Mixture & Alligation

1.The ratio, in which tea costing Rs. 192 per kg is to be mixed with tea costing Rs. 150 per kg so that the mixed tea when sold for Rs. 194.40 per kg, gives a profit of 20%.

1. 2:5
2. 3:5
3. 5:3
4. 5:2

Answer:Option 1

Explanation:

Solution:

CP of first tea = Rs. 192 per kg.
CP of Second tea = Rs. 150 per kg.
Mixture is to be sold in Rs. 194.40 per kg, which has included 20% profit. So,
SP of Mixture = Rs. 194.40 per kg.
Let the CP of Mixture be Rs. X per kg. Therefore,
X + 20% of X = SP
6X /5 = 194.40
6X = 194.40 *5
X = Rs. 162 per kg.
Let N kg of first tea and M kg of second tea to be added.
Now, Using Alligation,
We get,

N/M = 12/30
N : M = 2 : 5.

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2.An alloy contains zinc, copper and tin in the ratio 2:3:1 and another contains copper, tin and lead in the ratio 5:4:3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:

1. 1/2 kg
2. 1/8 kg
3. 3/14 kg
4. 7/9 kg

Answer:Option 1

Explanation:

Solution:

Ratio of Zinc, Copper and Tin is given as,
Z : C : T = 2 : 3 : 1.
Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.)
Alloys are mixed together to form third alloy. Then the ratio of content in it,
Z : C : T : L = 4 : (6+5) : (2+4) : 3
Weight of third alloy = 12+12 = 24 Kg.
So, weight of the Lead = 3/24 = 1/8 kg.

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3.In a 729 litres mixture of milk and water, the ratio of milk to water is 7:2. to get a new mixture containing milk and water in the ratio 7:3, the amount of water to be added is:

1. 81 litres
2. 71 litres
3. 56 litres
4. 50 litres

Answer:Option 1

Explanation:

Solution:

Quantity of milk in 729 litre of mixture
= 7*729/9 = 567 litre
Quantity of water
= 729-567 = 162 litre.
Let x litre of water be added to become ratio 7:3.

Or, 7/3 = 567/(162+x)
Or, 162*7 +7x = 567*3
Or, 7x = 1701-1134 = 567
Or, x = 567/7 = 81 litre water is to be added.

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4.To gain 10% on selling sample of milk at the cost price of pure milk, the quantity of water to be mixed with 50 kg. of pure milk is:

1. 2.5 kg
2. 5 kg
3. 7.5 kg
4. 10 kg

Answer:Option 1

Explanation:

Solution:

1^st Method (Method of Alligation):
Let the quantity of water to be mixed is x kg.
Let cost of milk be Rs. 1 per kg. Then SP of 50 kg of milk with gain 10% = Rs. 55 [As Cost of 50 kg milk = Rs. 50, then SP = (50 +10% of 50)]

Then, water to be mixed is 5 kg, As the selling price of the milk is Rs. 1 per kg. Seller has to have 10% gain on 50 kg milk, he must have to add 5 kg water to 50 kg milk.

2^nd Method (Simple Method):

Let the quantity of water mixed be x kg.
Let CP of 1 kg of pure milk = Rs. 1.
Hence,
% gain = x*100/50
10 = 100x/50
Or, 2x = 10
or, x = 5 kg.

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5.From a container, full of pure milk, 20% is replaced by water and this process is repeated three times. At the end of third operation, the quantity of pure milk reduces to:

1. 40.0%
2. 50.0%
3. 51.2%
4. 58.8%

Answer:Option 1

Explanation:

Solution:

Let pure milk was 100L. So,
Water is replaced 20% in per process = 20% of 100 = 20L.
Now, we use short-cut formula for it.
Quantity of Milk reduced to,
= X* [1 - (Y/X)]ⁿ
= 100 * [1- (20/100)]³
= (100* 64)/(125)
= 51.2 L.

Here,
X = Initial quantity of milk.
Y = Replaced water in per process.
n = No. of process repeated.

Note:
The formula used in above problem is quite similar to depreciation formula or Compound interest formula.

Alternatively,
Let pure milk be 100 litres initially.
After third operation, milk will be
100==20%↓(- 20L)==> 80==20%↓(- 16L)==>64==20%↓(-12.8L)==>51.2 L.

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6.Three types of wheat of Rs. 1.27, Rs. 1.29 and Rs. 1.32 per kg are mixed together to be sold at Rs. 1.30 per kg. In what ratio should this wheat be mixed?

1. 1:2:3
2. 2:2:3
3. 2:3:1
4. 1:1:2

Answer:Option 1

Explanation:

Solution:

Average cost: Rs. 1.30 per kg.

And,

Hence, ratio of 1.27, 1.29 and 1.32 = 2:2:(3+1),
= 1:1:2.

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7.In a zoo, there are deers and ducks. If the heads are counted, there are 180, while the legs are 448. What will be the number of deers in the zoo?

1. 136
2. 68
3. 44
4. 22

Answer:Option 1

Explanation:

Solution:

Average legs per head = 448/180.

Deers: Ducks = 88: 272
Hence,the number of Deers = (180*88)/360 = 44.

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8.A trader has 50 kg of pulses, part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4%. How much quantity is sold at 14% profit and that at 6% loss?

1. 5 kg, 45 kg
2. 15 kg, 35 kg
3. 10kg, 40 kg
4. None of these

Answer:Option 1

Explanation:

Solution:

Hence, ratio of quantity sold at 14% profit and 6% loss,
= 2:18 = 1:9
Hence, pulses sold at 14% profit,
= (50*1)/10 = 5 kg
At 6% Wheat sold at loss = 45 kg.

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9.Weights of two friends Ram and Shyam are in the ratio 4:5. If Ram's weight is increased by 10% and total weight of Ram and Shyam become 82.8 kg, with an increases of 15%. By what percent did the weight of Shyam has to be increased?

1. 19%
2. 10%
3. 21%
4. 16%

Answer:Option 1

Explanation:

Solution:

Now, given ratio of Ram and Shayam's weight = 4:5
Hence, (x-15)/(15-10) = 4/5
Or, x = 19%.

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10.A lump of two metals weighing 18 g is worth Rs. 87 but if their weight is interchanged, it would be worth Rs. 78.60. If the price of one metal be Rs. 6.70 per gram, find the weight of the other metal in the mixture.

1. 8g
2. 12g
3. 15g
4. 18g

Answer:Option 1

Explanation:

Solution:

Cost of (18 g of 1^st metal + 18 g of 2^nd metal) = Rs. 165.60
Cost of (1 g of 1^st metal + 1 g metal of 2^nd metal) = Rs. 9.20
Hence cost of 1 g of 2nd metal,
= 9.20 - 6.70
= Rs. 2.5
Mean price = Rs. 87/18