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1. If 20 liters of water is added to a tub already filled with 120 liters of diluted milk, the concentration of milk in the tub will be 80%. What is the concentration of milk in the tub before adding water to the tub?

A. 88.66%

B. 90%

C. 93.33%

D. 96%

**Answer : ****93.33%**

Let concentration of milk be "k".

therefore, Concentration of 120 liters of diluted milk = 120k

Concentration of 20 liters of water = 20k = 20 x 0 = 0 (ஃConcentration of milk is zero in water)

Given,

(Concentration of 120 liters of diluted milk + Concentration of 20 liters of water) / total liters = 80

[120k + (20 x 0)] / 140 = 80

120 k = 140 x 80

k = 93.33 %

2. The ratio of milk and water in a mixture is 2 : 3 and the ratio of water to milk in another mixture is 3 : 4. If 70 liters of each type of milk is added to 100 lts of pure milk, what will be the ratio of milk and water now ?

A. 3 : 4

B. 4 : 5

C. 5 : 2

D. 7 : 3

**Answer : ****7 : 3**

70 liters of first type of mixture = ( 2 / 5 ) x 70 = 28 liters of milk.

70 liters of 2^{nd} type of mixture = ( 4 / 7 ) x 70 = 40 liters of milk.

Total liters from both the types = 68 liters. Now, it is added with 100 liters of pure milk.

Total amount of milk now = 168 liters.

Similary,

Water from the first type = 70 - 28 = 42 liters

Water from the second type = 70 - 40 = 30 liters

Total amount of water now = 72 liters

The ratio of milk and water now = 168 / 72 = 7/3 --> 7 : 3

3. An alloy contains zinc and tin in the ratio 3 : 4. Another alloy contains zinc and silver in the ratio 4 : 3. If both the alloys are melted and mixed in equal ratio, what will be the ratio of zinc and tin in the new alloy ?

A. 4 : 3

B. 7 : 4

C. 1 : 1

D. 3 : 1

**Answer : ****7 : 4**

Let,

7 kg of the first alloy contains 3 kg of zinc and 4 kg of Tin.

7 kg of the second alloy contains 4 kg of zinc and 3 kg of *silver*.

If both these alloys are melted and mixed then we will get 7 kg of zinc and 4 kg of Tin.

The ratio of zinc and tin in the new alloy = 7 : 4.

**(Attention : Silver is given here instead of Tin in the second alloy)**

4. A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?

A. 26.34 litres

B. 27.36 litres

C. 28 litres

D. 29.16 litres

**Answer : ****29.16 litres**

**Formula:**

Amount remaining after taken out

= [ total amount [ 1 - (amount taken out) / (total amount) ] ^{times}]

Therefore, Amount of milk left after 3 operations = [40 [1 - ( 4/40 )]^{3}] liters.

= [40 x (9/10) x (9/10) x (9/10)] = 29.16

5. There are two bottles A and B, B is twice of A size. A is half full of milk while B is one quarter full of milk. Both bottles are filled with water and the contents are poured into a third bottle C. What portion of bottle C mixture is milk?

A. 1/2

B. 1/3

C. 1/4

D. 2/3

**Answer : ****1/3**

Let the size of A be 2X

Since B is two times larger than A, size of B = 4X

Given that A is half full of milk.

i.e Amount of Milk in bottle A = (1/2) x 2X = X

Amount of Water in bottle A = Size of A - Amount of Milk in A

= 2X - X = X

Given that B is one quarter full of milk.

i.e Amount of Milk in bottle B = (1/4) x 4X = X

Amount of Water in bottle B = Size of B - Amount of Milk in B

= 4X - X = 3X

Now, contents of A and B are poured into C

Water in C = Water that was present in A + Water that was present in B = X + 3X = 4X

Milk in C = Milk that was present in A + Milk that was present in B = X + X = 2X

Total Mixture in C = Milk in C + Water in C = 4X + 2X = 6X

So portion of milk in C = Milk in C / Total Mixture in C = 2X/6X = 1/3 part.

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