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GIVEN:

The ratio of two numbers = 3 : 5

HCF of two numbers = 25

CONCEPT:

Product of two numbers = LCM × HCF

CALCULATION:

Suppose two numbers are 3x and 5x, respectively.

HCF = 25

∴ First number = 3x = 3(25) = 75

Second number = 5x = 5(25) = 125

⇒ 75 × 125 = 25 × LCM

⇒ LCM = 3 × 125

⇒ LCM = 375

GIVEN:

N = 4200

CONCEPT:

Let there be a composite number N and its prime factors be a, b, c, d, … etc. and p, q, r, s, … etc. be the indices (or powers) of the a, b, c, d, … etc respectively i.e., if N can be expressed as –

N = a^p. b^q.c^r.d^s…..

∴ The sum of the factors of N is \(\frac{{\left( {{{\rm{a}}^{{\rm{p\;}} + {\rm{\;}}1}} - 1} \right)\left( {{{\rm{b}}^{{\rm{q\;}} + {\rm{\;}}1}} - 1} \right)\left( {{{\rm{c}}^{{\rm{r\;}} + {\rm{\;}}1}} - 1} \right)\left( {{{\rm{d}}^{{\rm{s\;}} + {\rm{\;}}1}} - 1} \right)}}{{\left( {{\rm{a}} - 1} \right)\left( {{\rm{b}} - 1} \right)\left( {{\rm{c}} - 1} \right)\left( {{\rm{d}} - 1} \right)}}\)

CALCULATION:

The factor of 4200 is-

⇒ 4200 = 2³ × 3¹ × 5² × 7¹

∴ Sum of factors of 4200 = \(\frac{{\left( {{2^{3\; + \;1}} - 1} \right)\left( {{3^{1\; + \;1}} - 1} \right)\left( {{5^{2\; + \;1}} - 1} \right)\left( {{7^{1\; + \;1}} - 1} \right)}}{{\left( {2 - 1} \right)\left( {3 - 1} \right)\left( {5 - 1} \right)\left( {7 - 1} \right)}}\)

⇒ \(\frac{{\left( {16 - 1} \right)\left( {9 - 1} \right)\left( {125 - 1} \right)\left( {49 - 1} \right)}}{{\left( 1 \right)\left( 2 \right)\left( 4 \right)\left( 6 \right)}}\)

⇒ \(\frac{{15\; \times \;8\; \times \;124\; \times \;48}}{{1\; \times \;2\; \times \;4\; \times \;6}}\)

⇒ 15 × 992

⇒ 14880

Correct answer is (B) tan^-13/4

Correct answer is (B) P(A) . P(B/A)

Correct answer is (B) P(A'B') = [1 – P(A)] [1 – P(B)]

K_j, j and u represent the ages of Kiara, Juhi and Udita respectively.

Since, Kiara is twice as old as Juhi.

\(\therefore\) K = 2j

⇒ k  - 2j = 0---(1)

Since, Kiara is thrice as old as Udita.

\(\therefore\) K = 3u

⇒ k - 3u = 0---(2)

Since, sum of their ages is 33 years.

Matrix equation which represent all given scenario.

(or equation (1), (2) and (3)) is

\(\begin{bmatrix}1&-2&0\\1&0&-3\\1&1&1\end{bmatrix}\)\(\begin{bmatrix}k\\j\\u\end{bmatrix}=\begin{bmatrix}0\\0\\3\end{bmatrix}\)

or Ax = b, where A = \(\begin{bmatrix}1&-2&0\\1&0&-3\\1&1&1\end{bmatrix}\), x = \(\begin{bmatrix}k\\j\\u\end{bmatrix}\) b = \(\begin{bmatrix}0\\0\\3\end{bmatrix}\) 

Let \(\theta = 18°\)

\(5\theta = 18° \times 5 = 90°\)

⇒ \(2\theta + 3\theta = 90°\)

⇒ \(3\theta = 90° - 2\theta\)

⇒ \(sin3\theta = sin(90° - 2\theta ) = cos2\theta\)

⇒ \(3sin \theta - 4sin^3\theta = 1 - 2sin^2 \theta\)

⇒ \(4sin^3\theta - 2sin^2\theta - 3sin\theta + 1 = 0\)

Let \(sin\theta = x\)

\(\therefore 4x^3 - 2x^2 - 3x + 1 = 0\)

\((x - 1) (4x^2 + 2x - 1) = 0\)

\(\therefore x = 1 , \frac {-2 \pm \sqrt{4 + 16} }{2\times 4}\)

\(= 1, \frac{-2\pm 2\sqrt 5}8\)

\(= 1, \frac{-1\pm \sqrt 5}4\)

\(\because\) sin 18° is a positive value and not equal to 1.

\(\therefore sin 18° = x = \frac {-1 + \sqrt 5}4\).

W net =W mg +W air = 1 2 m(v f ^ 2 -v i ^ 2 )

Rightarrow0+W air = 1 2 *0.1[(6)^ 2 -(10)^ 2 ]=- - 3.2

From work energy thergy theorem,

W_net=W_mg + W_{air resistance}

According to Work-Energy theorem,

W_net = change in K.E.

 So, W_mg = 0 and W_air = 1/2m(v_f ² - v_i ² )

  =1/2×0.1(6² - 10² ) = -3.2 J.

\(\frac{d}{dx}sec^4x\) = 4 sec³x \(\frac{d}{dx}sec\,x\)

 = 4 sec³ x sec x tan x

 = 4 sec⁴ x tan x.

\(\int \frac{5}{\sin^2x\, \cos^2x}dx\)

\(=\int \frac{20}{4\sin^2x\, \cos^2x}dx\)

\(=\int \frac{20}{\sin^2 2x}dx\) (∵ 2 sin x cos x = sin 2x)

\(= \int\) 20 cosec² 2x dx

\(= \frac{-20\, \cot x}{2} + c\)

= -10 cot x + c

Penicillin is discovered by Fleming.

d. Vancomycin 

Correct option:  c. both.

Synergism & Antaginism

 The drug of choice in anaphylactic shock is Epinephrine.

d) Epinephrine

Correct answer is (b) 45°

Correct answer is (b) 0.83

Correct answer is (d) A - Z

(b) Law of conservation of energy

Correct answer is (b) (h/2π)Js

• Planning is the process of thinking about the activities required to achieve a desired goal. 
• It is the first and foremost activity to achieve the desired results. 
• It involves the creation and maintenance of a plan, such as psychological aspects that require conceptual skills.
• four types of planning used by managers, including strategic, tactical, operational and contingency planning. 
• Terms, such as single-use plans, continuing plans, policy, procedure, and rule, 

Correct answer is (c) 10²⁰

Answer: (d) Six month

b. 24.74 kbps 

Correct option is c. 48 Mbps 

Let the 3 parts be x, y and z 

Given 3x = 5y = 62 

Let 3x = 6z and 5y = 6/5 = z 

x = 2z, y = 2 

∴ x : y : z = 2z : 6/5z : z= 10:6:5 

Sum of the ratios = 10 + 6 + 5 = 21

∴ X = 10/21 X 1800 = 6000/7 = 857.14; Y = 6/21 X 1800 = 3600/7 = 514.29; Z = 5/21 X 1800 = 3000/7

Given :

Sachin takes three times longer than Sourav to do a work 

To complete the same work, Sourav takes three times as long as Yuvraj

To complete the same work Yuvraj takes three times as long as Dhoni

Concept used :

Person's per day work = Total work/Number of days he takes to complete the work 

Calculations :

Let the time taken by Dhoni be 't'

The time taken by Yuvraj will be '3t' 

The time taken by Sourav will be '9t' 

The time taken by Sachin will be '27t'

Now, 

Total work = LCM of t, 3t, 9t, and 27t 

⇒ 27t units 

Sachin's per day work = 27t/27t = 1 unit

Sourav's per day work = 27t/9t = 3 units 

Yuvraj's per day work = 27t/9t = 9 units 

Dhoni's per day work = 27t/t = 27 units 

One group can complete the work in 13 days and another group can complete in 31 days so,

The efficiency of the 1st group : Efficiency of 2nd group = 1/13 : 1/31 

⇒ 31 : 13 

We can see here 

Efficiency (Dhoni, Sourav and Sachin) : Efficiency (Sachin, Sourav and Yuvraj) = 31 : 13 

∴ Dhoni, Saurav, and Sachin group can complete the work in 13 days 

Given :

p / q = 1 / 3, q / r = 2, r /s = 1 / 2, s / t = 3 and t / u = 1 / 4

Concept used :

If p/q = a/b, q/r = b/c, r/s = c/d, s/t = d/e and t/u = e/f, then,

p : q : r : s : t : u = a.b.c.d.e : b.c.d.e.b : c.d.e.b.c : d.e.b.c.d : e.b.c.d.e : b.c.d.e.f 

Calculations :

p : q : r : s : t : u = 1.2.1.3.1 : 2.1.3.1.3 : 1.3.1.3.1 :  3.1.3.1.2 : 1.3.1.2.1 :  3.1.2.1.4

⇒ 6 : 18 : 9 : 18 : 6 : 24 or 6x : 18x : 9x : 18x : 6x : 24x 

Now for [pqr / str]3 put value of pqrstu in this equation 

[pqr / str]³ = [(6x × 18x × 9x)/18x × 6x × 24x)]³

⇒ (3/8)²

⇒ 27/512 

∴ 27/512 will be the correct answer 

Quantity A:

Simple interest = (principal × rate × time)/100;

Simple interest = Amount – principal

Simple interest = Rs. 2,640 – Rs. 2,200

⇒ Simple Interest = Rs. 440

⇒ 440 = (2,200 × 5 × time)/100

⇒ Time = 4 years

∴ Required time = 4 years

Quantity B:

8 years

∴ Quantity A < Quantity B

Given:

Hari capital = 5 (Om’s capital); Om’s capital = 2 (Jagdish’s capital)

Calculation:

Let the Jagdish’s capital be Rs x

Then, Om’s capital = 2x

And, Hari’s capital = 5 (2x) = 10 x

Ratio of capital of Hari, Om and Jagdish = 10x : 2x : x

⇒ 10 : 2 : 1

∴ The required ratio is 10 : 2 : 1

Quantity A:

S.P. = M.P. × (100 – d1/100) (100 – d2/100)

SP, MP, d1 and d2 are selling price, market price, first discount and second discount respectively.

⇒ 240 = MP × (100 – 20/100) (100 – 20/100)

⇒ 240 = MP × (80/100) (80/100)

⇒ (240 × 100 × 100/80 × 80) = MP

⇒ Rs 375 = MP

∴ Market Price of the article is Rs. 375

Quantity B:

Rs. 400

∴ Quantity A < Quantity B

Perimeter of rectangle = 40 m

Length = 15 m

Breadth = ?

Perimeter of rectangle = 2(l + b)

Perimeter of rectangle = 2(l + b) = 40 m

So, l + b = 20 m

l = 15 m

So b = 20 -15 = 5 m

x − y = 50...(1)

Substituting the value of x = 200, in equation (1), we get

200 − y = 50 ⇒ y = 200 − 50 = 150.

Hence, the breadth of the rectangular field is y = 150 m.

GIVEN:

Three statements.

CONCEPT:

Mensuration

FORMULA USED:

Area of rhombus = (D₁ × D₂) / 2

Area of an equilateral triangle = √3a² / 4

CALCULATION:

A:

Let diagonals of rhombus are ‘3x’ and ‘4x’ respectively.

Area of rhombus = (3x × 4x) / 2

96 = 6x²

⇒ x = 4

Side of rhombus = √[6² + 8²] = 10 m

Perimeter of rhombus = 10 × 4 = 40 cm

Cost of fencing = 40 × 24 = Rs. 960

B:

Let side of triangle = ‘a’

Now,

√3a² / 4 = 36√3

⇒ a = 12

Let length of perpendicular from a vertex = ‘x’

Now,

36√3 = (1 / 2) × x × 12

⇒ x = 6√3 cm

C:

Let side of triangle = ‘a’

Now,

√3a² / 4 = 49√3

⇒ a = 14

Perimeter of triangle = 3a

= 42 cm

Given that C is centre of the circle and AB is a diameter of the circle where 

coordinates of A and B are (–2, 9) and (6, 3) respectively. 

∴ C is midpoint of points A and B. 

(∵ Centre is always mid-point of diameter in a circle)

∴ the coordinates of centre C is \((\frac{-2+6}2,\frac{9+3}2)\) = \((\frac{4}2,\frac{12}2)\) = (2,6)

Hence, the co-ordinates of the centre of the circle is (2, 6).

x − y = 50...(1)

2x + y = 550...(2)

Adding equations (1) and (2), we get

x − y + 2x + y = 50 + 550 ⇒ 3x = 600 ⇒ x = 200.

Hence, the length of the rectangular field is x = 200 m.

The faces of die shows the letters as A, B, C, A, D, A. 

Total numbers of A’s = 3. 

Total number of D’s = 1. 

Total number of faces = 6. 

(i) Probability of getting A = \(\frac{Total\,number\,of\,face\,having\,A}{Total\,number\,of\,faces}\) = \(\frac{3}6\) = \(\frac{1}2\).

(ii) And probability of getting D = \(\frac{Total\,number\,of\,faces\,having\,D}{Total\,number\,of\,faces}\) = \(\frac{1}6\).

Total number of English alphabets is 26. 

Total consonants in English alphabet are 21. 

One letter of English alphabet is chosen at random. 

∴ the probability that the chosen letter is a consonant

= \(\frac{Total\,consonant\,in\,english\,alphabet}{Total\,number\,of\,english\,alphabets}\) = \(\frac{21}{26}\).

Hence, the probability that the chosen letter is a constant is \(\frac{21}{26}\).

(c) P(A) + P(B) - P(A ∩ B) 

Given that 3 tan A = 4 sin A

⇒ \(\frac{3}{sin A}=\frac4{tan A}\) (On dividing both sides by Sin A tan A)

⇒ 3 cosec A = 4 cot A (\(\because\) \(\frac1{sin A}=cosec A \,and\,\frac{1}{tan A}=cot A\))

which is required relation between cosec A and cot A

Let cos⁻¹(√3/2​​) = x, where x ∈ [0,π].

Then, cosx = √3/2 ​​= cos π/6 ​⇒ x = π/6​.

cosec²(90°- θ) - tan²θ

= sec² θ − tan² θ 

= 1

Probability of getting a doublet = \(\frac{1}{6}\)

πR²H = 12 x \(\frac{4}{3}πr^3\)

1 x 1 x 16 = \(\frac{22}{7}\) x r³ x 12 

r³ = 1 

r = 1 

d = 2cm