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(a)?/84= 189/?

or  ?² = 84 × 189
or  ?² = 21 × 4 × 21 × 9
or  ?² = 21² × 2² × 3²
 ? = 21 × 2 × 3 = 126

Given:

The given number = 3481

Concept:

Calculation skills trick

Calculation:

Step(1):

The unit digit in this number is 1, which can be a unit digit of its square root number such as 1 or 9. Because 1 x 1 is 1 and 9 x 9 is 81.

Step(2):

Now let us consider the first two digits that is 34 which comes between the squares of 5 and 6 because 25 < 34 < 36

Step(3):

We can assume that the ten’s digit of the square root of 3481 is the lowest among the two numbers i.e. 5

Step(4):

Now, we need to find among 51 and 59 which is the square root of 3481.

Step(5):

Since the ten’s digit is 4 and the next number is 5, we need to multiply both the numbers like 5 x 6 = 30 and since 30 is smaller than 34.

Step(6):

Square root of 3481 will be the greater number among 51 and 59 i.e. 59.

∴ √3481 = 59

Given:

The given number = 1764

Concept:

Calculation skills trick

Calculation:

Step(1):

The unit digit in this number is 4, which can be a unit digit of its square root number such as 2 or 8. Because 2 x 2 is 4 and 8 x 8 is 64.

Step(2):

Now let us consider the first two digits that is 17 which comes between the squares of 4 and 5 because 16 < 17 < 25

Step(3):

We can assume that the ten’s digit of the square root of 1764 is the lowest among the two numbers i.e. 4

Step(4):

Now, we need to find among 42 and 48 which is the square root of 1764.

Step(5):

Since the ten’s digit is 4 and the next number is 5, we need to multiply both the numbers like 4 x 5 = 20 and since 20 is greater than 17.

Step(6):

Square root of 1764 will be the smaller number among 42 and 48 i.e. 42.

∴ √1764 = 42

Calculation:

Let the investment of Chikku be x.

Investment of Beena = x – 300

Investment of Arjun = x – 300 + 600

Total investment = 7200

⇒ x + x – 300 + x – 300 + 600 = 7200

⇒ 3x = 7200

⇒ x = 2400

Investment of Chikku = Rs. 2400.

Investment of Beena = Rs. 2100

Investment of Arjun = Rs. 2700

∴ Ratio of Chikku : Beena : Arjun = 8 : 7 : 9

∴ Arjun’s Share = 864 × 9/24

⇒ Rs. 324

Investment of Arjun in saving scheme = 25% of 324

⇒ Rs. 81

Interest earned by Arjun = 81 × 15 × 1/100

⇒ Rs. 12.15

The interest earned by Arjun from the saving scheme in one year is Rs. 12.15.

Let the third proportional to 16 and 36 be x. 

Then, 16 : 36 : : 36 : x 

16 x x = 36 x 36 

x=(36 x 36)/16 =81 

Third proportional to 16 and 36 is 81.

Answer is (b) (b⁶ - a⁶)/6

Answer is (b) x² - y² = k

Answer is (b) e^{tan x}

Answer is (a) 1

Answer is (d) 4

a:b=5:9 and b:c=4:7= (4X9/4): (7x9/4) = 9:63/4 

a:b:c = 5:9:63/4 =20:36:63.

Let the fourth proportional to 4, 9, 12 be x. 

Then, 4 : 9 : : 12 : x 

4 x x=9 x12 

X=(9 x 12)/14=27; 

Fourth proportional to 4, 9, 12 is 27.

Answer is (d) vector(i + 2k)

Answer is (b) 3√6

Correct option:

(A) 12π 

Correct option:

(D) 2

Given:

A and B can do the work together in 18 days. A is three times as efficient as B.

Concept Used:

Time ∝ 1/efficiencey

If a person can complete a work in x days then in 1 day he will do 1/x part of the work

Calculation:

A is three times as efficient as B which imply the rate of efficiency of A and B is 3 : 1

Required time ratio of A and B to complete the work alone is 1/3 : 1

⇒ 1 : 3

Let, A can do the work in x day and B can do the same in 3x day

In 1 day A can do 1/x part of the work

In 1 day B can do 1/3x part of the work

Together in 1 day they can do (1/x + 1/3x) part of the work

⇒ 4/3x part of the work

A and B can do the work together in 18 days.

In 1 day they can do 1/18 part of the work together

4/3x = 1/18

⇒ x = 24

B alone take time to complete the work (3 × 24)

⇒ 72 days

∴ In 72 days B can complete the work alone.

Given:

A positive integer ‘n’ is divisible by 3, 5, and 7.

Concept:

The LCM of two or more numbers is the least number which is exactly divisible by each of them.

Calculation:

The LCM of (3, 5, and 7) = 105

So next larger integer = n + 105

∴ The next larger integer divisible by 3, 5, and 7 is n + 105.

GIVEN:

Number = (14 + 6√5)

CALCULATION:

Let square root of (14 + 6√5) = (a + b√5)

Now,

(a + b√5)² = 14 + 6√5

a² + 5b² + 2√5ab = 14 + 6√5

After comparing :

a² + 5b² = 14      ….. (1)

ab = 3         ..… (2)

From (1) and (2) :

a = 3 and b = 1

Hence, required square root = (a + b√5)

= (3 + √5)

Correct option:

(C) my 

(D) log x/(1 + log x)²

Correct option:

(C)  y/(1 - y)

(B) –sin(sinx)· cosx 

correct option:

(C) 3/4

Answer is : (a) Otto Von Bismarck

Given,
The sum of the two angles of a quadrilateral is 180°

Let the sum of the remaining two angles be x+y.

Concept used:
The sum of the angles of a quadrilateral is given
=360°

Apply the above concept, we get
180°+(x+y)° =360° 
 (x+y)°=360° −180°
 (x+y)° =180°

Correct option:

(A) 0 

Every point in a number line represents a unique number.

Correct option:

(C) -√y/√x 

Require number = \(\frac{2/3+7/4}{2}\)

= \(\frac{8+21}{2\times 21}\)

= \(\frac{29}{24}\).

As per the question Cos²a + Cos²b = 2
We know that the maximum value of cos x is 1 and the minimum value is -1.
so by taking 1 and -1 value, we can obtain the sum 2. 

cos function gives this value at (2n-1)π/2 where sin has value 0 at this place.

so, tan²A.tan²B = sin²A.sin²B / cos²A.cos²B = 0/1 = 0

As per the question Cos²a + Cos²b = 2
We know that the maximum value of cos x is 1 and the minimum value is -1.
so by taking 1 and -1 value, we can obtain the sum 2. 

cos function gives this value at (2n-1)π/2 where sin has value 0 at this place.

so, tan²A.tan²B = sin²A.sin²B / cos²A.cos²B = 0/1 = 0

It is given that a/b=4

so, a= 4b

now put this value of a in main equation:

a+b/a-b

= 4b+b/4b-b

= 5b/3b

= 5/3

(i) (3²)³ × (2 × 3⁵)⁻² × 18² 

= 3⁶ x 2^-2 x 3^-10 x (-2 x 3²)²

= 2^-2 x 3^6-10 x 2² x 3⁴

= 2^{-2 + 2} x 3^{-4 + 4}

= 2⁰ x 3⁰ 

= 1 x 1 = 1

(ii) \(\cfrac{9^2\times7^3\times2^5}{84^3}\) = \(\cfrac{(3^2)^2\times7^3\times2^5}{(2^2\times3\times7)^3}\)

= \(\cfrac{3^4\times7^3\times2^5}{2^6\times3^3\times7^3}\) = \(\cfrac{3^{4-3}}{2^{6-5}}\)

= \(\cfrac32\) = 1.5

(iii) \(\cfrac{2^8\times2187}{3^5\times32}\) = \(\cfrac{2^8\times3^7}{3^5\times2^5}\)

= 2^8-5 x 3^7-5

= 2³ x 3² = 8 x 9  

= 72

BODMAS :- While simplifying an expressions we can involve six operation in following orders.

B Stands for “BRACKET”

O Stands for “OF”

D Stands for “DIVISION”

M Stands for “MULTIPLICATION”

A Stands for “ADDITION”

S Stands for “SUBTRACTION”

Given:

The income of P and Q ratio = 5 : 6

And expenditure ratio = 5 : 8

Each saves = Rs. 500

Formula used:

Income = Expenditure + saving

Calculations:

Let Income = 5x and 6x

Income – saving = expenditure

According to the question

⇒ (5x – 500)/(6x - 500) = 5/8

⇒ 40x – 4000 = 30x – 2500

⇒ 10x = 1500

⇒ x = 150

P’s income = 5 × 150 = Rs. 750

Q’s income = 6 × 150 = Rs. 900

∴ P and Q’s income is Rs. 750 and Rs. 900 respectively.

Given

Income of A and B are in ratio = 5 : 3

expenses of A, B and C are in the ratio of 8 : 5 : 2

C spends 2000 and B saves ₹ 700

Concept used 

Ratio method 

Calculation

Let the income of A and B be Rs.5x and Rs.3x respectively.

Let the expenses of A, B and C be Rs.8y, Rs.5y and Rs.2y

⇒ 2y = 2000

⇒ y = 1000

⇒ B saves = Rs.700

⇒ 3x –5y = 700

⇒ 3x –5×1000 = 700

⇒ 3x = 700 + 5000

⇒ x = 5700/3 = 1900

⇒ A’s saving = (5x–8y)

⇒ (5×1900 – 8×1000) = (9500 – 8000)

⇒ Rs.1500

∴ A saves Rs.1500

x \(=\frac{2bt}{1+t^2}\)

∴ \(\frac{dx}{dt}\) \(\frac{(1+t^2)\times2b-2bt\times2t}{(1+t^2)^2}\) \(\left(\because\frac{d}{dt}\frac{u}{v}=\cfrac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^2}\right)\)

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

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

And y \(=\frac{a(1-t^2)}{1+t^2}\)

∴ \(\frac{dy}{dt}\) \(\frac{(1+t^2)\times-2at-a(1-t^2)\times2t}{(1+t^2)^2}\) (By u/v formula)

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

\(=\frac{-4at}{(1+t^2)^2}\)

Now, \(\frac{dy}{dx}\) \(=\cfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\) \(=\cfrac{\frac{-4at}{(1+t^2)^2}}{\frac{2b(1-t^2)}{(1+t^2)^2}}\) \(=\frac{-2at}{b(1-t^2)}\)

= -a/b x \(\frac{2bt}{b(1+t^2)}\times\frac{a(1+t^2)}{a(1-t^2)}\)

\(=-\frac{a^2}{b^2}\times\frac{2bt}{1+t^2}\times\frac{1+t^2}{a(1-t^2)}\)

\(=\frac{-a^2}{b^2}\times x\times\frac1y=-\frac{a^2x}{b^2y}\)

u \(=\frac1{1+x^2}\)

∴ \(\frac{du}{dx}\) \(=\frac{-1}{(1+x^2)^2}\) d/dx(1+x²) (By chain rule and d/dx (1/x) = -1/x²)

\(=\frac{-2x}{(1+x^2)^2}\) (∵ d/dx (1+x²) = d/dx 1 + d/dx x² = 2x)

And V = tan^-1x

∵ \(\frac{dV}{dx}=\frac1{1+x^2}\)

Now, \(\frac{du}{dV}=\cfrac{\frac{du}{dx}}{\frac{dV}{dx}}=\cfrac{\frac{-2x}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{-2x}{1+x^2}\)

(A) The time T_AO = T_OA 

(B) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope. 

(D) The velocity of any pulse along the rope is independent of its frequency and wavelength.

Speed of transverse pulse at the point = \(\sqrt{\frac{Tension in ropeat the point}{Linear mass density of rope}}\)

So, T_AO = T_OA

Wavelength becomes longer when speed of the pulse increases.

(a) (i) Zone refining is the method which is used to obtain the semiconductors like Si, Ge and Ga in high purity. 

(ii) This method is applicable for metals, such as Sn, Pb and Bi, which have low melting points as compared to their impurities.

(b) Cu₂S+\(\frac{3}{2}\)O₂→CU₂O+SO₂ ↑

Cu₂S+2Cu₂O→6Cu(I)+SO₂ ↑

I = 2Cos(t) mA

\(R=5K\Omega\)

[Hint : Zinc reacts at faster rate as compared with copper, further zinc is cheaper
than copper.]

It will lies on straight line.

Explanation:

Any solution to the equation zⁿ = (z+1)ⁿ satisfies |z| = |z+1| i.e. the point is equidistant from (0,0) and (-1,0).

This means it lies on the perpendicular bisector of the line joining (0,0) and (-1,0) i.e. x = -1/2 or 2x+1 = 0.

[Hint : Molten Al₂O₃ + Na₃AlF6 or CaF₂]