All questions of Mixtures and Alligations for Banking Exams Exam

Let the time taken be x
.•.6x+18x= 75
.•. 24x = 75
.•.x= 3.125
.•. distance travelled by scooter=3.125×18
=56.25

50 liters of milk from the jar. How much milk is left in the jar?

There are two ways to approach this problem:

Method 1: Subtraction
To find the amount of milk left in the jar, we can simply subtract the amount stolen from the total amount:
200 - 50 = 150 liters
Therefore, there are 150 liters of milk left in the jar.

Method 2: Proportion
We can also use proportions to find the amount of milk left in the jar. Let x be the amount of milk left in the jar. Then we can set up a proportion:
200/100 = x/50
This proportion states that the initial amount of milk (200 liters) is 100% of the jar, and the stolen amount (50 liters) is 50% of the jar. We can solve for x by cross-multiplying:
200 * 50 = 100 * x
10,000 = 100x
x = 100 liters
Therefore, there are 100 liters of milk left in the jar.

Initial Amount of Milk in the Container: 80 liters

To solve this problem, let's break it down step by step:

Step 1: Understanding the Problem
- A thief steals 10 liters of milk from a container.
- The thief replaces the stolen milk with the same quantity of water.
- This process is repeated three times.
- After these replacements, the ratio of milk to water in the container is 343:169.

Step 2: Setting up the Equation
- Let's assume the initial amount of milk in the container is x liters.
- After the first replacement, the amount of milk in the container becomes x - 10 liters.
- After the second replacement, the amount of milk in the container becomes (x - 10) - 10 liters, which can be simplified to x - 20 liters.
- After the third replacement, the amount of milk in the container becomes (x - 20) - 10 liters, which can be simplified to x - 30 liters.

Step 3: Calculating the Amount of Water
- In the final state, the amount of water in the container is the same as the amount of milk.
- So, after the three replacements, the amount of water in the container is also x - 30 liters.

Step 4: Setting up the Ratio Equation
- The ratio of milk to water in the container after the three replacements is given as 343:169.
- This can be written as (x - 30): (x - 30) = 343:169.

Step 5: Solving the Equation
- Cross-multiplying, we get (x - 30) * 169 = (x - 30) * 343.
- Expanding the equation, we have 169x - 5070 = 343x - 10290.
- Simplifying the equation, we get 174x = 5220.
- Dividing both sides by 174, we find x = 30.

Step 6: Calculating the Initial Amount of Milk
- The initial amount of milk in the container is given as x liters, which is equal to 30 liters.

Therefore, the correct answer is option A) 30 liters.

Solution:

Let's assume that the cost price of the part sold at 20% profit is x.

According to the question,

The part sold at 20% profit = x

The part sold at 15% loss = 560 - x

Now, as per the question, the total profit earned is 10%.

So, the selling price of the part sold at 20% profit = x + 0.2x = 1.2x

The selling price of the part sold at 15% loss = (560 - x) - 0.15(560 - x) = 476.5 - 0.85x

The total selling price = (1.2x) + (476.5 - 0.85x) = 1.15x + 476.5

As per the question, the total profit earned is 10%.

So, (Total Selling Price) - (Total Cost Price) = 10% of Total Cost Price

Or, (1.15x + 476.5) - 560 = 0.1(560)

Or, 1.15x - 83.5 = 56

Or, 1.15x = 139.5

Or, x = 121

Therefore, the cost price of the part sold at 20% profit is Rs 121.

Hence, the correct answer is option C (Rs 400).

Given:
Cost of variety-1 rice = Rs. 42/kg
Cost of variety-2 rice = Rs. 24/kg
Selling price = Rs. 40/kg
Profit required = 25%

Let the quantity of variety-1 rice to be mixed be x kgs.

Total cost price of the mixture = cost of variety-1 rice + cost of variety-2 rice
= 42x + 24(25) = 42x + 600

Total selling price of the mixture = selling price * total quantity
= 40(x+25)

Profit = selling price - cost price
= 40(x+25) - (42x + 600)
= 40x + 1000 - 42x - 600
= -2x + 400

As profit is required to be 25% of cost price, we have:

Profit/Cost Price * 100 = 25
-2x + 400 / (42x + 600) * 100 = 25
-2x + 400 = 10.5x + 150
12.5x = 250
x = 20

Therefore, the shopkeeper should mix 20 kgs of variety-1 rice with 25 kgs of variety-2 rice to make a profit of 25% on selling the mixture at Rs.40/kg.

Given:
Fresh fruit contains 75% water
Dry fruit contains 20% water

To find:
How much dry fruit can be obtained from 150 kg of fresh fruit

Solution:
Let the weight of dry fruit obtained be x kg.
From the given information, we can say that:

Weight of water in fresh fruit = 75% of weight of fresh fruit
Weight of dry matter in fresh fruit = 100% - 75% = 25% of weight of fresh fruit

Weight of water in dry fruit = 20% of weight of dry fruit
Weight of dry matter in dry fruit = 100% - 20% = 80% of weight of dry fruit

Now, we can form the following equation:

Weight of dry matter in fresh fruit = Weight of dry matter in dry fruit
0.25 * 150 = 0.8x
37.5 = 0.8x
x = 46.875

Therefore, the weight of dry fruit obtained from 150 kg of fresh fruit is 46.875 kg, which is closest to option C (47).

The ratio of final mixture is m:w = 2:3. So W= 3/5. 

2:5

Given:
- Quantity of milk = 10 liters
- Cost of milk = Rs 50 per liter
- Selling price of mixture = Rs 44 per kg

To find:
- Quantity of water to be added

Solution:
1. Let's assume that x liters of water is added to 10 liters of milk to make the mixture.
2. So, the total quantity of the mixture will be (10 + x) liters.
3. Let's find the cost price of the mixture.
- The cost of 10 liters of milk = 10 x 50 = Rs 500
- The cost of x liters of water = 0 (as water is free)
- So, the cost price of (10 + x) liters of mixture = Rs 500
4. Let's find the weight of (10 + x) liters of mixture in kg.
- 1 liter of milk weighs 1 kg (approx.)
- 1 liter of water weighs 1 kg (exactly)
- So, (10 + x) liters of mixture will weigh (10 + x) kg.
5. We know that the selling price of the mixture is Rs 44 per kg.
- So, the selling price of (10 + x) kg of mixture = (10 + x) x 44 = 440 + 44x Rs
6. We can equate the cost price and selling price of the mixture to find the value of x.
- Cost price of mixture = Selling price of mixture
- Rs 500 = 440 + 44x
- 44x = 60
- x = 60/44 = 1 4/11 liters

Therefore, the quantity of water to be added is 1 4/11 liters, which is option D.

To solve this problem, we need to set up a system of equations based on the given information. Let's denote the quantity of milk in the first mixture as 3x and the quantity of water as 2x. Similarly, for the second mixture, let's denote the quantity of milk as 4y and the quantity of water as 5y.

We are given that 3 litres of the first mixture are mixed with some quantity of the second mixture. Let's assume that the quantity of the second mixture added is z litres. So, we have:

Milk in the resulting mixture = 3x + 4y
Water in the resulting mixture = 2x + 5y

Since the resulting mixture must contain equal quantities of milk and water, we can set up the following equation:

3x + 4y = 2x + 5y

Simplifying this equation, we get:

x = y

This means that the ratio of milk to water in both mixtures is the same, and we can equate the ratios:

3x/2x = 4y/5y

Cross-multiplying, we get:

15x = 8x

Simplifying this equation, we get:

7x = 0

Since x cannot be equal to 0, this equation has no solution. Therefore, it is not possible to mix the two given mixtures in such a way that the resulting mixture contains equal quantities of milk and water.

Hence, the correct answer is option (e) None of these.

Given:
- The container is filled with a liquid mixture consisting of 4 parts water and 6 parts milk.
- We need to draw off some of the mixture and fill it with water in such a way that the resulting mixture contains equal parts of water and milk.

Approach:
Let's assume that the container initially contains 10 parts of the mixture (4 parts water + 6 parts milk).
We need to find out how much of the mixture needs to be drawn off and replaced with water to obtain a 1:1 ratio of water and milk in the final mixture.

Solution:
Step 1: Assume the total quantity of the mixture in the container as 10 units (to simplify calculations).

Step 2: Calculate the initial quantities of water and milk in the container:
- Water in the mixture = (4/10) * 10 = 4 units
- Milk in the mixture = (6/10) * 10 = 6 units

Step 3: Let's assume x units of the mixture is drawn off and replaced with water.

Step 4: After drawing off x units of the mixture, the remaining quantities of water and milk in the container will be:
- Remaining water = 4 - (4/10) * x
- Remaining milk = 6 - (6/10) * x

Step 5: Now, we need to calculate how much water needs to be added to make the final mixture contain equal parts of water and milk. Let's assume y units of water are added.

Step 6: After adding y units of water, the final quantities of water and milk in the container will be:
- Final water = 4 - (4/10) * x + y
- Final milk = 6 - (6/10) * x

Step 7: According to the problem, the final mixture should contain equal parts of water and milk. So, the final water and milk quantities should be equal:
Final water = Final milk

Step 8: Set up the equation:
4 - (4/10) * x + y = 6 - (6/10) * x

Step 9: Simplify the equation:
2/10 * x = y - 2

Step 10: We know that the final mixture should contain equal parts of water and milk. So, the quantity of water added should be equal to the quantity of milk removed from the mixture.
Therefore, y = (4/10) * x

Step 11: Substitute the value of y in the equation from Step 9:
2/10 * x = (4/10) * x - 2

Step 12: Simplify the equation:
2x = 4x - 20
2x - 4x = -20
-2x = -20
x = -20 / -2
x = 10

Step 13: The quantity of the mixture that needs to be drawn off and replaced with water is 10 units.

Step 14: The total quantity of the mixture in the container is 10 units (as assumed in Step 1). And we need to draw off 10 units of the mixture.

Step 15: The fraction of the mixture that needs to be drawn off is:
Fraction = (10 units) / (10 units) = 1/