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805+ (Hard)|   Mixture Problems|         
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findingmyself
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A milk vendor mixes water with milk and sells the mixture at the same 05 Jul 2025, 12:48
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Hi braddouglas10

The answer provided by Harshvardhan in this chat is the best answer and explains well. Share your specific doubts if any.

braddouglas10
Bunuel,

Since you're an incredible and competent teacher, can you please do your version of this one?
PReciSioN
Should be D, but actually both statements give different answers for X. A gives 22.72 while B gives 25. This should not happen with an official question. Right? Bunuel
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Pr4n
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A milk vendor mixes water with milk and sells the mixture at the same 08 Jul 2025, 07:25
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Everything aside, I don't understand how is this question labeled as easy in the DI Review 2024-2025. When I looked it and wrote the equations i was already like 90s in the questions.
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A milk vendor mixes water with milk and sells the mixture at the same 03 Aug 2025, 09:59
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CapnSal
A milk vendor mixes water with milk and sells the mixture at the same price per liter as if it were undiluted milk. The selling price per liter of the mixture is the vendor’s cost per liter of the milk plus a markup of x %. The water costs the vendor nothing. If the vendor gets a 50% profit on the sale of the mixture, what is the value of x ?

(1) If the vendor mixes half the intended quantity of water and sells every liter of the mixture at the cost price per liter of the undiluted milk, the vendor will get a 10% profit.

(2) The concentration of milk in the mixture after adding water is 5/6.­


ID: 700325

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Unable to add tags but this is from Official Online Question Bank for DI Review 2024-25­
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Points to Note :
Mixture(M) sold at same price as milk(m)
CP=CP of milk
SP=CP(1+x/100)
P=50%
SP-CP=0.5CP
SP=1.5CP
CP(1+x/100)M=1.5CP*m
m/M=1+x/100 ...... (i)
x = ?

Statement 1
So,
SP=1.1CP
M=2m
CP(1+x/100)2m=1.1CP*m
(1+x/100)2 = 1.1
We can find x from here.
SUFFICIENT.

Statement 2
m=5/6M
If we substitute this in i, we can get x.
SUFFICIENT.

Ans D
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A milk vendor mixes water with milk and sells the mixture at the same 20 Jun 2026, 16:12
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SP per liter of mix = CP per liter of milk + x% mkup (Notice per liter here) -----> (1)

Entire mixture (milk + water) is sold at the price of the milk (assuming water is also milk) - There is a 50% profit on the sale of Mixture

Let's assume we started with 100 liters of milk and CP per liter of milk is $1

Stmt 1: If the vendor mixes half the intended quantity of water and sells every liter of the mixture at the cost price per liter of the undiluted milk, the vendor will get a 10%
profit.

Let water be y liters
100 liters + y/2 water sold at the cost per liter of milk (no water) gives 10% profit. Since water has no cost, CP of mix = $100, SP of mix = $110 (10 liters of water added)
100 liters + y water sold at the cost per liter of milk (no water) gives 20% profit. Since water has no cost, CP of mix = $100, SP of mix = $120 (20 liters of water added)

Now actual SP of mix with markup is $150
SP of 120 liters of mix is $150
SP of 1 liter of mix = (150/120) = $1.25
Now go to equation 1:
SP per liter of mix = CP per liter of milk + x% mkup
$1.25 = $1 + $0.25
$0.25 markup on $1 means a 25% markup . So, x = 25% --> SUFFICIENT

Stmt 2: The concentration of milk in the mixture after adding water is 5/6.
Again assuming 100 liters of milk, w liters of water
100/(100 + w) = 5/6, gives w = 20 liters, same as what we got in statement 1. So, by same calculation as statement 1, x = 25%. --> SUFFICIENT

Hence, D is the answer. Each statement alone is sufficient.
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