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Given:

a3 = 1 + 7      ----(i)

33 = 1 + 7 + b      ----(ii)

43 = 1 + 7 + c      ----(iii)

Calculations:

Solving the equation (i)

⇒ a3 = 1 + 7

⇒ a3 = 8

⇒ a = 2

Solving the equation (ii)

⇒ 33 = 1 + 7 + b

⇒ 27 = 8 + b

⇒ b = 19

Solving the equation (iii)

⇒ 43 = 1 + 7 + c

⇒ 64 = 8 + c

⇒ c = 56

Using the values of a, b, and c,

⇒ a + b + c = 2 + 19 + 56

⇒ a + b + c = 77

∴ The value of a + b + c is 77

Formula used:

a³ + b³ = (a + b) × (a² + b² – a × b)

Calculation:

[(4.35)³ + (3.25)³]/[(43.5)² + (32.5)² – 43.5 × 32.5]

⇒ [(4.35 + 3.25) × {(4.35)² +(3.25)² - (4.35 × 3.25)}]/[100 × {(4.35)² + (3.25)² – (4.35 × 3.25)}]

⇒ [4.35 + 3.25]/100

⇒ 7.6/100

⇒ 0.076

∴ The answer is 0.076 

Given:

a2 + b2 = 14

a + b = 4

Formula Used:

(a + b)² = a² + b² + 2ab

(a - b)2 = a2 + b2 - 2ab

a³ - b³ = (a - b)(a² + b² + ab)

Calculation:

(4)² = 14 + 2ab

⇒ 16 - 14 = 2ab

⇒ ab = 2/2 = 1

(a - b)² = 14 - 2 × 1

⇒ (a - b) = 2√3

a3 - b3 = (2√3)(14 + 1)

⇒ a3 - b3 = 30√3

∴ The value of  a3 - b³ is 30√3.

Given:

x + y = 25 and xy = 35

We have to find the value of 1/x + 1/y

Calculation:

1/x + 1/y

⇒ (y + x)/xy     [Using the given values]

⇒ 25/35

⇒ 5/7

∴ Required value of (1/x + 1/y) is (5/7).

Identity used:

(a + b)³ = a³ + b³ + 3ab(a + b)  

Calculation:

⇒ (.98 + .02)³ = (.98)³ + (.02)³ + 3 × .98 × .02(.98 + .02)

⇒ 1 = (.98)3 + (.02)3 + 3 × .98 × .02 

⇒ (.98)3 + (.02)3 + 3 × .98 × .02 - 1 = 0

∴ Solution is 0.

Given:

a + 3b = 12

ab = 9

Calculation:

∵ a + 3b = 12

⇒ a = 12 - 3b      ------(1)

∵ ab = 9

Putting the value of 'a' from (1);

⇒ (12 - 3b) × b = 9

⇒ 12b - 3b² = 9

⇒ 4b - b² = 3

⇒ b² - 4b + 3 = 0

⇒ b² - b - 3b + 3 = 0

⇒ b(b - 1) - 3(b - 1) = 0 

⇒ (b - 1) (b - 3) = 0 

∴ b - 1 = 0

⇒ b = 1

∴ b - 3 = 0

⇒ b = 3

∴ When b = 3 then a = 3

a - 3b = 3 - (3 × 3) = 3 - 9 = (-6)

When b = 1 then a = 9

a - 3b = 9 - (3 × 1) = 9 - 3 = 6

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

⇒ (x + 2)² = ?

⇒ x² + 4 + 2 × x × 2 = ?

⇒ x² + 4 + 4x = ?

∴ The value of ? is x² + 4x + 4

Given:

Total student in group A, B, and C = 56

B consists of twice the number of students in group C

Group A has half the number of students than group C

Calculation:

Let group C have x candidates

B consists of twice the number of students in group C

⇒ B = 2x

A has half the number of students than group C

⇒ A = x/2

According to the question:

x + 2x + x/2 = 56

⇒ x = 112/7 = 16

∴ Number of student in group C = x = 16 students

Check:

Number of student in group C = x = 16 students

Given:

\({\rm{x}} + \frac{1}{{\rm{x}}} = 5\)

Formula used:

\({\rm{a}} - \frac{1}{{\rm{a}}} = {\rm{\;}}\sqrt {{{\left( {{\rm{a}} + \frac{1}{{\rm{a}}}} \right)}^2} - 4}\)

Calculation:

\({\rm{x}} - \frac{1}{{\rm{x}}} = {\rm{\;}}\sqrt {{{\left( {{\rm{x}} + \frac{1}{{\rm{x}}}} \right)}^2} - 4}\)

⇒ \({\rm{x}} - \frac{1}{{\rm{x}}} = {\rm{\;}}\sqrt {{25} - 4}\)

⇒ \({\rm{x}} - \frac{1}{{\rm{x}}} = {\rm{\;}}\sqrt {{21}}\)

\({\rm{x}} + \frac{1}{{\rm{x}}} = 5\)

⇒ x² + 1 = 5x

⇒ 5x – x2 = 1

\(\frac{{{{\left( {{\rm{x}} - {{\rm{x}}^{ - 1}}} \right)}^2}}}{{5{\rm{x}} - {{\rm{x}}^2} + 6}}\)

⇒ \(\frac{{{{\left( {{\rm{x}} - \frac{1}{{\rm{x}}}} \right)}^2}}}{{5{\rm{x}} - {{\rm{x}}^2} + 6}}\)

⇒ {21/(1 + 6)}

⇒ 21/7

⇒ 3

∴ Required value is 3

Given:

a + b = 19

ab = 88

Formula Used:

a3 + b3  = (a + b)³ – 3ab(a + b)

Calculation:

substituting the given values in the formula 

a3 + b3  = (19)3 – 3 × 88(19) 

a3 + b3 = 6859 - 5016 = 1843

∴ The value of a3 + b3 is 1843

Given:

If x² – 7x = -12, then find the value of x?

Calculation:

⇒ x² – 7x + 12 = 0

⇒ x² – 3x – 4x + 12 = 0

⇒ x( x – 3 ) – 4( x – 3 ) = 0

⇒ ( x – 3 )( x – 4 ) = 0

∴ x = 3,4

Given:

Total money = Rs. 8,200

A's share = Rs. 500 more than B

C's share = Rs. 300 more than A

Calculations:

Let share of A, B and C are A, B and C respectively.

A = 500 + B

⇒ B = A - 500

C = 300 + A

⇒ C = 300 + A

A + B + C = 8,200

⇒ A + A - 500 + 300 + A = 8,200

⇒ 3A - 200 = 8,200

⇒ 3A = 8,400

⇒ A = 2,800

∴ A's share was Rs. 2,800

Calculation:

3X + Y = 81      ----(i)

81X – Y = 3      ----(ii)

Adding equation (i) and (ii)

84X = 84

⇒ X = 1

∴ The value of X is 1

Given :

The pair of linear equation x + y = 0 and 2x + 2y = 0

Calculation :

x + y = 0      ----(1)equation

2x + 2y =0      ----(2)equation

Dividing equation (2) by 2, we get

⇒ x + y = 0 

∴ The two lines represent coincident lines So they have infinitely many solutions.

Given:

\(2\sqrt 2 {x^2} + 4x + \sqrt 2 = 0\)

Formula used :

D = b² - 4ac   

Calculation:

D = 4² - 4 × 2√2 × √2 = 16 - 16 = 0

⇒ D = 0, means equal roots 

∴ The equation has equal roots.

Concept:

Peacock has 2 foots

Calculation:

Number of heads = 35

So, the total number of peacocks and four-footed animals = 35

Let there be X four-footed animals and Y peacock

Total number of feet = total foot of four-footed animals + total foot of peacock

4X + 2Y = 98       ----(i)

And Total head = X + Y = 35       ----(ii)

Multiply equation (ii) by 2

⇒ 2X + 2Y = 70      ----(iii)

(i) - (iii)

⇒ 2X = 28 or X = 14

Put X = 14 in equation (i)

56 + 2Y = 98

∴ Y = 42/2 = 21

∴ Number of peacocks = Y = 21

Given:

Number = 1452058

Calculation:

Place value of first 5 = 5 × 10000 = 50000

Place value of last 5 = 5 × 10 = 50

Difference between the place values = 50000 - 50 = 49950.

∴ The required difference is 49950.

Given:

⇒ p(x) = x³ - x² + 2

A rational zero theorem tells that if p/q is a zero then p = factor of leading term and q = factor of constant.

⇒ p/q for p(x) = +_1 and +_1/2

So, dividing p(x) with (x + 1) gives p(x) = 0, then (-1) is the only real zero function of given polynomial.

∴ For given polynomial, there is only one real zero.

  - A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0.

Given:

The product of two numbers is 228 and one number is 7 more than the second number

Concept Used:

Basic concept of arithmetic

Calculation:

Let the first number be x then second number will be (x + 7) and their product is 228

∴ x × (x + 7) = 228

⇒ x² + 7x – 228 = 0

⇒ x² + 19x – 12x – 228 = 0

⇒ (x + 19) × (x - 12) = 0

⇒ x = -19 and 12

So, the greatest number is 19

Hence, option (4) is correct