- 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-3
Congratulations - you have completed Arithmetic: Alligation and Mixture Test-3.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 |
A container has 30 litres of water. If 3 litres of water is replaced by 3 litres of spirit and this operation is repeated twice, what will be the quantity of water in the new mixture?
24 litres | |
23 litres | |
24.3 litres | |
23.3 litres |
Question 1 Explanation:
After 1st operation: Water=27litres, Sprit=3litres
Now water in 3 litres= 27/30×3=2.7,
This amount would be replaced by sprit
=27-2.7=24.3
This amount would be replaced by sprit
=27-2.7=24.3
Question 2 |
The wheat sold by a grocer contained 10% low quality wheat. What quantity of good quality wheat should be added to 150 kgs of wheat so that the percentage of low quality wheat becomes 5%?
150 kgs | |
135 kgs | |
50 kgs | |
85 kgs |
Question 2 Explanation:
Bad quality wheat in 150 kgs=15kg
Let the good quantity wheat added be x kgs
Then We want,{15/(150+x)}×100=5
X=150
Let the good quantity wheat added be x kgs
Then We want,{15/(150+x)}×100=5
X=150
Question 3 |
A man purchased 35 kg of rice at the rate of Rs.9.50 per kg and 30 kg at the rate of Rs.10.50 per kg. He missed the two. Approximately, at what price (in Rupees) per kg should he sell the mixture to make 35 per cent profit in the transaction?
12 | |
12.50 | |
13 | |
13.50 |
Question 3 Explanation:
Total cost of the bought rice=35×9.50+30×10.5=647.50
Total amount he should earn for a profit of
35%=1.35×647.50=874.125
So the SP per kg is 874.125/65=13.45≅13.5
Total amount he should earn for a profit of
35%=1.35×647.50=874.125
So the SP per kg is 874.125/65=13.45≅13.5
Question 4 |
When one litre of water is added to a mixture of acid and water, the new mixture contains 20% acid. When one litre of acid is added to the new mixture, then the resulting mixture contains $ \displaystyle 33\frac{1}{3}%$ acid. The percentage of acid in the original mixture was
20% | |
22% | |
24% | |
23% |
Question 4 Explanation:
$ \displaystyle \begin{array}{l}Let\text{ }the\text{ }quantity\text{ }of\text{ }water\text{ }in\text{ }\\mixture\text{ }be\text{ }x\text{ }and\text{ }acid\text{ }be\text{ }y\\so,\frac{y}{x+y+1}=\frac{20}{100}\\4y=x+1...............a\\Also\,\,\frac{y+1}{x+y+1}=\frac{100}{300}\\2y+2=x...............b\\By\,\,a\,\,and\,\,b\,\,we\,\,get\\y=1.5\\x=5\\Acid\,in\,original\,mixture\,=\frac{1.5}{6.5}\times 100=23\end{array}$
Question 5 |
A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7: 2 and 7: 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be:
5: 7 | |
5: 9 | |
7: 5 | |
9: 5 |
Question 5 Explanation:
Let the amount added of both be 1 kg
Amount of gold in A=7/9
Amount of copper in A=2/9
Amount of gold in B=7/18
Amount of copper in B=11/18
SO ratio of gold and copper is
$ \frac{\frac{7}{9}+\frac{7}{18}}{\frac{2}{9}+\frac{11}{18}}=\frac{7}{5}$
Amount of gold in A=7/9
Amount of copper in A=2/9
Amount of gold in B=7/18
Amount of copper in B=11/18
SO ratio of gold and copper is
$ \frac{\frac{7}{9}+\frac{7}{18}}{\frac{2}{9}+\frac{11}{18}}=\frac{7}{5}$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
