cost Price) x 100

Loss percentage = (Loss/Cost price) x 100

Calculation:

Ramesh bought 200 spoons at a total price of Rs. 2400

Selling price = Rs. (150 × 14) = Rs. 2100

Amount remains for no profit no loss = Rs. (2400 – 2100) = Rs. 300

To gain a overall 10% the selling price = Rs. (2400 × 11/10) = Rs. 2640

⇒ 2640 – 2100 = 540

Rate = 540/50 = 10.8

∴ The rate is 10.8%

x = Δx/Δ  = 1

y = Δy/Δ = 1

Let the sum be Rs. x. Then, 

C.I. = x ( 1 + ( 10 /100 ))² - x = 21x / 100 , 

S.I. = (( x * 10 * 2) / 100) = x / 5 

(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100

(x/100) = 632

x = 63100

Hence, the sum is Rs.63,100.

Let the rate be R% p.a. then, 

[ 18000 ( 1 + ( R / 100 )² ) - 18000 ] - ((18000 * R * 2) / 100 ) = 405 

18000 [ ( 100 + (R / 100 )² / 10000) - 1 - (2R / 100 ) ] = 405 

18000[( (100 + R )² - 10000 - 200R) / 10000 ] = 405 

9R² / 5 = 405 

R² =((405 * 5 ) / 9) = 225 

R = 15. 

Rate = 15%.

Let (X, Y) be the new coordinates of (x, y)

So,  x = X cosα – Y sinα, y = X sinα + Y cosα

The transformed equation is (X cosα – Y sinα) cosα + (X sinα + Y cosα) sinα = p

=> X cos²α – Y sinα cosα + Y sin²α + Y cosα sinα = p 

=> X(cos²α + sin²α) = p

=>X = p.

We know that Mass = Volume x density

m = (A x H).\(\rho\) 

m = 200 x 10^-4 x 5 x 997

m = 997 x 10^-1

m = 99.7 kg

Given the diameter of the wheel is 2r = 40 cm. 

∴ The radius of the wheel is r = 20 cm. 

∴ The circumference of the wheel is 2\(\pi\)r = 40\(\pi\) cm = 40 × 3.14 cm = 125.6 cm. 

So, distance covered in 1 revolution = 125.6 cm. 

∴ No. of revolution to covering distance 125.6 cm is 1. 

∴ No. of revolution to covering distance 1 cm is \(\frac{1}{1256}\).

∴ No. of revolution to covering distance 17600 cm is \(\frac{1}{125.6} \times 17600 = 140.12\).

Hence, total 141 revolution needed for a wheel of diameter 40 cm to cover a distance of 176 m in travelling.

Diameter of wheel = 40 cm = 0.4 m   Radius of wheel = D/2  = 0.4 / 2       = 0.2m

Circumference of wheel = 2πr  = 2 × 22/7 × 0.2  =  1.256m .

Therefore , distance covered in one revolution = 1.256m 

No. Of revolutions to cover 176 m   = Total Distance / Distance     covered in one revolution

= 176 / 1.256  = 140.12

Hence , No of revolutions required to cover the distance of 176 m is 141 revolutions .

Concept:

\(\rm \left(\vec a + \vec b\right) \cdot \left(\vec a - \vec b\right) = \left|\vec a\right|^2-\left|\vec b\right|^2\)

If \(\rm \vec u\) is a unit vector, then \(\rm \left|\vec u\right|=1\).

Calculation:

It is given that \((\rm \vec x + \rm 2\vec a) \cdot (\rm \vec x - \rm 2\vec a) = 12\).

⇒ \(\rm \left|\vec x\right|^2 - 4\left|\vec a\right|^2 = 12\)

Since \(\rm \vec a\) is a unit vector, we get:

⇒ \(\rm \left|\vec x\right|^2 - 4= 12\)

⇒ \(\rm \left|\vec x\right|^2 =16\)

⇒ \(|\rm \vec x |\) = 4

Let each installment be Rs.x.  

Then,(P.W. of Rs.x due 1 year hence)+(P>W of Rs.x due 2 years hence)+(P.W of Rs. X due 3 years hence)=7620. 

x/(1+(50/3*100))+ x/(1+(50/3*100))² + x/(1+(50/3*100))³ =7620 

(6x/7)+(936x/49)+(216x/343)=7620. 

294x+252x+216x=7620*343. 

x=(7620*343/762)=3430. 

Amount of each installment=Rs.3430.

Direct and inverse proportion:

Let us consider some examples to understand direct and inverse proportion:

1. More you ride your bicycle the more distance you will cover.
2. More the packets of biscuits more the number of biscuits.
3. If there are many shops of the same kind then less will be the crowd.
4. If there is more advertisement in newspaper then less will be the news.

From the above examples, it is clear that change in one quantity leads to change in the other quantity.

Types of proportions:

There are two proportions i.e. direct and inverse proportion.

1. Direct proportion:

• We can say two quantities x and y in direct proportion if they increase (decrease) together in the same manner.
• Also, two quantities x and y in direct proportion if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant. That is if \(\frac{x}{y}\) = k [k is a positive number], then x and y are said to vary directly.
• In such a case if y₁, y₂ are the values of y corresponding to the values x₁ , x₂ of x respectively then \(\frac{x_{1}}{y_{1}}\) = \(\frac{x_{2}}{y_{2}}\)
• If x is directly proportional to y, we can write it as x ∝ y.

Variables increasing (or decreasing) together need not always be in direct proportion. For example:

• In human beings, physical changes occur with time but not necessarily in a predetermined ratio.
• Among individuals when we talk about changes in weight and height are not in any known proportion.
• In natural phenomena such as there is no direct relationship or ratio between the height of a tree and the number of leaves growing on its branches.

2. Inverse proportion:

• We can say two quantities x and y in direct proportion if they increase (decrease) together in the same manner.
• Also, two quantities x and y are said to be in inverse proportion if an increase in x causes a proportional decrease in y (and vice-versa) in such a manner that the product of their corresponding values remains constant. That is, if xy = k, then x and y are said to vary inversely.
• In such a case if y₁, y₂ are the values of y corresponding to the values x₁ , x₂ of x respectively then \(\frac{x_{1}}{x_{2}}\) = \(\frac{y_{2}}{y_{1}}\)
• If x is inversly proportional to y, we can write it as x ∝ \(\frac{1}{y}\).

Direct and inverse proportion examples:

1. An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.

Solution:

More the height of an object, the more would be the length of its shadow.
Thus, x1 : x2 = y1 : y2
14 : x = 10 : 15
10 × x = 15 × 14
thus, x = 21

2. 6 pipes are required to fill a tank in 1 hour 20 minutes. How long will it take if we use only 5 pipes of the same type?
   Solution:

This is a case of inverse proportion as less number of pipes, it will require more time to fill the tank.
Hence, 80 × 6 = x × 5  as [x₁y₁ = x₂ y₂]
or, \(\frac{80x6}{5}\)
x = 96
Thus, time taken to fill the tank by 5 pipes is 96 minutes or 1 hour 36 minutes.

Given:

Anant invested Rs. 81000 in a business for 8 months and then 4000 more for 4 months. B invested certain amount for 4 months and then increased the invested amount by 5000 for the remaining 8 months. The ratio of their profits was 247 : 130.

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Let the amounts invested by B for 4 and 8 months respectively be x and x + 5000 respectively.

(81000 × 8 + 85000 × 4)/(x × 4 + (x + 5000) × 8) = 247/130

(988000)/(12x + 40000) = 247/130

⇒ 4000/(12x + 40000) = 1/130

⇒ 1000/(3x + 10000) = 1/130

⇒ 130000 = 3x + 10000

⇒ x = 40000

The amount invested by B in the last 8 months = x + 5000 = 40000 + 5000 = 45000

Most common petrocrops  belongs to family Euphorbiaceae which have property of converting large amount of their photosynthesis  into latex with hydrocarbons . their hydrocarbons contents can be increased by genetic manipulations .

Correct answer is (C)

Given:

Total amount = Rs 1120

Ratio of coins of Rs 1, 50 paisa, and 25 paisa = 1 ∶ 1/2 ∶ 1/3

Formula Used:

Value of coins = Number of coins × per coin value

Calculation:

We know that

Rs 1 = 100 paisa

Ratio of value per coin = 100 ∶ 50 ∶ 25

⇒ 4 ∶ 2 ∶ 1

Ratio of number of coins = 1 ∶ 1/2 ∶ 1/3

⇒ 6 ∶ 3 ∶ 2

Total value of coins = 24 ∶ 6 ∶ 2

⇒ 12 ∶ 3 ∶ 1 = 16 units

⇒ 16 units = 1120

⇒ 1 units = 70

Value of 25 paisa coins = 1 units

⇒ 1 × 70

⇒ Rs. 70

∴ The total number of 25 paisa coins is Rs. 70.

Given:

Total amount = Rs 1120

Ratio of coins of Rs 1, 50 paisa, and 25 paisa = 1 ∶ 1/2 ∶ 1/3

Formula Used:

Value of coins = Number of coins × per coin value

Calculation:

We know that

Rs 1 = 100 paisa

Ratio of value per coin = 100 ∶ 50 ∶ 25

⇒ 4 ∶ 2 ∶ 1

Ratio of number of coins = 1 ∶ 1/2 ∶ 1/3

⇒ 6 ∶ 3 ∶ 2

Total value of coins = 24 ∶ 6 ∶ 2

⇒ 12 ∶ 3 ∶ 1 = 16 units
⇒ 16 units = 1120

⇒ 1 units = 70

Value of 50 paisa coins = 3 units

⇒ 3 × 70

⇒ Rs 210

Given:

Total amount = Rs 1120

Ratio of coins of Rs 1, 50 paisa, and 25 paisa = 1 ∶ 1/2 ∶ 1/3

Formula Used:

Value of coins = Number of coins × per coin value 

Calculation:

We know that

Rs 1 = 100 paisa

Ratio of value per coin = 100 ∶ 50 ∶ 25

⇒ 4 ∶ 2 ∶ 1

Ratio of number of coins = 1 ∶ 1/2 ∶ 1/3

⇒ 6 ∶ 3 ∶ 2

Total value of coins = 24 ∶ 6 ∶ 2

⇒ 12 ∶ 3 ∶ 1 = 16 units
⇒ 16 units = 1120

⇒ 1 units = 70

Value of Rs 1 coins = 12 units

⇒ 12 × 70

⇒ Rs 840

∴ The total number of Rs 1 coins is Rs 840.  

Have you ever found yourself waiting impatiently for the online release of a product, one that you’re eagerly waiting to purchase? You keep refreshing the page, waiting for that moment when the product will go live. Then, as you press F5 for the last time, the page shows an error: “Service Unavailable.” The server must be overloaded! 

There are indeed cases like these where a website’s server gets overloaded with traffic and simply crashes, sometimes when a news story breaks. But more commonly, this is what happens to a website during a DoS attack, or denial-of-service, a malicious traffic overload that occurs when attackers over flood a website with traffic. When a website has too much traffic, it’s unable to serve its content to visitors. 

A DoS attack is performed by one machine and its internet connection, by flooding a website with packets and making it impossible for legitimate users to access the content of flooded website. Fortunately, you can’t really overload a server with a single other server or a PC anymore. In the past years it hasn’t been that common if anything, then by flaws in the protocol. 

A DDoS attack, or distributed denial-of-service attack, is similar to DoS, but is more forceful. It’s harder to overcome a DDoS attack. It’s launched from several computers, and the number of computers involved can range from just a couple of them to thousands or even more. 

Since it’s likely that not all of those machines belong to the attacker, they are compromised and added to the attacker’s network by malware. These computers can be distributed around the entire globe, and that network of compromised computers is called botnet.

Since the attack comes from so many different IP addresses simultaneously, a DDoS attack is much more difficult for the victim to locate and defend against.