A petrol distributor adulterates petrol with kerosene and naphthalene....
The correct option is Option B.
Let the capacity of Container 1 be ‘a’ litres and that of Container 2 be ‘6’ litres.
∴ Petrol in container 1 = 0.3a
Petrol in container 2 = 0.456
Similarly, petrol in the new container = 0.35(a + b)
∴ 035 a + 0.45 b = 0.35(a + b)
∴ a = 2b
Since a = 8, b = 4.
Hence, the capacity of container 2 is 8 litres.
Hence, the answer is 4
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A petrol distributor adulterates petrol with kerosene and naphthalene....
To solve this problem, we can use the concept of weighted averages. Let's break down the problem into smaller steps:
Step 1: Calculate the ratio of petrol, kerosene, and naphthalene in container 1:
- Container 1 contains 30% petrol, which means it contains 70% of the adulterants (kerosene and naphthalene).
- Let's assume the capacity of container 1 is 8 liters. So, container 1 contains 8 * 30/100 = 2.4 liters of petrol.
- Therefore, container 1 contains 8 - 2.4 = 5.6 liters of the adulterants.
Step 2: Calculate the ratio of petrol, kerosene, and naphthalene in container 2:
- Container 2 contains 45% petrol, which means it contains 55% of the adulterants (kerosene and naphthalene).
- Let's assume the capacity of container 2 is x liters. So, container 2 contains x * 45/100 = 0.45x liters of petrol.
- Therefore, container 2 contains x - 0.45x = 0.55x liters of the adulterants.
Step 3: Calculate the ratio of petrol, kerosene, and naphthalene in the larger container:
- Let's assume the capacity of the larger container is C liters.
- The kerosene in the larger container is 25% of C, which means it is 0.25C liters.
- The naphthalene in the larger container is 40% of C, which means it is 0.4C liters.
- Therefore, the petrol in the larger container is C - 0.25C - 0.4C = 0.35C liters.
Step 4: Set up the equation:
- Since we know the amount of petrol in each container, we can set up the equation as follows:
2.4 + 0.45x = 0.35C
Step 5: Solve for x:
- Substitute the value of C from the equation above into the equation for container 2:
2.4 + 0.45x = 0.35(8 + x)
- Simplify the equation:
2.4 + 0.45x = 2.8 + 0.35x
- Subtract 0.35x from both sides:
0.1x = 0.4
- Divide both sides by 0.1:
x = 4
Therefore, the capacity of container 2 is 4 liters, which is option B.
Note: It is important to understand the concept of weighted averages to solve this problem. The weighted average takes into account both the quantity and quality (percentage) of the components in each container to determine the overall ratio in the larger container.
A petrol distributor adulterates petrol with kerosene and naphthalene....
To solve this problem, let's assume the capacity of container 2 is 'x' litres.
Given:
Container 1 capacity = 8 litres
Container 1 petrol percentage = 30%
Container 2 petrol percentage = 45%
Kerosene percentage in the mixture = 25%
Naphthalene percentage in the mixture = 40%
Let's calculate the quantities of petrol, kerosene, and naphthalene in each container.
Container 1:
Petrol quantity in container 1 = (30/100) * 8 = 2.4 litres
Kerosene quantity in container 1 = 8 - 2.4 = 5.6 litres
Naphthalene quantity in container 1 = 0 (as it is not mentioned)
Container 2:
Petrol quantity in container 2 = (45/100) * x = 0.45x litres
Kerosene quantity in container 2 = 0 (as it is not mentioned)
Naphthalene quantity in container 2 = 0 (as it is not mentioned)
Now, let's calculate the total quantities of petrol, kerosene, and naphthalene in the mixture.
Total petrol quantity = Petrol quantity in container 1 + Petrol quantity in container 2 = 2.4 + 0.45x
Total kerosene quantity = Kerosene quantity in container 1 + Kerosene quantity in container 2 = 5.6 + 0
Total naphthalene quantity = Naphthalene quantity in container 1 + Naphthalene quantity in container 2 = 0 + 0
Given:
Kerosene percentage in the mixture = 25%
Naphthalene percentage in the mixture = 40%
Using these percentages, we can equate the quantities of kerosene and naphthalene in the mixture to their respective percentages of the total mixture.
Total kerosene quantity = (25/100) * (8 + x)
Total naphthalene quantity = (40/100) * (8 + x)
Now, let's equate the quantities of kerosene and naphthalene in the mixture.
(25/100) * (8 + x) = 5.6
(40/100) * (8 + x) = 0
Simplifying these equations, we get:
2 + 0.1x = 5.6
3.2 + 0.16x = 0
Solving these equations, we find:
x = 40/0.1 = 4
Therefore, the capacity of container 2 is 4 litres.
Hence, option B is the correct answer.
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