Shortcut tricks on ratio are one of the most important topics in exams. Time is the main factor in competitive exams. If you know how to manage time then you will surely do great in your exam. Most of us skip that part. Few examples on ratio shortcuts is given in this page below. These shortcut tricks cover all sorts of tricks on Ratio. 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. Write down twenty math problems related to this topic on a page. Using basic math formula do first ten maths of that page. You also need to keep track of timing. Write down the time taken by you to solve those questions. Now practice our shortcut tricks on ratio and read examples carefully. After doing this go back to the remaining ten questions and solve those using shortcut methods. Again keep track of Timing. The timing will be surely improved this time. But this is not all you need. You need to practice more to improve your timing more.

You all know that math portion is very much important in competitive exams. That doesn’t mean that other topics are less important. But only math portion can leads you to a good score. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and you can do this 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 other peoples 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 it possible we miss any. We appreciate if you share that with us. Your little help will help so many needy.

**Ration Example 3 :**

Ration based problem are very important for Competitive here is some problems which are given in exams that is in bag contain some coins like 50 p, 25 p, 10 p, in a ratio and total amounting is given and we need to find each type of coins in a bag,

Now we discuss this example in ratio example 3. This type of problem are given in Quantitative Aptitude which is a very essential paper in banking exam. Under below given some more example for your better practice.

**Example 1:**

A money bag contains 50 p, 25 p, and 10 p coins in the ratio 5 : 9 : 4, and the total amounting to Rs.206.

Find the individual number of coins of each type.

**Answer :**

**Step 1:** Let the number of 50 p ,25 p, and 10 p coins be 5x, 9x, 4x respectively.

we can say two 50 paise = 1 rupee that’s why we take 2

we can say four 25 paise = 1 rupee that’s why we take 4

we can say Ten 10 paise = 1 rupee that’s why we take 10

Then, 5x / 2 + 9x / 4 + 4x / 10 = 206

= 50x + 45x + 8x = 4120

= 103x = 4120

= x = 40.

**Step 2:** Number of 50 p coins is **( 5 x 40 = 200 ),**

Number of 25 p coins is**( 9 x 40 = 360 ),**

Number of 10 p coins **( 4 x 40 = 160 ),**

**Example 2:**

On a self there are 4 books on Economics, 3 books on Management and 4 books on Statistics. In how many different ways can be the books be arranged so that the books on Economics are kept together ?

**Answer :**

( 4 books on Statistics ! + 3 books on Management ! + 1 x 4 books on Economics ! )

Total ways = 8! x 4!

= ( 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) x ( 4 x 3 x 2 x 1 )

= 40320 x 24

=967680.

**So , we can 967680 way be the books be arranged.**

**Example 3:**

How many bags are required for filling 1824 kg of wheat if each bag filled with 152 kg of wheat ?

**Answer :**

Each bag filled means 1 bag filled with 152 kg .

Total kg of wheat is 1824 kg , So we divide it by each bag filled with 152 kg

**Number of bags = 1824 / 152**

**= 12.**

**Example 4:**

An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles. If four marbles are drawn at random, what is the probability that two are blue and two are red ?

**Answer :**

Required probability = ^{5}C_{2} x ^{2}C_{2} / ^{14}C_{4}

= 10 / 1001.

**Example 5:**

An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles. If eight marbles are drawn at random, what is the probability that there are equal number of marbles of each color ?

**Answer :**

Required probability = ^{4}C_{2} x ^{2}C_{2} x ^{3}C_{2} / ^{14}C_{8}

= 18 / 3003.

**Example 6:**

An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles. If two marbles are drawn at random, what is the probability that both are red or at least one is red ?

**Answer :**

Required probability = ^{2}C_{2} / ^{14}C_{2} + [ 1 – ^{12}C_{2} / ^{14}C_{2}]

= 1 / 91 + [ – 66 / 91 ]

= 1 / 91 + 25 / 91

= 26 / 91.

**Example 7:**

An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles. If three marbles are drawn at random, what is the probability that at least one is yellow ?

**Answer :**

Required probability = 1 – ^{11}C_{3} / ^{14}C_{3}

= 1 – 165 / 364

= 199 / 364.

**Example 8:**

An urn contains 4 green , 5 blue , 2 red and 3 yellow marbles. If three marbles are drawn at random, what is the probability that none is green ?

**Answer :**

Required probability = ^{10}C_{3} / ^{14}C_{3}

= 30 / 91.

**Example 9:** In a cotton Bag has 50 paise and 25 paise coins, and the cotton bag contains total 60 coins which is sum of Rs.24.25. In cotton bag How many 25 paise coins are present ?

**Answer:** X x 0.50 + ( 60 – X ) x 0.25 = 24.25

0.50X + 15 – 0.25X = 24.25

0.25X = 24.25 – 15

0.25X = 9.25

X = 925 x 100 / 25 x 100

X = 37

Total no of coins is 60. So, ( 60 – X ) = ( 60 – 37 ) = 23

**So, 25 paise coins are present in bag is 23.**

**Example 10:**

If each one plastic bag is filled with 160 Kgs of rice, than for 2240 Kgs of rice how many plastic bags are required ?

**Answer :**Total rice is 2240, So required bags is 2240 / 160 = 14.

So plastic bags are required 14.

- Ratio Example 1
- Ratio Example 2
- Ratio Example 4
- Ratio Hard Example 1
- << Go back to Ratio and Proportion main page

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Dear tutor,

Please explai the following:

A money bag contains 50 p, 25 p, and 10 p coins in the ratio 5 : 9 : 4, and the total amounting to Rs.206.

Find the individual number of coins of each type.

Answer :

Step 1: Let the number of 50 p ,25 p, and 10 p coins be 5x, 9x, 4x respectively.

Then, 5x / 2 + 9x / 4 + 4x / 10 = 206

= 50x + 45x + 8x = 4120

= 103x = 4120

= x = 40.

Step 2: Number of 50 p coins is ( 5 x 40 = 200 ),

Number of 25 p coins is( 9 x 40 = 360 ),

Number of 10 p coins ( 4 x 40 = 160 ),

which line u don’t understand ,

i will explain….

how 50x+45x+8x came

LCM of denominators is found which is 20 and it is multiplied in both sides of the equation to eliminate denominator

pis explain ex-2

Example 2 complete plz reffer in a simple way

pls explain ex-3

Sir I don’t understand how 103 x come

103x comes from addition of 50x + 45x + 8x = 103x

I can’t understand ex1 how 5x/2,9x/4,4x/10 came

we can say two 50 paise = 1 rupee thats why we take 2

we can say four 25 paise = 1 rupee thats why we take 4

we can say Ten 10 paise = 1 rupee thats why we take 10

if a:c=2/3 and b:c=8/9 find a:b:c

ANS-A:B=2/3,B:C=8/9 FIND A:B:C

SO,

=(B*B/B):(C*B/B)

=(8*3/8):(9*3/8)

=3:27/8

=2::3:27/8

=16:24:27

Superb!!! Man

Plz Explain example 2, how the total ways = 8! * 4!

where as total no. of books are 11

Also economics books should kept together

Here economic books considered as 1 set as place together and remaining management and statistics are 4+3 as 7 …. so total books 7+1=8 … now these can be arranged as 8! Ways and we use 4! Bcz these economic books can be internally changed in the set so the ans was 8!*4!

Why is 5C2 excluded in example 5 plz explain