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A milkman sells milk after adding some water in it. Quantity 30 Jul 2021, 09:58
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

59% (02:45) correct 41% (02:53) wrong based on 335 sessions

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A milkman sells milk after adding some water in it. Quantity of water is 74% of quantity of milk in mixture. If milkman sells the mixture (milk+ water) at 16.66% discount (16.66% less price than CP of pure milk) find profit % of milkman?


a)37%
b)40%
c)45%
d)50%
e)57.5%

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C
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A milkman sells milk after adding some water in it. Quantity 30 Jul 2021, 10:09
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Explanation:

Let milk is 100 litre
Water added 74% = 100x0.74 = 74 litre
Total mixture = 174 litre
Let rate per litre is 1$
Discounted price for 174 litre of milk mixture = 174x(1-0.166)
= 145

Profit % = 45%

IMO-C

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A milkman sells milk after adding some water in it. Quantity 31 Jul 2021, 00:40
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Solution:

Since the answer options are in percentage, we can take numbers and work to get our asnwer.

Let us assume the milkman bought \(100\) liters of milk in \($100\) (His Cost Price)

Now we know that the quantity of water is \(74\)% of the quantity of milk in the mixture. This means that in this \(100\) liters of milk, he added \(\frac{74}{100}\times 100=74\) liters of water.

So the total mixture \(= 100+74=174\) liters

Now ideally he should sell this \(174\) liters mixture in \($174\). But he gives a \(16.66\)% discount.

So, the Selling Price (SP) \(= 174-(\frac{16.66}{100}\times 174)=174-(\frac{1}{6}\times 174)=174-29=145\)

Thus, the milkman bought milk for \($100\) and sold it for \($145\). Hence profit \(= 145-100=45\).

Profit percent \(= \frac{45}{100}\times 100=45\)%

Hence the right answer is Option C.
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A milkman sells milk after adding some water in it. Quantity 31 Jul 2021, 06:36
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Solution:

Since the answer options are in percentage, we can take numbers and work to get our asnwer.

Let us assume the milkman bought \(100\) liters of milk in \($100\) (His Cost Price)

Now we know that the quantity of water is \(74\)% of the quantity of milk in the mixture. This means that in this \(100\) liters of milk, he added \(\frac{74}{100}\times 100=74\) liters of water.

So the total mixture \(= 100+74=174\) liters

Now ideally he should sell this \(174\) liters mixture in \($174\). But he gives a \(16.66\)% discount.

So, the Selling Price (SP) \(= 174-(\frac{16.66}{100}\times 174)=174-(\frac{1}{6}\times 174)=174-29=145\)

Thus, the milkman bought milk for \($100\) and sold it for \($145\). Hence profit \(= 145-100=45\).

Profit percent \(= \frac{45}{100}\times 100=45\)%

Hence the right answer is Option C.
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Thank You, but the question says (16.66% less price than CP of pure milk)...then shouldn’t i discount on 100 other than 174 ?

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A milkman sells milk after adding some water in it. Quantity 31 Jul 2021, 10:00
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Mugdho
A milkman sells milk after adding some water in it. Quantity of water is 74% of quantity of milk in mixture. If milkman sells the mixture (milk+ water) at 16.66% discount (16.66% less price than CP of pure milk) find profit % of milkman?


a)37%
b)40%
c)45%
d)50%
e)57.5%

Posted from my mobile device
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Knowing 16.66% is better written as 50/3 or 100/6 will ease calculation.
\(\frac{100}{6}\)% discount means 1/6 less or 5/6 of original price.
Let us take price be 6 of pure milk, so that price of mix becomes 5.

Let milk be 100 L, so water is 74 L.

So CP is 100*6 or 600, while SP is 174*5 or 870
Profit = 870-600=270.
Profit % = \(\frac{270}{600}*100=45\)%

Logical
Money gained on 74 L at the rate of 5perL is restricted by loss on 100L at rate of 1 per liter.
Profit =74*5-100*1=270.
As a %=100*270/600=45%

C
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