Answer / kumar prashant

PORK/CHOP = C
P=5 ; O=4 ; R=9 ; K=0 SO PORK=5490
C=2 ; H=7 ; O=4 ; P=5 SO CHOP=2745
5490/2745=2


ATOM + BOMB = BINGO (I AM SOLVING IT FOR A=9; G=5)

A=9 ; T=4 ; O=3 ; M=2 SO ATOM = 9432
B=1 ; O=3 ; M=2 ; B=1 SO BOMB = 1321
B=1 ; I=0 ; N=7 ; G=5 ; O=3 SO BINGO = 10753
HENCE ATOM + BOMB = BINGO

Answer / amit

@Vian: Pork/chop=c where c=2
so we need to divide 'pork' by a number which stars with 2 since in CHOP(most significant digit c=2)....and also result is 2(given since c=2).... so P CAN BE EITHER 4 0R 5
WHEN U WILL USE P=4 AND APPLY HIT & TRAIL U WILL GET STUCK AT SOME POINT...SO P MUST BE 5 NOW WE HAVE 5ORK/2HO5=2 (SIMPLY SUBSTITUTING VALUES)NOW '2HO5' contains 5 as least significant bit so k should be 0 to get completely divisible now we have
5OR0/2HO5=2(DONT GET CONFUSED BY ZERO AND 'O')NOW WE CAN SEE THAT VALUE OF H SHOULD BE GREATER THAN 5 COZ WE NNEED 2 AS RESULT TRYING IT WID 6 AND 7 YOU WILL FOULD 7 SUITABLE AND IN THAT CASE O WILL BE 4 ... I HOPE ITS HELPFUL

Answer / vian

kumar or amit can u guys plz tell me how did u get the ans....how is p=5 o=4 and al......plz explain

Answer / amit

@Kiran: plz dont post incorrect answers.

acc to kiran: chop should be 2+7+14+15=38 and not 28.

so answer posted by kiran is incorrect.

BOTH QUESTIONS ARE CORRECT.
GOOD WORK KUMAR...KEEP IT UP.

Answer / kiran

as c=2
so we conclude that c=3 (original acc. to position)
3-1=2
so we subtract 1 from each alphabet position
so p=15,0=14,r=17,k=10,c=2,h=7
so pork/chop=56/28=2
hence proved



and 2nd part is wrong cn't be solved acc to some rule

Answer / elprince

C is chop so pork=4 & chop=2
Atom=7 & bomb=2 & bingo=9

3 boys ,a,b,c . have 3 chocklets .. a ate the chocklet.b thrown away the chocklet and c ate half chocklet .now how many choclect each have in their hand.

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