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.
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.
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.
Anamika said:
1 decade ago
Where it is given that C.P. is rupee one?
Would you please explain.
Would you please explain.
Ted said:
1 decade ago
You are assuming CP. AS rupee one for easy calculation. Thats all!
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%.
DHANANJAY SINGH said:
1 decade ago
CP = 100 RS FOR 100 LITERS.
BUT ACTUALLY IT HAS 125 LITERS (25 LITERS WATER).
SO CP OF 100 LITERS BE RS 100.
CP OF 125 LITERS BE RS 125.
PROFIT = 125-100/125*100=20%.
BUT ACTUALLY IT HAS 125 LITERS (25 LITERS WATER).
SO CP OF 100 LITERS BE RS 100.
CP OF 125 LITERS BE RS 125.
PROFIT = 125-100/125*100=20%.
Thamaraikannan said:
1 decade ago
100:gain%.
So 100:25 (this meant for 100% we get 25% profit).
Now 4:1 (for every 4l milk 1l water to be added).
Ratio wise 1l. Then percentage wise (1/5)*100 = 20%.
So 100:25 (this meant for 100% we get 25% profit).
Now 4:1 (for every 4l milk 1l water to be added).
Ratio wise 1l. Then percentage wise (1/5)*100 = 20%.
Vishnu said:
10 years ago
For example he sells 1 liter of mixture =0.8 liter milk 0.2 liter water.
I liter is equal to 125% (100% cp + profit 25 %).
Thus cost price achieved by selling 0.8 liters of milk.
Profit by 0.2 liters which is 25 %.
125% = 0.2/0.8*100.
I liter is equal to 125% (100% cp + profit 25 %).
Thus cost price achieved by selling 0.8 liters of milk.
Profit by 0.2 liters which is 25 %.
125% = 0.2/0.8*100.
Sreyansh joshi said:
10 years ago
I am still not getting this according to alligation rule the cheaper material should be used and in this solution they have considered milk as the cheaper material.
Parvy Govil said:
10 years ago
Let the original amount of milk be 100 and the cost price is Rs. 1/L.
CP1 = 100.
SP1 = 100.
Case 2 : When the milk is milk with x liters of water milk remaining is 100-x.
But now the SP is 100.
Given gain% = 25.
Thus, (100-(100-x))/100-x *100 = 25.
x = 20 liters.
Therefore, 20 L of water is added to 80 liters of milk so that the quantities remain the same however the selling price now makes all the difference!
CP1 = 100.
SP1 = 100.
Case 2 : When the milk is milk with x liters of water milk remaining is 100-x.
But now the SP is 100.
Given gain% = 25.
Thus, (100-(100-x))/100-x *100 = 25.
x = 20 liters.
Therefore, 20 L of water is added to 80 liters of milk so that the quantities remain the same however the selling price now makes all the difference!
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