Dinalal divides his property among his four sons after donating RS.20,000 and 10% of his remaining property. The amounts received by the last three sons are in arithmetic progression and the amount received by the fourth son is equal to the total amount donated. The first son receives as his share RS.20,000 more than the share of the second son. The last son received RS.1 lakh less than the eldest son.
10. Find the share of the third son.
option
a) RS.80,000
b) RS.1,00,000
c) RS.1,20,000
d) RS.1,50,000

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• 11 years agoHelpfull: Yes(22) No(8)
• let d shares are a,b,c,d.
  and let property be x.
  so d=20000+0.1(x-20000) .. equation 1
  
  now let d differences b/w last three sons be n.
  c=d-n
  b=d-2n
  a=b+20000=d-2n+20000
  
  so a+b+c+2d = x
  d-2n+20000+d-2n+d-n+2d = x
  5d-5n+20000 = x
  putting the value of d from equation 1 we get.
  5(20000+0.1(x-20000))-5n+20000 = x
  100000+0.5x-10000-5n+20000 = x ... equation 2
  as last son received 100000 less than eldest son.
  so d-2n+20000 = d+100000
  n = -40000 .... equation 3
  by putting value of equation 3 in equation 2 we get,
  100000+0.5x-10000-5(-40000)+20000 = x
  solving this we get x = 620000.
  now putting d value of x in equation 1 we get.
  d=20000+0.1(620000-20000)
  = 20000+0.1(600000) = 80000
  therefore c = d-n = 80000-(-40000) = 120000 answer. option (c).
• 11 years agoHelpfull: Yes(19) No(0)
• ans is (C). Let the total property be P. Consider the four brothers as a,b,c and d. d receives a share equal to the donation which is {20,000 + 1/10(P-20,000)}. Convert all the shares in terms using any one of the brother's share and substitute in a+b+c+2d = P. I converted everything in terms of d.
• 11 years agoHelpfull: Yes(8) No(0)
• consider four sons are a1,a2,a3,a4
  according to problem
  a4=20,000+(1/10)y
  property is divided to four sons
  y=a1+a2+a3+a4
  a1=a2+20,000
  a4=a1-1,00,000
  so
  a4=a2-80,000
  since its a ap it should a2,a2-40000,a2-2(40,000)
  a3=a2-40,000
  rearrange the equations
  a1=a4+1,00,000
  a2=a4+80,000
  a3=a4+40,000
  substitute to equation y=a1+a2+a3+a4
  y=2,20,000+4(a4)
  substitute the given a4=20,000+(1/10)y
  soultion is y=5,00,000
  a4=70,000
  a3=1,10,000
  a2=1,50,000
  a1=1,70,000
  
• 11 years agoHelpfull: Yes(7) No(9)
• @METASPOIT Oh God No :(
• 11 years agoHelpfull: Yes(5) No(1)
• @tanu-is it tcs 2014 q???????
• 11 years agoHelpfull: Yes(4) No(1)
• Assume the amounts received by the 2nd, 3rd, and 4th sons are a+d, a, a-d (as they are in AP)
  Now Eldest son received Rs.20,000 more than the 2nd son. So He gets a+d+20,000
  Last son received 1 lakh less than the eldest son. So (a+d+20,000) - (a-d) = 1,00,000 ⇒ 2d = 80,000 ⇒ d = 40,000
  So Amounts received by the 4 sons are a + 60,000, a+40,000, a, a - 40,000.
  Assume His property = K rupees.
  It was given that the youngest son's share is equal to 20,000 + 110(K−20,000)
  Then 20,000 + 1/10(K−20,000) = a - 40,000 ...........(1)
  and the Remaining property = Sum of the properties received by all the four son's together.
  Remaining property = 9/10(K-20,000)
  ⇒9/10(K-20,000) = ( a + 60,000 ) + (a+40,000) + a +( a - 40,000) ..(2)
  
  solve equation:-
  9/10(k-20,000)=4a+60,000
  1/10(k-20,000)=a-60,000
  -----------------------------------
  k-20,000=5a
  put k=20,000+5a in eq (1)
  a=1,20,000 so,ans is c.
• 9 years agoHelpfull: Yes(3) No(0)
• C is the correct option
• 11 years agoHelpfull: Yes(1) No(2)
• (a)Rs80,000
  A,B,C,D are sons
  D get 20,000
  B,C,D has common diffrence
  A,B has diffrence of 20,000
  A,D had diffrence of 1,00,000
  so. B=20,000+2(common diffrence)=80,000+D
• 11 years agoHelpfull: Yes(1) No(0)
• i think c
  
• 9 years agoHelpfull: Yes(1) No(1)
• hi frds i think i made mistake can anyone please help me to spot out
• 11 years agoHelpfull: Yes(0) No(2)
• Assume the amounts received by the 2nd, 3rd, and 4th sons are a+d, a, a-d (as they are in AP)
  Now Eldest son received Rs.20,000 more than the 2nd son. So He gets a+d+20,000
  Last son received 1 lakh less than the eldest son. So (a+d+20,000) - (a-d) = 1,00,000 ⇒⇒ 2d = 80,000 ⇒⇒ d = 40,000
  So Amounts received by the 4 sons are a + 60,000, a+40,000, a, a - 40,000.
  Assume His property = K rupees.
  It was given that the youngest son's share is equal to 20,000 + 110(K−20,000)110(K−20,000)
  Then 20,000 + 110(K−20,000)110(K−20,000) = a - 40,000 ...........(1)
  and the Remaining property = Sum of the properties received by all the four son's together.
  Remaining property = 910910(K-20,000)
  ⇒910⇒910(K-20,000) = ( a + 60,000 ) + (a+40,000) + a +( a - 40,000) ..(2)
  Solving We get K = 40,000 and a = 1,20,000
  So third son got Rs.1,20,000
  
• 8 years agoHelpfull: Yes(0) No(0)