Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 6)
6.
A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:
4%
6
%
%20%
25%
Answer: Option
Explanation:
Let C.P. of 1 litre milk be Re. 1
Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
| C.P. of 1 litre mixture = Re. |  |
100 | x 1 |  |
= | 4 |
| 125 | 5 |
By the rule of alligation, we have:
| C.P. of 1 litre of milk C.P. of 1 litre of water | |||||
| Re. 1 | Mean Price
| 0 | |||
|
|
||||
 Ratio of milk to water = |
4 | : | 1 | = 4 : 1. |
| 5 | 5 |
| Hence, percentage of water in the mixture = | |
1 | x 100 | % |
= 20%. |
| 5 |
Discussion:
65 comments Page 2 of 7.
Karan said:
7 years ago
If we apply alligation on % profit i.e. if he sell the milk at CP, he gains 0% and if he sell water at CP, he gains 100%. Thus by rule of alligation the ratio comes out to be 3:1.
& in that case % water in mixture becomes 25%. Now, by solving it by the method given and the method used by me, I m getting confused what is the perfect answer. Can anyone help?
& in that case % water in mixture becomes 25%. Now, by solving it by the method given and the method used by me, I m getting confused what is the perfect answer. Can anyone help?
Prabhat Ranjan said:
1 decade ago
I have very simple solution for the above problem, which is as follow:
Suppose we have 100 kg of milk @ Rs.1 per kg Cost price.
Total cost price = Rs.100.
Profit is given 25%, therefore SP of 100kg milk = Rs. 125.
But according to question, he sells at CP. therefore he has to mix 25 litres of water.
Therefore % of water in mixture = 25*100/125 = 20%.
Suppose we have 100 kg of milk @ Rs.1 per kg Cost price.
Total cost price = Rs.100.
Profit is given 25%, therefore SP of 100kg milk = Rs. 125.
But according to question, he sells at CP. therefore he has to mix 25 litres of water.
Therefore % of water in mixture = 25*100/125 = 20%.
Naux said:
8 years ago
Say the vendor buys L liters of milk and the cost of this is X.
Now, the vendor mixes some amount of water so the quantity is becoming say (L+Z) the cost of this is X(original cost)+25/100(X) that would be 5X/4.
Cost---->quantity
X---->L
5X/4--->?
So, ? = 5L/4,
So (L+Z) = 5L/4,
Z =L/4,
So, that would be 25% right but the answer is 20%.
Now, the vendor mixes some amount of water so the quantity is becoming say (L+Z) the cost of this is X(original cost)+25/100(X) that would be 5X/4.
Cost---->quantity
X---->L
5X/4--->?
So, ? = 5L/4,
So (L+Z) = 5L/4,
Z =L/4,
So, that would be 25% right but the answer is 20%.
Mukund said:
9 years ago
@Srinithi.
Profit is 25% its because he is not selling pure milk which Worth's 1Rs (i.e 100%) he is adding water to it .It does mean that milk is no more pure and also its cost reduced. But even after he selling it with the same price.
Hence he is getting profit.
Wkt if SP is given the formula is CP = SP *(100/(GAIN + 100)).
Profit is 25% its because he is not selling pure milk which Worth's 1Rs (i.e 100%) he is adding water to it .It does mean that milk is no more pure and also its cost reduced. But even after he selling it with the same price.
Hence he is getting profit.
Wkt if SP is given the formula is CP = SP *(100/(GAIN + 100)).
Thuder said:
7 years ago
Suppose 1 litre is 1rs.
Then 0.5 litres is 0.5 right.
x litres cost x rupee,
So gain = ((s.p-c.p)/c.p)*100;
So he actually sells it for 1 rupee as 1 litre of milk but he has less than 1 litre of milk, say x,
So x litres cost x rupee,
So 25=((1-x)/x)*100;,
x=4/5;,
So water is 1/5 which is 20%,
Hope it helps.
Then 0.5 litres is 0.5 right.
x litres cost x rupee,
So gain = ((s.p-c.p)/c.p)*100;
So he actually sells it for 1 rupee as 1 litre of milk but he has less than 1 litre of milk, say x,
So x litres cost x rupee,
So 25=((1-x)/x)*100;,
x=4/5;,
So water is 1/5 which is 20%,
Hope it helps.
Vishal said:
1 decade ago
Let the price of the milk is 100% and he is getting the profit of 25% so the selling price must be 125%.
Now according the formula:
[(x/x+100)*100]%
{if price is 25% more than original then the milk is [25/100+25*100]% less than}
So,
[25/125*100]% = 20%.
He is selling the milk mixed with 20% of water.
Now according the formula:
[(x/x+100)*100]%
{if price is 25% more than original then the milk is [25/100+25*100]% less than}
So,
[25/125*100]% = 20%.
He is selling the milk mixed with 20% of water.
Ramesh said:
7 years ago
CP of milk = X.
SP of Milk = X.
He did not get any profit so he mixed some water and sold it. Then he got a profit of 25%
SP of mixture=1.25X.
1.25X = X + 0.25X.
x = milk and 0.25X = water.
we need to calculate how much water is there in the mixture.
0.25X/1.25X = 0.2.
---->20% water.
SP of Milk = X.
He did not get any profit so he mixed some water and sold it. Then he got a profit of 25%
SP of mixture=1.25X.
1.25X = X + 0.25X.
x = milk and 0.25X = water.
we need to calculate how much water is there in the mixture.
0.25X/1.25X = 0.2.
---->20% water.
Vikatakavee said:
1 decade ago
Let us consider C.P of the milk = Rs. 100 (milk is 100%).
--> S.P also = 100.
If you are getting 25% profit S.P = 1.25 C.P.
--> 100 = 1.25 C.P.
--> C.P = 80.
C.P 80 means 80% of milk. So remaining 20% water.
--> S.P also = 100.
If you are getting 25% profit S.P = 1.25 C.P.
--> 100 = 1.25 C.P.
--> C.P = 80.
C.P 80 means 80% of milk. So remaining 20% water.
Target spotted said:
9 years ago
A dishonest milkman dilutes milk and water with water and then sells the diluted milk at a price that is 20% more than the price at which he purchased the milk. If he makes a profit of 50%in this manner. How many ml of water does he add to every liter of milk?
Sujit said:
9 years ago
S.P. = C.P. (given) ...... (i).
Gain = 25%.
Then,
S.P. = (125/100)C.P. = (5/4)C.P.
ie., S.P/C.P = 5/4.
ie., 4S.P = 5C.P.
For them to b equal 1should be removed from 5 C.P.
i.e., 1/5 is removed.
Hence percentage removed is 1/5*100 = 20%.
Gain = 25%.
Then,
S.P. = (125/100)C.P. = (5/4)C.P.
ie., S.P/C.P = 5/4.
ie., 4S.P = 5C.P.
For them to b equal 1should be removed from 5 C.P.
i.e., 1/5 is removed.
Hence percentage removed is 1/5*100 = 20%.
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