**Explanation :**Let the number be x. Then, x - 36 = 86 - x => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.

**Explanation :**Let the number be x. Then, 7x - 15 = 2x + 10 => 5x = 25 =>x = 5. Hence, the required number is 5.

**Explanation :**Let the number be x.

Then, x + (1/x) = 13/6 ( x * x + 1)/x = 13/6 6x2 – 13x + 6 = 0 6x2 – 9x – 4x + 6 = 0 (3x – 2) (2x – 3) = 0 x = 2/3 or x = 3/2

Hence the required number is 2/3 or 3/2.

**Explanation :**Let the numbers be x and (184 - x). Then, (X/3) - ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72. So, the numbers are 72 and 112. Hence, smaller number = 72.

**Explanation :**Let the number be x and y. Then,

x – y = 11 ----(i) and 1/5 (x + y) = 9 => x + y = 45 (ii) Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17. Hence, the numbers are 28 and 17.

**Explanation :**Let the numbers be x and y. Then, x + y = 42 and xy = 437 x - y = (sqrt(x + y)^2) - 4xy)) = \(\sqrt((42)2 - 4 * 437 )\) = \(\sqrt(1764 – 1748)\) = \(\sqrt(16)\) = 4. Required difference = 4.

**Explanation :**Let the numbers be x and (15 - x). Then,

x2 + (15 - x)2 = 113 x2 + 225 + X2 - 30x = 113 2x2 - 30x + 112 = 0 x2 - 15x + 56 = 0 (x - 7) (x - 8) = 0 x = 7 or x = 8. So, the numbers are 7 and 8.

**Explanation :**Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6. Then, sum of these numbers = (27 x 4) = 108. So, x + (x + 2) + (x + 4) + (x + 6) = 108 or 4x = 96 or x = 24. Largest number = (x + 6) = 30.

**Explanation :**Let the numbers be x, x + 2 and x + 4. Then, \(X^2\) + \((x + 2)^2\) + \((x + 4)^2\) = 2531 => 3\(x^2\) + 12x - 2511 = 0 => \(X^2\) + 4x - 837 = 0 => (x - 27) (x + 31) = 0 => x = 27.

Hence, the required numbers are 27, 29 and 31.

**Explanation :**Let the numbers be x and y, such that x > y

Then, 3x - 4y = 5 ...(i)

and (x + y) - 6 (x - y) = 6 => -5x + 7y = 6 …(ii)

Solving (i) and (ii), we get: x = 59 and y = 43. Hence, the required numbers are 59 and 43.

**Explanation :**Let the ten's digit be x. Then, unit's digit = (x + 3). Sum of the digits = x + (x + 3) = 2x + 3. Number = l0x + (x + 3) = llx + 3. 11x+3 / 2x + 3 = 4 / 1 => 1lx + 3 = 4 (2x + 3) => 3x = 9 => x = 3. Hence, required number = 11x + 3 = 36.

**Explanation :**Let the ten's digit be x. Then, unit's digit = (9 - x). Number = l0x + (9 - x) = 9x + 9. Number obtained by reversing the digits = 10 (9 - x) + x = 90 - 9x. therefore, (9x + 9) - 63 = 90 - 9x => 18x = 144 => x = 8. So, ten's digit = 8 and unit's digit = 1. Hence, the required number is 81.

**Explanation :**Let the required fraction be x/y. Then, \(\frac{x+1 } {y+1}\) = \(\frac{2}{3}\)=> 3x – 2y = - 1 …(i)

and x – 1 / y – 1 = 1 / 2= 2x – y = 1 …(ii)

Solving (i) and (ii), we get : x = 3 , y = 5 therefore, Required fraction = \(\frac {3}{5}\).

**Explanation :**Let the two parts be x and (50 - x). Then,\(\frac {1} {x}\) + \(\frac {1} {(50 – x)}\) = \(\frac {1} {12}\) => \(\frac {(50 – x + x)} {x ( 50 – x)}\) = \(\frac {1} {12}\) => \(x^2\) – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20. So, the parts are 30 and 20.

**Explanation :**Let the numbers be x, y and z. Then, x+ y = 10 ...(i)

y + z = 19 ...(ii)

x + z = 21 …(iii)

Adding (i) ,(ii) and (iii), we get: 2 (x + y + z ) = 50 or (x + y + z) = 25. Thus, x= (25 - 19) = 6; y = (25 - 21) = 4; z = (25 - 10) = 15. Hence, the required numbers are 6, 4 and 15.