## Alligations and Mixtures Part 1

Problem 1: In what ratio of two varieties of milk costing Rs.16 and Rs.18 per liter respectively mixed, so the mixture cost RS.16.60 per liter?

Solution:

Basic metod:

16——->x——–>16x

18——->y———>18y

16x+18y = 16.60(x+y)

16x+16y = 16.60x+16.60y

0.60x = 1.40y

$\frac{x}{y}=\frac{14}{6}=7:3$

Shortcut:

Problem 2: The average weight of 30 students in a class is 45 kg. The average weight of girls is 37 kg and that of the boy is 49 kg. Find the number of boys in the class?

Solution:

Basic method:

Boys =x

Girls = (30-x)

x*49 + (30-x)*37 = 30*45

Solving this equation is time taking process.

Shortcut:

2+1=3

3——->30

1———>?$\frac{30}{3}=10$

2———->?2*10=20 BOYS

Problem 3: A shopkeeper sold 40kg of goods partly at 7% profit and rest at 5% loss, suffering 2% loss on the whole transactions. What is the quantity sold at profit?

Solution:

CP = 100

1+3=4

4——–>40

1———>?$\frac{40}{4}=10$ KG sold for profit

3———->?3*10 = 30

## Alligation and Mixtures Part 2

Problem 1: A mixture of 70 liters of wine and water contains 10% water. How much water should be added to the mixture to make 37% water in the resulting mixture?

Solution:

Basic method

Wine              Water

90%                  10%    ——>  ( 9   :   1)*7 —–>63  :  7

63%                  37%  ———>  (63 : 37)*1 —–> 63   :  37

63 + 7 =70

30 liter of water is added

Shortcut:

Mixture        Water

7                    3

Wine            Water

4                        3

7——->70

1———>?$\frac{70}{7}$ = 10

3——->?3*10 =30

Problem 2: In a zoo, there are rabbits and pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there?

Solution:

Pigeons = x

Rabbit = y

x+y=200

2x+4y = 580

Solve this equation

Shortcut:

$\frac{580}{200}=2.9&space;legs/heads$

Pigeon =11

Rabbit = 9

$\frac{11}{20}\times&space;200=110$

Problem 3: In what ratio must a grocer mix tea at RS. 60 a kg and 65 a kg so that by selling a mixture at RS. 68.20 a kg, he may gain 10%?

Solution:

110%——>68.20

100%——->?$\frac{6820\times&space;100}{110}=62$

## Alligation and Mixture Part 3

Problem 1: The cost of Type 1 material is RS.15 per kg and Type 2 material is RS 20 per kg. If both type 1 and Type 2 are mixed in the ratio of 2:3, then what is the price per kg of a mixed variety of material?

Solution:

Problem 2: In an examination, a students gets 3 marks for every correct answer and loses 1 mark for every wrong answer. If he scores 120 marks in a paper of 100 questions, how many of his answers were correct assuming that he attempts all the questions?

Solution:

Let the number of correct answers = x

number of wrong answers = 100-x

3x + (100-x)-1 = 100

Solve the equation

Shortcut:

$\frac{120}{3}=40$

$\frac{1}{4}\times&space;60=15$

40+15=55

Problem 3: A man possessing Rs. 8400 lent a part of it at 8% simple interest and the remaining at 20/3% simple interest. His total income after 1 ½ years was Rs. 882. Find the sum lent at 8% rates?
Solution:
$S.I=\frac{PTR}{100}$
$882=\frac{8400\times&space;R\times&space;3}{2\times&space;100}=7$%
4————->8400
1————–>?$\frac{8400}{4}=2100$

## Alligations and Mixture Part 4

Problem1: Some amount out of Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent at 6% per annum was?

Solution:

$S.I=\frac{PTR}{100}$

$1600=\frac{7000\times&space;5\times&space;R}{100}$

$R=\frac{32}{7}$

2  :  5

$\frac{2}{7}\times&space;7000=2000$

Problem 2: The blowing average of a cricketer was 12.4. he improves his bowling average by 0.2 points when he takes 5 wickets for 26 runs in his last match. The number of wickets taken by him before the last match was?

Solution:

$Blowing&space;Average&space;=\frac{Runs}{Wicket}$

Let the number of wickets = x

Runs=12.4X

$12.2=\frac{12.4x+26}{x+5}$

Solve the equation

Shortcut:

$\frac{26}{5}=5.2$

7  :  0.2—–>70  :  2—–>35  :  1

1——–>5

35——>? 35*5 = 175

Problem 3: Two mixtures A and B contain milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and a water in the ratio 9:4?

Solution:

$\frac{2}{91}:\frac{1\times&space;7}{13\times&space;7}$

$\frac{2}{91}:\frac{7}{91}$

2  :  7

## Alligations and Mixture Part 5

Problem 1: There are two mixtures A and B, both contain milk and water. In the mixture A ratio of milk and water is 5:4 whereas in mixture B ratio of milk and water is 5:7. What part of mixture B should be taken out and replaced with mixture A so that in mixture B quantity of milk and water becomes equal?

Solution:

Answer is $\frac{2}{3}$

Problem 2: Milk and water are in can A as 4:1, and in can B as 3:2. For Can C, if one takes equal quantities from A and B, find the ratio of milk to water in can C?

Solution:

$\frac{x-\frac{3}{5}}{\frac{4}{5}-x}=\frac{1}{1}$

$x-\frac{3}{5}=\frac{4}{5}-x$

$2x=\frac{7}{5}$

$x=\frac{7}{10}$

Here 7 is milk and 3 parts are water

7  :  3 is the answer.

Problem 3: A jar contains a mixture of two liquids A and B in the ratio 4:1. When 10 liters of the mixture is taken out and 10 liters of liquid B is poured into the jar, the ratio becomes 2:3. How many liters of liquid A was contained in the jar?

Solution:

A  :  B = 4X  :  X——->5X

In 10 liters 8 parts is milk and 2 parts is water

$\frac{4x-3}{x-2+1}=\frac{2}{3}$

12x-24=2x+16

10x=40

x=4

4*4=16

Al ligation method

1  :  1

1——->10

1——-> Is remaining mixture

1———->10+10=20

$20\times&space;\frac{4}{5}=16$

$20\times&space;\frac{1}{5}=4$

Shortcut

A  :  B

4  :  1—————>4  :  1

2  :  3 *2———–>4  :   6

6-1=5

5——>10

1——->?2

10+10=20

$20\times&space;\frac{4}{5}=16$

$20\times&space;\frac{1}{5}=4$