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On the set Z of all integers * is defined by a * b = a + b – 5. If 2* (x * 3) = 5, then x is equal to

1) 0 2) 3 3) 5 4) 10 Solution: Given a * b = a + b - 5 2* (x * 3) = 5 => 2* (x + 3 - 5) = 5 => 2* (x - 2) = 5 => 2 + x... View Article

1) 137 2) 173 3) 170 4) None of these Solution: (10101101)2 => 27×1 + 26×0 + 25×1 + 24×0 + 23×1 +... View Article

1) identity law 2) inverse law 3) closure law 4) associative law Solution: A = {-1, 0, 1} Check the options. (1) 1 ∈ A. So... View Article

1) 1 2) 3 3) 4 4) 2 Solution: Zn will have only 2 subgroups. Hence option (4) is the answer.... View Article

1) one 2) two 3) three 4) None of the above Solution: The vertex connectivity of any tree is one. Hence option (1) is the answer.... View Article

1) 3ex 2) 3(h - 1)ex 3) 3(eh - 1)2ex 4) None of these Solution: ∆ (3ex ) = 3(ex+h - ex) = 3ex(eh - 1) ∆2 (3ex) = ∆(3ex(eh -... View Article

1) ∆ -∇ 2) ∆ +∇ 3) ∇ - ∆ 4) None of these Solution: In usual notation the value of ∆ ∇ is equal to ∆ -... View Article

1) 9 2) 30 3) 32 4) 182 Solution: 10110110 = 27×1 + 26×0 + 25×1 + 24×1 + 23×0 + 22×1 +... View Article

1) (970)10 2) (973)10 3) (975)10 4) None of these Solution: (1111001101)2 => 29×1 + 28×1 + 27×1 + 26×1 +... View Article

1) 37 2) 36 3) 34 4) 32 Solution: Given binary number is 100101 (Multiply each digit of the binary number by the... View Article

1) 100 2) 1000 3) 10000 4) None of the above Solution: (101100 - 1) = (1 + 100)100 - 1 (1+ x)n = nC0x0 + nC1x1 + nC2x2 + …+ nCnxn... View Article

1) 145 2) 135 3) 225 4) 125 Solution: 1080 = 23 × 33 × 5 (Prime factorisation) 675 = 33 × 52 Greatest common... View Article

1) a + b 2) a - b 3) ab 4) 1 Solution: (a2 - b2) = (a + b)(a - b) Given (a2 - b2) is prime. So (a2 - b2) is divisible by 1 or the... View Article

1) 0 2) 1 3) 3 4) 5 Solution: 51 = 5 52 = 25 53 = 125 54 = 625 We know that the unit’s place of 5k where... View Article

1) 6 2) 7 3) 8 4) 9 Solution: 599 = 5. 598 = 5.(25)49 = 5 (26 - 1)49 = 5 [ 49C0 (26)49 - 49C1(26)48 + …..+ 49C48 (26)1 - 49C49... View Article

1) 156 2) 242 3) 342 4) 399 Solution: 242 = 2 × 112 Sum of divisors = (20 + 21)(110 + 111 + 112) = 3 × 133 = 399 Sum of all... View Article

1) 51! 2) 51! - 1 3) 51! + 1 4) 2 × 51! Solution: 1 × 1! + 2 × 2! + ...... + 50 × 50! = (2 - 1)1! + (3 -... View Article

1) 9 2) 7 3) 3 4) 1 Solution: Last digit in 71 = 7 Last digit in 72 = 9 Similarly 73 → 3 74 → 1 75 → 7 76 →... View Article

1) 4 2) 120 3) 121 4) 130 Solution: Since 123 = -1 (mod 124) 125 = 1 (mod 124) 127 = 3 (mod 124) So 123 × 125 × 127 =... View Article

1) 29351 2) 23951 3) 33851 4) None of these Solution: 9900 = 22 × 32 × 52 × 11 Sum of proper divisors of 9900 =... View Article