Ratio And Proportion Formula, Questiosn, problems And Examples

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Mathematics is one of the most important that is asked in the government recruitment exams. Generally, there are questions asked related to basic concepts and formulas of trigonometry. To let you make the most of the Mathematics section, we are providing important facts related to Ratio And Proportion. Also, RRB NTPC CBT 2, CGL Tier 2, and State exams are nearby with bunches of posts for the interested candidates in which Mathematics is a major part. We have covered important notes and questions focusing on these prestigious exams. We wish you all the best of luck to come over the fear of the Mathematics section.

Ratio And Proportion

Ratio and Proportion is an important concept while dealing with various competitive examinations. The ratio is the relation between two amounts showing how one value is dependent on another. While proportion can be defined as a part or portion of number in comparison to the whole number. This article consists of study notes on Ratio & proportion containing the ratio and proportion formula that must be applied to solve the questions. The examples with solutions have also been provided with respect to Ratio and Proportion problems.

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Ratio & Proportion

a : b ∷ c : d

Product of Means = Product of Extremes

a×d = b×c

Fourth Proportion

a : b ∷ c : x

x → Fourth Proportion

x=(b×c)/a

Ratio and proportion questions

Example. Find the fourth proportion to the numbers 4, 10, 12.

Sol. Fourth Proportion

=(12×10)/4

= 30

Third Proportion→

a : b ∷ b : x

x → Third Proportion

Third Proportion of a, b = b²/a

Now let’s look at some more ratio and proportion problems

Find the third proportion to the numbers 4, 12.

Sol. Third Proportion

=(12×12)/4

= 36

Mean Proportional

a : x ∷ x : b

x → Mean Proportion

Mean Proportion of ab is given by = √ab

Ratio and Proportion Example. Find the mean proportion of 4, 16?

Sol. Mean proportion = √(4×16)

=√64

= 8

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If two numbers are in the ratio a : b and there sum is x, then these numbers will be

ax/(a+b) & bx/(a+b)

If three numbers are in the ratio of a : b : c and there sum is x then the numbers are

ax/(a+b+c) , bx/(a+b+c) & cx/(a+b+c)

If a : b = n₁ : d₁ & b : c = n₂ : d₂

then a : b : c = n₁ × n₂ : n₂ × d₁ : d₁ × d₂

Example. If A : B = 3 : 5 & B : C = 9 : 10, find A : B : C.

Sol. A : B = 3 : 5

B : C = 9 : 10

A : B : C = 3 × 9 : 9 × 5 : 5 × 10

= 27 : 45 : 50

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If a : b = n₁ : d₁ , b : c = n₂ : d₂ , c : d = n₃ : d₃

a : b : c : d = n₁ × n₂ × n₃ : d₁ × n₂ × n₃ : d₁ × d₂ × n₃ : d₁ × d₂ × d₃

Example. If A : B = 2 : 3, B : C = 4 : 5, C : D = 6 : 7. Find A : B : C : D.

Sol. A : B : C : D = 2 × 4 × 6 : 3 × 4 × 6 : 3 × 5 × 6 : 3 × 5 × 7

= 48 : 72 : 90 : 105

= 16 : 24 : 30 : 35

If the ratio between two numbers is a : b & x is added to both of them then the ratio becomes c : d. Then the two numbers are given by:

ax(c-d)/(ad-bc) & bx (c-d)/(ad-bc)

Example. If two numbers are in the ratio of 3 : 4. If 8 is added to both the number, the ratio becomes 5 : 6. Find the numbers.

Sol. 1st Number

=(3×8 (5-6))/(3×8-5×4)

=(24 (-1))/(-2)=12

Number=(4×8 (5-6))/(3×6-5×4)

=(32×(-1))/((-2))=16

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If the ratio of two numbers is a : b, then the number that should be added to each numbers to make the ratio c : d is given by

(ad-bc)/(c-d)

Example. Find the number that should be added to the numbers in ratio 11 : 29, to make it equal to 11 : 20?

Sol. Number =(ad – bc)/(c – d)

=(11×20-29×11)/(11-20)=11

The incomes of two persons are in the ratio → a : b and their expenditures are is the ratio → c : d. It saving of each person is S, then their incomes are.

aS(d-c)/(ad-bc) & bS(d-c)/(ad-bc)

And their expenditures are given by

cS(b-a)/(ad-bc) & dS(b-a)/(ad-bc)

Example. The annual salary of A & B are in the ratio of 5 : 4 and their annual expenses bear a ratio of 4 : 3. If each of them saves Rs. 800 at the end of year. Find their Incomes.

Sol. A’s Incomes

=(5×500(3-4))/(15-16)b

= 2500 Rs.

B’s Income

=(4×500(3-4))/(15-16)=2000 Rs.

When two ingredients A & B of quantities q₁ & q₂ with cost price/unit c₁ & c₂ respectively are mixed to get a mixture c having cost price cm/unit then.

(a) Ratio in which A & B are mixed

(q₁)/q₂ =(c₂-cm)/(cm-c₁ )

(b) Cost of the mixture

cm = (c₁×q₁+c₂×q₂)/(q₁+q₂)

Example. In what ratio two kinds of tea must be mixed together into one at Rs. 9/kg and another at Rs. 15/kg, so that the mixture may cost Rs. 10.2/kg?

Sol.

q₁/q₂ =(15-10.2)/(10.2-9)=4.8/1.2

= 4 : 1

Example. In a mixture of two types of oils O₁ & O₂ the ratio O₁ : O₂ is 3 : 2. If the cost of oil O₁ is Rs. 4/L and that of oil O₂ is Rs. 9/L. Then find the cost of Resulting mixture?

Sol.

c_m=(c₁×q₁+c₂×q₂)/(q₁+q₂ )

=(4×3+9×2)/(3+2)

=(12+18)/5=30/5=Rs.6

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