A shopkeeper cheats to the extent of $10$ % while buying as well as selling, by using false weights. His total gain is

10

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20

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Solution

The correct option is C $20$ %
While buying, let the shopkeeper cheats $0.1 k g$

for each $1 k g,$ so he gets $1.1 k g$ as the price of $1 k g .$

Again while selling he sells $0.9$ kg at the price of $1 k g .$

So, his total gain per kg is $(1.1 - 0.9) k g$

$= 0.2. k g$

i.e; $20 %$

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A shopkeeper cheats to the extent of 10 % while buying as well as selling, by using false weights. His total gain is

The correct option is C 20%While buying, let the shopkeeper cheats 0.1kgfor each 1kg, so he gets 1.1kg as the price of 1kg.Again while selling he sells 0.9 kg at the price of 1kg.So, his total gain per kg is (1.1−0.9)kg=0.2.kgi.e; 20%

A shopkeeper cheats to the extent of $10$ % while buying as well as selling, by using false weights. His total gain is

The correct option is C $20$ %
While buying, let the shopkeeper cheats $0.1 k g$

for each $1 k g,$ so he gets $1.1 k g$ as the price of $1 k g .$

Again while selling he sells $0.9$ kg at the price of $1 k g .$

So, his total gain per kg is $(1.1 - 0.9) k g$

$= 0.2. k g$

i.e; $20 %$