2.5 Mixture Sold at Profit/Loss (Link with Profit & Loss)

A very common SSC pattern: a trader mixes two grades of a commodity and sells the mixture at a stated price, making a given profit or loss. You must first convert the selling price and profit/loss percentage into the mixture’s cost price (mean price), because alligation always works on cost prices, never on selling prices directly.

Mean CP of mixture = (SP of mixture)/(1 + (Profit%)/(100)) (profit case), (SP)/(1-(Loss%)/(100)) (loss case)

Worked Example 2.5.1: A trader mixes 26 kg of rice at Rs 20/kg with 30 kg of rice at Rs 36/kg and sells the mixture at Rs 30/kg. Find his gain percentage.

Solution: Total CP = 26(20)+30(36) = 520+1080=1600. Total quantity =56 kg. CP per kg = (1600)/(56) = (200)/(7) ≈ 28.57. Gain% = (30-(200)/(7))/((200)/(7)) × 100 = ((10)/(7))/((200)/(7))× 100 = (10)/(200)×100 = 5%

Worked Example 2.5.2 (dishonest dealer / “free” ingredient): A dishonest dealer mixes chalk powder (which costs him nothing) with rice costing Rs 1/kg, and sells the mixture at the price of pure rice, thereby gaining 25%. Find the ratio of rice to chalk powder in the mixture.

Solution: SP per kg = Rs 1 (price of pure rice). Since gain =25%: Mean CP = (1)/(1.25) = 0.8 Alligation with rice (Rs 1) and chalk (Rs 0): Rice : Chalk = (0-0.8):(0.8-1) ⇒ taking magnitudes 0.8:0.2 = 4:1 So rice and chalk powder are mixed in the ratio 4 : 1.

Worked Example 2.5.3: In what ratio must water be mixed with milk costing Rs 20/L so that, after selling the mixture at Rs 20/L (the cost price of pure milk), the seller gains 25%?

Solution: Mean CP of mixture = (20)/(1.25) = 16. Milk : Water = (0-16):(16-20) ⇒ 16:4 = 4:1

Worked Example 2.5.4 (loss case): A shopkeeper mixes 40 kg of rice costing Rs 25/kg with 60 kg of rice costing Rs 15/kg, and sells the mixture at Rs 15.20/kg. Find his loss percentage.

Solution: Total CP =40(25)+60(15)=1000+900=1900. Total quantity =100 kg, so CP per kg =19. Loss per kg =19-15.20=3.80. Loss% = (3.80)/(19)×100 = 20% Note the mirror-image relationship with the profit formula in Section 2.5’s box — for a loss, SP is below CP, so use Mean CP = (SP)/(1-Loss%/100) whenever you are working backward from a stated selling price and loss percentage.

Worked Example 2.5.5 (find the total quantity, given the ratio, both costs, overall profit%, and total revenue): A trader mixes two varieties of pulses costing Rs 45/kg and Rs 60/kg in the ratio 3 : 2, and sells the entire mixture for Rs 3060, making an overall profit of 20%. Find the total quantity of the mixture sold.

Solution: Mean CP =(45(3)+60(2))/(5)=(135+120)/(5)=(255)/(5)=51 per kg. SP per kg with 20% profit =51×1.20=61.20. Total quantity = (Total SP)/(SP per kg) = (3060)/(61.20) = 50 kg Check: Total CP =51×50=2550; SP at 20% profit =2550×1.20=3060 ✓. Total quantity sold: 50 kg.

Worked Example 2.5.6 (dishonest dealer with a fractional gain percentage — tests comfort with mixed numbers): A milkman sells pure milk mixed with water at the cost price of pure milk, and thereby gains 11(1)/(9)%. Find the percentage of water in the mixture.

Solution: Let CP of pure milk =1 (per litre); water costs 0. Gain =11(1)/(9)% = (100)/(9)%. Since he sells at the cost price of pure milk, SP =1, so: Mean CP = (SP)/(1+Gain%/100) = (1)/(1+(100/9)/(100)) = (1)/(1+(1)/(9)) = (1)/((10)/(9)) = (9)/(10) = 0.9 Milk : Water = (0-0.9):(0.9-1) → magnitudes 0.9:0.1 = 9:1. Water’s share of the mixture =(1)/(9+1)=(1)/(10)= 10%.