# TRICKS TO SOLVE PROBLEMS ON PROFIT AND LOSS PART 3 -LEARN SERIES

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## TYPE 1:

### PROBLEMS BASED ON FINDING  HOW MANY NUMBERS ARE THERE:

1.A trader  buys oranges at 7 for a rupee and sells them at 40% Profit.How many oranges does he sell for a Rupee?

Explanation

Cost Price of 1 apple=1/7

Selling Price  of 1 apple at 40% Profit is equal to 140% of its cost Price .

Selling Price of 1 apple=(1/7) * 140% =(1/7) * (140/100)=1/5

Therefore 5 Oranges are sold for 1 Rupee

## TYPE 2:

### Chain Rule:

1.A Manufacturer sells a scooter at 10% Profit to wholesaler who in turn sells it to a retailer at 20% Profit. If the Price Paid by the retailer is Rs.13200,How much the scooter costs to the manufacturer?

Explanation

Let the manufacturer bought the Scooter  for x Rupees

Given, x*(110/100) * (120/100)=13200

X=(13200*100*100)/(110*120) =10000

The Scooter costs Rs.10000 to the manufacturer

## TYPE 3:

### FALSE WEIGHT:

To find Profit/Loss Percentage=[Error /(True Value –False Value)]*100

Error=True Value – False Value

1.A dishonest dealer to sell his goods at cost price,but he uses a weight of 960grams for the 1Kg weight,then the Percentage of gainis?

Explanation

Error=1000-960=40grams

Profit%=[40/(1000-40)]*100 ⇒ (40 * 100)/960⇒4(1/6)%

2.A grocer sells rice at a profit of 10%  and uses  a weight which is 20% less .Find his total percentage Gain?

Explanation

Let the Original weight=1000 gm

Here the grocer sells 20% less weight=200 gm, Therefore he sold only 800 gm with 10% Profit

Profit% when he sells rice 20% less than the original quantity

(200/800)*100  =25%

To find Successive Profit/Loss%=x+y+(xy/100)

Here x=10%   and   y=25%  if it is Profit  its sign is ‘+’ve if it is Loss then its sign is ‘-‘ve

Profit%=10+25+ (250/100)⇒ 35+2.5=37.5gm