## Ratios and Proportions Part 6

Problem 1: Two vessels of 48 lit and 42 lit are filled with a mixture of milk and water, the proportion of two vessels being respectively 13:7 and 18:17. if the contents of two vessels are mixed and 20 lit of water be added to the whole what will be the proportion of milk and water in the resulting mixture?

Solution:

M=$48\times&space;\frac{13}{20}+42\times&space;\frac{18}{15}=\frac{264}{5}$

w=$48\times&space;\frac{7}{20}+42\times&space;\frac{17}{35}+20=\frac{286}{5}$

264 : 286

132 : 143

12   :   13

Problem 2: A vessel contains some liter of pure milk. if 25 lit of water is added to the vessel then ratio of milk and water becomes 12 : 5. 17 liters of  mixture is drawn from the vessel and 10 lit of water is added to the mixture. Find the new ratio between milk and water?

Solution:

Milk        Water

12         :        5

5——->25

12——>?

$\frac{12\times&space;25}{5}=60$

in 17 liters 12 liters are milk and 5 liters are water is drawn

Milk=60-12=48

Water=25-5=20

and 10 liters of water is added so 20+10=30

48 : 30

8:5

Problem 3: 18 lit of pure water was added to the vessel contains 80 lit of pure milk. 49 lit of the resultant mixture was then sold and some more quantity of pure milk and water was added to the vessel in the ratio of 2:1. if the resultant ratio of  Milk and water in the vessel was 4:1 what was the quantity of pure milk added in the vessel?

Solution:

W          M

18   +   80——>98 lit

9     :      40

49 lit of mixture was sold out

98-49=49 lit

Some more quantity of pure milk and water was added to the vessel in the ratio of 2:1

Milk=$49\times&space;\frac{2}{3}=40$

Water=49-40=9

$\frac{40+2x}{9+x}=\frac{4}{1}$

40+2x=36+4x

2x=4

x=2

Problem 4: A vessel contains a mixture of milk and water in the ratio 14:3. 25.5 lit of mixture is taken out from the vessel and 2.5 lit of pure water and 5 lit of milk is added to the mixture. If the resultant mixture contains 20% water. What was the initial quantity of mixture in the vessel before the replacement?

Solution:

Milk         Water

14x     :        3x

14x+3x = 17x

Milk = $\frac{25.5\times&space;14}{17}=21$

Water = 25.5-21 = 4.5

$\frac{14x-21+5}{3x-4.5+2.5}=\frac{80}{20}$

14x-21+5=12x-1.8+2.5

x=4

17x=17*4=68

## Ratios and Proportions Part 5

Problem 1: An alloy contains A, B, and C in the ratio 5:2:1 another alloy contains 12:5:7. If equal quantities of these two alloys are melted together to form a third alloy, then what will be the weight of C per kg in new alloy?

Solution:

A:B:C=5:2:1  ————> 8*3 ———> 15:6:3  ——> 24

A:B:C=12:5:7———–>24———->12:5:7——>24

3+7=10            24+24=48

$\frac{10}{48}=\frac{5}{24}$

Problem 2: An alloy contains the copper and aluminum in the ratio of  7:4 while making the weapons from this alloy 12% of alloy gets destroyed. If there is 12kg of aluminum in the weapon, then the weight of the alloy required will be?

Solution:

Copper       :         aluminum

7                   :                   4

4————>12

1——–>?$\frac{12}{4}=3$

11——->?11*3=33

$33\times&space;\frac{100}{88}=37.5$

Problem 3: In a college, there are two courses B.Tech and M.Tech and the ratio of boys and girls in the ratio is 5:4. If in M.Tech no.of boys and girls is equal then find the ratio of boys and girls in Be.ch. The no.of B.Tech students are 25% more than the no. of M.Tech students?

Solution:

Boys     :        Girls

5             :            4

The no.of B.Tech students are 25% more than the no. of M.Tech students

25%=$\frac{1}{4}$

B.Tech = 4+1 = 5

M.Tech = 4

Assume there are 400 students in M.Tech. In M.tech, the no.of boys and girls are equal.

Boys     :        Girls

2             :            2

Assume there are 500 students in B.tech

Boys     :        Girls

3             :            2

## Ratios and Proportions Part 4

Problem 1: Seat of arts, commerce, and science are in the ratio of 3:5:8 respectively. If the number of students studying arts, commerce and science is increased by 20%, 40%, 25% respectively. What will be the new ratio of seats for students in arts, commerce, and science respectively?

Solution:

20%=$\frac{1}{5}$

40%=$\frac{2}{5}$

25%=$\frac{1}{4}$

$3\times&space;\frac{6}{5}:5\times&space;\frac{7}{5}:8\times&space;\frac{5}{4}$

18 : 35 : 50

Problem 2: Three containers have their volume in the ratio 1:4:7 they are full of the mixture of milk and water. The mixture contains milk and water in the ratio 4:1, 3:1, 5:2 respectively. The contents of all these containers are poured into the fourth container. Find the ratio of milk and water in the fourth container is?

Solution:

Milk           :             Water

4                 :                1

3                :                  1

5                :                  2

Water=$1\times&space;\frac{1}{5}+4\times&space;\frac{1}{4}+7\times&space;\frac{2}{7}$

=$\frac{1}{5}+1+2$

1+5+10=16

Milk= $1\times&space;\frac{4}{5}+4\times&space;\frac{3}{4}+&space;7\times&space;\frac{5}{7}$

$\frac{4}{5}+3+5$

$4+15+25$=44

Milk            :            Water

44                :             16

11               :                 4

Problem 3: The ratio of birds in two branches of a tree is 2:1. When 45 birds from the first branch fly to the second branch, the new ratio of birds become 1:5. Find the total number of birds on the tree.

Solution:

1st Branch  :  2nd Branch

2                  :              1

$\frac{2x-45}{x+45}=\frac{1}{5}$

10x-225=x+45

2x=270

x=30

## Ratios and Proportions Part 3

Problem 1: The quantity of water in 120 lit of milk and water is only 25% the milk man sold 20 liters of the mixture in this mixture and added 16.2 liters of pure milk and 3.8 liters of water in the remaining mixture. What is the percentage of water in the new mixture?

Solution: Total quantity=120

25%=$\frac{1}{4}$

$120\times&space;\frac{1}{4}=30$

120-30=90

W        :           M

30                   90

1           :           3

A milk man sold 20 liters of mixture so 120-20=100

Water =$100\times&space;\frac{1}{4}=25$

100-25=75

3.8 liters of water is added to remaining mixture=25+3.8=28.8

16.2 liters of water is added to remaining mixture=16.2+75=91.2

$\frac{288}{120}\times&space;100=24$%

Problem 2: In a bag there are 25p, 50p and 1r coins are there in the ratio 2:1:3. If their total is Rs.240 in all, how many 25p, 50p and 1 RS coins respectively are there in the bag?

Solution:

25            50         1

2        :       1    :      3

50p           50p        3RS

50P+50P+3RS=4

4———->240

2———–>?   $\frac{240\times&space;2}{4}=120$

1————>?  $\frac{240}{4}=60$

3————->? $\frac{240\times&space;3}{4}=180$

Problem 3: Zinc and copper are in the ratio 2:1 in 75 kg of an alloy. The amount of copper to be further added to the alloy so as to make the ratio of  zinc and copper 1:2?

Solution:

$75\times&space;\frac{2}{3}=50$

75-50=25

Zinc         :           Copper

50            :           25     +        75

50            :         100

1              :           2

100-25=75

75 kg of copper is added to get 1:2 ratio

Problem 4: A mixture contains alcohol and water in the ratio of 4:3 if 5 liters of water is added to the mixture the ratio becomes 4:5. Find the quantity of alcohol in the given mixture?

Solution:

Alcohol        :            water

4                     :                   3

4                     :                   5

2—————->5

4—————–>?

$\frac{5\times&space;4}{2}=10$

Problem 5: An perfume production company prepares perfume by mixing it water. The new mixture of 340 liters contains perfume and water in the ratio 27:7 later the company decides to liquefy the mixture to perform anew mixture containing perfume and water in the ratio of 3:1. How much water must they have to add in the previous mixture?

Solution:

Perfume=$340\times&space;\frac{27}{34}=270$

340-270=70

Perfume           :              Water

270                     :               70  +  20

270                     :               90

3                          :                1

20 liters of water is added to get the mixture of 3:1

## Ratios and Proportions Part 2

Problem 1: A container of 108 lit capacity is filled with pure milk from this 18 liters is taken out and replaced with water if this process is replaced two more times find the quantity of water in a container?

Solution:

Basic method: $\frac{18}{108}=\frac{1}{6}$

Milk=$108\times&space;\frac{5}{6}\times&space;\frac{5}{6}\times&space;\frac{5}{6}=&space;\frac{125}{2}=62.5$

Water=108-62.5=45.5 liters

Formula: $x(1-\frac{y}{x})^{n}$

= $108(1-\frac{18}{108})^{3}$

=$108\times&space;\frac{5}{6}\times&space;\frac{5}{6}\times&space;\frac{5}{6}=&space;\frac{125}{2}=62.5$

108-62.5 = 45.5 liters

Problem 2: A vessel contains 100 liters of pure milk 10 liters of milk is taken out from the vessel and is replaced by 10 lit of water. Noe 20 liters of the mixture is taken out from the vessel and is replaced by water. What is the ratio of the quantity of milk left and water?

Solution:

$\frac{10}{100}=\frac{1}{10}$

$\frac{20}{100}=\frac{1}{5}$

m=$100\times&space;\frac{9}{10}\times&space;\frac{4}{5}=72$

w = 100-72 = 28

Problem 3: 9 liters of spirit is drawn from a vessel full of spirit and is filled with water, this operation is performed one more time. The ratio of the quantity of spirit now left in cask to that of water is 16:9. Find the capacity of the vessel?

Solution:

S : W

16X : 9X

16+9 = 25

S=$x(1-\frac{y}{x})^{n}$

$\frac{16}{25}X=X(1-\frac{9}{X})^{2}$

$\frac{4}{5}=1-\frac{9}{X}$

$\frac{9}{X}=1-\frac{4}{5}$

$\frac{9}{X}=\frac{1}{5}$

X=45

Problem 4: A jar is filled with milk. A person replaces 25% of milk with water. He repeats the process 5 times and as a result, there are only 972 ml of milk left in the jar, the remaining part of the jar is filled with water. The initial quantity of milk in the jar was?

Solution: 1 lit = 1000 ml

25%=$\frac{1}{4}$

$x\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}\times&space;\frac{3}{4}=972$

x=$972000\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}\times&space;\frac{4}{3}=68,259.84$