(a) Rahul contributed =rs(2x + 80)
Sunil contributed = rs(8x + 20)
The total amount=rs[(2x+80)+(8x+20)]
=2100
2x + 8x + 80+ 20 = 2100
10x=2000
x=200
Thus, Rahul's contribution
= 2x + 80= 2 x 200 + 80
= 400 + 80= rs480
and Sunil's contributuib
= 8x + 20 = 8 x 200 + 20
= 1600 + 20 = rs1620
Thus, Rahul's and Sunil contributed rs480 and rs1620 respectively
(b) Rahul and Sunil are very cooperative by nature and they literally proved that "A friend in need is a friend indeed".

Given,51²-49²=100p
2601-2401=100p
200 =100p
\(p=\frac { 200 }{ 100 } =2\)
So, it is true
Alternate Method
51²- 49² =100p
(51- 49)(51 + 49) =100p
2 x100 =100p
200 =100p
\(p=\frac { 200 }{ 100 } =2\)

(a) We know that, for a square
(Side)² = Area
(x + 4)² = 289 or (x + 4)² = (17)²
x+4=17
x=17-4=13
The value of x is 13 m.
(b) Given, side of the square shaped garden is
(x+4) =(13+4)m.
:. Side of the garden = 17
(c) With the garden in our surrounding, we get fresh air which is basic need for our body

(a) (52)² = (50 + 2)²
= (50)² + (2)² + 2 x 50 x 2
[.: (a + b)² =a² + b² + 2ab]
= 2500 + 4 + 200= 2704
(b) (1005)² =(1000+ 5)²
= (1000)²+ (5)² + 2 x1000 x 5
=1000000 + 25 + 10000=1010025
(c) (9.9)² =(10.0-0.1)²
=(10)² + (0.1)² - 2 x10 x 0.1
=100 + 0.01- 2 = 98.01
(d)10.1 x10.2 = (10 + 0.1) (10+ 0.2)
=(10)² + (0.1 + 0.2) x 10 + 0.1 x 0.2
=100 + 0.3x 10+ 0.02
=100 + 3 + 0.02= 103.02

Given,\(\ x+\frac { 1 }{ x } =5\)
\(\ \left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }={ x }^{ 2 }+2\times x\times \frac { 1 }{ x } +\left( \frac { 1 }{ x } \right) ^{ 2 }\quad \)
(a+ b)² =a² + 2ab + b². Here, a = x and \(b=\frac { 1 }{ x } \)
\(={ x }^{ 2 }+2+\left( \frac { 1 }{ { x }^{ 2 } } \right) ={ x }^{ 2 }+\left( \frac { 1 }{ { x }^{ 2 } } \right) +2\)
\(\left( x+\frac { 1 }{ x } \right) ^{ 2 }=25\)
\(\quad { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2=25\)
\({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =25-2=23\)

(i)(48)²=(50-2)²
Since, (a - b)² = a² - 2ab + b²
(50-2)²=(50)²-2x50x2+(2)²
= 2500 - 200 + 4   (a-b)2=a²-2ab+b²
= 2504 - 200 = 2304
(ii) 181² -19² =(181-19)(181+ 19)
where, a = 50 and b = 2
= 162 x 200 = 32400
(iii) 497 x 505 =(500 - 3)(500 + 5)
= 5002 + (-3 + 5) x 500 + (-3)(5)
[.: (x + a)(x + b) = x² + (a +. b) x + ab]
= 250000 + 1000 -15 = 250985
(iv) 2.07 x1.93 = (2 + 0.07)(2 - 0.07)
= 2² -(0.07)²
[.: where, a = 50 and b = 2]
= 3.9951

(5x+3)²

39601

\(\frac { { x }^{ 2 }-{ y }^{ 2 } }{ 4 } \)

(48)²=(50-2)²
Since, (a - b)² = a² - 2ab + b²
(50-2)²=(50)²-2x50x2+(2)²
= 2500 - 200 + 4
[(a-b)²=a²-2ab+b²]
= 2504 - 200 = 2304
(ii) 181² -19² =(181-19)(181 + 19)
where, a = 50 and b = 2
= 162 x 200 = 32400
(iii) 497 x 505 =(500 - 3)(500 + 5)
= 500² + (-3 + 5) x 500 + (-3)(5)
[.: (x + a)(x + b) = x² + (a +b) x + ab]
= 250000 + 1000 -15 = 250985
(iv) 2.07 x1.93 = (2 + 0.07)(2 - 0.07)
= 2² -(0.07)²
[.: where, a = 50 and b = 2]
= 3.9951

Monomial, binomial, trinomial

(i) 4X
(ii) 1