### 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.

**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.

**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.

**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.

**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.

**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.

**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.

**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.

**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.

**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.

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