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22 Sep 2025, 21:53
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:19) correct
42%
(02:37)
wrong
based on 131
sessions
History
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Time
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Not Attempted Yet
Ram mixed milk and freely available water in a ratio of 3:2, and sold it at a profit of 30%. What will be the profit/loss percentage, if the milk and water were mixed in the ratio 4:1 and the selling price remains unchanged?
A. 2.5% profit
B. 2.5% loss
C. 5% profit
D. 5% loss
E. None of the above
Screenshot 2025-09-23 10.24.33 AM.png [ 69.28 KiB | Viewed 1017 times ]
A. 2.5% profit
B. 2.5% loss
C. 5% profit
D. 5% loss
E. None of the above
Attachment:
Screenshot 2025-09-23 10.24.33 AM.png [ 69.28 KiB | Viewed 1017 times ]
27 Sep 2025, 01:52
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Multiplication factor (MF)
= Selling Price ÷ Cost Price.
Case 1: Ratio 3:2
Profit = 30% ⇒ MF = 1.3.
Cost of mixture = 3 parts milk × 1 + 2 parts water × 0 = 3.
Selling price
= 3 × 1.3
= 3.9
= 39/10
So MF of mixture
= (39/10) ÷ 3 = 13/10
Case 2: Ratio 4:1
Cost price = 4 parts milk × 1 + 1 part water × 0 = 4.
Selling price (using same MF of 1.3) would have been 4 × 13/10 = 52/10.
But actual selling price remains 3.9 (same as case 1).
So, new MF
= (39/10) ÷ 4 = 39/40.
Profit/Loss % = (MF − 1) ×100
= ((39/40) − 1) ×100
= (-1/40)*100
= −2.5%
👉 Hence, Loss = 2.5%.
Long way to solve:
Let cost price of milk = ₹1 per liter, cost of water = ₹0.
Case 1: Ratio 3:2
Total = 5 liters → 3 liters milk + 2 liters water.
Cost price = 3×1 + 2×0 = 3.
Profit = 30%, so Selling price = 3 × 1.3 = 3.9.
Selling price per liter = 3.9/5 = 0.78.
Case 2: Ratio 4:1
Total = 5 liters → 4 liters milk + 1 liter water.
Cost price = 4×1 + 1×0 = 4.
Selling price (unchanged per liter) = 5 × 0.78 = 3.9.
Profit/Loss = (SP − CP)/CP ×100 = (3.9 − 4)/4 ×100 = −2.5%.
So, there is a 2.5% loss.
= Selling Price ÷ Cost Price.
Case 1: Ratio 3:2
Profit = 30% ⇒ MF = 1.3.
Cost of mixture = 3 parts milk × 1 + 2 parts water × 0 = 3.
Selling price
= 3 × 1.3
= 3.9
= 39/10
So MF of mixture
= (39/10) ÷ 3 = 13/10
Case 2: Ratio 4:1
Cost price = 4 parts milk × 1 + 1 part water × 0 = 4.
Selling price (using same MF of 1.3) would have been 4 × 13/10 = 52/10.
But actual selling price remains 3.9 (same as case 1).
So, new MF
= (39/10) ÷ 4 = 39/40.
Profit/Loss % = (MF − 1) ×100
= ((39/40) − 1) ×100
= (-1/40)*100
= −2.5%
👉 Hence, Loss = 2.5%.
Long way to solve:
Let cost price of milk = ₹1 per liter, cost of water = ₹0.
Case 1: Ratio 3:2
Total = 5 liters → 3 liters milk + 2 liters water.
Cost price = 3×1 + 2×0 = 3.
Profit = 30%, so Selling price = 3 × 1.3 = 3.9.
Selling price per liter = 3.9/5 = 0.78.
Case 2: Ratio 4:1
Total = 5 liters → 4 liters milk + 1 liter water.
Cost price = 4×1 + 1×0 = 4.
Selling price (unchanged per liter) = 5 × 0.78 = 3.9.
Profit/Loss = (SP − CP)/CP ×100 = (3.9 − 4)/4 ×100 = −2.5%.
So, there is a 2.5% loss.
27 Sep 2025, 02:12
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okay since milk will the only cost & water is free-> if the ratio changes to 4:1-> i.e. more milk in the mixture the price would increase & the profit would decrease.
let's take the value as 10 for milk & calculate.
When it was 3:2-> 1liter would cost-> 6
& the profit would be 1.8. so sp=7.8
similarly for 4:1-> 1l would cost 8.
that's a 0.2 loss which equals to 2.5%. (0.2/8*100)
let's take the value as 10 for milk & calculate.
When it was 3:2-> 1liter would cost-> 6
& the profit would be 1.8. so sp=7.8
similarly for 4:1-> 1l would cost 8.
that's a 0.2 loss which equals to 2.5%. (0.2/8*100)
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