Arranging the data in ascending order, we have
5, 7, 10, 12, 14, 19, 23, 26.
Here, n = 8 , which is even.
Therefore, Median = \(\cfrac { 1 }{ 2 } \left\{ \left( \cfrac { 8 }{ 2 } \right) ^{ th }term+\left( \cfrac { 8 }{ 2 } +1 \right) ^{ th }term \right\} \) 

= \(\cfrac { 1 }{ 2 } \left\{ { 4 }^{ th }term+5^{ th }term \right\} \) 
= \(\cfrac { 1 }{ 2 } \left\{ 12+14 \right\} \) 
= \(\cfrac { 26 }{ 2 } =13\) 
Therefore, the Median is 13.

Arranging the golf scores in ascending order, we have,
65, 68, 70, 73, 75, 78, 79
Here n = 7 , which is odd.
Therefore, Median = \(\left( \cfrac { n+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 7+1 }{ 2 } \right) ^{ th }term\) 
= \(\left( \cfrac { 8 }{ 2 } \right) ^{ th }term\) 
4^th term = 73
Hence, the Median is 73.

\((a-b)^{2}=a^{2}-2 a b+b^{2}\)
put a = xyz, b = 1 inthe above identity,
\( (x y z-1)^{2} =(x y z)^{2}-2 \times x y z \times 1+1^{2} \)
\(\\ =x^{2} y^{2} z^{2}-2 x y z+1 \)

\( (a-b)^{2}=a^{2}-2 a b+b^{2}\)
\(\\ put \ a=b, b=7 \)
\(\\\text {in the above identity,} \)
\(\\ (b-7)^{2} =b^{2}-2 \times b \times 7+7^{2}\)
\( \\ =b^{2}-14 b+49 \)

put x = pq, a = 8,b = 7 in the above formula, we get
\( (p q+8)(p q+7) =(p q)^{2}+(p q)(8+7)+(8 \times 7) \)
\(\\ =p^{2} q^{2}+15 p q+56 \)

(x + a)(x + b)= x² + x(a + b) + ab
put x = 6A, a = 9, b = 5 in the above formula,
\( (6 a+9)(6 a-5) =(6 a) 2+6 a(9-5)-(9 \times 5) \)
\(\\ =36 a^{2}+6 a(4)-(45) \)
\(\\ =36 a^{2}+24 a-45 \)

x² −2xy+ y² −z² = (x² −2xy + y² )−z² = (x− y)² −z² [by identity-3]
Let, (x − y) = a and z = b.
Therefore, (x− y)² −z² = a² −b²
We know that, a² -b² = (a + b)(a − b) [Identity-4]
Hence, x² −2xy+ y² −z² = (x − y + z)(x − y − z).
Therefore, the factors of x² −2xy+ y² −z² are (x − y + z) and (x − y − z).

9x² + 30xy + 25y² = (3² x x² )+(2 x 3 x 5)x xy +(5²xy² )
= (3x)² + 2x(3x) x (5y) + (5y)²
This expression is in the form of identity a² + 2ab + b² = (a + b)².
Hence, (3x)² + 2x(3x) x (5y) + (5y)² = (3x + 5y)²
= (3x + 5y)(3x + 5y).
Therefore, the factors of 9x² + 30xy + 25y² are (3x + 5y) and (3x + 5y).

(i) ab² = a x b x b .
(ii) −3pq³ = −3 x p x q x q x q .
(iii) 12 m²n²p = 4 x 3 x m x m x n x n x p = 2 x 2 x 3 x m x m x n x n x p.

(i) (3x - 5y)²
Put a = 3x and b = 5y in the identity (a−b)² = a² −2ab + b² , we get,
(3x - 5y)² = (3x)² −2x (3x) x (5y) + (5y)²
= 32x² −(2 x 3 x 5)xy +(5² x y²)
= 9x² −30xy +25y².
(ii) 47² = (50−3)² , substituting a = 50 and b = 3 in the identity (a−b)²  = a² − 2ab + b², we get,
(50 − 3)² =  50² −2 x 50 x 3 + 3²
= 2500 − 300 + 9
= 2509 − 300 = 2209.

(x + 3)(x + 5)
Let us represent the expression geometrically, as shown in the Figure
In the rectangle with length (x + 5) and breadth (x + 3), we get,

Area of bigger rectangle = Area of a square + Area of three rectangles
Therefore,
(x + 3)(x + 5) = x² + 5x + 3x +15
= x² + (5 + 3)x +15
= x² + 8x + 15.
Area of bigger rectangle = Area of a square + Area of three rectangles
Therefore,
(x + 3)(x + 5) = x² + 5x + 3x + 15
= x² + (5 + 3)x + 15
 = x² + 8x + 15 .
(ii) (y + 6)(y + 8)
Let us represent the expression geometrically, as shown in Figure. In the rectangle with length (y + 6) and breadth (y + 8) units, we get, 
Area of bigger rectangle = Area of square + Area of three rectangles

(iii) 43 x 36 = (40 + 3) x (40 − 4)
We know the identity
(x + a)(x + b) = x² + x(a + b) + ab
Taking, x = 40, a = 3 and b =  –4, we get
(40 + 3)(40 − 4) = 40² + 40(3 − 4) + 3(−4)
= 1600 + 40(−1) − 1²
= 1600 − 40 − 1²
= 1600 − 52
Therefore, 43 x 36 = 1548.

Cost price of T-shirt is Rs.110 and Selling price is Rs.90. So, the loss is Rs.20
Hence, for Rs. 100 the loss is \(\cfrac { 20 }{ 110 } \times 100=\cfrac { 200 }{ 11 } =18\cfrac { 2 }{ 11 }\% \) 

Original amount = the number of literate persons initially = 5 lakhs
Amount of change = increase in the number of literate persons = 8 – 5 = 3 lakhs
Therefore, the percentage of increase = \(\cfrac { Amount\quad of\quad change }{ Original\quad amount } \times 100\) 
= \(\cfrac { 3 }{ 5 } \times 100=60\%\)

The total number of parts of the work is 1+ 2+ 3 = 6
That is, the work is divided into 3 portions as \(\cfrac { 1 }{ 6 } ,\cfrac { 2 }{ 6 } \) and \(\cfrac { 3 }{ 6 } \) .
Thus, the percentage of \(\quad \cfrac { { 1 }^{ th } }{ 6 } \quad \cfrac { 1 }{ 6 } \times 100\%=\cfrac { 100 }{ 6 }\% =16\cfrac { 2 }{ 3 }\% \) 
Similarly, the percentage of \(\cfrac { { 2 }^{ th } }{ 6 } \) portion of work would be \(\cfrac { 2 }{ 6 } \times 100\%=\cfrac { 200 }{ 6 }\% =33\cfrac { 1 }{ 3 }\% \) 
Similarly, the percentage of \(\cfrac { { 3 }^{ th } }{ 6 } \) portion of work would be \(\quad \cfrac { 3 }{ 6 } \times 100\%=\cfrac { 300 }{ 6 } =50\%\)

Let the quantity of alloy required be Q g
Then 26 % of Q = 260 g
\(\cfrac { 26 }{ 100 } \times Q=260g\) 
\(Q=\cfrac { 260\times 100 }{ 26 } g\) 
Q = 1000 g
Therefore, the required quantity of alloy is 1000 g.

Number of students absent on a particular day = 14% of 50
= \(\cfrac { 14 }{ 100 } \times 50=7\) 
Therefore, the number of students present = 50 − 7 = 43 students.

Let us find the percentage of boys and girls. It is given in the form of table below.

To find the percentage of boys and girls we can also use unitary method or multiply both numerator and denominator by a same number which makes denominator 100.

Fraction = \(\cfrac { 1 }{ 5 } \)
Percentage = \(\cfrac { 1 }{ 5 } \times 100\%=20\%\)

\(\cfrac { 15 }{ 10 }\% =\cfrac { \frac { 15 }{ 10 } }{ 100 } =\cfrac { \frac { 3 }{ 2 } }{ 100 } =\cfrac { 3 }{ 200 } \)

We have \(\cfrac { 23 }{ 30 } =\cfrac { 23 }{ 30 } \times \cfrac { 100 }{ 100 } =\cfrac { 23 }{ 30 } \times 100\%=76\cfrac { 2 }{ 3 }\% \) 
From these examples we see that the percentage of proper fractions are less than 100 and that of improper fractions are more than 100.

We have \(\\ \cfrac { 7 }{ 4 } =\cfrac { 7 }{ 4 } \times \cfrac { 100 }{ 100 } =\cfrac { 7 }{ 4 } \times 100\%=\cfrac { 700 }{ 4 }\% =175\%\)

Cost of one ticket = Rs. 12.50
Amount to be paid for 20 students = 12.50 x 20 = Rs. 250
We have already discussed about the multiplication of decimal numbers by 10, 100 and 1000. In the same way we can find patterns for multiplying decimal numbers by 0.1, 0.01 and 0.001. Observe the following.
\(12.3\times 0.1=\cfrac { 123 }{ 10 } \times \cfrac { 1 }{ 10 } =\cfrac { 123 }{ 100 } =1.23\) 
\(12.3\times 0.01=\cfrac { 123 }{ 10 } \times \cfrac { 1 }{ 100 } =\cfrac { 123 }{ 1000 } =0.123\) 
\(12.3\times 0.001=\cfrac { 123 }{ 10 } \times \cfrac { 1 }{ 1000 } =\cfrac { 123 }{ 10000 } =0.0123\)
From the above multiplication we can conclude that, when multiplying by
• 0.1, the decimal point moves one place left.
• 0.01, the decimal point moves two places left.
• 0.001, the decimal point moves three places left.
Zeros may be needed after the decimal point as they are needed before the decimal
point of numbers when multiplied 0.1, 0.01 and 0.0001.

Area of a rectangle = l × b sq. units
Here, l = 23.5 cm, b = 1.5 cm
Area of the rectangle = 23.5 x 1.5
= 35.25 sq.cm.

Round 52.6583 upto 2 places of decimal means round to the nearest hundredths place.
Underline the digit in the hundredth place of 52.6583 gives 52.6583.
We observe that the digit after the hundredth place value is 8 which is more than 5.
Therefore, we should add 1 to the underlined digit. Hence, we get 52.66.
So the rounded value of 52.6583 upto 2 places of decimal is 52.66.

Underline the digit to be rounded - 2.367
Since the digit next to the underlined digit is 3 which is less than 5, the underlined digit 2 remains the same.
Also look at the number line shown below

On the number line, we observe that 2.3 is closer to 2.0 than 3.0
Hence, the rounded value of 2.367 to the nearest whole number is 2.