Ratio Example 4

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.

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 Example 4 :
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: Rs. 75840/- are divided between P and Q in the ratio of 3 : 7. What would be the diffderence between thrice the share of P and twice the share of Q ?
Answer : P : Q = 3 : 7
75840 x 3 / 10 = Rs. 22752
75840 x 7 / 10 = 53088
22752 x 3 = 106176
53088 x 2 = 68256
( 106176 – 68256 ) = 37920

 

 

Example 2: In the ratio of 8 : 12 Copper and silver are melted together. What is the weight of melted mixture if 24 kg of silver has been consumed in it ?
Answer: The mixture melted of copper and silver = (12 + 8)kg = 20 kg.
for 24 kg silver, so mixture melted is = 20 x 24 / 8 = 60 kg.

 

 


Example 3:

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 ?
Answer :
Share of A = 75,500 x 1  / 1 + 3 = 75,500 x 1 / 4 = 18875 .
Share of B = 75,500 x 3 / 1 + 3 = 75,500 x 3 /4  = 56625 .
Difference between thrice the share of A and twice the share of B is
= 2B – 3A
= 2 x 56625 – 3 x 18875
= 113250 – 56625
= 56625.

 

 

Example 4:
Milk and water in the ratio 5 : 3 is contain in a 20 litres of mixture. If 4 litres of this mixture be replaced by 4 litres of milk, the ratio of milk to water in the new mixture would be ?
Answer :
Quantity of milk in the mixture after taking out 4 liters of the mixture = ( 16 x 5 / 8 ) = 10 litres.
Now we add 4 liters of milk in this mixture.
quantity of milk in 20 litres of new mix = ( 10 + 4 ) = 14 litres.
quantity of water in it ( 20 – 14 ) = 6 litres.
Ratio of milk and water in the new ratio mix is = 14 : 6 = 7 : 3.

 

 

Example 5:
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 ?
Answer :
Share of A = 75,500 x 1  / 1 + 3 = 75,500 x 1 / 4 = 18875 .
Share of B = 75,500 x 3 / 1 + 3 = 75,500 x 3 /4  = 56625 .
Difference between thrice the share of A and twice the share of B is
= 2B – 3A
= 2 x 56625 – 3 x 18875
= 113250 – 56625
= 56625.

 

 

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 the quantity of water to be further added ?
Answer :
Step 1: Quantity of Milk ( 80 x 3 / 5 ) = 48 liters, So Quantity of water in it  ( 80 – 48 ) = 32 liters.
Step 2: New Ratio required 2 : 3, Let x water to be added, Then Milk : Water is = 48 : (32+x)
=48 / (32 + x).
Step 3: Now 48 / (32 + x) = 2 : 3
48 / (32 + x) = 2 / 3
2x = 144 – 64
x = 80/2
=40 liters.

 

 

Example 7:
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?
Answer :
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.

 

 

Example 8: In a liquid mixture of 60 litres , 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 :
Answer :
So Quantity of milk = ( 60 x 2 / 3 ) = 40 litres.
Water in it = ( 60 – 40 ) = 20 litres .
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 lires
So, 60 litres of water to be added further.

 

 

Example 9:
20% alcohol present in a 15 litres mixture and rest of water. If 3 litres of water be mixed with it, the percentage of alcohol in the new mixture would be  ?
Answer :
20% present in a 15 litres mixture
So, 20 x 15 / 100 = 3 litres.
Water in it = ( 15 – 3 ) 12 litres.
New quantity of mixture = ( 15 + 3 ) = 18 litres.
Percentage of alcohol in new mix is = ( 3 x 100 / 18 ) % = 50 / 3.

 

 

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 ?
Answer :
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.

 

 

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.
Answer : Total mixture is 80 liters and ratio is 3 : 2
So, milk is 80 x 3 / 5 = 240 / 5 = 48
Water is ( 80 – 48 ) = 32
48 / ( 32 + x ) = 32
2x = 80
x = 40
Quantity of water to be added more is 40 litres.

 

 

Example 12: In a mixture of 25 litres has contains 40% milk and remaining is water, and if 5 litrs of water mix with it, then what would be the new percentage of milk in mixture ?
Answer :  40% milk means 60% water
So, milk in mixture is = 25 x 40 / 100 = 10 litres.
and water in mixure is ( 25 – 10 ) = 15 litres.
new mixture quantity = ( 25 + 5 ) = 30 litres.
10 x 100 / 30 = 100 / 3 litres.
new percentage of milk in mixture is 100 / 3 litres.

 

 

 

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15 comments

    • bala says:

      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

  1. vatsal says:

    Hi,

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

  2. gurpreet says:

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

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