In a certain code language, DISABLE is written as HRL20. How is ENABLE written in that code?... pls explain
Answers
Answered by
33
Let us code the
Each alphabets equal to natural naumber given below:
A=1
B=2
C=3
D=4
E=5
F=6
G=7
H=8
I=9
J=10
K=11
L=12
M=13
N=14
O=15
P=16
Q=17
R=18
S=19
T=20
U=21
V=22
W=23
X=24
Y=25
Z=26
Now decode the given words
1)DISABLE
D=4 decoded as 2×D=2×4=8=H
I=9 decoded as 2×I=2×9=18=R
S=19 decoded as 2×S=2×19=38 we doesn't have 38 letters then we take
Cyclic way =38=26+12=12=L
[A+B+L+E=1+2+12+5=20]
Therefore
DISABLE : HRL20
Now required
E=5 decoded as 2×E=2×5=10=J
N=14 decoded as 2×N=2×14=28=(26+2)=2=B
[A+B+L+E=1+2+12+5=20]
Therefore
ENABLE:JB20
Each alphabets equal to natural naumber given below:
A=1
B=2
C=3
D=4
E=5
F=6
G=7
H=8
I=9
J=10
K=11
L=12
M=13
N=14
O=15
P=16
Q=17
R=18
S=19
T=20
U=21
V=22
W=23
X=24
Y=25
Z=26
Now decode the given words
1)DISABLE
D=4 decoded as 2×D=2×4=8=H
I=9 decoded as 2×I=2×9=18=R
S=19 decoded as 2×S=2×19=38 we doesn't have 38 letters then we take
Cyclic way =38=26+12=12=L
[A+B+L+E=1+2+12+5=20]
Therefore
DISABLE : HRL20
Now required
E=5 decoded as 2×E=2×5=10=J
N=14 decoded as 2×N=2×14=28=(26+2)=2=B
[A+B+L+E=1+2+12+5=20]
Therefore
ENABLE:JB20
Answered by
0
ENABLE = JB20
This is a multiple-choice question from BBET 2018 sample paper, so its better you solve it on your own.
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