Shortcut tricks on ratio are one of the most important topics in exams. Time takes a huge part in competitive exams. If you know time management then everything will be easier for you. Most of us skip that part. Few examples on ratio shortcuts is given in this page below. All tricks on ratio are provided here. We request all visitors to read all examples carefully. You can understand shortcut tricks on Ratio by these examples.

First of all do a practice set on math of any exam. Then find out twenty math problems related to this topic and write those on a paper. Solve first ten math problems according to basic math formula. You also need to keep track of the time. After solving all ten math questions write down total time taken by you to solve those questions. Now read our examples on ratio shortcut tricks and practice few questions. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of the time. The timing will be surely improved this time. But this is not enough. If you need to improve your timing more then you need to practice more.

Few Important things to Remember

Math section in a competitive exam is the most important part of the exam. That doesn’t mean that other sections are not so important. But only math portion can leads you to a good score. A good score comes with practice and practice. You should do your math problems within time with correctness, and this can be achieved only by using shortcut tricks. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may do math problems within time without using any shortcut tricks. You may have that potential.

But, so many other people may not do the same. So Ratio shortcut tricks here for those people. We try our level best to put together all types of shortcut methods here. But if you see any tricks are missing from the list then please inform us. Your little help will help others.

Ratio

Ratio based problem are very important for Competitive exams. Here is some problems we provide you which are given in exams that is, some item are divide into persons and find the amount of one or two persons, we discuss this example in ration example 4. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example and ratio shortcut tricks for your better practice.

Example #1 – Ratio Examples

Rs.75840/- is divided between P and Q in the ratio of 3 : 7. What would be the difference between thrice the share of P and twice the share of Q?

Rs.37920/-

Rs.39720/-

Rs.32790/-

Rs.30792/-

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve P : Q = 3 : 7

P’s share: 75840 x 3 / 10 = 22752 Q’s share: 75840 x 7 / 10 = 53088

Thrice of the P’s share: 22752 x 3 = 106176 Twice of the Q’s share: 53088 x 2 = 68256

Difference: 106176 – 68256 = 37920

Rough Workspace

Example #2 – Ratio Examples

In the ratio of 8 : 12 Copper and Silver are melted together. What is the weight of melted mixture if 24 kg of copper has been consumed in it?

52 Kg.

65 Kg.

56 Kg.

60 Kg.

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (D)

How to Solve The mixture melted of Copper and Silver = (12 + 8) = 20 kg. For 24 kg of copper, mixture melted is = 24 x 20 / 8 = 60 kg.

Rough Workspace

Example #3 – Ratio Examples

Rs.75,500/- are divided between A and B in the ratio 1 : 3. what is the difference between thrice the share of A and twice the share of B?

Rs.50025/-

Rs.52526/-

Rs.56625/-

Rs.66525/-

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (C)

How to Solve Share of A = 75500 x 1 / 1 + 3 = 75500 x 1 / 4 = 18875. Share of B = 75500 x 3 / 1 + 3 = 75500 x 3 /4 = 56625.

Difference between thrice the share of A and twice the share of B is: 2B – 3A or, 2 x 56625 – 3 x 18875 or, 113250 – 56625 finally 56625

Rough Workspace

Example #4 – Ratio Examples

Milk and Water in the ratio 5 : 3 is contain in a 20 liters of mixture. If 4 liters of this mixture be replaced by 4 liters of milk, the ratio of milk to water in the new mixture would be?

7 : 3

3 : 7

11 : 9

8 : 7

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve Quantity of milk in the mixture after taking out 4 liters of the mixture = ( 16 x 5 / 8 ) = 10 liters. Now we add 4 liters of milk in this mixture. quantity of milk in 20 liters of new mix = ( 10 + 4 ) = 14 liters. quantity of water in it ( 20 – 14 ) = 6 liters. Ratio of milk and water in the new ratio mix is = 14 : 6 = 7 : 3.

Rough Workspace

Example #5 – Ratio Examples

Rs.75,500/- are divided between A and B in the ratio 1 : 3. what is the difference between thrice the share of A and twice the share of B?

Rs.56625/-

Rs.65625/-

Rs.59525/-

Rs.58500/-

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve Share of A = 75500 x 1 / 1 + 3 = 75500 x 1 / 4 = 18875. Share of B = 75500 x 3 / 1 + 3 = 75500 x 3 /4 = 56625.

Difference between thrice the share of A and twice the share of B is: = 2B – 3A or, 2 x 56625 – 3 x 18875 or, 113250 – 56625 finally, 56625

Rough Workspace

Example #6

In a bottle mixture of 80 liters and the ratio of milk and water is 3 : 2. If this mixture ratio is to be 2 : 3. What is the quantity of water to be further added?

30 Liters

36 Liters

40 Liters

42 Liters

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (C)

How to Solve Quantity of Milk ( 80 x 3 / 5 ) = 48 liters. So, quantity of Water in it ( 80 – 48 ) = 32 liters.

New Ratio required 2 : 3. Let, X water to be added Now, Milk : Water is = 48 : (32 + X) = 48 / (32 + X)

The ratio between the number of Men and Women in a society is 31 : 23. When 75 more women are added in the society, this ratio becomes 124 : 107. How many more women should be added in the society in order to make the number of men and women be equal?

82

85

88

95

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve 31x / 23x + 75 = 124 / 107 3317x = 2852x + 9300 465x = 9300 x = 20 .

So, the number of women in society after added = 20 x 23 + 75 = 535.

So, the number of men in society after added = 31 x 20 = 620.

Number of more women is = 620 – 535 = 85.

Rough Workspace

Example #8

In a liquid mixture of 60 liters, the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is?

50 Liters

60 Liters

62 Liters

69 Liters

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve So, Quantity of milk = ( 60 x 2 / 3 ) = 40 liters. Water in it = ( 60 – 40 ) = 20 liters.

New ratio required = 1 : 2 Let, quantity of water to be added further be X liters. Milk : Water = 40 / ( 20 + x ) Now, 40 / ( 20 + x ) = 1 / 2 20 + x = 80 x = 60 liters

So, 60 liters of water to be added further.

Rough Workspace

Example #9

20% alcohol present in a 15 liters mixture and rest of water. If 3 liters of water be mixed with it, the percentage of alcohol in the new mixture would be?

50/3%

40/3%

55/3%

68/3%

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve 20% present in a 15 liters mixture So, 20 x 15 / 100 = 3 liters of Alcohol. Water in it ( 15 – 3 ) = 12 liters.

New quantity of mixture = ( 15 + 3 ) = 18 liters. Percentage of alcohol in new mix is = ( 3 x 100 / 18 ) % = 50 / 3.

Rough Workspace

Example #10

A liquid mixture contain alcohol and water in the ratio of 4 : 3. If 5 liters of water is added to the mixture the ration becomes 4 : 5. Find the quantities of alcohol in the given mixture?

7 Liters

8 Liters

9 Liters

10 Liters

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (D)

How to Solve Let, the quantity of alcohol and water be 4x and 3x liters. Then, 4x / 3x + 5 = 4 / 5 20x = 4(3x + 5) 8x = 20 x = 2.5

Quantity of alcohol = ( 4 x 2.5 ) = 10 liters.

Rough Workspace

Example #11

If the ratio of milk and water is 3 : 2 in a mixture of 80 liters. If this ratio later to be 2 : 3, Then what quantity of water to be added more?

36 Liters

40 Liters

42 Liters

50 Liters

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (B)

How to Solve Total mixture is 80 liters and ratio is 3 : 2 So, Milk is 80 x 3 / 5 = 240 / 5 = 48 liters. Water is ( 80 – 48 ) = 32 liters.

48 / ( 32 + x ) = 32 2x = 80 x = 40

Quantity of water to be added more is 40 liters.

Rough Workspace

Example #12

In a mixture of 25 liters has contains 40% milk and remaining is water. If 5 liters of water mix with it, then what would be the new percentage of milk in mixture?

100/3%

101/3%

110/3%

111/3%

Show AnswerShow How to SolveOpen Rough Workspace

Answer: Option (A)

How to Solve 40% milk means 60% water So, milk in mixture is = 25 x 40 / 100 = 10 liters. and water in mixture is ( 25 – 10 ) = 15 liters.

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18 comments

md ahamad says:

This site very best for competative exam. I am satisfied this site.

first b:c and c:d value take it, 4:5 and 6:7 then common value of =c to l.c.m 5&6=30 so convert the value of c 4*6:5*6and6*5:7*5 24:30and 30:35, so c is common value=30 so, b:c:d=24:30:35 then,take it a:b and b:c:d 2:3 and 24:30:35 then, common b value 3 and 24 l.c.m in 24, so, 2*8:3*8and 24*1:30*1:35*1 then,16:24 and 24:30:35 common value of b=24, ans,a:b:c:d=16:24:30:35

5*1:3*2:1*4 = 5:6:4 => this is ratio of no of coins each have……………..now add 5+6+4=15…………..divide 15 with total no of coins that is 465. We get 31 now multiply 31 to each ratios that is 5*31:6*31*4*31= 155:186:124 this is no. Of individual coins of 1 rs,50p, and 25p

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This site very best for competative exam. I am satisfied this site.

thank u

sir inQ3 ratio of cu: zn is 9:3 so weight of cu must be more than Zn

pls explain it

sir plz tell me how to solve if a:b=2:3, b:c=4:5 c:d=6:7 what is the ratio of a:b:c:d with short method

A:B=m:n

B:C=p:q

C:D=r:s

A:B:C:D=mpr:npr:nqr:nqs

hi just arrange the words horizontally as a:b:c:d

1. a:b – 2:3

2. b:c – 4:5

3. c:d – 6:7

a b c d

2 3

4 5

6 7

a = 2*4*6 = 48

b = 3*4*6 = 72

c = 3*5*6 = 90

d = 3 *5*7 = 105

a:b:c:d = 48:72:90:105 after simplifying = 16;24;30;35

first b:c and c:d value take it,

4:5 and 6:7

then common value of =c to l.c.m

5&6=30 so convert the value of c

4*6:5*6and6*5:7*5

24:30and 30:35,

so c is common value=30

so, b:c:d=24:30:35

then,take it a:b and b:c:d

2:3 and 24:30:35

then, common b value 3 and 24 l.c.m in 24,

so, 2*8:3*8and 24*1:30*1:35*1

then,16:24 and 24:30:35

common value of b=24,

ans,a:b:c:d=16:24:30:35

Hi,

Can you please explain the logic if quantity of liquid is replaced instead of added so that the proportion of mixture gets inverse.

there are 465 coins.1rupee,50paise,25paise.there values are in ratio 5:3:1.number of each coins respectively are?

5*1:3*2:1*4 = 5:6:4 => this is ratio of no of coins each have……………..now add 5+6+4=15…………..divide 15 with total no of coins that is 465. We get 31 now multiply 31 to each ratios that is 5*31:6*31*4*31= 155:186:124 this is no. Of individual coins of 1 rs,50p, and 25p

thanks sir,pls explain

Very good help to those who are trying for best – thank you sir.

thank u jagdamba prasad

sir copper and silver question.

please explain throughly.

sir,

please explain Q1

This helped me lot, thanku.

pls…expalin 1st question