Allegations and Mixture Part 6

Problem 1: A jar contains a mixture of two liquids A and B in the ratio 3:1. When 15 liters of the mixture is taken out and 9 liters of liquid B is poured into the jar, the ratio becomes 3: 4. How many liters of liquid A was contained in the jar?

Solution:

A  :  B

3  :  1

3  :  4

1-4=3

3——->9

1——–>?\frac{9}{3}=3

4——->?4*3=12

12+15=27

27\times \frac{3}{4}=20.25

Problem 2: A vessel is filled with liquid, which is 3 parts water and 5 parts milk. How much of the liquid should be drawn off and replaced by water to make it half water and half milk?

Solution:

w   :   m

3    :    5————-> 3    :   5

1    :    1*5———–>5   :   5

\frac{2}{8+2}=\frac{2}{10}=\frac{1}{5}

Allegations and mixture

\frac{1}{5} is answer

Problem 3: In three classes X, Y, and Z an algebra test is conducted. The average score of class X is 83. The average score of class Y is 76 and the average score of class Z is 85. The average score of class X & Y is 79 and the average score of class Y & Z is 81. What is the average score of the classes X, Y, and Z?

Solution:

x : y : z= 3 : 4 : 5
The average score of a class
\frac{3\times 83+4\times 76+85\times 5}{3+4+5}=\frac{978}{12}=81.5

 

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 \frac{4}{5}=16

20\times \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 \frac{4}{5}=16

20\times \frac{1}{5}=4

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 5\times R}{100}

R=\frac{32}{7}

2  :  5

\frac{2}{7}\times 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 Average =\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 7}{13\times 7}

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

2  :  7

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 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 R\times 3}{2\times 100}=7%
4————->8400
1————–>?\frac{8400}{4}=2100

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 legs/heads

Pigeon =11

Rabbit = 9

\frac{11}{20}\times 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 100}{110}=62