Alligation or Mixture -> Formula/Rule 6

Direct Formula / Rule 6 :
• Theorem: There are 'N' students in a class. Rs X are distributed among them so that each boy gets Rs x and each girl gets Rs y. Then the ratio of boys to the girls is given by

  =	(	X - Ny	)	and
  		Nx - X

  the no. of boys =	(	X - Ny	)	and
  		x - y

  the no. of girls are =	(	Nx - X	)	and
  		x - y
• Example :
There are 65 students in a class, 39 rupees are distributed among them so that each boy gets 80 P and each girl gets 30 P. Find the number of boys and girls in that class.
• Detail Method : 
  Let the ratio of boys to the girls in the

  and the no. of girls =	65 X a
  	a + b

  No. of boys =	65 X b
  	a + b

  or,	65 X a	X 80 +	65 X b	X 30 = 3900
  	a + b		a + b

  or, (5200 - 3900)a = (3900 - 1950)b

  or,	a	=	1950	=	3
  	b		1300		2

  ∴ a : b = 3 : 2

  ∴ the no. of boys =	65 X 3	= 39 and
  	5

  the no. of girls =	65 X 2	= 26
  	5

• Ailigation Method : 
  Here alligation is applicable for "money per boy or girl."

  Mean value of money per student =	3900	= 60 P
  	65

  Boys

  Girls

  80		60		30
  30				20

  ∴ Boys : Girls = 3 : 2
  

  ∴ Number of boys =	65	X 3 = 39
  	3 + 2

  and number of girls = 65 - 39 = 26.
• Quicker Method : Here you can use direct formula : 
  Applying the above theorem, we have
  N = 65 
  X = Rs 39 =3900 P
  x= 80 P 
  y = 30 P

  No. of boys =	3900 - 65 X 30	=	1950	= 39
  	80 - 30		50

  No. of girls =	65 X 80 - 3900	=	1300	= 26
  	80 - 30		50

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Exercise :
1. There are 60 students in a class, 120 rupees are distributed among them so that each boy gets Rs 2.50 and each girl gets 50 P. Find the ratio of boys to the girls.
2. There are 75 students in a class, 48 rupees are distributed among them so that each boy gets Re 1 and each girl gets 40 P. Find the number of boys and girls in that class.
3. There are 50 students in a class, 32 rupees are distributed among them so that each boy gets Re 1 and each girl gets 50 P. Find the number of girls -and boys in that class. 
Answers : 1 = 3 : 5,   2 = 30, 45,   3 = 36 girls, 14 boys

Alligation or Mixture -> Formula/Rule 5

Direct Formula / Rule 5 :
• Theorem: n gm of sugar solution has x% sugar in it. The quantity of sugar should be added to make it y% in the solution is given by

  = n	(	y - x	)	gm.
  		100 - y

  or Quantity of sugar added

  =	(	Solution (required % value - present % value)	)
  		(100 - required % value)
• Example :
300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?
• Detail Method : 
  300 X 40/100 = 120 gm
  300 - 120 = 180 gm
  180gm = 120 + 60 gm
  so 60 gm 
• Ailigation Method : 
  The existing solution has 40% sugar. And sugar is to be mixed; so the other solution has 100% sugar. So by alligation method:

  40%		50%		100%
  50%				10%

  The two mixtures should be added in the ratio 5 : 1.
  

  Therefore, required sugar =	300	X 1 = 60 gm.
  	5
• Quicker Method : Here you can use direct formula : 

  quantity of sugar added =	300 (50 - 40)	= 60 gm.
  	100 - 50

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Exercise :
1. A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make 20% water in the new mixture?
2. A petrol pump owner mixed leaded and unleaded petrol in such a way that the mixture contains 10% unleaded petrol. What quantity of leaded petrol should be added to 1 litre mixture so that the percentage of unleaded petrol becomes 5%.
3. 15 litres of a mixture contains 20% alcohol and the rest water. If 3 litres of water be mixed in it, the percentage of alcohol in the new mixture will be: 
Answers : 1 = 5 litres,   2 = 1000 ml,   3 = 16 2/3%

Alligation or Mixture -> Formula/Rule 4

Direct Formula / Rule 4 :
• Theorem: The proportion in which water must be mixed with spirit to gain or to lose x% by selling it at cost price is given by

  =	(	x	)
  		100
• Example :
In what proportion must water be mixed with spirit to gain 16 ²/₃% by selling it at cost price?
• Detail Method : Let the required proportion of water to spirit be a : b and the cost price of spirit be Rs x per litre.
  As per the question,
  Selling price of the mixture = Rs x per litre.
  Cost price of the mixture  

  = x X	100
  	x X 100 100

  = Rs	6x	= per litre.
  	7

  Now, assume that the cost price of water = Rs 0 per litre.

  ∴ (a X 0 + b X x) = (a + b)	6x
  	7

  or, bx = (a + b) =	6	x
  	7

  or, b	(	1 -	6	)	=	6a
  			7			7

  or,	b	=	6a
  	7		7

  or,	a	=	1
  	b		6

  ∴ required ratio = 1: 6 
• Ailigation Method : 
  Let CP of sprit be Re 1 per litre.
  Then SP of 1 litre o f mixture = Rs 1.

  Gain = 16	2	%
  	3

  CP of 1 litre of mixture = Rs	(	100 X 3 X 1	)	= Rs	6
  		350			7

  CP of 1 litre water

  CP of 1 litre pure spirit

  (Rs 0)		Mean price    Rs 6/7		(Rs 1)
  1/7				6/7

  Quantity of water	=	1/7	=	1
  Quantity of spirit		6/7		6

  or Ratio of water and spirit = 1 : 6.
  
• Quicker Method : Here you can use direct formula : 

  the required proportion =	50	= 1 : 6
  	300

---

Exercise :
1. In what proportion must water be mixed with spirit to gain 33 ¹/₃%by selling it at cost price?
2. In what proportion must water be mixed with spirit to gain 16% by selling it at cost price?
3. In what proportion must water be mixed with spirit to gain 25% by selling it at cost price? 
Answers : 1 = 1 : 3,   2 = 4 : 25,   3 = 1 : 4

Alligation or Mixture -> Formula/Rule 3

Direct Formula / Rule 3 : Theorem: A mixture of a certain quantity of milk with 'L' litres of water is worth Rs x per litre. If pure milk be worth Rs y per litre, then the quantity of milk is given by : = L ( x ) litres. y - x Example : A mixture of a certain quantity of milk with 16 litres of water is worth 90 P per litre. If pure milk be worth 108 P per litre how much milk is there in the mixture? Detail Method : Let the quantity of milk be x litres.  (x + 16) 90 = x X 108 + 16 X 0 [because the price of water is 0 P]  Because Selling price of the mixture = 90 per kg given  or, 90x + 16 X 90 = 108x  or, 18x = 16 X 90  or, x = 80 litres.  ∴ The quantity of milk = 80 litres. Ailigation Method :  The mean value is 90 P and the price of water is 0 P. milk water 108 Mean price      90 0 90 - 0 108 - 90 = 90 : 18 By the Alligation Rule, milk and water are in the ratio of 5 : 1. ∴ quantity of milk in the mixture = 5 X 16= 80 litres. Quicker Method : Here you can use direct formula :  Quantity of milk in the mixture = 16 ( 90 )  = 16 X 5 = 80 litres. 108 - 90 Exercise : A mixture of a certain quantity of milk with 25 litres of water is worth Rs 2 per litre. If pure milk be worth Rs 12 per litre how much milk is there in the mixture? A mixture of a certain quantity of milk with 16 litres of water is worth Rs 3 per litre. If pure milk be worth Rs 7 per litre how much milk is there in the mixture? A mixture of a certain quantity of milk with 32 litres of water is worth Rs 1.50 per litre. If pure milk be worth Rs 4.50 per litre how much milk is there in the mixture?  Answers : 1 = 5 litres,   2 = 12 litres,   3 = 16 litres

Alligation or Mixture -> Formula/Rule 2

Direct Formula / Rule 2 :
• Theorem: The quantity of salt at Rs x per kg that a man must mix with n kg of salt at Rs y per kg, so that he may, on selling the mixture at Rs z per kg, gain p% on the outlay is given by
  = [ ^{100z - y(100 + p)}/_{x(100 + p) - 100z} ] X n
  
  Note : if we suppose that the quantity of salt at Rs x be m, then we have.
• Example :
How many kg of salt at 42 P per kg must a man mix with 25 kg of salt at 24 P per kg, so that he may, on selling the mixture at 40 per kg, gain 25% on the outlay?
• Detail Method : Let the required amount of salt be x kg
   As per Question,
   Because Selling price of the mixture = 40 per kg given
   so Cost price,of the mixture = 40 X ¹⁰⁰/₁₂₅ X (x + 25)
   or, 42x + 24 x 25 = 32x + 32 x 25
   or, 10x = 25 x 8 
   so x = 20 kg.
• Ailigation Method : 
  Cost price,of the mixture = 40 + ¹⁰⁰/₁₂₅ P = 32 p per kg..
  By the rule of fraction :

  42		Mean price      32		24
  8				10

  Ratio = 4 : 5
  Thus for every 5 kg of salt at 24 P, 4 kg of salt at 42 P is used.
  the required no. of kg = 25 x ⁴/₅ = 20
  
• Quicker Method : Here you can use direct formula : 
  = [ ^{100z - y(100 + p)}/_{x(100 + p) - 100z} ] X n
  = [ ^{100 x 40 - 24(100 + 25)}/_{42 x (100 + 25) - 100 x 40} ] X 25
  = [ ^{4000 - 3000}/_{5250 - 4000} ] X 25
  = [ ¹⁰⁰⁰/₁₂₅₀ ] X 25 = 20 kg.

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Exercise :
1. Jaydeep purchased 25 kg of rice at the rate of Rs 16.50 per kg and 35 kg ofrice at the rate of Rs 25.50 per kg. He mixed the two and sold the mixture. Approximately, at What price per kg did he sell the mixture to make 25 per cent profit?   [Hint : here you need to find z as per formula]
2. Jagtar purchases 30 kg of wheat at the rate of Rs 11.50 per kg and 20 kg ofwheat at the rate of Rs 14.25 per kg. He mixed the two and sold the mixture. Approximately at what price per kg should he sell the mixture to make 30 per cent profit?   [Hint : here you need to find z as per formula]
3. In what proportion must a grocer mix one kind of wheat at Rs 4.50 per kg with another at Rs 4 per kg in order that by selling the mixture at Rs 5.20 per kg he may make a profit of 20 per cent?
Answers : 1 = Rs 26.50,   2 = Rs 16.30,   3 = 2 : 1

Alligation or Mixture -> Formula/Rule 1

Direct Formula / Rule 1 :
• Theorem: The proportion in which sugar at Rs. x per kg must be mixed with rice at Rs. y per kg, so that the mixture be worth Rs. z a kg, is given by

  =	(	y - z	)
  		z - x

• Example :
The proportion in which sugar at Rs. 3.10 per kg must be mixed with sugar at Rs. 3.60 per kg, so that the mixture be worth Rs. 3.25 a kg ?
• Detail Method : Let the required ratio be x. y.
   As per Question,
   310x + 360y = 325(x+y)
   or 310x + 360y = 325x + 325y
   325x - 310x = 360y - 325y
   150x = 350y
   15x = 35y
   3x = 7y

  =	x	:	7
  	y		3

   so x : y = 7 : 3    [here y goes to left side & 3 goes to right side]
• Ailigation Method : By the rule of alligation we have :

  C.P.of 1 kg cheaper sugar

  C.P.of 1 kg dearer sugar

  (310 paise)		Mean price (325 paise)		(360 paise)
  35				15

  So, ratio of 1st and 2nd quantities = 35 : 15 = 7 : 3 
  so, they must be mixed in the ratio of = 7 : 3
• Quicker Method : Here you can use direct formula : 

  =	(	y - z	)
  		z - x

  =	(	CP of dearer - Mean Price	)
  		Mean Price - CP of cheaper

  =	(	3.60 - 3.25	)	=	(	0.35	)
  		3.25 - 3.10				0.15

  so, they must be mixed in the ratio of = 7 : 3

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Exercise :
1. In what proportion must rice at Rs 21 per kg be mixed with rice at Rs 28 per kg, so.that the mixture be worth Rs. 25 a kg ? 
2. In what ratio, must coffee at Rs. 62 per kg Is mixed With coffee at Rs. 72 per kg so that the mixture must be worth RS. 64.50 per kg ? 
3. In what ratio must water be mixed with oil costing Rs. 12 per litre to obtain a mixture worth of Rs. 8 per litre ? 
Answers :
1 = 5 : 4,   2 = 3 : 1,   3 = 1 : 2

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