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# Alligation or Mixture Questions

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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:

ARs. 175.50

BRs. 180

CRs. 170

DRs. 169.50

Explanation:

126*1 135*1 2*x/1 1 2=153

126 135 2x=153*4

612-261=2x

351=2x

x=175.50

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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?

A9:1

B7:2

C8:1

D5:03

Explanation:

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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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?

A7:9

B9:7

C3:8

D8:9

Explanation:

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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In what ratio must water be added to 10 liters of milk at Rs.20 per liter so that cost of mixture is Rs.16 per liter?

A3:2

B1:4

C2:3

D4:1

Explanation:

10 lit milk cost 20 Rs For 1Rs, Milk = 1/2 lit
For 16Rs, Milk = (1/2)*16 = 8 lit (Milk)
Total quantity should be 10 litres.
so 10 - 8 = 2 lit ( water have to add)
Now ratio = 2(water) : 8(Milk) = 1:4

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A 10 Liter mixture of milk and water contains 30 percent water. Two liters of this mixture is taken away. How many liters of water should now be added so that the amount of milk in the mixture is double that of water?

A1.4

B0.8

C0.4

D0.7

Explanation:

Two liters were taken away So we have the only 8L of the mixture.
Amount of milk in 8 L of mixture = 8 * 70% = 5.6 liters
Amount of water in 8 L of mix = (8 - 5.6) = 2.4 L.
Half of milk i.e half of 5.6 = 2.8 L.
We need (2.8 - 2.4)L water more = 0.4 L

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An alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 16 kg of the alloy so that the ratio of the metal may be 3 : 1 is:

A2 Kg.

B4 Kg.

C8 Kg.

D3 Kg.

Explanation:

Lets assume quantity of zinc to added = x kg.
zinc quantity in alloy = 5/8*16=10 Kg.
and copper quantity = 3/8*16=6 kg.
Alloy new ratio of zinc and copper = 3:1
Zinc quantity in alloy => (10+x)/6=3/1 => 10 + x = 18 => x = 8 kg.

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A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

A1/4

B1/3

C1/2

D2/3

Explanation:

Consider a solution = 1 ltr
X is the certain quantity which has to be replaced
Now,
40%of (1-x)+(25%of x)=35/100
=> 40/100 * (1-x) + 25x/100 = 35/100
=> 40/100 - 40x/100 + 25x/100 = 35/100
=> 15x/100 = 5/100
=> x = 1/3

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