Let number of persons needed to fit the windows in one day be x.
Then, we have the following table:

Here, lesser the number of days, more will be the number of persons needed to fit the windows. Therefore, it is a case of inverse proportion.
3 x 2 = 1X x => x = 6
Here, 6 persons would be needed to fit the windows in one day.

Here, we can take two factors 4 and 5 such that,
4 X 5 = 20 and 4 + 5 = 9.
Now, put these value in given expression
x² +(4+5)x + 4 X5=x² + 4x+5x + 4 X 5=x(x + 4) +5 (x+ 4)=(x + 4)(x +5).

The given number is 294.
Sum of digits of 294 = 2 + 9 + 4 = 15
Now, on dividing this sum by 3, we get
15 \(\div\) 3 = 5 and remainder = 0
which is divisible by 3.
Hence, 294 is divisible by 3.

We have, 0.0000000000085
=\(\frac{85}{10^{13}}=\frac{8.5\times 10}{10^{13}}=8.5\times 10 \times 10^{-13}\)      \([\because \frac{1}{a^{m}}=a^{-m}]\)
= \(8.5\times 10^{(1-13)}=8.5\times 10^{-12} \quad [\because a^{m}\times a^{n}=a^{m+n}]\)
which is the required standard form.

We have \((3^{0}+4^{-1})\times 2^{2}\)
=\((1+\frac{1}{4})\times 4\)     [∵ a⁰=1 and a^-m=\(\frac{1}{a^{m}}\)]
=\((\frac{4+1}{4})\times 4=\frac{5}{4}\times 4=5\)

A number is said to be in a generalised form, if it is expressed as the sum of the products of its digits with their respective place values as ab = a x 10 + b and abc = a x 100 + b x 10 + c.
129 = 100 x 1 + 10 x 2 + 1 x 9   [ \(\because\)abc =  100a + 10b + c ]

Given,length (I) = 6 cm, breadth (b) = 4 cm and heigh (h) = 2 cm
∴ Total surface area of cuboid = 2(lb+bh+hl)
 = 2(6\(\times\)4 + 4\(\times\)2+2\(\times\)6)
= 2(24 + 8 + 12) = 2\(\times\)44 = 88 m²

Here, 102 can be written as
102 = 100 + 2
(102)²= (100 + 2)²
= (100)² + 2(100)(2) + (2)²
= 10000 + 400 + 4 = 10404

We have, (x + 2) x (2x + 13)= x x (2x + 13) + 2 x (2x + 13)
= (x x 2x) + (x x 13) + (2 x 2x) + (2 x 13) = 2x² + 13x + 4x + 26
= 2x² + 17x + 26

The given solid and its top view, front view and side view are as follow:

22a + 30ab + 6b + 8
13a -18ab + 3b + 16
(-)     (+)     (-)      (-)
__________________
9a + 48ab + 3b - 8
___________________

Sum of the digits of 27x
= 2 + 7 + x = 9 + x
Since, 27x is a multiple of 3,
∴ (9 + x) is divisible by 3.
⇒ 9 + x must be equal to 9 or 12 or 15
When 9 + x = 9 ⇒  x = 0
When 9 + x = 12 ⇒  x = 3
When 9 + x = 15 ⇒  x = 6
When 9 + x = 18 ⇒  x = 9
When 9 + x = 21 ⇒  x = 12, which is not possible. [∵ x is a digit]
∴ x can equal to 0, 3, 6 or 9.

24 (x²yz + xy²z + xyz²)
\(=2 \times 2 \times 2 \times 3 \times[(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)] \)
\(=2 \times 2 \times 2 \times 3 \times x \times y \times z \times(x+y+z)=8 \times 3 \times x y z \times(x+y+z)\)
Therefore, 24 (x²yz + xy²z + xyz²) ÷ 8xyz
\(=\frac{8 \times 3 \times x y z \times(x+y+z)}{8 \times x y z}=3 \times(x+y+z)=3(x+y+z)\)
Alternately,24(x²yz + xy²z + xyz²) ÷ 8xyz
= \(\frac{24 x^{2} y z}{8 x y z}+\frac{24 x y^{2} z}{8 x y z}+\frac{24 x y z^{2}}{8 x y z}\)
= 3x + 3y + 3z = 3(x + y + z)

LHS=(4pq + 3q)² - (4pq - 3q)²
= [(4pq)² + 2(4pq)(3q) + (3q)²] - [(4pq)² - 2(4pq)(3q) + (3q)²]
= 16p²q² + 24pq² + 9q²-[16p²q²-24pq²+ 9q²]
=  16p²q² + 24pq² + 9q²-16p²q²-24pq²- 9q²
= (16 - 16)p²q² + (24 + 24)pq² + (9 - 9)q²
= (0)p²q² + 48pq² + (0)q² = 48pq² = RHS
Since, LHS = RHS
∴ (4pq + 3q)² - (4pq - 3q)²  = 48pq²

40 m²

Radius of cylindrical pillar, r = 28 cm = 0.28 m height, h = 4 m
curved surface area of a cylinder = 2prh
curved surface area of a pillar = \(2 \times \frac{22}{7} \times 0.28 \times 4=7.04 \mathrm{~m}^{2}\)
curved surface area of 24 such pillar = 7.04 x 24 = 168.96 m²
cost of painting an area of 1 m² = RS. 8
Therefore, cost of painting 1689.6 m² = 168.96 x 8 = RS. 1351.68

Since 125 = 5 \(\times\)5\(\times\)5=5³
6^-5=(2\(\times\)3)^-5 = 2^-5 \(\times\) 3^-5
\(\therefore \frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times 125 }{ { 5 }^{ -4 }\times { 6 }^{ -5 } } =\frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times { 5 }^{ 3 } }{ { 5 }^{ -4 }\times { 2 }^{ -5 }\times { 3 }^{ -5 } } \)
\(=\frac { { 2 }^{ -5 } }{ { 2 }^{ -5 } } \times \frac { { 3 }^{ -5 } }{ { 3 }^{ -5 } } \times { 5 }^{ 3+4 }=1\times { 5 }^{ 7 }\)
=5 \(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5 = 78125.

Let the number of boxes filled be x.
Then, we have the following table:

Here, more the number of bottles in a box, less would be the number of boxes. Therefore, it is a case of inverse proportion.
So, \(12\times 25=20\times x\Rightarrow x=\frac { 12\times 25 }{ 20 } =\frac { 12\times 5 }{ 4 } \)
[dividing numerator and denominator by 5]
x = 15
Hence, 15 boxes would be filled, if the same batch is packed using 20 bottles in each box.

Given, road roller is in cylindrical shape.
Length of road roller = 1m =100 cm  [∵1 m= 100 cm]
 and diameter of road roller = 84 cm
∴ Radius of road roller =\(\frac { 84 }{ 2 } \) = 42 cm
The curved surface area of cylindrical road roller = 2\(\pi\)rh
= \(2\times \frac { 22 }{ 7 } \times 42\times 10\) =44\(\times\)6\(\times\)100 = 26400 cm²
∵ Area covered by road roller in 1 revolution = Curved surface area of road roller = 26400 cm² = \(\frac { 26400 }{ 100\times 100 } { m }^{ 2 }\)
∴  Area covered by road roller in 750 revolutions
 = 750\(\times\)2.64 = 1980 m²
Hence, the required area of road is 1980 m²