"Tricks to solve ratio and proportion problems pdf" is the much needed stuff to the people who prepare for ACT, SAT and other competitive exams. Because problems on ratio and proportion play a major role
quantitative aptitude test. There is no competitive exam without the
questions from this topic.
a/b is the ratio of a to b . That is a:b
When two ratios are equal, they are said to be in proportion.
If a:b = c:d, then a,b,c & d are proportion.
Cross product rule in proportion:
Product of extremes = Product of means.
Let us consider the proportion a:b = c:d
Extremes = a & d, means = b & c
Then, as per the cross product rule, we have
ad = bc
b:a is the inverse ratio of a:b and vice versa.
That is, a:b & b:a are the two ratios inverse to each other.
Verification of inverse ratios:
If two ratios are inverse to each other, then their product must be 1.
That is, a:b & b:a are two ratios inverse to each other.
Then, (a:b)X(b:a) = (a/b)X(b/a) = ab/ab = 1
If the ratio of two quantities is given and we want to get the original quantities, we have to multiply both the terms of the ratio by some constant, say “x”.
The ratio of earnings of two persons is 3:4.
the ratio of the first person = 3x
the ratio of the second person = 4x
If we want compare any two ratios, first we have to express the given ratios as fractions.
Then, we have to make them to be like fractions.
That is, we have to convert the fractions to have same denominators.
Compare: 3:5 and 4:7.
First, let us write the ratios 3:5 and 4:7 as fractions.
That is 3/5 and 4/7.
The above two fractions do not have the same denominators. Let us make them to be same.
For that, we have to find L.C.M of the denominators (5,7).
That is, 5X7 = 35. We have to make each denominator as 35.
Then the fractions will be 21/35 and 20/35.
Now compare the numerators 21 and 20.
21 is greater.
So the first fraction is greater.
Hence the first ratio 3:5 is greater than 4:7.
If two ratios P:Q and Q:R are given and we want to find the ratio P:Q:R, we have to do the following steps.
First find the common tern in the given two ratios P:Q and Q:R. That is Q.
In both the ratios try to get the same value for “Q”.
After having done the above step, take the values corresponding to P, Q, R in the above ratios and form the ratio P:Q:R.
If P:Q = 2:3 and Q:R = 4:7, find the ratio P:Q:R.
In the above two ratios, we find “Q” in common.
The value corresponding to Q in the first ratio is 3 and in the second ratio is 4.
L.C.M of (3, 4) = 12.
So, if multiply the first ratio by 4 and second by 3,
we get P:Q = 8:12 and Q:R = 12:21
Now we have same value (12) for “Q” in both the ratios.
Now the values corresponding to P, Q & R are 8, 12 & 21.
Hence the ratio P:Q:R = 8:12:21
If the ratio of speeds of two vehicles in the ratio a:b, then time taken ratio of the two vehicles would be b:a.
The ratio of speeds of two vehicles is 2:3. Then time taken ratio of the two vehicles to cover the same distance would be 3:2.
If the ratio of speeds of two vehicles in the ratio a:b, then the distance covered ratio in the same amount of time would also be a:b.
The ratio of speeds of two vehicles is 2:3. Each vehicle is given one hour time. Then, the distance covered by the two vehicles would be in the ratio 2:3.
If A is twice as good as B, then the work completed ratio of A and B in the same amount of time would be 2:1.
A is twice as good as B and each given 1 hour time. If A completes 2 unit of work in 1 hour, then B will complete 1 unit of work in one hour.
If A is twice as good as B, then the tame taken ratio of A and B to do the same work would be 1:2.
A is twice as good as B and each given the same amount of work to complete. If A takes 1 hour to complete the work, then B will take 2 hours to complete the same work.
If “m” kg of one kind costing $a per kg is mixed with “n” kg of another kind costing $b per kg, then the price of the mixture would be $ (ma+nb)/(m+n) per kg.
If one quantity increased or decreases in the ratio a:b,
then the new quantity is = “b” of the original quantity/a
More clearly, new quantity = (“b” X original quantity) / a
David weighs 56 kg. If he reduces his weight in the ratio 7:6, find his new weight.
New weight = (6 X 56) / 7 = 48 kg.
Hence, David’s new weight = 48 kg.
Students who are preparing to improve their aptitude skills
and those who are preparing for the competitive exams like ACT, SAT must need this tricks to solve ratio and proportion problems pdf in order to have better score. Because, today there is no competitive
exam without questions from the topic ratio and proportion . Whether a person
is going to write placement exam to get placed or a students is going to write
a competitive exam in order to get admission in university, they are in need of the tricks to solve ratio and proportion problems pdf.
As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to know the ideas given in this tricks to solve ratio and proportion problems pdf. Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes
Students have to learn few basic operations in this topic ratio and proportion. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff "tricks to solve ratio and proportion problems pdf" is the one much needed to solve the problems in a very short time.