Explore topic-wise InterviewSolutions in .

Biological Oxygen Demand (BOD), represents the amount of oxygen needed by aerobic bacteria to decompose the waste.

Oxygen is found in the elemental form in the atmosphere to the extent of nearly 21 %

Oxygen is also an essential component of most biological molecules like carbohydrates, proteins, nucleic acids, and fats.

Nitrification: It is the conversion of nitrates into ammonia by the denitrifying bacteria in the soil.

Nitrogen fixation: It is the conversion of atmospheric nitrogen gas into ammonium and nitrates.

Nitro gen cycle: In this cycle nitrogen is converted from its inert atmospheric molecular form (N₂) into a form that is useful in biological processes.

Water cycle or Hydrological cycle: It ¡s a process of constantly recycling water.

The Montreal agreement in 1987 was to control and phase out the production and supply of ozone-depleting chemicals specifically chlorofluorocarbons (CFCs).

A) 1-b, 2-c, 3-a

पोलैण्ड में ‘लेनिन जहाज कारखाना’ के मजदूरों ने 14 अगस्त, 1980 ई. को हड़ताल की।

एक विशेष आयु प्राप्त करने पर किसी देश में सभी नागरिकों को बिना किसी प्रकार के भेदभाव के मतदान (वोट) करने का अधिकार दे दिया जाए तो उस प्रणाली को सार्वभौमिक वयस्क मताधिकार कहते हैं।

जब किसी सरकार को अचानक गैर-कानूनी ढंग से हटा दिया जाता है तो उसे कूप कहते हैं।

As given,

a² + 3a – 88

= a² + 3a – 88 

= a² + 11a – 8a – 88

= a (a + 11) – 8 (a + 11)

= (a – 8) (a + 11)

We have,

a² – 14a – 51

= a² – 14a – 51 

= a² + 3a – 17a – 51

= a (a + 3) – 17 (a + 3)

= (a – 17) (a + 3)

Given, 

x² – y² - 4xz + 4z² = x² – 4xz +4z² – y² 

= (x)² - 2×x×2z+(2z)² – y² 

= (x-2z)² – y² 

= (x - 2z - y) (x – 2z + y) 

= (x – y – 2z) (x + y – 2z)

Given, 

x⁴+x²y²+y⁴ = x⁴+2x²y²+y⁴ – x²y² 

= (x²y²)² – (xy)² 

= (x²+ y² – xy) (x²+ y² + xy)

Given, 

4(x-y)² -12(x-y)(x+y)+9(x+y)² = 4(x²-2xy+y 2)-12(x² – y²)+9(x²+y²+2xy) 

= 4x²-8xy+4y²-12x²+12y²+9x²+9y+18xy 

= x²+25y²+10xy 

= (x)²+(5y)²+2×x×5y 

= (x+5y)²

Given, 

x⁴+x²y²+y⁴ = x⁴+2x²y²+y⁴ – x²y² 

= (x²y²)² – (xy)² 

= (x²+ y² – xy) (x²+ y² + xy)

Given, 

a² +4b²-4ab-4c² = (a)²+ (2b)² – 2×a×2b-4^c2 

= (a-2b)² – (2c)² 

= (a-2b-2c)(a-2b+2c)

We know that,

Area of rectangle = length × breadth

Given,

35y² + 13y – 12 = 35y² + 28y – 15y -12

= 7y(5y+4) -3 (5y + 4)

= (7y – 3) (5y + 4)

Thus,

Length = (7y – 3), then breadth = (5y + 4)

Length = (5y + 4), then breadth = (7y – 3)

Given, 

a²-b²+2ab-c² = a²-(b²-2bc+c²) 

= a – (b-c)² 

= (a + (b-c)) (a-(b-c)) 

= (a+b-c) (a-b+c)

Given,

64a³ - b³,

= (4a)³ - (b)³

= (4a - b) (16a² + b² + 4ab)

Given,

8x³ y³ +27a 3,

= (2xy)³ + (3a)³

= (2xy + 3a) (4x² y² + 9a² – 6axy)

Given,

1 – 27a³,

= 1 – (3a)³

= (1 – 3a) (1+9a²+3a)

Given,

y³+125,

= y³ + (5)³ [∵ a³+b³ = (a+b)(a² – 2ab +b²)

= (y + 5) (y² – 5y + 25)

Given,

P³+27,

= p³ + (3)³ [∵ a³+b³ = (a+b)(a² – 2ab +b²)]

= (p + 3) (p² + 9 – 3p)

Given, 

6ab-b²+12ac-2bc = b(6a-b)+2c(6a-b) 

= (b+2c)(6a-b)

Given, 

8x²y³ – x⁵ 

= x² (8y³ – x³) 

= x² { (2y)² – (x)³} 

= x² (2y - x) (4y² + x² + 2xy)

Given,

(x + 2)³ + (x - 2)³ = (x + 2 + x - 2) { (x + 2)² + (x - 2)² – (x + 2)(x - 2)}

= 2x (x² + 4 + 4x + x² + 4 – 4x – x² + 4)

= 2x (x² +12)

Given,

(a + b)³ – {2(a – b)}³

= {(a + b) –2(a - b)} { (a+b)² +4(a-b)² +2(a+b)(a - b)} [ By using: x³ - y³ = (x - y)(x² + y² +xy)

= (a + b – 2a + 2b)(a² + b² + 2ab + 4a² +4b² – 8ab + 2a² – 2b²)

= (3b – a) (7a 2 +3b 2 - 6ab)

Given,

32a³+108b³,

= 4 (8a³+27b³)

= 4 { (2a)³ + (3b)³ }

= 4 (2a + 3b) (4a² +9b² – 6ab)

Given, 

54x⁶ y + 2x³ y⁴, 

= 2x³ y (27x³ + y³) 

= 2x³ y {(3x)³ + (y)³} 

= 2x³ y (3x + y) (9x² + y² – 3xy)

As given,

4(xy + 1)² – 9(x – 1)²

= [2x (xy + 1)]² – [3 (x – 1)]²

= (2xy + 2 + 3x – 3) (2xy + 2 – 3x + 3)

= (2xy + 3x – 1) (2xy – 3x + 5)

We have,

a² + 2a – 3

a² + 2a – 3 

= a² + 3a – a – 3

= a (a + 3) – 1 (a + 3)

= (a – 1) (a + 3)

We have,

x² – 11x – 42

= x² – 11x – 42 

= x² + 3x – 14x – 42

= x (x + 3) – 14 (x + 3)

= (x – 14) (x + 3)

Given,

\(\frac{\text{x}^{3}}{216}-8\text{y}^{3}=(\frac{\text{x}}{6})^{3}\,-(2\text{y})^{3}\)

= \((\frac{\text{x}}{6}-2\text{y})(\frac{\text{x}^{2}}{36}+4\text{y}^{2}+\frac{\text{xy}}{3})\) 

We have,

\(=a^{2}-1-2\text{x}-\text{x}^{2}\)

\(=a^{2}-(1+2\text{x}+\text{x}^{2})\)

\(=a^{2}-(1+\text{x})^{2}\)

\(=(a-1-\text{x})(a+1+\text{x})\)

\([a^{2}-b^{2}=(a+b)(a-b)]\)

Thus, the factors of

\(a^{2}-1-2\text{x}-\text{x}^{2}\,are \,(a-1-\text{x})(a+1+\text{x}).\)

Given, 

(x+2) (x²+25) - 10x² - 20x = (x+2) (x²+25)-10x (x+2) 

= (x+2) (x²+25-10x) 

= (x+2) (x-5)²

10x⁴ y – 10xy⁴ 

= 10xy(x³ − y³)   [a³ – b³ = (a – b)(a² + ab + b²)] 

= 10xy (x−y)(x² + xy + y²) 

Therefore, 10x⁴y – 10xy⁴ = 10xy (x−y)(x² + xy + y²)

(x + 2) (x² + 25) – 10x (x + 2) 

Take (x + 2 ) as common factor;

= (x + 2)(x² + 25 – 10x) 

= (x + 2) (x² – 10x + 25) 

Expanding the middle term of (x² – 10x + 25 ) 

= (x + 2) (x² – 5x – 5x + 25)

= (x + 2){x(x – 5) – 5( x – 5)} 

= (x + 2)(x – 5)(x – 5) 

= (x + 2)(x – 5)² 

Therefore, (x + 2) (x² + 25) – 10x(x + 2) = (x + 2)(x – 5)².

As given,

x² – 4x – 21

By considering, p+q = -4 and pq = -21

So we can replace -4x by 3x – 7x

-21 by 3 × -7

x² + 4x – 21

= x² + 3x – 7x – 21

= x (x + 3) – 7 (x + 3)

= (x – 7) (x + 3)

 (i) 6xy + 6 – 9y – 4x

= 6xy – 4x – 9y + 6

= 2x (3y – 2) – 3 (3y – 2)

= (2x – 3) (3y – 2)

(ii) x² – 2ax – 2ab + bx

= x² + bx – 2ax – 2ab

= x (x + b) – 2a (x + b)

= (x – 2a) (x + b)

(iii) x³ – 2x²y + 3xy² – 6y³

= x³ + 3xy² – 2x²y – 6y³

= x (x² + 3y²) – 2y (x² + 3y²)

= (x – 2y) (x² + 3y²)

(iv) abx² + (ay – b) x – y

= abx² – ayx – bx – y

= abx² – bx – ayx – y

= bx (ax – 1) + y (ax – 1)

= (bx + y) (ax – 1)

(i) a (a + b – c) – bc

= a² + ab – ac – bc

= a (a + b) – c (a + b)

= (a + b) (a – c)

(ii) x² – 11xy – x + 11y

= x² – x – 11xy + 11y

= x (x – 1) – 11y (x – 1)

= (x – 11y) (x – 1)

(iii) ab – a – b + 1

= a (b – 1) – 1 (b – 1)

= (a – 1) (b – 1)

Given,

a²x² + (ax²+1)x + a = a²x² + ax³ + x + a

= ax²(a + x) + 1(x + a)

= (ax² + 1)(x + a)

(i) (ax + by)² + (bx – ay)²

= a²x² + b²y² + 2axby + b²x² + a²y² – 2axby

= a²x² + b²y² + b²x² + a²y²

= a²x² + a²y² + b²y² + b²x²

= a² (x² + y²) + b² (x² + y²)

= (a² + b²) (x² + y²)

(ii) 16 (a – b)³ – 24 (a – b)²

= 8 (a – b)² [2 (a – b) – 3]

= 8 (a – b)² (2a – 2b – 3)

(iii) ab (x² + 1) + x (a² + b²)

= abx² + ab + xa² + xb²

= abx² + xa² + ab + xb²

= ax (bx + a) + b (bx + a)

= (ax + b) (bx + a)

(iv) a²x² + (ax² + 1)x + a

= a²x² + ax³ + x + a

= ax² (a + x) + 1 (x + a)

= (x + a) (ax² + 1)

(v) a (a – 2b – c) + 2bc

= a² – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)

We have,

ax + ay – bx – by

= a(x + y) –b (x + y)

= (a – b) (x + y)

Given,

ax²y + bxy² + cxyz

= xy (ax + by + cz)

a²x² + (ax² + 1)x + a 

= a²x² + a + (ax² + 1)x

= a(ax² + 1) + x(ax² + 1) 

= (ax² + 1) (a + x)

Radius of first circle = r₁ = 15 cm

Radius of second circle = r₂ = 18 cm

∴ Circumference of first circle = 2πr₁ = 30π cm

Circumference of second circle = 2πr₂ = 36π cm

Let the radius of the circle = R

According to the question,

Circumference of circle = Circumference of first circle + Circumference of second circle

2πR= 2πr₁+ 2πr₂

⇒ 2πR = 30π + 36π

⇒ 66π ⇒ R = 33

⇒ Radius = 33 cm

Hence, required radius of a circle is 33 cm.

True

We are given that

Circumference of circle with radius R₁ = Circumference of circle with radius R₂

⇒ 2πR₁= 2πR₂

⇒ R₁= R₂

⇒ π(R₁)² = π(R₂)²

and hence the areas are also equal.