Question

A = 5, B = 0, C = 2, D = 10, E = 2. What is then AB + EE – (ED)powerB +
(AC)powerE = ?
1. 113
2. 103
3. 93
4. 111

Answer 1

Given A = 5, B = 0, C = 2, D = 10, E = 2

AB + EE – (ED)^B + (AC)^E = (5 x 0) + (2 x 2) - (2 x10)^0 + (5 x 2)^2

AB + EE – (ED)^B + (AC)^E = 0 + 4 - 1 + 100

AB + EE – (ED)^B + (AC)^E = 103

Answer: (2) 103

Answer 2

Given :

A = 5 , B = 0 , C = 2 , D = 10 , E = 2.

$AB \: + EE \: + {ED}^{B} + {AC}^{E}$

$(5 \times 0)+(2 \times 2)-{(2 \times 10})^{0} + {(5 \times 2)}^{2}$

=0 + 4 - 1 + 100

= 3 + 100

= 103 .

Hence : Option "2" is the answer .

$\bold{Answer: \: is \: 103}$