The cost of a precious stone varies directly with the cube of its weight. The stone is broken into three pieces, whose weights are in the ratio 4 : 5 : 6. If the stone had been broken into three pieces of equal weights, then there would have been a further loss of र24,000. What is the cost price of the unbroken stone?
a) 250,000
b) 300,000
c) 4,500,000
d) 2,700,000

• Let V be the value of a precious stone of weight W.
  
  Then, from the condition of the problem, we have,
  
  V ∝ W3
  
  V ∝ kW3 ……..... (1) [k = constant of variation]
  
  Since the weights of the three broken pieces are proportional to 4 : 5: 6, we assume their weights as 4w, 5w and 6w respectively.
  Hence, the weight of the unbroken piece of stone = 4w + 5w + 6w = 15w
  
  v=k((15w)^3)
  v= 3375k(w^3)...............(2)
  k(w^3)=v/3375.
  Let, v1, v2 and v3 be the values of the 3 pieces of weights 4w, 5w and 6w respectively.
  v1=64k(w^3)
  v2=125k(w^3)
  v3=216k(w^3)
  
  Therefore the total value of the 3 pieces
  =v1+v2+v3
  =405k(w^3)
  from eq 2..........
  the total value of the 3 pieces= 405*v/3375
  loss = v - (405*v/3375)........................(3)
  
  similarly
  second time it breaks in three equal parts i.e 1:1:1
  in this case total value equals = 3*v/27............(4)
  loss = v-(3*v/27...............................(5)
  acc to question
  eq(5)-eq(3)=24000
  (8v/9)-(2970v/3375)=24000
  v=2,700,000
  
• 11 years agoHelpfull: Yes(14) No(1)
• Let V be the value of a precious stone of weight W.
  
  Then, from the condition of the problem, we have,
  
  V ∝ W^3
  
  V ∝ kW^3 ……..... (1) [k = constant of variation]
  
  Since the weights of the three broken pieces are proportional to 4 : 5: 6, we assume their weights as 4w, 5w and 6w respectively.
  Hence, the weight of the unbroken piece of stone = 4w + 5w + 6w = 15w
  
  v=k((15w)^3)
  v= 3375k(w^3)...............(2)
  k(w^3)=v/3375.
  Let, v1, v2 and v3 be the values of the 3 pieces of weights 4w, 5w and 6w respectively.
  v1=64k(w^3)
  v2=125k(w^3)
  v3=216k(w^3)
  
  Therefore the total value of the 3 pieces
  =v1+v2+v3
  =405k(w^3)
  from eq 2..........
  the total value of the 3 pieces= 405*v/3375
  loss = v - (405*v/3375)........................(3)
  
  similarly
  second time it breaks in three equal parts i.e 1:1:1
  in this case total value equals = 3*v/27............(4)
  loss = v-(3*v/27...............................(5)
  acc to question
  eq(5)-eq(3)=24000
  (8v/9)-(2970v/3375)=24000
  v=2,700,000
• 11 years agoHelpfull: Yes(5) No(2)
• ans. (d)2700000
• 11 years agoHelpfull: Yes(0) No(3)