Alligation:
It is the specific process of bonding that helps us to find the ratio in which two or more commodities at the given price must be mixed to produce a mixture of the desired price. The rule of allegation is:
(d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
(d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Ques. 1.
A mixture of two liquids A and B is in the ratio of 7: 5. When 9 liters of mixture is taken out and the can is filled with liquid B, the ratio of A and B becomes 7: 9. How many liters of liquid A was there in the container initially?
Solution: Just multiply the three figures given: 7x5x9 = 108.
Let the initial quantity of mixture = 108 K liters.
So A liquid = 108x7/12 = 63 K liters.
And B liquid = 108 x5/12 = 45 K liters.
According to question 9 liters of the mixture as per old ratio has been taken out and 9 liters of B liquid is added.
A liquid taken out = 9 x 7/12 = 63/12 liters OR 21/4 liters.
B liquid taken out = 9 x 5/12 = 45/12 liters OR 15/4 liters.
Now A liquid = 63 K – 21/4 liters.
Now B liquid = 45 K – 15/4 + 9 liters OR 45 K + 21/4 liters.
New Ratio = 7/9 Given.
OR (63 K – 21/4) 9 = 7(45K + 21/4)
OR 567 K – 189/4 = 315 K + 147/4
OR 567 K – 315 K = 189/4 + 147/4
252 K = 84
K = 84/252
K = 1/3
Original Quantity of A liquid = 63 x 1/3 = 21 Litres Answer.
Let the initial quantity of mixture = 108 K liters.
So A liquid = 108x7/12 = 63 K liters.
And B liquid = 108 x5/12 = 45 K liters.
According to question 9 liters of the mixture as per old ratio has been taken out and 9 liters of B liquid is added.
A liquid taken out = 9 x 7/12 = 63/12 liters OR 21/4 liters.
B liquid taken out = 9 x 5/12 = 45/12 liters OR 15/4 liters.
Now A liquid = 63 K – 21/4 liters.
Now B liquid = 45 K – 15/4 + 9 liters OR 45 K + 21/4 liters.
New Ratio = 7/9 Given.
OR (63 K – 21/4) 9 = 7(45K + 21/4)
OR 567 K – 189/4 = 315 K + 147/4
OR 567 K – 315 K = 189/4 + 147/4
252 K = 84
K = 84/252
K = 1/3
Original Quantity of A liquid = 63 x 1/3 = 21 Litres Answer.
Ques 2.
A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:
Solution: Let the cost price of milk = Rs. 100 per litre.
Gain = 25%
Sale Price of adulterated milk = Rs. 100 per litre.
So cost price of adulterated milk 100 x 100/125 OR 100 x 4/5 Rs.80
Thus there is 100-80 = 20% water in the adulterated milk – Answer.
Gain = 25%
Sale Price of adulterated milk = Rs. 100 per litre.
So cost price of adulterated milk 100 x 100/125 OR 100 x 4/5 Rs.80
Thus there is 100-80 = 20% water in the adulterated milk – Answer.
Ques 3.
The cost of Type-1 rice is Rs. 30 per kg and Type-2 rice is Rs. 40 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2: 3, then the price per kg of the mixed variety of rice is:
Solution: (d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Let mean price m = x per kg. Given Ratio = 2/3
2/3 = (40 – x)/ (x – 30)
3 (40 – x) = 2 (x – 30)
120 – 3 x = 2 x – 60
120 + 60 = 2 x + 3 x
5 x = 180
X = Rs. 36 per kg Answer.
Simple Trick = 30 x 2 + 40 x 3 = 60 + 120 = 180 for 5 (2+3) kgs i.e. 180/5 Rs. 36 per kg.
Let mean price m = x per kg. Given Ratio = 2/3
2/3 = (40 – x)/ (x – 30)
3 (40 – x) = 2 (x – 30)
120 – 3 x = 2 x – 60
120 + 60 = 2 x + 3 x
5 x = 180
X = Rs. 36 per kg Answer.
Simple Trick = 30 x 2 + 40 x 3 = 60 + 120 = 180 for 5 (2+3) kgs i.e. 180/5 Rs. 36 per kg.
Ques 4.
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:
Solution:
(d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
(18 – 14)/ (14-8) = 4/6 OR 2:3 Answer.
(18 – 14)/ (14-8) = 4/6 OR 2:3 Answer.
Ques 5.
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?
Solution: (d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Let the price of pure milk = Rs. 100 per litre.
So here cost of I can of milk = 100 – 25 = Rs. 75 per litre.
Cost of II can of milk = 100 – 50 = Rs. 50 per litre.
Desired Ratio = 3/5 i.e. 3 litres of water and 5 litres of pure milk costing Rs. 5x100 = Rs.500
Desired Price of the mixed milk = 500/ (3+5) = 500/8
Desired Ratio of milk from I can: II can
(75-500/8): (500/8- 50) = 100/8: 100/8 or 1:1
So milk from each container = 12 x ½ = 6 litres Answer.
Let the price of pure milk = Rs. 100 per litre.
So here cost of I can of milk = 100 – 25 = Rs. 75 per litre.
Cost of II can of milk = 100 – 50 = Rs. 50 per litre.
Desired Ratio = 3/5 i.e. 3 litres of water and 5 litres of pure milk costing Rs. 5x100 = Rs.500
Desired Price of the mixed milk = 500/ (3+5) = 500/8
Desired Ratio of milk from I can: II can
(75-500/8): (500/8- 50) = 100/8: 100/8 or 1:1
So milk from each container = 12 x ½ = 6 litres Answer.
Ques 6.
In what ratio must a grocer mix two varieties of pulses costing Rs. 150 and Rs. 200 per kg respectively so as to get a mixture worth Rs. 165 kg?
Solution: (d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Desired Ratio: (200-165)/ (165-150) OR 35/15 OR 7/3
7:3 is the Answer.
Desired Ratio: (200-165)/ (165-150) OR 35/15 OR 7/3
7:3 is the Answer.
Ques 7.
A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?
Solution: Original mixture contains 20% water or it contains 80% pure wine.
Wine content in 150 litres = 150x80/100 = 120 litres.
120 litres – now to become 75%; so 100% or the total new mixture = 120 x 100/75 = 120 x 4/3 = 160 litres.
Hence 160 – 150 = 10 litres water is added Answer.
Wine content in 150 litres = 150x80/100 = 120 litres.
120 litres – now to become 75%; so 100% or the total new mixture = 120 x 100/75 = 120 x 4/3 = 160 litres.
Hence 160 – 150 = 10 litres water is added Answer.
Ques 8.
In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg?
Solution: (d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Desired Ratio: (24-20)/ (20 - 18) OR 4/2 OR 2/1
2:1 is the Answer.
Desired Ratio: (24-20)/ (20 - 18) OR 4/2 OR 2/1
2:1 is the Answer.
Ques 9.
How many liters of oil at Rs.40 per litre should be mixed with 240 liters of a second variety of oil at Rs.60 per litre so as to get a mixture whose cost is Rs.52 per litre?
Solution: (d - m): (m - c) where d is the dearer value, m is the mean value and c is the cheaper value.
Desired Ratio: (60-52)/ (52-40) OR 8/12 OR 2/3
If 240 litres is the second quantity, then the first quantity should be 240 x 2/3 =
160 litres Answer.
Desired Ratio: (60-52)/ (52-40) OR 8/12 OR 2/3
If 240 litres is the second quantity, then the first quantity should be 240 x 2/3 =
160 litres Answer.
