3AB=DE
5BC=LM
7CD=??

• 7CD = XY
  
  e>xplanation-
  3AB = DE => 3,1,2 = 4,5
  5BC = LM => 5,2,3 = 12, 13
  We can observe,first number of RHS = first number of LHS * first number eqv to digit + first number eqv to digit
  similarly,
  second number of RHS = first number of LHS * first number eqv to digit + Second number eqv to digit
  so,
  7CD = 7*3+3,7*3+4 = 24, 25 = XY
  
• 11 years agoHelpfull: Yes(4) No(1)
• Answer is TU
• 11 years agoHelpfull: Yes(1) No(1)
• ANSWER IS XY.
  AS EXPLAINED BY CHANDAN...
• 11 years agoHelpfull: Yes(1) No(0)
• 7CD=XY
  
  3AB=3*(1+2)=9=(4+5)=DE
  5BC=5*(2+3)=25=(12+13)=LM
  7CD=7*(3+4)=49=(24+25)=XY
  
  
• 11 years agoHelpfull: Yes(1) No(0)
• Answer is TU
  solution:
  let A = 1 B = 2 C = 3 and so on...........
  now see the L.H.S the numbers are increasing in a perticular sequence.
  then see the second and third, the alphabets are also increasing in a perticular sequnce.
  Now see the R.H.S
  Like LHS the letters must increase with a perticular sequence
  i.e. value of L = 12 and value of D = 4. so L - D = 12 - 4 = 8
  =>(value of L)+8 = 20 = T
  then, value of M = 13 and value of E = 5. so M - E = 8
  =>(value of M) + 8= 21 = U
  so the answer is TU
• 11 years agoHelpfull: Yes(0) No(1)