If 1111=R, 2222=T, 3333=E, 4444=N Then 5555=?

Question

If
1111=FR,
2222=ET,
3333=TE,
4444=SN
Then
6666=??

Answer 1

1111=4=fouR 
2222=8=eighT 
3333=12=twelvE 
4444=16=sixteeN 
5555=20=twentY

So answer is Y

Comments

By addition also we can solve the same

Answer 2

1*4=4 fouR
2*4=eighT
3*4=twelvE
4*4=sixteeN
5*4=twentY

Answer 3

1111=r
2222=t
3333=e
4444=n
5555=y

Answer 4

Y........................................................

Answer 5

Answer is Y
5+5+5+5=20=twentY

Answer 6

1111=1+1+1+1=FOU(R)
similarly
,2222=EIGH(T)=T,
3333=TWELV(E)=E,
4444=SIXTEE(N)=N
Then 5555TWENT(Y)=Y.

Answer 7

answer is Y.
ENTRY (jumbled words R T E N and Y)
Y is the missing alphabet.

Answer 8

if 1+1+1+1=4 in word FOUR last word id R
and 2+2+2+2=8 in word EIGHT LAST WORD IS T
3+3+3+3=12 IN WORD TWELVE LAST WORD IS E
4+4+4+4=16 IN WORD SIXTEEN LAST WORD IS N
5+5+5+5=20 IN WORD TWENTY LAST WORD IS Y
SO ANSWER IS Y

Answer 9

Y.
1111=4, whose last letter is R
2222=8, whose last letter is T
and so on.

Answer 10

Its Y .................so simple