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Bunuel
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Nidzo
Lets say that the person buys 1l of juice concentrate for $100.

On day 1 he mixes 1l of juice concentrate and 3l of water, to get 4l in total. He then sells all of it for $160, making 60% profit. Each litre sells for \(\frac{160}{4} = $40\).

On day 2 he mixes 1l of juice concentrate with 4l of water this time, and has 5l mix in total. He sells each litre for same price as the previous day, meaning he makes 40*5 = $200.

Percentage of profit formula is: \(\frac{revenue - cost }{ cost}* 100\)

\(\frac{200 - 100}{ 100}* 100\)

\(= 100%\)

Answer A

yes, you are right. My error.
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On the first day, a person mixed juice concentrate with freely available water in the ratio 1:3. On the second day, he mixed the same in the ratio 1:4 and sold the drink at the same price as the first day. if on day 1, the profit on investment on the juice concentrate was 60% what was the value on day 2?

A. 100%
B. 75%
C. 80%
D. 60%
E. 48%


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Lets assume cost price of Juice = Rs: 100/-
Water -- Free of Cost (Rs: 0 /-)
100 0
\ /
X(Cost Price)
/ \
1 : 3

(100+(3*0)) / (1+3)= Rs:25 = COST PRICE of X

100 0
\ /
Y (New CP)
/ \
1 : 4
therefore , the new CP will be 20.

But SP is 60% more of CP = 25*160/100 = 40

new profit = (40 -20) /20 *100
= 100%
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Profit is B) minus A)

Method 1:


First day:
A) you pay 1 to get 4 liters of a certain quantity x --> 1x/4
B) selling price is: selling price - cost price = profit --> selling price = cost price + profit --> selling price = 1x/4 * 160/100 = 2x/5

Second day:
A) you now pay 1 to get 5 liters of a ceratin quantity x --> 1x/5
B) price is still 2x/5

2x/5 - 1x/5 = 1x/5 which is 100% of cost

Method 2:

First day:
A) cost is 100 for concentration equal to 1/4 (you get 4 liters of mixture but pay only for 1 liter which is the juice)
B) selling price is: selling price - cost price = profit --> selling price = cost price + profit --> selling price = 100 * 160/100 = 160 for 1/4 of concentration

Second day:
A) you now have 1/5 of concentration, so you buy 5 liters and still pay for 1 only --> 100*4/5 = 80
B) price is still 160 for 1/5 of concentration

160 - 80 = 80 which is 100% of cost
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Simple and best explanation. Thank you
Bunuel
Bunuel
On the first day, a person mixed juice concentrate with freely available water in the ratio 1:3. On the second day, he mixed the same in the ratio 1:4 and sold the drink at the same price as the first day. if on day 1, the profit on investment on the juice concentrate was 60% what was the value on day 2?

A. 100%
B. 75%
C. 80%
D. 60%
E. 48%


EXPERT'S GLOBAL OFFICIAL VIDEO EXPLANATION



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On the first day, a person mixed juice concentrate with freely available water in the ratio 1:3. On the second day, he mixed the same in the ratio 1:4 and sold the drink at the same price as the first day.

If on day 1, the profit on investment on the juice concentrate was 60% what was the value on day 2?

Let the cost price for 1 unit of juice concentrate = c

First day: -
Cost price of 1 unit of juice concentrate (1 unit) = c
Cost price of 3 units of water = 0
Cost price of 4 units of mixture = c
Profit = 60%
Selling price of 4 units of mixture = 1.6c
Selling price of 1 unit of mixture = .4c


Second day: -
Cost price of 1 unit of juice concentrate (1 unit) = c
Cost price of 4 units of water = 0
Cost price of 5 units of mixture = c
Selling price of 5 units of mixture = .4c*5 = 2c
Profit = (2c-c)/c*100% = 100%

IMO A
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Is this correct ?

Assuming price = 100 of 1 liter orange

160% * 100 (1) + 3(0) = x (4)
so x = 40

now,
40* 4 = 100 (4)*1/5 + (0) * 4 * (4/5) + y
y = 80

s0 100% profit on investment
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rak08
Is this correct ?

Assuming price = 100 of 1 liter orange

160% * 100 (1) + 3(0) = x (4)
so x = 40

now,
40* 4 = 100 (4)*1/5 + (0) * 4 * (4/5) + y
y = 80

s0 100% profit on investment
Your answer of 100% profit is correct! However, let me show you a cleaner approach that will make these mixture profit problems more systematic and easier to follow.

Step 1: Set up the problem with clear variables

Let's say the cost of juice concentrate = \(C\) per unit Water is free (cost = 0)

Step 2: Analyze Day 1
  • Mixture ratio = 1 (concentrate)
  • Total cost = \(C \times 1 + 0 \times 3 = C\)
  • Total volume = 1 + 3 = 4 units
  • Profit = 60%, so selling price = \(1.6C\)
  • Price per unit of drink = \(\frac{1.6C}{4} = 0.4C\)

Step 3: Analyze Day 2
  • Mixture ratio = 1 (concentrate)
  • Total cost = \(C \times 1 + 0 \times 4 = C\)
  • Total volume = 1 + 4 = 5 units
  • Same price per unit as Day 1 = \(0.4C\)
  • Total selling price = \(5 \times 0.4C = 2C\)
  • Profit = \(2C - C = C\)
  • Profit percentage = \(\frac{C}{C} \times 100% = 100%\)

Answer: A. 100%

What went right in your solution:

✓ You correctly identified that the selling price per unit stays constant
✓ You arrived at the correct answer of 100%

Key Insight for Similar Problems: When the price per unit of final product remains constant but you're diluting more (adding more free water), your profit margin increases because you're selling more volume at the same unit price while your cost base remains the same.

Quick Check Method:
Day 1: Spend \(C\), get \(1.6C\) from 4 units
Day 2: Spend \(C\), get \(2C\) from 5 units (since \(5 \times \frac{1.6C}{4} = 2C\))

Profit doubles from \(0.6C\) to \(C\), giving 100% profit!
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