3 mangoes and 4 apples costs Rs 85. 5 apples and 6 peaches costs Rs. 122. 6 mangoes and 2 peaches cost Rs.114. what is the combined price of 1 apple, 1 peach and 1 mango?
option
(a) 37 Rs
(b) 39 Rs
(c) 35 Rs
(d) 36 Rs

• 3m+4a=85...(i)
  5a+6p=122....(ii)
  6m+2p=114....(iii)
  after solving (i) & (ii),we get the eqn in terms of m & p i.e
  15m-24p=-63....(iv)
  now, we solve eqn(iii) & eqn(iv),we get
  m=15,a=10,p=12
• 10 years agoHelpfull: Yes(35) No(8)
• I think there is typing mistake in qn.
  
  If statement is
  3 mangoes and 4 apples costs Rs 85. 5 apples and 6 peaches costs Rs. 122. 6 mangoes and '3' peaches cost Rs.114. what is the combined price of 1 apple, 1peach and 1 mango?
  
  then
  3M+4A+0P=85
  0M+5A+6P=122
  6M+0A+3P=144
  adding these eqns
  9(M+A+P)= 351
  M+A+P= 351/9=39 ... option (b)
  
• 10 years agoHelpfull: Yes(19) No(40)
• ans 37.
  3m+4a=85....(1)
  5a+6P=122....(2)
  6m+2p=114....(3)
  solving i and 2 and 3 we'll get 37
• 10 years agoHelpfull: Yes(13) No(3)
• 3m+4a=85...(i)
  5a+6p=122....(ii)
  6m+2p=114....(iii)
  Adding (i) (ii) (iii) we get
  9a+9m+8p= 321
  9a+9m+9p= 321+p
  a+m+p=(321+p)/9------(iv) It must have to be a integer number.
  To make (iv) as an integer p must have to be either 3 or (3+9) or (3+18)or.....
  Let check it out..
  if p=3 then
  5a+18=122
  5a=104
  a!= an integer so p=3 is false
  
  If a=12, then
  by(ii)
  5a=122-72
  a=10
  
  by(i)
  3m=85=40
  a=15
  a+m+p= 15+10+12=37.
  
• 8 years agoHelpfull: Yes(5) No(0)
• 10 + 15 +12 =37
  
• 10 years agoHelpfull: Yes(3) No(3)
• Let x b mango cost, y b apple cost, z b peaches cost.
  3x+4y=85----eq1
  5y+6z=122-----eq2
  6x+2z=144----eq3
  by solvin above equations v get x=15,y=10,z=12
  so combined price--> 15+10+12=37
  ans==> 37
• 10 years agoHelpfull: Yes(3) No(0)
• from the question
  3m+4a=85---->1
  5a+6p=122---->2
  6m+2p=114----->3
  solve 1 and 3
  we get 8a-2p=56.let it be 4th eq.
  now solve 4 and 2.
  29a=290
  a=10.from 4,by substituting a,we get p=12.
  substitute a,p in eq3.we get m=15
  now a+b+c=10+12+15=37.
  
• 10 years agoHelpfull: Yes(3) No(2)
• 37..solve the equations we get A=10 P=12..now we have by adding all the given eqs 9A+9M+8P=321.9A+9M+9P=321+12...so A+P+M=37
• 10 years agoHelpfull: Yes(2) No(2)
• answer is 37
• 10 years agoHelpfull: Yes(2) No(2)
• 37
  3m+4A=85
  5a+6p=122
  6m+2p=114
  solving these three we'll get the answer
• 10 years agoHelpfull: Yes(1) No(0)
• 3x+4y=85
  5y+6z=122
  6x+2z=114
  solving three equations,,x+y+z=37
• 10 years agoHelpfull: Yes(1) No(2)
• Solve equation get answer !!!
• 10 years agoHelpfull: Yes(1) No(2)
• 3m + 4a = 85 ..(1)
  5a + 6p = 122 ..(2)
  6m + 2p = 144 ..(3)
  (1) x 2 => 6m + 8a = 170
  4 x (3) => 6m + 2p = 144
  Solving we get 8a - 2p = 56 ..(4)
  
  (2) => 5a + 6p = 122
  3 x (4) = 24a - 6p = 168
  Solving we get a = 10, p = 12, m = 15
  So a + p + m = 37
• 9 years agoHelpfull: Yes(1) No(0)
• 3M+4A=85 RS...(1)
  6P+5A=122 RS...(2)
  6M+2P=114 RS....(3)
  FROM EQUATIION 1 DETERMINE VALUE A AND PUT IN 2 EQUAT.. AND SOLVE 2AND 3 EQUATION...
  M=15.A=10,P=12
• 9 years agoHelpfull: Yes(1) No(1)
• a)37
  as on solving we get
  3M+4A+0P=85
  0M+5A+6P=122
  6M+0A+2P=144
  on further solving we get:
  m=15,a=10 and p=12
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• answer is c
• 10 years agoHelpfull: Yes(0) No(0)
• 1+15+12=37
  
• 10 years agoHelpfull: Yes(0) No(0)
• @shubham kumar please solve the equations because after solving the equations u will not get these values.
• 10 years agoHelpfull: Yes(0) No(1)
• 3M+4A=85...(i)
  5A+6P=122...9(ii)
  6M+2P=114...(iii)
  after solving the eqn we get M=15,A=10 and P=12
  so,oprion (a) is correct
• 8 years agoHelpfull: Yes(0) No(0)
• 37
  answer
  
  by solving
• 4 years agoHelpfull: Yes(0) No(0)
• 3M + 4A = 85....eq1
  5A + 6P = 122...eq2
  6M + 2P = 114...eq3
  sub eq3 by 2*eq1
  => 8A - 2P = 56....eq4
  sub eq2 by 3 * eq4
  29A = 290
  => A = 10
  put value of A in 1st and 2nd eq
  M = 15
  P = 12
  now, A + P + M = 10 + 12 + 15 = 37 answer
• 2 years agoHelpfull: Yes(0) No(0)