Ratio Example 1

Shortcut Tricks are very important things in competitive exam. Time takes a huge part in competitive exams. If you manage your time then you can do well in those exams. Most of us miss that part. We provide examples on Ratio shortcut tricks here in this page below. All tricks on ratio are provided here. Visitors are requested to carefully read all shortcut examples. These examples here will help you to better understand shortcut tricks on ratio.

Before doing anything we recommend you to do a math practice set. Then find out twenty math problems related to this topic and write those on a paper. Do first ten maths using basic formula of this math topic. You also need to keep track of the time. After finish write down total time taken by you to solve those ten maths. Now practice our shortcut tricks on ratio and read examples carefully. After finishing this do remaining questions using Ratio shortcut tricks. Again keep track of Timing. This time you will surely see improvement in your timing. But this is not all you need. You need more practice to improve your timing more.

We all know that the most important thing in competitive exams is Mathematics. That doesn’t mean that other sections are not so important. You can get a good score only if you get a good score in math section. You can get good score only by practicing more and more. 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 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 1:

Ratio Example 1 shortcut tricks and formula based problem are very important for Competitive exams here is some problems which are given in exams that is some item or product are divide into persons and find the number of item that the person should have, we discuss this example in ratio example 1.

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: If A : B = 7 : 9, C : D = 6 : 13 then A : B : C is
Answer : A : B : C = (7 x 6) : (9 x 6) : (9 x 13) = 42 : 54 : 117
Example 2:

If P : Q : R = 2 : 3 : 4, Then P / Q : Q / R : R / P = ?
Answer :
P : Q : R = 2 : 3 : 4.
Let P = 2k,
Q = 3k,
R = 4k.
Then,
P / Q = 2k / 3k = 2 / 3 ,
Q / R = 3k / 4k = 3 / 4
R / P = 4k / 2k = 2 / 1.

At first we do LCM of 2, 3, 4 is 12, now we multiplied with ratio numbers ,
Like, 2 x 12 / 3 = 8
3 x 12 / 4 = 9
2 x 12 / 1 = 24
So,the ratio of P / Q : Q / R : R / P is 8 : 9 : 24.

 

 

Example 3:
If a : b = 3 : 7 and b : c = 5 : 9, Find a : b : c.
Answer: 
A : B = 3 : 7,
B : C = 5 : 9
= ( 5 x 7 / 5 ) : ( 9 x 7 / 5 ) = 7 : 63 / 5.
= A : B : C = 3 : 7 : 63 / 5 = 15 : 35 : 63 .

 

 

Example 4:
If P : Q = 2 : 3 and Q : R = 4 : 5, then R : P = ?
Answer :
P / R = ( P / Q x Q / R ) = ( 2 / 3 x 4 / 5 ) = 8 / 15
=R / P = 15 / 8 = R : P = 15 : 8.

 

 

Example 5:
If P : Q = 2 : 3, Q : R = 4 : 5 and R : S = 6 : 7, Then P : S = ?
Answer :
P / S = ( P / Q x Q / R x R / S ) = ( 2 /3 x 4 / 5 x 6 / 7 ) = 16 / 35 = P : S = 16 : 35.

 

 

Example 6:
If P : Q = 4 : 3 , Q : R = 3 : 5 and R : S = 10 : 9, Then P : Q : R : S = ?
Answer :
P : Q = 4 : 3
Q : R = 3 : 5
R : S =10 : 9

= 4 x 3 x 10 : 3 x 3 x 10 : 3 x 5 x 10 : 3 x 5 x 9
P : Q : R : S = 120 : 90 : 150 : 135 = 8 : 6 : 10 : 9.

 

 

Example 7: what would be the 3rd proportional to 0.25 to 0.38 ?
Answer : 0.25 : 0.38 :: 0.38 : X
X = 0.38 x 0.38 / 0.25 = 0.5776

 

 

Eaxmple 8: Divide the Rs, 520 in the ratio of 6 : 4 in between ramesh and suresh, How much amount would be both are getting ?
Answer: Sum of ratio = ( 6 + 4 )= 10,
Ramesh got his amount = 520 x 6 / 10 = 312.
Suresh got his amount = 520 x 4 / 10 = 208.

 

 

Example 9:
What is the smallest part, If 75 is divided into three parts proportional to 3, 5, 8, 9.
Answer :
ratio is = 3 : 5 : 8 : 9, sum of ratio terms is = 25.
So the smallest part is = ( 75 x 3 / 25 ) = 9.

 

 

Example 10:
Rama gives his pencils between his four friends Rakesh, Rahul, Ranjan, and Rohit in the ratio 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5. What would be the minimum number of pencils Rama should have?
Answer :
Rakesh : Rahul : Ranjan : Rohit = 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5
Step 1: At First we need to do is LCM of 2,3,4 and 5 is 60.
Step 2: Then pencils are given in ratio among friends,
Rakesh = ( 1 / 2 x 60 ) = 30.
Rahul = ( 1 / 3 x 60 ) = 20.
Ranjan = ( 1 / 4 x 60 ) = 15.
Rohit = ( 1 / 5 x 60 ) = 12.
Step 3: Total number of pencils are ( 30 x + 20 x + 15 x + 12 x) = 77 x.
The minimum number of pencils are  x = 1.
Rama should have only 77 pencils.

 

 

Example 11:
Two numbers are respectively 40% and 60% more than third number. What would be the ratio of two numbers ?
Answer :
Step 1: Let the third number is A
Then first number is 140% of A = 140 x A / 100 = 7A / 5 and second number is 160% of B = 160 x B / 100 = 8B / 5.
Step 2: now ratio of first and second number is 7A / 5 : 8B / 5 = 35A : 40B = 7 : 8.

 

 

Example 12: A sum money is divided among P, Q, R, S in the ratio of 2 : 3 : 4 : 7 respectively. If the share of R is Rs.9872 more than the share of P, then what is the total amount of money of Q and S together ?
Answer: Let P, Q, R and S money respectively is 2x, 3x, 4x, 7x
So, Share of R is Rs. 9872 more than share of P
4x = 2x + 9872
2x = 9872
x = 4936
Amount of Share of Q = 4936 x 3 = 14808.
Amount of Share of S = 4936 x 7 = 34552.

 

 

Example 13:
If A : B = 4 : 9 and B : C = 3 : 6 , Then find A : C is :
Answer : (A / B = 4 / 9, B / C = 3 / 6 )
= A / C = ( A /B x B / C) = ( 4 / 9 x 3 / 6 ) = 2 / 9 = A : C = 2 : 9.

 

 

 

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54 comments

  1. karthik says:

    in ratio example 1, question no.3
    ratio given is 3,5,8,9
    but substituted value is 2 instead of 9. am i right? and,

    regarding the same question. 90 is divided into 3 parts but given is 4 ratios..

    clarify..
    and your short cuts are very useful..

    • Admin says:

      At first we do LCM of 2, 3, 4 is 12. now we multiplied with ratio numbers ,
      Like, 2 x 12 / 3 = 8
      3 x 12 / 4 = 9
      2 x 12 / 1 = 24
      So,the ratio of P / Q : Q / R : R / P is 8 : 9 : 24

  2. G C rana says:

    Example 2:
    If 2P = 3Q = 4R, Then P : Q : R = ? is incorrect This in very simple terms should be as under; no matter you take k or 1
    By taking 2P = 3Q = 4R = 1, we have P = ½ Q= 1/3 and R = ¼
    So ratio = ½ : 1/3 : ¼ = 6: 4 : 3 (taking LCM as 12). Ans

    • Admin says:

      pls check it ….

      At first we do LCM of 2, 3, 4 is 12. now we multiplied with ratio numbers ,
      Like, 2 x 12 / 3 = 8
      3 x 12 / 4 = 9
      2 x 12 / 1 = 24
      So,the ratio of P / Q : Q / R : R / P is 8 : 9 : 24

    • Sarath says:

      sry bro you understood the question wrongly.The first number is 40% greater than A, since the percentage of A should be inside 100%, u took it as 140%.

  3. suji says:

    I GUESS SOMETHING IS WRONG WITH EXAMPLE 12. Q’s share is 9872 which should be taken as 3x=9872 and hence x=9872/3 am i right? thus share of P and S together is 9x=9*9872/3

    if im wrong. plz explain this example sir…

  4. suji says:

    I have a problem with example 12.
    Q’s share is 9872 which should be taken as 3x=9872
    hence x=9872/3
    P and S ‘s share together = 2x+7x=9x=9*9872/3=29616

    If im wrong plz do explain sir… waiting for ur reply…

  5. Soumi says:

    I have a ques…
    A and B have marbles 5:7. If B gives 3 marbles to A , then the number of marbles each is same. How many marbles did A have initially???

    • Tejaswi says:

      A will be having 15 marbels
      Initially A=5k
      B=7k
      Now when B gave 3marbles it has 7k-3 and A has 5k+3..….. as both values same equate them
      7k-3=5k+3 so k=3 ….so 5k =15

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