Latest TCS Aptitude Question SOLUTION: 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 bro

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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

Asked In TCS Tanu (11 years ago)

Solved Robin Read Solution (3)

Is this Puzzle helpful?   (14)   (6) Submit Your Solution Arithmetic

Read Solution (3) :

Robin  (11 years ago)

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

Is this solution Helpfull? Yes (14) | No (1)  

Robin  (11 years ago)

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

Is this solution Helpfull? Yes (5) | No (2)  

Neema Melkani  (11 years ago)

ans. (d)2700000

Is this solution Helpfull? Yes | No (3)  

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