Ratio and Proportion Methods shortcut tricks
Shortcut Tricks are very important things in competitive exam. Competitive
exams are all about time. If you manage your time then you can do well in those
exams. Most of us miss that part. We provide examples on Ratio and
Proportion shortcut tricks here in this page below. These shortcut
tricks cover all sorts of tricks on Ratio and Proportion. Visitors are
requested to carefully read all shortcut examples. These examples here will
help you to better understand shortcut tricks on ratio and proportion.First of all do a practice set on math of any exam. Write down twenty math problems related to this topic on a page. Do first ten maths using basic formula of this math topic. 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 go through our page for ratio and proportion shortcut trick. After this do remaining ten questions and apply shortcut formula on those math problems. Again keep track of the time. This time you will surely see improvement in your timing. But this is not enough. You need more practice to improve your timing more.
You all know that math portion is very much important in competitive exams. 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. All you need to do is to do math problems correctly within time, and only shortcut tricks can give you that success. But it doesn’t mean that without using shortcut tricks you can’t do any math problems. You may have that potential that you may do maths within time without using any shortcut tricks. But so many people can’t do this. Here we prepared ratio and proportion shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Ratio and Proportion. But we may miss few of them. If you know anything else rather than this please do share with us. Your little help will help so many needy.
Now we will discuss some basic ideas of Ratio and Proportion. On the basis of these ideas we will learn trick and tips of shortcut ratio and proportion. If you think that how to solve ratio and proportion questions using ratio and proportion shortcut tricks, then further studies will help you to do so.
What is Ratio?: A ratio is a relationship
between two numbers by division of the same kind. The ration of a to b is
written as a : b = a / b, In ratio a : b, we can say that a as the first
term or antecedent and b the second term or consequent.
Example : The ratio 4 : 9 we can represent
as 4 / 9 after this 4 is a antecedent and, consequent = 9
Rule of ratio : In ratio
multiplication or division of each and every term of a ratio by the same non-
zero number does not affect the ratio.
Different type of ratio problem are given in Quantitative Aptitude which is
a very essential topic in banking exam. Under below given some more example for
your better practice.Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks and formula are comes into action.
What is Proportion?
The idea of proportions is that two ratios are like equal.
If a : b = c : d, we write a : b : : c : d,
Ex. 3 / 15 = 1 / 5
a and d called extremes, where as b and c called mean terms.
Proportion of quantities
the four quantities like a, b, c, d we can say proportion then we can express it
a : b = c : d
Then a : b : : c : d <–> ( a x d ) = ( b x c )
product of means = product of extremes.
If there is given three quantities like a, d, c of same like then we can say it proportion of continued.
a : d = d : c , d is called mean term. a and c are called extremes.
Example of Ratio and Proportion method :
Example Set -1
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 : 1 : 7 respectively. If the share of Q is Rs.9872, than what would be the total amount of P and S together ?
Answer: Share of Q = 9872, P, Q, R, S in the ratio of 2 : 3 : 1 : 7
Total amount of P ans S together is = 9872 x 9 = 88848.
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.
Example Set -2
Example 1: In a town, 20% of the mans are same in numbers as 1 / 4th of the women. What would be the ratio of mans and woman in that town ?
Answer : 20% mans
= 1 / 4 womens<=> 20man / 100
So, mans = 5 / 4 womens
mans / womens = 5 / 4
mans : womens = 5 : 4
Example 2: In a school, the number of ratio of boys and girls is 4 : 9, after inclusion of 32 new girls, the ratio become 4 : 17. How many boys were present at starting in this school ?
Answer : 4x / 9x + 32 = 4 / 17
68x = 36x + 128
x = 4
So, the number of boys in the school is (4 x 4) = 16.
Example 3: Ina school ratio of number of boys and girls is 9 : 6 and if there present 180 boys, than find the total number of students in the school.
Answer: Let number of boys and girls 9x and 6x.
Then 9x = 180 = x = 20.
So, Total number of students is = 15x = (15 x 20) = 300.
Example 4:
Share Rs.4200 among joy, sanjay and bijoy in the ration 2 : 4 : 6.Find the amount received by sanjay.
Answer :
Amount received by sanjay.
4 / 12 X 4200 = 1400= ( related ratio / sum of ratio ) x Total amount
So, the Amount received by sanjay is 1400.
Example 5:
Find the mean proportional between given two number that is 64 and 49.
Answer :
The mean proportion of two numbers is
Root of 64 and 49 is √8 x √ 7 = 8 x 7 = 56.
So, the mean proportional is 56.
Example 6:
Rs. 385 were divided among P , Q , R in such a way that P had Rs 20 more than Q and R had Rs 15 more than P . How much was R’s share ?
Answer :
Let Q gets Rs x. Then We can say P gets Rs (x + 20 ) and R gets Rs ( x + 35) .
x + 20 + x + x + 35 = 385
3x = 330
x = 110 .
R’s share = Rs ( 110 + 35 ) = Rs 145.
Example 7:
Rs 1210 were divided among three person P, Q, R so that P : Q = 5 : 4 and Q : R = 9 : 10. Then R gets the amount.
Answer :
P : Q = 5 : 4, Q : R = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9.
So, P : Q : R = 5 : 4 : 40 /9 = 45 : 36 : 40
Sum of ratio terms is = ( 45 + 36 + 40 ) =121.
R share of amount is Rs (1210 x 40 / 121) = Rs. 400 .
Example 8:
Rs 64000 are divided among three friends in the ratio 3 / 5 : 2 / 1 : 5 / 3 . The share of the second friend is :
Answer :
Here the given ratio is = 3 / 5 : 2 / 1 : 5 / 3 = 9 : 30 : 25 .
So the Share of the second friend = Rs ( 64000 x 30 / 64) = 30000.
Example 9:
Which of the following ratio is the greatest ?
7 : 15 , 15 : 23 , 17 : 25 , 21 : 29
Answer :
7 / 15 = 0.466
15 / 23 = 0.652
17 / 25 = 0.68
21 / 29 = 0.724
So, 0.724 is greatest and therefore, 21 : 29.
Example 10:
What number has to be added to each term of 3 : 5 to make the ratio 5 : 6 :
Answer :
Let the number to be added x , Then
3 + x / 5 + x = 5 / 6
6 ( 3 + x ) = 5 ( 5 + x )
x = ( 25 – 18 ) = 7
So , the number to be added is 7.
Example 11:
On dividing a sum of Rs 832 between Paul and john in the ratio 3 : 5 , their shares are
Answer :
So the given ratio is = 3 : 5
So , the 1st part = Rs ( 832 x 5 / 8 ) = Rs 520 , 2nd part = Rs ( 832 x 3 / 8 ) = Rs 312
So , their shares are : 312 and 520.
Example 12:
A certain amount was divided between A and B in the ratio 4 : 3 . If B’s share was Rs, 4800, the total amount was ?
Answer :
If B ‘s share is Rs, 3, total amount = Rs, 7
If B’s share is Rs, 4800, total amount = Rs, (7 / 3 x 4800) = 11200.
Example 13:
The Salary of Three friend A, B, C are divided into ratio 5 : 6 : 8. If the increment has given of 10%, 20%, 25%, Find the new ratio of three friend salary ?
Answer :
Step 1: We assume ration as 5x, 6x, 8x
now the increment of new salary of A is 10% = 110 / 100, B is 20% = 120 / 100, C is 25% = 125 / 100.
Step 2: A,s new salary is 110 x 5X / 100 = 55X / 10.
B,s new salary is 120 x 6X / 100 = 36X / 5.
C,s new salary is 125 x 8X / 100 = 10.
Step 3: New ratio is 55X / 10 : 36X / 5 : 40X / 4.
Example Set -3
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 = 5C2 x 2C2 / 14C4
= 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 = 4C2 x 2C2 x 3C2 / 14C8
= 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 = 2C2 / 14C2 + [ 1 – 12C2 / 14C2]
= 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 – 11C3 / 14C3
= 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 = 10C3 / 14C3
= 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.
Example Set -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: 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 24kg silver, so mixture melted is = 20 x 24 / 8 = 60 kg.
Example 2:
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 3:
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 4:
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 5:
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 6:
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 7: 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 8:
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 9:
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 10: 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 liters.
Example 11: 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.
Now, you can solve the following Questions out of Test Paper No168-E23 and submit answers in the form of 1a,2b,3c,4d,5e,6b in the Comments Section. You will get your scorecard, answer key, and detailed explanation of each question. Please submit your email address, WhatsApp# to download full Practice Test Paper and similar other practice test papers. You can also download Lessons and Test Papers from https://www.facebook.com/groups/NextGenCareers/
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