In this question we are asked to find the given integer x is a positive number or a negative number or a zero. For that, first we have to assign the value of x and check whether x is greater than 0. If ‘yes’, we will print “x is a Positive number”. If ‘no’ then we will check again if the value of x is less than 0. If ‘yes’, we will print “x is a Negative number”. If ‘no’, then we will print “x = zero”. Step by Step process:

Arranging the data in ascending order, we have,
365, 365, 547, 547, 730, 912, 1095, 1460, 1500, 1825, 2000.
The number of observations are 11, which is odd.
\(Median=\left( \cfrac { n+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 11+1 }{ 2 } \right) ^{ th }term\) 
= 6 ^th term = 912
Therefore, the median is 912.
Hence, the median lifetime of the LED bulb is 912 days.

Arranging the weights in ascending order, we have, 20, 24, 25, 29, 32, 35, 38, 40, 42, 45
Here, n = 10 , which is even.
Therefore, median weight = \(\cfrac { 1 }{ 2 } \left\{ \left( \cfrac { n }{ 2 } \right) ^{ th }term+\left( \cfrac { n }{ 2 } +1 \right) ^{ th }term \right\} \) 
= \(\cfrac { 1 }{ 2 } \left\{ \left( \cfrac { 10 }{ 2 } \right) ^{ th }term+\left( \cfrac { 10 }{ 2 } +1 \right) ^{ th }term \right\} \) 
= \(\cfrac { 1 }{ 2 } \left\{ { 5 }^{ th }tern+{ 6 }^{ th }term \right\} \) 
= \(\cfrac { 1 }{ 2 } \left\{ 32+35 \right\} kg=\cfrac { 67 }{ 2 } =33.5kg\) 
Hence, Median is 33.5 kg. 

\(Arthemetic\ Mean=\cfrac { Sum\ all\ observation }{ Number\ of\ observation } \) 
Thus, \(24=\cfrac { Sum\ of\ all\ observations }{ 9 } \) 
Sum of all observations = 9 x 24 = 216

(i) Reflection
(ii) Rotation
(iii) Translation

(i) 90˚ counter clockwise

(ii) 180^o

(x + a)(x + b) = x² + x(a + b) + ab
put x = 4x, a = 3y, b = 5y is the above formula, we get
\( (4 x+3 y)(4 x+5 y) =(4 x)^{2}+(4 x)(3 y+5 y)+(3 y \times 5 y) \)
\(\\ =16 x^{2}+(4 x)(8 y)+15 y^{2} \)
\(\\ =16 x^{2}+32 x y+15 y^{2} \)

(i) a² −1 = a² −12 = (a + 1)(a − 1) [since, 12 = 1 x 1 = 1]
Therefore, the factors of a² -1 are (a + 1) and (a − 1).
(ii) 9k² − 25 = (32 x k² ) − 5² = (3k)² −5² [since, amxbm =(axb)m ]
= (3k + 5)(3k − 5) |since,a²- b² = (a + b)(a - b)|
Therefore, the factors of 9k² -25 are (3k + 5) and (3k − 5).
(iii) 64x² −81y² = (82xx² ) − (92xy² ) = (8x)² −(9y)²
= (8x + 9y)(8x − 9y)
Therefore, the factors of 64x² -81y² are (8x + 9y) and (8x – 9y).

(i) (3x + 4)(3x − 4)
Substitute, a = 3x and b = 4 in the identity (a+b) x (a − b) = a² −b², we get,
(3x + 4)(3x − 4) = (3x)² −4²
(3² x x² )−16 = 9x² −16
(ii) 53 x 47 = (50 + 3) x (50 − 3).
Take, a = 50 and b = 3,
Substituting the values of ‘a’ and ‘b’ in the identity (a + b)(a – b) = a² – b²), we get,
53 x 47 = 50² −3²
= 2500 − 9
= 2491.

(i) (2x + 5)²

Let the side of the square be 2x + 5 units.
Then its area is side´side , that is (2x + 5)²
The geometrical representation of the given expression is as shown in Figure.
Area of the bigger square = Area of two squares+ Area of two rectangles.
(2x + 5)² = 4x² + 25 + 10x + 10x
= 4x²  + (10 + 10)x  + 25 (Adding like terms)
= 4x² + 20x + 25 .
(ii) 21²
Let the side of the square be 21. Hence, its area is 212 .
Consider 21² as (20 + 1)² which is one of the way to represent it geometrically as shown in Figure.

Now,
Area of the bigger square = Area of two squares+ Area of two rectangles.
21² = 400 + 1 + 20 + 20 = 441.
Aliter method:
We know the identity,(a + b)² = (a + b)(a + b)
= a² + 2ab + b²

To find the value of 21²
We take,21² = (20 + 1)²
= (2 0 + 1)(20 + 1)
Here, a = 20 and b = 1.
Therefore,
\( a^{2}+2 a b+b^{2}=20^{2}+2 \times 20 \times 1+1^{2} \)
\(\\ =400+40+1=441 \)

To find the cost price,
Loss per cent = \(\cfrac { Loss }{ C.P } \times 100\) 
\(4\%=\cfrac { Loss }{ C.P } \times 100\) 
\(\\ 4\%=\cfrac { Loss }{ 200 } \times 100\) 
Loss
C.P = S.P + Loss
= 200 + 8
= 208
Hence the cost price of the item is Rs.208.

\(Profit\quad percent=\cfrac { Profit }{ C.P } \times 100\) 
= \(\cfrac { 25 }{ 325 } \times 100=\cfrac { 100 }{ 13 } =7\cfrac { 9 }{ 13 } \%\)

Original price = the price of the shirt before Aadi month
Amount of change = the decrease in the price = 90 – 50 = Rs. 40
Therefore, the percentage of decrease = \(\cfrac { Amount\ of\ change }{ Original\ amount } \times 100\) 
= \(\cfrac { 40 }{ 90 } \times 100=\cfrac { 400 }{ 9 } \) 
= \(44\cfrac { 4 }{ 9 }\%\)

Representing ingredients used in the batter as ratio, we get, rice : urad dhall = 4 :1
Now, the total number of parts is 4 +1 = 5 .
That is \(\cfrac { 4 }{ 5 } \) portion of rice is mixed with \(\cfrac { 1 }{ 5 } \) portion of urad dhall.
Thus, the percentage of rice would be \(\cfrac { 4 }{ 5 } \times 100\%=\cfrac { 400 }{ 5 } =80\%\) 
The percentage of urad dhall would be \(\cfrac { 1 }{ 5 } \times 100\%=\cfrac { 100 }{ 5 }\% =20\%\)

Let the price of the raincoat (in Rs. ) be P. So 20% of P = 25
\(\cfrac { 20 }{ 100 } \times P=25\) 
\(P=\cfrac { 25\times 100 }{ 20 } =125\) 
Therefore, the original price of the raincoat is Rs.125.

Total number of students = 560
Number of boys = 320
Number of girls = 560 – 320 = 240
\(Percentage=\cfrac { 240 }{ 560 } \times 100\%=\cfrac { 24 }{ 56 } \times 100\%\) 
= \(\cfrac { 3 }{ 7 } \times 100\%=\cfrac { 300 }{ 7 } \%\) 
= 42.86%

Total number of students = 75
Number of students declared passed = 72
Percentage = \(\cfrac { 72 }{ 75 } \times 100\%\) 
= \(\cfrac { 24 }{ 25 } \times 100\%\) 
= 24 x 4%
= 96%

\(\cfrac { 3 }{ 5 } =\cfrac { \frac { 3 }{ 5 } }{ 100 } =\cfrac { 3 }{ 500 } \)

\(60\%=\cfrac { 60 }{ 100 } =\cfrac { 6 }{ 10 } =\cfrac { 3 }{ 5 } \)

We have \(\cfrac { 5 }{ 20 } =\cfrac { 5 }{ 20 } \times \cfrac { 100 }{ 100 } =\cfrac { 5 }{ 20 } \times 100\%=\cfrac { 500 }{ 20 }\% =25\%\)

We have \(\cfrac { 1 }{ 5 } =\cfrac { 1 }{ 5 } \times \cfrac { 100 }{ 100 } =\cfrac { 1 }{ 5 }\% \times 100=\cfrac { 100 }{ 5 }\% =20\%\)

(i) 3.26 x 10 = 32.6
(ii) 3.26 x 100 = 326.0
(iii) 3.26 x 1000 = 3260.0
(iv) 7.01 x 10 = 70.1
(v) 7.01 x 100 = 701.0
(vi) 7.01 x 1000 = 7010.0

Cost of churidhar material = Rs. 62.50 per metre
Length of churidhar material = 3.75 m
Amount to be paid = 3.75 x 62.50
= Rs. 234.3750
= Rs.234.38 (rounded to two decimals)

Round 189.0007 upto 3 places of decimal means rounding to the nearest thousandth place.
Underline the digit in the thousandth place of 189.0007 gives 189.0007 In 189.0007 we observe that the digit next to the thousandth place value is 7, which is greater than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 189.001. So the rounded value of 189.0007 upto 3 places of decimal is 189.001.
So the rounded value of 189.0007 upto 3 places of decimal is 189.001

Underline the digit to be rounded 99.95.
Since the digit right to the tenth place is 5, we add 1 to the tenth place (underlined digit) of 99.95 and we get 100.0.
Look at the number line shown below

So, the rounded value of 99.95 is 100.0.