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 3 of 7.
Bathi said:
7 years ago
x liters of milk ------- re.1.
(x+y) liters of mixture -------- re.1.25 where y is for water.
(x+y)/x = 1.25/1.
y/x = 1/4.
y = 1/5 * 100 = 20%.
(x+y) liters of mixture -------- re.1.25 where y is for water.
(x+y)/x = 1.25/1.
y/x = 1/4.
y = 1/5 * 100 = 20%.
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.
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?
Palash said:
6 years ago
Let 1 lt milk price is rs1 and x lt water mixed.
Then CP of (1-x)lt milk is rs(1-x) and SP is rs 1.
So, x/ (1-x) = 25% implies X= 20%.
Then CP of (1-x)lt milk is rs(1-x) and SP is rs 1.
So, x/ (1-x) = 25% implies X= 20%.
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.
Vihan said:
6 years ago
If the price of milk is 15% more than that of water than the price of water is how much percent less than that of milk? Can anyone solve this?
Siva said:
8 years ago
Simple.
Gain=25.
Total milk =125.
Then 25/125=20%.
Gain=25.
Total milk =125.
Then 25/125=20%.
Karthi said:
6 years ago
Excellent @Amir.
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.
Shikha said:
1 decade ago
CP of 1 liter of milk = Re1.
SP of 1 liter of milk = Re1(given).
Gain = 25%(for the mixture [M+W]).
CP of milk(mixture) = (SP*(100/100+gain)) = 1*100/125 = 4/5.
Hence by allegation rule,
Milk(old) water
CP = Re1(dearer) CP = 0(for milk)(cheaper)
Mixture
CP = 4/5(mean).
Cheaper:Dearer = (1-4/5)/(4/5-0) = 1:4.
Water : milk = 1:4.
Water% = 1/5*100 = 20%.
SP of 1 liter of milk = Re1(given).
Gain = 25%(for the mixture [M+W]).
CP of milk(mixture) = (SP*(100/100+gain)) = 1*100/125 = 4/5.
Hence by allegation rule,
Milk(old) water
CP = Re1(dearer) CP = 0(for milk)(cheaper)
Mixture
CP = 4/5(mean).
Cheaper:Dearer = (1-4/5)/(4/5-0) = 1:4.
Water : milk = 1:4.
Water% = 1/5*100 = 20%.
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