Railway Exams Maths Ratio and Proportion: Concepts, Formulas and Solved Questions

Ratio and Proportion is one of the most important topics in the Arithmetic section of Railway Exams and other competitive examinations. Questions from this topic frequently test a candidate's ability to compare quantities, distribute values, and solve proportional relationships quickly. 

Understanding the fundamental concepts of ratio, direct proportion, mean proportion, third proportion, and fourth proportion can significantly improve problem-solving speed and accuracy. Here, you will learn the main concepts, formulas, and problem-solving methods used in ratio and proportion questions.

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Railway Exams Maths Arithmetic Ratio & Proportion Overview

Railway Exams Maths Arithmetic Ratio & Proportion is a key topic in competitive exam preparation. It helps candidates solve questions related to comparisons, distribution, and proportional relationships. This guide covers important concepts, formulas, and problem-solving techniques commonly used in railway exam questions.

Combining Ratios (A:B, B:C to A:B:C)

This section focuses on combining individual ratios like A:B and B:C to derive a combined ratio A:B:C.

Method 1: Option-Based Approach

This method is often quick. You identify an option where A:B matches the given ratio (e.g., 11:7). If multiple options fit, check the B:C ratio (e.g., 5:19) in the remaining options to find the correct answer.

Method 2: Systematic Calculation

This is a fundamental, step-by-step method ensuring correctness.

1. Align Ratios: Write down the variables A, B, C.

• A:B = 11:7

• B:C = 5:19

• Represent as:
  A B C 11 7 _ _ 5 19

2. Fill Blanks: Fill empty spaces with the adjacent values.
   A B C 11 7 **7** **5** 5 19

3. Multiply Columns: Multiply values in each column to get the combined ratio.

• (11 * 5) : (7 * 5) : (7 * 19)

• 55 : 35 : 133

Comparative Analysis of Methods

Ratios in "Equal-Equal Form" (e.g., xa = yb = zc)

When terms are presented in an "equal-equal form" like 7a = 4b = 14c, there are methods to find the combined ratio A:B:C.

Method 1: Sequential Ratio Conversion

1. Derive A:B from the first two terms: 7a = 4b => A/B = 4/7.

2. Derive B:C from the second and third terms: 4b = 14c => B/C = 14/4 = 7/2.

3. Combine A:B and B:C using the systematic calculation method.

Method 2: The "Hiding Method"

To find A:B:C:

1. To find A's part: Conceal the coefficient of A (e.g., 7) and multiply the coefficients of B and C (e.g., 4 * 14 = 56).

2. To find B's part: Conceal the coefficient of B (e.g., 4) and multiply the coefficients of A and C (e.g., 7 * 14 = 98).

3. To find C's part: Conceal the coefficient of C (e.g., 14) and multiply the coefficients of A and B (e.g., 7 * 4 = 28).

• This results in: A : B : C = 56 : 98 : 28.

4. Simplify the ratio by dividing by a common factor (e.g., 14).

• 56/14 : 98/14 : 28/14 = 4 : 7 : 2.

Ratio Application: Distribution and Differences

Consider a sum 'X' divided among A, B, and C. Given A:C, B:C, and the difference between A and B, find the total sum 'X'

.

Solution Steps:

1. Establish Combined Ratio A:B:C:

• Given A:C = 8:7 and B:C = 3:2.

• To combine, equate the common element, C. The LCM of 7 and 2 is 14.

• Multiply A:C by 2: (8*2) : (7*2) = 16:14 (A:C)

• Multiply B:C by 7: (3*7) : (2*7) = 21:14 (B:C)

• The combined ratio A:B:C is 16:21:14.

2. Calculate Value of One Ratio Unit:

• If the difference between A and B is 240, then from the combined ratio, A-B = 21 - 16 = 5 ratio units.

• Thus, 5 ratio units = 240, so 1 ratio unit = 240 / 5 = 48.

3. Calculate the Total Sum (X):

• Total sum X = sum of all ratio units = 16 + 21 + 14 = 51 ratio units.

• Total Sum X = 51 * 48.

Direct Proportionality

When a variable A is directly proportional to B (A ∝ B), it means A is equal to a constant (k) multiplied by B: A = kB.

Problem: Given A = 52 when B = 91, find A when B = 126.

Solution Steps:

1. Determine the Constant of Proportionality (k):

• Using the first set of values: 52 = k * 91

• k = 52 / 91. Simplify by dividing by 13: k = 4/7.

2. Calculate the new value of A:

• Using the derived constant and the new B value: A = (4/7) * 126

• A = 4 * 18 = 72.

• (Do not resort to guessing or shortcuts without understanding the underlying concept. Honest effort in solving problems leads to reliable results.)

Combining Multiple Ratios (e.g., P:Q, Q:R, R:S)

For combining multiple ratios into a single P:Q:R:S, the "By Option" method is often the most efficient.

Steps:

1. Check the first ratio (e.g., P:Q) in each given option.

2. Eliminate options that do not match.

3. If multiple options remain, check the next ratio (e.g., Q:R) in the remaining options.

4. Continue until only one option correctly satisfies all given sub-ratios.

• (When using the "By Option" method, avoid haste. Do not select an answer based on only the first part of the ratio. Always verify all given sub-ratios within the chosen option to ensure accuracy and save time during exams.)

Mean Proportion, Third Proportion, and Fourth Proportion: Basic Concepts

These concepts are based on a continuous proportion: A : B = B : C, meaning A/B = B/C.

1. Mean Proportion (मध्य अनुपात):

• B is the mean proportion between A and C.

• Formula: B² = A * C => B = √(A * C)

2. Third Proportion (तृतीय अनुपात):

• C is the third proportion to A and B.

• Formula: C = B² / A

Problem Example (Mean Proportion):

Find the mean proportion of terms (a+b)³(a-b) and (a+b)(a-b)³.

Solution:

Using B = √(A * C):

B = √[ ((a+b)³(a-b)) * ((a+b)(a-b)³) ]

B = √[ (a+b)⁴ * (a-b)⁴ ]

B = (a+b)²(a-b)²

(It is essential to understand the basic concept and formulas for mean, third, and fourth proportions, as this knowledge underpins solving related problems effectively.)

Third and Fourth Proportions

1. Third Proportion:

• Definition: If A, B, and C are in continuous proportion (A:B = B:C), then C is the third proportion to A and B.

• Formula: C = B² / A

• Problem Example: Find the third proportion of A and (B²/4A).

• Here, 'A' is the first term and 'B²/4A' is the second term.

• Third Proportion = (Second Term)² / (First Term)

• C = (B²/4A)² / A = (B⁴ / 16A²) / A = B⁴ / 16A³

2. Fourth Proportion:

• Definition: If A, B, C, and D are in proportion (A:B = C:D), then D is the fourth proportion to A, B, and C.

• Formula: A/B = C/D => AD = BC => D = (B * C) / A

Combined Problem on Third and Fourth Proportions:

Problem: The third proportion of 8 and 20 is P. The fourth proportion of 3, 5, and 24 is Q. Find the value of 2P + Q.

Solution Steps:

1. Calculate P (Third Proportion):

• A = 8, B = 20

• P = B² / A = 20² / 8 = 400 / 8 = 50.

2. Calculate Q (Fourth Proportion):

• A = 3, B = 5, C = 24

• Q = (B * C) / A = (5 * 24) / 3 = 5 * 8 = 40.

3. Calculate 2P + Q:

• 2P + Q = 2 * 50 + 40 = 100 + 40 = 140.

Applications of Mean and Third Proportions (Cont.)

Problem: Find the ratio of the mean proportion of 1.6 and 3.6 to the third proportion of 5 and 8.

Solution Steps:

1. Calculate Mean Proportion (MP) of 1.6 and 3.6:

• MP = √(1.6 * 3.6) = √((16/10) * (36/10)) = √((16 * 36) / 100)

• MP = (√16 * √36) / √100 = (4 * 6) / 10 = 24 / 10 = 2.4

2. Calculate Third Proportion (TP) of 5 and 8:

• TP = B² / A = 8² / 5 = 64 / 5 = 12.8

3. Find the Ratio MP : TP:

• 2.4 : 12.8

• Multiply by 10 to remove decimals: 24 : 128

• Divide by common factor 8: 3 : 16

Ratio Equalities and Expression Evaluation

Concept: If a/x = b/y = c/z, then the ratio a:b:c is x:y:z. This allows for direct substitution (e.g., a=x, b=y, c=z) for calculations.

Problem: Given a/2 = b/3 = c/7, find the ratio of (4a + 3b - c) to (a + b + c).

Solution Steps:

1. Direct Substitution: Assume a=2, b=3, and c=7.

2. Evaluate first expression (4a + 3b - c):

• 4(2) + 3(3) - 7 = 8 + 9 - 7 = 17 - 7 = 10.

3. Evaluate second expression (a + b + c):

• 2 + 3 + 7 = 12.

4. Form and simplify ratio:

• 10 : 12 = 5 : 6.

Ratio Applications (Word Problems)

Problem: Marks Score Distribution

Given the ratio of marks obtained by Rajesh, Rakesh, and Ramesh as 2 : 4 : 9, and Rajesh scored 30 marks. Calculate individual marks for Rakesh and Ramesh.

Example:

• Rajesh's marks (2 ratio units) = 30 marks.

• Therefore, 1 ratio unit = 30 / 2 = 15 marks.

• Rakesh's marks = 4 * 15 = 60.

• Ramesh's marks = 9 * 15 = 135.

Advanced Ratio Equalities and Expression Evaluation

Problem: Given k²/m = m/16, find the value of (k² + 2m²) : (k² + m²).

Solution Steps:

1. Establish a relationship between k and m:

• k²/m = m/16 => 16k² = m²

• Solve for m: m = √(16k²) = 4k.

2. Substitute 'm' into the expression:

• (k² + 2(4k)²) : (k² + (4k)²)

3. Simplify the expression:

• (k² + 2 * 16k²) : (k² + 16k²)

• (k² + 32k²) : (k² + 16k²)

• 33k² : 17k²

• Divide by k² (assuming k ≠ 0): 33 : 17.

Chained Ratios and Simplification

Problem: Given C = (2/3)D and D = (3/8)E, find the ratio of C to E (C:E).

Solution Steps:

1. Express given relations as ratios:

• C = (2/3)D => C:D = 2:3

• D = (3/8)E => D:E = 3:8

2. Combine Ratios (C:D:E):

• The common variable 'D' already has the same ratio value (3) in both expressions.

• Therefore, C:D:E = 2:3:8.

3. Extract the required ratio (C:E):

• From C:D:E = 2:3:8, C=2 and E=8.

• The ratio C:E = 2:8.

• Simplify: 1:4.

• (When combining chained ratios, ensure the common element's value is consistent across all ratios. If not, adjust the ratios by multiplication. Always double-check the final ratio's direction, e.g., 1:4 versus 4:1.)

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