• This is an assessment test.
  • To draw maximum benefit, study the concepts for the topic concerned.
  • Kindly take the tests in this series with a pre-defined schedule.

Arithmetic: Alligation and Mixture Test-4

Congratulations - you have completed Arithmetic: Alligation and Mixture Test-4. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
Your answers are highlighted below.
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}$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect. Get Results
There are 4 questions to complete.
List
Return
Shaded items are complete.
1234End
Return

Want to explore more Arithmetic Tests?

Explore Our Arithmetic Tests

Get Posts Like This Sent to your Email
Updates for Free Live sessions and offers are sent on mail. Don't worry: we do not send too many emails..:)
Get Posts Like This Sent to your Email
Updates for Free Live sessions and offers are sent on mail. Don't worry: we do not send too many emails..:)
Join Our Newsletter
Get the latest updates from our side, including offers and free live updates, on email.
Join Our Newsletter
Leverage agile frameworks to provide a robust synopsis for high level overviews.
Join our Free TELEGRAM GROUP for exclusive content and updates
Join our Free TELEGRAM GROUP for exclusive content and updates

Pin It on Pinterest