Numerical Computation Quiz-1B

Solution:

Let A = 12 cos^{2}\alpha + 3 sec^{2}\alpha

⇒ A =  3(4 cos^{2}\alpha+ 3 sec^{2}\alpha)

⇒ A = 3[(2 cos \alpha– sec\alpha)^{2}+ 4]

⇒ A = 3(2cos \alpha– sec\alpha)^{2}+ 12

For the minimum value of A, 2cos \alpha – sec\alpha = 0

2cos \alpha = sec\alpha

Hence, the minimum value of the expression = 12

Solution:

cos X = m cos Y

\frac{cos X }{cos Y} = \frac{m}{1}

By componendo and dividend,

\frac{(cos X + cos Y)}{(cos X – cos Y)} =\frac{ (m + 1)}{(m – 1)}

{2 cos (X + Y / 2) cos (X – Y / 2)} / {–2 sin (X – Y / 2) sin (X – Y / 2)} = (m + 1) / (m – 1)

{cos(X + Y / 2) cos (Y – X / 2 )} / {sin (X + Y / 2) sin (Y – X / 2)} = (m + 1) / (m – 1)

Since, cos (– X) = cos X whereas sin (– X) = – sin X

cot (X + Y / 2) × cot (Y – X / 2) = m + 1 / m –1

cot (X + Y / 2) = [(m + 1) / (m – 1)] × tan (Y – X / 2)

since, cot \Theta = (1 / tan\Theta )

Solution:

5x + 2(x + 7) = 14x – 7

5x+2x+14=14x-7

7x+14=14x-7

7x=21

x=3

Solution:

 x^{3} + y^{3} = (x + y) × (x^{2} − xy + y^{2}) 



Putting given values of  x^{3} + y^{3} and (x + y) 

 9 = 3\times ((x+y)^{2} − 3xy) 

= 3\times(9 − 3xy) 

= 27 − 9xy

9xy = 18

xy = 2



 x^{4}+y^{4} = (x^{2} + y^{2})^{2} - 2x^{2}y^{2}    

=  (x^{2} + y^{2})^{2}  - 2\times 4 

[Putting value of xy]                           

 = ((x + y)^{2} - 2xy)^{2} - 2\times 4 

 [Putting values of (x+y) and xy]

  =  (9 - 4)^{2} - 2\times 4  

  = 17

Solution:

log_{b}x=y

b^{y}=x

Here, log_{2}64

2^{y}=64

2^{6}=64

Therefore,y=6

Solution:

In the year 2000, exports were 60 while the imports were 50. So 10% of 50 is 5. And thus the total will be 55. Here 60 > 55. Similarly in 2001, 110 = 10% of 100.

In 2002, 120 < 130. For 2003, 120 > 10% of 100. For 2004, 150 < 10 % of 140.

In 2005, 190 < 10 % of 180. So there are years i.e. 2000 and 2003 where exports were 10 % greater than the imports.

Solution:

Here from the figure the average of the exports will be: \frac{(60 + 110 + 120 + 120 + 150 + 140)}{6} = 116.67.

Solution:

Imports done in 2001 = 110

Imports done in 2005 = 190

So, the annual growth = \frac{(190 – 110)}{110 \times 100} = 72.72%

Now, the average growth will be \frac{72.72}{ 5 }= 14.5 .

Solution:

From the figure, we can determine that imports grew only in 2001, 2002, 2004, and 2005.

For 2001, \frac{(50 \times 100) }{ 60} = 83.33

For 2002, \frac{(10 \times 100) }{ 110} = 9.09

For 2004, \frac{(30\times 100) }{ 120}= 25

For 2005, \frac{(40 \times 100) }{ 150} = 26.66

Solution:

Out of all the years given in the figure, the imports and exports as compared to the previous are increased in every year except 2000 and 2003. Out of the four in which imports and exports have increased, in no year does the percentage increase in export is greater than the percentage increase in imports.