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Arithmetic: Alligation and Mixture Test-4

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Question 1
In a mixture of milk and water the proportion of water by weight was 75%. If in the 60 gms mixture 15 gms water was added, what would be the percentage of water in the new mixture?
A
75%
B
88%
C
90%
D
None
Question 1 Explanation: 
Water in 60 gms=75/100×60=45
Milk in 60 gms=60-45=15
Water in new mixture=45+15=60
So percentage of water=60/75×100=80%
Question 2
In 1 kg mixture of sand and iron, 20% is iron. How much sand should be added, so that the proportion of iron becomes 5%?
A
3 kg
B
4 kg
C
5 kg
D
6 kg
Question 2 Explanation: 
Iron present in 1 kg mixture= 200gms
Let the quantity of sand added be x kgs
Then We want, $ \frac{0.2}{1+x}\times 100=5$ x=3
Question 3
A grocer purchased 2 kg. of rice at the rate of Rs.15 per kg and 3 kg of rice at the rate of Rs.13 per kg. At what price per kg should he sell the mixture to earn$ \displaystyle 33\frac{1}{3}%$ profit on the cost price?
A
Rs.28.00
B
Rs.20.00
C
Rs.18.40
D
Rs.17.40
Question 3 Explanation: 
Total cost of the bought rice=15×2+13×3=69
Total amount he should earn for a profit of $ \frac{100}{3}%=\frac{4}{3}\times 69=92$
So the SP per kg is 92/4=18.40
Question 4
To x litres of an x% solution of acid, y litres of water is added to get (x –10)% solution of acid. If x > 20, then value of y is
A
$ \displaystyle \frac{{{x}^{2}}}{100}$
B
$ \displaystyle \frac{10x}{x-10}$
C
$ \displaystyle \frac{10x}{x+10}$
D
$ \displaystyle \frac{10{{x}^{2}}}{x-10}$
Question 4 Explanation: 
$ \begin{array}{l}Acid\text{ }present\text{ }in\text{ }x\text{ }litres=\frac{{{x}^{2}}}{100}\\After\text{ }adding\text{ }y\text{ }litres\text{ }of\text{ }water\\Acid\text{ }percentage=\frac{\frac{{{x}^{2}}}{100}}{x+y}\times 100=x-10\\{{x}^{2}}={{x}^{2}}-10x+xy-10y\\Y=\frac{10x}{x-10}\\Y=\frac{10x}{x-10}\end{array}$
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