The row sum of elements of a Pascal’s Triangle are shown below:

First row = 21 − 1 = 1
Second row = 22 − 1 = 2 x 1 = 2
Third row = 23 − 1 = 2 x 2 = 4
Fourth row = 24 − 1 = 2 x 2 x 2 = 8
Fifth row = 25 − 1 = 2 × 2 × 2 × 2 = 16
Sixth row = 26 − 1 = 2 × 2 × 2 × 2 × 2 = 32
Seventh row = 27 − 1 = 2 × 2 × 2 × 2 × 2 × 2 = 64
Eighth row = 28 − 1 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 128
Here x denotes the row and y denotes the corresponding row sum. The values of x and y can be tabulated as follows:

The relationship between x and y is y = 2^{x – 1.}

Observe the table carefully. To verify the relationship between x and y, let us substitute the values of x and get the values of y.
If x = 1, y = \(\frac { 1(1+1) }{ 2 } =\frac { 2 }{ 2 } =1\)
If x = 2 , y = \(\frac { 2(2+1) }{ 2 } =\frac { 6 }{ 2 } =3\)
If x = 3, y = \(\frac { 3(3+1) }{ 2 } =\frac { 12 }{ 2 } =6\) 
If x = 4 , y = \(\frac { 4(4+1) }{ 2 } =\frac { 20 }{ 2 } =10\)
If x = 5, y =  \(\frac { 5(5+1) }{ 2 } =\frac { 30 }{ 2 } =15\)
Hence, y = \(\frac { x(x+1) }{ 2 } \) is verified.

9¹²
Unit digit of base 9 is 9 and power is 12 (even power).
Therefore, unit digit of 9¹² is 1.

64¹⁰
Unit digit of base 64 is 4 and power is 10 (even power). Therefore, unit digit of 64¹⁰ is 6.

4⁷
Unit digit of base 4 is 4 and power is 7 (odd power). Therefore, unit digit of 4⁷ is 4.

46³¹
Unit digit of base 46 is 6 and power is 31
Thus, the unit digit of 46³¹ is 6.

81¹⁰⁰
Unit digit of base 81 is 1 and power is 100
Thus, the unit digit of 81¹⁰⁰ is 1.

25²³
Unit digit of base 25 is 5 and power is 23
Thus, the unit digit of 25²³ is 5.

The total area of the garden
Total area of the garden = Area of rectangle + Area of 2 semicircle
= Area of rectangle + Area of the circle
Here, the area of the rectangle = l x b sq. units
= 16 x 8
= 128 m² ...(1)
Area of the circle = πr² sq.units
= 3.14 x 4 x 4
= 3.14 x 16
= 50.24 m² ...(2)
From (1) and (2), the total area of garden = 128 + 50.24
= 178.24 m²

The perimeter of the rectangular garden
Perimeter include two lengths each of 16 m and two semi circular arcs of diameter 8 m.
Circumference of the semicircle = \(\frac { \pi d }{ 2 } \) units
= \(\frac { \pi \times 8 }{ 2 } \) = 4π
= 4 x 3.14
= 12.56 m
Therefore, circumference of two semicircles = 2 x 12.56
= 25.12 m
Total perimeter = length + length + circumference of two semicircles 
= 16 +16 + 25.12
= 32 + 25.12
= 57.12 m

Given that, the area of the circle = 9856 sq.m
  πr² = 9856
   \(\frac { 22 }{ 7 } \)x r² = 9856
  r² = 9856 x \(\frac { 7 }{ 22 } \)
  r² = 448 x 7 = 3136
  
  r² = 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7
  r² = 8 x 8 x 7 x 7 = 8² x 7² = (8 x 7)²
  r = 8 x 7 = 56 m
Therefore the length of the rope should be 56 m.

The distance covered by the man is nothing but the circumference of the circle.
Given that the distance covered = 154 m
Therefore circumference of the circle = 154 m
That is, 2πr = 154
2 x \(\frac { 22 }{ 7 } \)x r = 154
r = 154 x \(\frac { 7 }{ 44 } \)
r = 3.5 x 7
= 24.5
Area of the park = πr² sq.units
= \(\frac { 22 }{ 7 } \)× 24.5 x 24.5
= 22 x 3.5 x 24.5
= 1886.5 m²
Cost of levelling the park per sq.m = Rs.25.
Cost of levelling the park of 1886.5 m² = 1886.5 x 25 = Rs.47,162.50

Given that the area of the circular region = 2464 cm ²
πr² = 2464
\(\frac { 22 }{ 7 } \)x r² = 2464
 r² = 2464 x  \(\frac { 7 }{ 22 } \)
  
 r² = 112 × 7 = 784
 r² = 2 x 2 x 2 x 2 x 7 x 7
 = 4 x 4 x 7 x 7
 = 4² x 7²
 r² = (4 x 7)² [r x r = (4 x 7) x (4 x 7)]
 r = 4 x 7
 = 28 cm
Diameter (d) = 2 x r = 2 x 28 = 56 cm.

Given the diameter (d) = 28 cm
 Radius (r) = \(\frac { 28 }{ 2 } \) = 14 cm
Area of a circle = πr² sq.units
 = \(\frac { 22 }{ 7 } \) x 14 x 14
So, the area of the circle = 616 cm².

Radius (r) = 21 cm
Area of a circle = πr² sq. units
 = 3.14 x 21 x 21
 = 1384.74
Area of the circle = 1384.74 cm²

4.5 = 4 + \(\frac { 5 }{ 10 } \)
  = 4\(\frac { 5 }{ 10 } \) = 4\(\frac { 1 }{ 2 } \)
A juice container has 4\(\frac { 1 }{ 2 } \) litres of mango juice.

Two and twenty five thousandths
  = 2 + \(\frac { 25 }{ 1000 } \)
  = 2 + \(\frac { 2 }{ 100 } \)+\(\frac { 5 }{ 1000 } \)  [ since, \(\frac { 25 }{ 1000 } \)=\(\frac { 20+5 }{ 1000 } \)=\(\frac { 20 }{ 1000 } \)+\(\frac { 5 }{ 1000 } \)=\(\frac { 2 }{ 100 } \)+\(\frac { 5 }{ 1000 } \)]
  = 2 + \(\frac { 0 }{ 10 } \)+\(\frac { 2 }{ 100 } \)+\(\frac { 5 }{ 1000 } \) = 2.025 [as there is no tenth we take it as 0 tenth]

23 + \(\frac { 6 }{ 10 } \) + \(\frac { 8 }{ 1000 } \) = 23 + 6 x \(\frac { 1 }{ 10 } \) + 0 x  \(\frac { 1 }{ 100 } \) + 8 x \(\frac { 1 }{ 1000 } \)
  = 23.608 (since hundredths place is not there, it is taken as '0')

999 + 99 + 9 + \(\frac { 9 }{ 10 } \) + \(\frac { 9 }{ 100 } \) = 1107 + 9 x \(\frac { 1 }{ 10 } \) + 9 x \(\frac { 1 }{ 100 } \)
  = 1107.99

We have to find a fraction equivalent to \(\frac { 1 }{ 50 } \) whose denominator is 100.
 \(\frac { 1 }{ 50 } \) = \(\frac { 1\times 2 }{ 50\times 2 } =\frac { 2 }{ 100 }\) = 0.02

In \(\frac { 9 }{ 1000 } \), tenth and hundredth place is zero.
 Therefore, \(\frac { 9 }{ 1000 } \) = 0.009

0.85 = 0 + \(\frac { 8 }{ 10 } \)+\(\frac { 5 }{ 100 } \)
  = \(\frac { 85 }{ 100 } \)=\(\frac { 17 }{ 20 } \)
 A capsule holds \(\frac { 17 }{ 20 } \) mg of medicine

Four hundred four and five hundredths.
  = 404 + \(\frac { 5 }{ 100 } \)
  = 404 + 0 x \(\frac { 1 }{ 10 } \) + 5 x \(\frac { 1 }{ 100 } \) = 404.05

153 + 96 + 7 + \(\frac { 5 }{ 10 } \) + \(\frac { 2 }{ 1000 } \) = 256 + 5 x \(\frac { 1 }{ 10 } \) + 0 x \(\frac { 1 }{ 100 } \) + 2 x \(\frac { 1 }{ 1000 } \)
  = 256.502 (since hundredths place is not there, it is taken as '0' )