Analogy varieties Question TCS (Part – II)

[(2^nd number × 6) - {(2^nd number - 1^st number) × 2}] = 3^rd number

In (12, 13, 76) :- [(13 × 6) - {(13 - 12) × 2}] = 76

Similarly,

(15, 17, 98) → [(17 × 6) - {(17 - 15) × 2}] = 98

Hence, (15, 17, 98) is the correct answer.

a = n and b = {(n + 1) × (n + 4)}
(8 : 108) → {(8 + 1) × (8 + 4)} ⇒ 9 × 12 = 108
Only,

(12 : 208) → {(12 + 1) × (12 + 4)} ⇒ 13 × 16 = 208

Hence, (12 : 208) is the correct answer.

(61 : 97) → (8² − 3) : (10² − 3)
Similarly, in option (c),

(46 : 78) → (7² − 3) : (9² − 3)

Hence, (46 : 78) is the correct answer.

(84 : 96) → (84 + 8 + 4) = 96
Similarly, in option (d)

(68 : 82) → (68 + 6 + 8) = 82

Hence, (68 : 82) is the correct answer.

(242 : 82) → {(242 + 4) ÷ 3} ⇒ 246 ÷ 3 = 82

Hence, (242 : 82) is the correct answer.

{10 × 2^nd number × (2^nd number + 2)} = 1^st number

(350 : 5) → {10 × 5 × (5 + 2)} = 350
Similarly, only in option (a)

(1950 : 13) → {10 × 13 × (13 + 2)} = 1950

Hence, (1950 : 13) is the correct answer.

The pattern followed here is :

{(1^st number)⁴ + 2)} = 2^nd number

(3 : 83) → (3⁴ + 2) = 83
Similarly,

(5 : 627) → (5⁴ + 2) = 627

Hence, 627 is the correct answer.

Hence, (36, 27, 77) is the correct answer.

2^nd number = {(1^st number)² + 4}

(21² + 4) = 445
Similarly,

(24² + 4) = 580

Hence, 580 is the correct answer.

Middle number = [{(1^st number)³ + (3^rd number)³} ÷ 2]
Given set is:
(4, 140, 6) → {(4³ + 6³) ÷ 2 } ⇒ (280 ÷ 2) = 140
Similarly,
(5, 234, 7) → {(5³ + 7³) ÷ 2} ⇒ (468 ÷ 2) = 234

Hence, (5, 234, 7) is the correct answer.