Z4     X3    V9     A6    C2     ?      T5    R4   P15 A. E10 B. S10 C. E12 D. S12 E. None

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Zab

  • Nov 14th, 2005
 

E12

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ji

  • Nov 16th, 2005
 

ans is 24

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ravi

  • Feb 7th, 2006
 

Ans-E10

Solution- Z4 X3 V9 A6 C2 ? T5 R4 P15

starting from Z .After that z ,x occur so differnce between z and x are one letter.And c is in the between of A and E So we find that E occurs.C2 is the middle of the questions .So we divide into 4 part.In first part three no occurs which always divisible by 3.So in next part three no always divisible by 5 so answer is 10.So finally answer is E10.

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shrikant

  • Feb 22nd, 2006
 

E10 and...............

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Bhagyashree

  • Jun 14th, 2006
 

E12

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ramya

  • Aug 19th, 2006
 

ans is E12 not E10 i am very much sure with my answer

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raja_letstalk

  • Jan 21st, 2008
 

CLEAR ANS..................................

OBSERVE -- Z4  (Y)  X3   (W)   V9

First --- Alternate from reverse alphabets missing
Second --- Again two ans. for this
                   i) 4+3=7  &  7+2=9 ( Here 2 is a normal number,taken)
                                  (or)
                   ii) 4x3=12 & 12-3=9  (Here 3 is the second alphabet's postfix i.e x3)

NOW ACCORDING TO THIS...............

                   A6  (B)  C2  (D)  E10


First --- Alternate from front alphabets missing
Second --- i) 6+2=8 & 8+2=10 (Here 2 is a normal number, taken)
                             (or)
                 ii) 6x2=12 & 12-2=10 ( Here 2 is the second alphabet's postfix i.e c2)

       -----------------------                    ENJOY                                -------------------------

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dhaliwalinderjot

  • Jun 22nd, 2008
 

i think its E12

exp:
           z4 x3   v9
           4+3+2=9
 
           a6 c2  e12
           6+2+4=12

           t5 r4 p15
           5+4+6=15


now pattern is add 2 to first grp to get 3rd no.
                        add 4 to second grp to get 3rd no.
                        add 6 to third grp to get 3rd no.

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