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

+4 votes
421,010 views
closed with the note: BEST ANSWER PICKED
posted Jul 27, 2014 by anonymous

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14 Solutions

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

So answer is Y

solution Jul 27, 2014 by Salil Agrawal
By addition also we can solve the same
+2 votes

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

solution Jul 31, 2014 by anonymous
+2 votes

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

solution Aug 22, 2014 by anonymous
+1 vote

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

solution Apr 8, 2015 by Varghese Anthony
+1 vote

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

solution Jul 27, 2015 by Pankaj Bora
+1 vote

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

solution Aug 3, 2015 by Hamza Khan
+1 vote

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

solution Sep 28, 2015 by anonymous
+1 vote

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

solution Mar 11, 2016 by Ashish Kumar Khanna
+1 vote

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

solution Mar 23, 2016 by Sundararajan Srinivasa Gopalan
0 votes

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

solution Oct 20, 2015 by Vikrant Mane
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