A vessel is filled with liquid, 3 parts of which are water and 5 parts 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?
A. | \(\frac{1}{3}\) | |
B. | \(\frac{1}{4}\) | |
C. | \(\frac{1}{5}\) | |
D. | \(\frac{1}{7}\) |
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = \(\left(3-\frac{3x}{8}+x\right) liters\)
Quantity of syrup in new mixture = \(\left(5-\frac{5x}{8}\right) liters\)
So, \(\left(3-\frac{3x}{8}+x\right)=\left(5-\frac{5x}{8}\right)\)
5x + 24 = 40 - 5x
10x = 16
x = \(\frac{8}{5}\)
So, part of the mixture replaced = \(\left(\frac{8}{5}\times\frac{1}{8}\right)=\frac{1}{5}\)
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More Questions Related to Quantitative Aptitude > Alligation :
- 3 gallons are drawn from a container full of wine. It is then filled with water. This process is repeated four times. The quantity of wine to water left in the container in now 6561/3439. Find the capacity of the container.
- 25 gallons
- 30 gallons
- 35 gallons
- 40 gallons
- A vessel contains a mixture of two liquids A and B in the proportion of 3 : 5. When 8 litres of mixture are drawn off and the vessel is filled with liquid B, the proportion of A and B becomes 1 : 3. How many litres of liquid was contained in the vessel initially?
- 24
- 28
- 32
- 36
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
- Rs. 169.50
- Rs. 170
- Rs. 175.50
- Rs. 180
Since first and second varieties are mixed in equal proportions.
So, their average price = Rs. = Rs.130.50
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:
Cost of 1 kg of 1st kindCost of 1 kg tea of 2nd kind | ||
Rs. 130.50 | Mean Price Rs. 153 |
Rs. x |
(x - 153) | 22.50 |
\(\frac{x-153}{22.50}=1\)
x - 153 = 22.50
x = 175.50
A can contains a mixture of two liquids A and B is the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
- 10
- 20
- 21
- 25
Suppose the can initially contains 7x and 5x of mixtures A and B respectively
Quantity of A in mixture left = \(\left(7x-\frac{7}{12}\times9\right) liters=\left(7x-\frac{21}{4}\right) liters\)
Quantity of B in mixture left = \(\left(5-\frac{5}{12}\times9\right) liters=\left(5x-\frac{15}{4}\right) liters\)
So, \(\frac{\left(7x-\frac{21}{4}\right)}{\left(5x-\frac{15}{4}\right)+9}=\frac{7}{9}\)
\(\frac{28x-21}{20x+21}=\frac{7}{9}\)
252x - 189 = 140x + 147
112x = 336
x = 3.
So, the can contained 21 litres of A.
A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
- 4 litres, 8 litres
- 6 litres, 6 litres
- 5 litres, 7 litres
- 7 litres, 5 litres
Let the cost of 1 litre milk be Re. 1
Milk in 1 litre mix. in 1st can = \(\frac{3}{4}\) litre, C.P. of 1 litre mix. in 1st can Re \(\frac{3}{4}\)
Milk in 1 litre mix. in 2nd can = \(\frac{1}{2}\) litre, C.P. of 1 litre mix. in 2nd can Re \(\frac{1}{2}\)
Milk in 1 litre of final mix.= \(\frac{5}{8}\) litre, Mean price = Re \(\frac{5}{8}\)
By the rule of alligation, we have:
C.P. of 1 litre mixture in 1st can = \(\frac{3}{4}\)
C.P. of mixture in 2nd can = \(\frac{1}{2}\)
Main price = \(\frac{5}{8}\)
So, Ratio of two mixtures = \(\frac{1}{8}:\frac{1}{8}=1:1\)
So, quantity of mixture taken from each can = \(\left(\frac{1}{2}\times12\right)= 6 litres\)