In a city there are few engineering, MBA and CA candidates. Sum of four times the engineering, three times the MBA and 5 times CA candidates is 3650. Also three times CA is equal to two times MBA and three times engineering is equal to two times CA. In total how many MBA candidates are there in the city?

• e = no. of engineering students, m = no. of MBA students and c = no. of CA students
  4e + 3m + 5c = 3650, therefore e = (3650-3m-5c)/4 --------(1)
  3c = 2m , therefore c = 2m/3
  3e = 2c replacing e with (1) we get,
  3(3650- 3m - 5c)/4 = 2c.....replacing c with 2m/3 and solving this equation we get m = 450
  So the number of MBA students is 450
  
• 9 years agoHelpfull: Yes(18) No(0)
• No of engg=e
  Mba=m
  Ca=c
  4e+3m+5c=3650
  3c=2m m=3c/2
  3e=2c e=2c/3
  put the value in equestionl
  4*2c/3+3*3c/2+5c=3650
  c=300
  m=3c/2
  m=450
• 9 years agoHelpfull: Yes(11) No(0)
• 1350 = mba
  4 e+3 m+ 5 c=3650, 3c=2m , 3e=2c
  8c/3 +9c/2 +c =3650, c=300
  3m= 9c/2=1350
  
• 9 years agoHelpfull: Yes(0) No(6)
• e = no. of engineering students, m = no. of MBA students and c = no. of CA students
  4e + 3m + 5c = 3650, therefore 3m=3650-5c-4e-----------1
  3c = 2m , therefore c = 2m/3
  3e = 2c ,threrfore e=2(2m/3)/3
  subtitute c and e values and slove eqa 1 you will get 730
  ans :730
  
• 9 years agoHelpfull: Yes(0) No(2)