A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?
Step 1 & 2: Understand Question and Draw Inference
Given:
To Find:
Step 3 : Analyze Statement 1 independent
⇒ x+y=350ml ...(Equation 2)
Step 4 : Analyze Statement 2 independent
Step 5: Analyze Both Statements Together (if needed)
Solution A contains equal amount of alcohol and water in it. It is heated till 50 percent of the water in solution A evaporates. Solution B, whose volume is equal to the reduced volume of water in Solution A, is then added to Solution A and the volume of alcohol in the resultant solution is equal to 12 liters. If solution B contains alcohol and water in the ratio 2:3, how many more liters of water should be added to the resultant solution to increase the concentration of water in the resultant solution to 50 percent?
Given:
Reduced volume of water in solution A = 0.5x liters, as 50% of the water evaporates from the original solution having x litres of water
To Find: Amount of water to be added to the resultant solution to increase the concentration of water to 50%?
Approach:
Working out:
2. Volume of Alcohol in the resultant solution = Volume of alcohol in solution A + volume of Alcohol in solution B
3. Volume of Water in the resultant solution = Volume of Water in solution A + volume of Water in solution B
4. Now the amount of water to be added is y liters
5. Hence 4 liters of water needs to be added to increase the volume of water to 50% of the solution
Answer : B
Solution A is made up of alcohol and water mixed in the ratio of 21:4 by volume; Solution B is made up of alcohol and water mixed in the ratio of 2:3 by volume. If Solution A and Solution B are mixed in the ratio of 5:6 by volume, what percent of the resultant mixture is alcohol?
Given:
To Find: % of alcohol in Resultant Solution
Approach:
We already know the ratio of And using our inferences in the Given section, we can easily find the values of C_{B} and C_{B} and thus, answer the question.
Working out:
Looking at the answer choices, we see that the correct answer is Option C
A 20 kilograms mixture of dry fruits consisting of x percent of almonds is formed by mixing a kilograms of mixture A and b kilograms of mixture B. If mixture A consists of y percent of almonds and mixture B consists of 20% of almonds, what is the weight of almonds in kilograms in mixture A? Assume that each mixture has a uniform composition, that is, samples taken from any part in a mixture will have the same composition as the overall mixture.
(1) The amount of almonds in mixture B is three-fifths of the total weight of mixture A
(2) If the weight of the final mixture is reduced by 25 percent, the almonds in the final mixture would reduce by 1.5 kilograms
Step 1 & 2: Understand Question and Draw Inference
To Find: We need to find weight of almonds in mixture A
Step 3 : Analyze Statement 1 independent
Insufficient to answer
Step 4 : Analyze Statement 2 independent
If the weight of the final mixture is reduced by 25 percent, the almonds in the
final mixture would reduce by 1.5 kilograms
Step 5: Analyze Both Statements Together (if needed)
a. Weight of mixture A = 5 kilograms
b. Weight of mixture B = 15 kilograms
2. Weight of almonds in final mixture = 6 kilograms
Now, Weight of almonds in final mixture = Weight of almonds in mixture A + Weight of almonds in mixture B
Answer : C
A salt-water solution that contains 8% salt by volume is kept in the sun till enough water evaporates from the solution such that the concentration of salt in the solution becomes 20%. If the volume of the remaining solution is 20 liters, what was the volume of the solution before it was kept in the sun?
Given:
To Find: X = ?
Approach:
Working out:
Looking at the answer choices, we see that the correct answer is option D
A container consists of x liters of milk. The milk in the container is adulterated by adding water equal to 50 percent of the milk in
the container. If a person takes out 6 liters of the milk solution from the container, how many liters of water does he take out?
Given:
To Find: The amount of water taken out from the milk solution?
Approach:
2. Using the same formula as above and putting the absolute value of total volume of milk solution (6 liters) and the calculated percentage, we will be ble to find the water taken out from the solution.
Working out:
1. Volume of milk = x liters
2. Volume of water = 0.5x liters
3. Total volume of milk solution = x + 0.5x = 1.5x liters
6. Hence, amount of water in 6 liters of milk solution is 2 liters
Answer : B
A solution contains ethanol, methanol and water in the ratio of 1:4:15 by volume. The composition of this solution is altered by adding ethanol, methanol and water such that the amount of ethanol per unit volume of methanol increases by 15 percentage points and the amount of water per unit volume of ethanol decreases by two-thirds. If the volume of the altered solution is 340 cubic units, which of the following is the volume, in cubic units, of methanol in the altered solution?
Given:
by 15 percent)
To Find : y = ?
Approach:
Working out:
Looking at the answer choices, we see that the correct answer is Option D
Packet A contains a mixture of salt and sugar and Packet B contains a mixture of salt and pepper. When the contents of the two packets are mixed in a bowl, the resultant mixture weighs 400 grams and contains 25% sugar. How many grams of salt are contained in Packet B?
(1) One-third of the mixture in Packet A is sugar
(2) 15% of the resultant mixture is pepper
Step 1 & 2: Understand Question and Draw Inference
Given:
Let packet A contain x grams of Salt and y grams of Sugar
Let packet B contain z grams of Salt and w grams of pepper
So, we can represent the 2 packets A and B visually as below:
To Find: the value of z
Step 3 : Analyze Statement 1 independent
Statement 1 says that One-third of the mixture in Packet A is sugar
By substituting (III) and (II) in (I), we get a linear equation with 2 unknowns: z and w.
Not sufficient to find a unique value of z.
Step 4 : Analyze Statement 2 independent
Statement 2 says that 15% of the resultant mixture is pepper
w = 15% of 400 = 60 grams . . . (IV)
Substituting (II) and (IV) in (I) gives us a linear equation with 2 unknowns: x and z.
Not sufficient to find a unique value of z.
Step 5: Analyze Both Statements Together (if needed)
Sufficient
Answer: C
A juice is made by mixing x liters of mango pulp with y liters of orange pulp and 14 liters of water where x and y are non-zero integers, mango pulp costs $6 per 10 liters and orange pulp costs $4 per 10 liters, is x >3?
(1) If water is available free of charge, the juice costs $1.6 per 10 liters
(2) If mango pulp had cost 20 cents more per liter, then the cost of the mango pulp used in the juice would have been more than $2 greater than the cost of the orange pulp used in the juice
Step 1 & 2: Understand Question and Draw Inference
Given:
To find: Is x > 3?
Step 3 : Analyze Statement 1 independent
Step 4 : Analyze Statement 2 independent
Statement 2 says that ‘If mango pulp had cost 20 cents more per liter, then the cost of the mango pulp used in the juice would have been more than $2 greater than the cost of the orange pulp used in the juice’
Sufficient.
Step 5: Analyze Both Statements Together (if needed)
Since we get a unique answer in each of Steps 3 and 4, this step is not required.
Answer: Option D
A tea company sells two brands of tea leaves, Brand A and Brand B. Both the brands of tea leaves consist of three types of tea leaves, types-I, II and III, but in different weights. The cost of a kilogram of a brand of tea leaves is given by the expression $(100p+80q+50r), where p, q and r are the weights of type-I, II and III tea leaves respectively in 1 kilogram of that brand of tea leaves. Which of the two brands of tea leaves costs more?
(1) The individual percentage weights of type-I and type-III tea leaves are greater in Brand A than in Brand B
(2) The percentage weight of type-I tea leaves in Brands A and B is 40 percent and 20 percent respectively
Step 1 & 2: Understand Question and Draw Inference
Given:
Step 3 : Analyze Statement 1 independent
The individual percentage weights of type-I and type-III tea leaves are greater
in Brand A than in Brand B.
Step 4 : Analyze Statement 2 independent
The percentage weight of type-I tea leaves in Brands A and B is 40 percent and
20 percent respectively
Information given about % weight of leaves of Type I
Note: Since, we are dealing with 1 kg of Tea of each brand
Step 5: Analyze Both Statements Together (if needed)
Hence the answer is E.
A solution contains water, milk and liquid chocolate in the ratio of 2:3:5 by volume. If y liters of water and milk each are added to this solution, the resultant solution would contain 25 percent of water by volume. If the volume of this resultant solution is 120 liters, what is the value of y in liters?
Given:
To Find: Value of y?
Approach :
a. Volume of water in the original solution =2x liters
b. Volume of water added to the original solution =y liters
c. So, Volume of water in the resultant solution = (2x+y) liters
d. The total volume of water in the new solution is also equal to 30 liters.
2. We will then find the total volume of the resultant solution to get another equation in x and y.
3. Using the two equations in x nd y, we will be able to find the value of y as asked in the question.
Working out:
4. So water to be added to the original solution is 10 liters.
5. So answer option B is correct.
A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
Let the quantity of the initial liquid in the vessel = 8 litre and
quantity of water in the initial liquid = 3 litre and
quantity of syrup in the initial liquid = 5 litre
Let x litre of the mixture is drawn off and replaced with waterQuantity of water in the new mixture
Quantity of syrup in the new mixture
Given that in the new mixture, quantity of water = quantity of syrup
Initially we assumed that the quantity of the initial liquid in the vessel = 8 litre for the ease of calculations. For that 8/5 litre of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup
Now, if the initial liquid in the vessel = 1 litre, quantity of the mixture to be drawn off and replaced with water so that the mixture may be half water and half syrup
It means 1/5 of the mixture has to be drawn off and replaced with water so that the mixture
may be half water and half syrup
A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
By the rule of alligation,we have
=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3
Total quantity is given as 1000Kg
So Quantity of part2 (Quantity sold at 18% profit)
John bought 20 kg of wheat at the rate of Rs.8.50 per kg and 35 kg at the rate of Rs.8.75 per kg. He mixed the two. Approximately at what price per kg should he sell the mixture to make 40% profit as the cost price?
A butler stole wine from a butt of sherry which contained 32% of spirit and then replaced what he stole by wine containing only 18% spirit. The butt was then of 24% strength only. How much of the butt had he stolen?
So, profit amount by selling remaining
2/3 part =600
cost price of 2/3 part =4000
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