An intelligent shopkeeper uses a special weighing balance in which one

Question

An intelligent shopkeeper uses a special weighing balance in which one pan is heavier than the other. While purchasing the material, he places a 1 kg weight on the heavier side but while selling the same product, he places 1 kg weight on the lighter side. If he decides to sell his goods at the cost price and gets an overall gain of 50 %, then find the value by which one side of the balance is heavier than the other?

A. 100 gm
B. 150 gm
C. 200 gm
D. 250 gm
E. 300 gm

A. 100 gm
B. 150 gm
C. 200 gm 
D. 250 gm
E. 300 gm

chetan2u · Accepted Answer

While purchase 
He purchases 1000+x g for the price of 1000g.
 While selling 
He sells 1000-x for the price of 1000g.

So what is his cost price - 1000-x
His profit x while purchasing and x while selling, so 2x

Thus,  \frac{2x}{1000-x}=\frac{50}{100}......4x=1000-x....5x=1000....x=200

C

TheNightKing · Answer

Trying options..

He purchases 1200g for the price of 1000g.
He sells 800g for the price of 1000g.

Cost Price=800
He gained 200g on each of his transaction (buying and selling)
so 400 over 800 is 50%.

Hence C

effatara · Answer

Let the weight of the heavier pan be 'x' kgs more than the lighter pan and the cost price of 1 kg of the merchandise be $'p'.
When appearing to buy 1 kg, since he puts the weights on the heavier pan and the merchandise on the lighter one, he actually gets (1+x) kg for $p. That is, buying 1 kg costs him $p/(1+x).
While appearing to sell 1 kg, since he puts the merchandise on the heavier pan and the weights on the lighter one, he actually sells (1-x) kg for $p.
That is, selling 1 kg fetches him $p/(1-x).
Since his overall gain is 50%  :
p/(1-x) = (3/2)*p/(1+x)....> (3/2)*(1-x) = 1+x....> x = (1/5) kg = 200 gms.
ANS: C

skmaan · Answer

wt of material bought = 1+x 
wt of material sold = 1-x

Cost for 1+x Kg is c 
Cost for 1 Kg is c/(1+x)*1

Selling price for 1-x Kg is c (Given)
Selling price for 1 Kg is c/(1-x)*1

Using Profit eqn; 
50% (cost of 1Kg) = 1kg wt * (SP of 1Kg) - 1Kg wt * (cost of 1 Kg)

0.5 (c/1+x) = 1* c/(1-x) - 1 * c/(1+x)

Solving, x = 0.2 Kg or 200 gm

An intelligent shopkeeper uses a special weighing balance in which one

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An intelligent shopkeeper uses a special weighing balance in which one

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