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:
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?
Milk in 1 litre mixture in 1st can = 3/4 litre.
Milk in 1 litre mixture in 2n d can = 1/2 litre.
Milk in 1 litre final mixture = 5/8 litre.
By rule of alligation we have required ratio 5 : 8
3/4 1/2
\ /
(5/8)
/ \
1/8 : 1/8
Therefore ratio of two mixtures =1/8:1/8=1:1
So, quantity of mixture taken from each can =(1/2)*12= 6 litres
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How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?
30 litres of the mixture has milk and water in the ratio 7 : 3. i.e. the solution has 21 litres of milk
and 9 litres of water.
When you add more water, the amount of milk in the mixture remains constant at 21 litres. In the
first case, before addition of further water, 21 litres of milk accounts for 70% by volume. After
water is added, the new mixture contains 60% milk and 40% water.
Therefore, the 21 litres of milk accounts for 60% by volume.
Hence, 100% volume =21/0.6 = 35 litres.
We started with 30 litres and ended up with 35 litres.
So, 5 litres of water was added.
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A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30 / kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?
lets assume x kg be rice A,
y kg be rice B
and 2 kgs is rice C
Given final CP = 25
So, 20x+24y+60=25(x+y+2) ------(1)
By Solving above eq. y=10-5x
Since , y cannot be zero or negative
Hence, x can only be 1 giving y = 5kg
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When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains 50% water, and the honey obtained from this nectar contains 15% water?
Flower-nectar contains 50% of non-water part.
In honey this non-water part constitutes 85% (100-15).
Therefore 0.5 X Amount of flower-nectar = 0.85 X Amount of honey = 0.85 X 1 kg
Therefore amount of flower-nectar needed = (0.85/0.5) * 1kg = 1.7 kg.
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A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?
The 20 litre mixture contains milk and water in the ratio of 3 : 2. Therefore, there will be 12 litres of milk in the mixture and 8 litres of water in the mixture.
Step 1: When 10 litres of the mixture is removed, 6 litres of milk is removed and 4 litres of water
is removed. Therefore, there will be 6 litres of milk and 4 litres of water left in the container. It is
then replaced with pure milk of 10 litres. Now the container will have 16 litres of milk and 4 litres
of water.
Step 2: When 10 litres of the new mixture is removed, 8 litres of milk and 2 litres of water is
removed. The container will have 8 litres of milk and 2 litres of water in it. Now 10 litres of pure
milk is added. Therefore, the container will have 18 litres of milk and 2 litres of water in it at the
end of the second step.
So, the ratio of milk and water is 18 : 2 or 9 : 1.
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There are two bottles A and B, each filled with milk and water in the ratio 5:3 and 1:2 respectively. A new mixture is formed by mixing the contents of A and B in the ratio 4:3.What is the ratio of composition of milk and water in the new mixture?
Bottle A mixture: milk = 5/8 & water=3/8
Bottle B mixture: milk = 1/3 & water=2/3
Given New mixture (Bottle A & B) ratio = 4:3
Now milk ratio = (5/8*4/7) + (1/3*3/7) = 1/2
and water = (3/8*4/7) + (2/3*3/7) = 1/2
So, New mixture ration milk : water = 1 : 1
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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:
Profit on 1st part Profit on 2nd part:
8% 18%
\ /
Mean Profi 14%
/ \
4 6
Ratio of 1st and 2nd parts = 4 : 6 = 2 : 3
Quantity of 2nd kind = (3/5 ) * 1000 kg = 600 kg
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Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8:5?
Let's assume C.P. of spirit = Re. 1 per litre.
Spirit in 1 litre mix. of A = 5/7 litre. So C.P of 1 litre mix in A = Re. 5/7.
Spirit in 1 litre mix. of B = 7/13 litre. So C.P of 1 litre mix in B = Re. 7/13.
Spirit in 1-litre mix. of C = 8/13 litre. So C.P. of 1 litre mix in C = Re. 8/13.
By rule of an allegation, we have required ratio X:Y.
(5/7) (7/13)
\ /
(8/13)
/ \
(1/13) (9/91)
So, required ratio = 1/13 : 9/91 = 7:9.
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A coffee seller has two types of coffee Brand A costing 5 bits per pound and Brand B costing 3 bits per pound. He mixes two brands to get a 40 pound mixture. he sold this at 6 bits per pound. the seller gets a profit of 33 1/2 percent. how much he has used Brand A in the mixture?
Let x and y quantities of A and B respectively are mixed.
=> x+y=40 --------- (1)
His total selling price = 40*6=240
His cost price 180 approx(as he makes 33.5 % profit)
5x+3y=180 ------------- (2)
On solving (1) and (2)
x=30 pounds.
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