## Ratios and Proportions Part 1

Problem 1: If A:B:C=2:3:4 find $\frac{A}{B}:\frac{B}{C}:\frac{C}{A}$

Solution:  $\frac{2}{3}:\frac{3}{4}:\frac{4}{2}$

$\frac{2}{3}:\frac{3}{4}:2$

L.C.M of 3 and 4 is 12

$12\times&space;\frac{2}{3}:12\frac{3}{4}:12\times&space;2$

8 : 9 : 24

Problem 2: A:B=2:3    B:C=4:5      C:D=6:7

Solution:

Problem 3: A : B= C : D= E : F = 1  :  2 then (PA+QC+RE) : (PB+QD+RF) is

Solution: A  : B = 1  :  2

$\frac{A}{B}=&space;\frac{C}{D}=&space;\frac{E}{F}=\frac{1}{2}$$\frac{(PA+QC+RE)}{(PB+QD+RF)}=\frac{3}{6}=\frac{1}{2}$

Problems 4: When a ball bounces, it raise to $\frac{2}{3}$ of the height from which it feels. if the ball is dropped from the height of 36 meters. How will height it rise at the third bounces?

Solution:

1st bounce =36$\times&space;\frac{2}{3}$=24

2nd bounce=24$\times&space;\frac{2}{3}=16$

3rd bounce=16$\times&space;\frac{2}{3}=\frac{32}{3}=10\tfrac{2}{3}$

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

## 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 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 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