ssc gd ratio and proportion

• Understanding Simple Ratio for SSC GD
• Comparing Ratios: Key to SSC GD Ratio and Proportion Questions
• What is Proportion?
• Direct Proportion in SSC GD Ratio and Proportion
• Inverse Proportion and SSC GD Ratio and Proportion Questions PDF
• Solving Word Problems: Ratio and Proportion SSC GD Questions

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Understanding Simple Ratio for SSC GD

Welcome! We are going to learn about ssc gd ratio and proportion. This is a very important topic for your exam.

What is a Ratio?

A ratio is a way to compare two numbers by dividing them. It tells you how much of one thing there is compared to another thing.

• Imagine you have 3 red apples and 2 green apples.
• The ratio of red apples to green apples is 3 to 2.
• We write this as $3:2$.
• The ratio $3:2$ is the same as the fraction $\frac{3}{2}$.

Key Points:

• The two numbers in a ratio are called terms.
• Ratios must always be in their simplest form (like simplifying a fraction).
• If you multiply or divide both parts of the ratio by the same number, the ratio stays the same. This is crucial when solving ssc gd ratio and proportion questions.

Solved Examples: Simple Ratio

Example 1: Simplifying a Ratio

Simplify the ratio $15:25$.

Solution:

1. Understand the Goal: We need to find the smallest whole numbers that represent this comparison.
2. Find the Common Factor: We look for the biggest number that can divide both 15 and 25. That number is 5.
3. Divide Both Terms:
   • $15÷5=3$
   • $25÷5=5$
4. Write the Answer: The simplified ratio is $3:5$.
5. So, the answer is $3:5$.

Example 2: Finding a Part of the Total

A class has 40 students. The ratio of boys to girls is $3:5$. How many girls are there?

Solution:

1. Understand the Ratio Parts: The ratio $3:5$ means 3 parts are boys and 5 parts are girls.
2. Find the Total Parts: We add the parts together: $3+5=8$ total parts.
3. Find the Value of One Part: The total number of students (40) is equal to 8 parts.
   • Value of 1 part = $\frac{\text{Total Students}}{\text{Total Parts}}$
   • Value of 1 part = $\frac{40}{8}=5$.
4. Calculate the Number of Girls: Girls have 5 parts in the ratio.
   • Number of girls = $5\times (\text{Value of 1 part})$
   • Number of girls = $5\times 5=25$.
5. So, the answer is 25 girls. These types of ratio and proportion ssc gd questions are very common.

Example 3: Combining Ratios

If $A:B=2:3$ and $B:C=4:5$, find the combined ratio $A:B:C$.

Solution:

1. Identify the Problem: The number for B is different in both ratios (3 and 4). We must make B the same.
2. Find the Common Multiple for B: We find the Least Common Multiple (LCM) of 3 and 4. The LCM is 12.
3. Adjust Ratio 1 ($A:B$): To make $B=12$, we multiply $3$ by $4$. We must multiply $A$ by 4 too.
   • $A:B=(2\times 4):(3\times 4)=8:12$.
4. Adjust Ratio 2 ($B:C$): To make $B=12$, we multiply $4$ by $3$. We must multiply $C$ by 3 too.
   • $B:C=(4\times 3):(5\times 3)=12:15$.
5. Combine the Ratios: Now B is 12 in both.
   • $A:B:C=8:12:15$.
6. So, the answer is $8:12:15$.

Example 4: Ratio of Different Units

Find the ratio of 500 grams to 2 kilograms.

Solution:

1. Rule Check: Ratios must compare the same units. We must change kilograms to grams.
2. Convert Units: We know that 1 kilogram = 1000 grams.
   • 2 kilograms = $2\times 1000=2000$ grams.
3. Write the Ratio: The ratio is 500 grams : 2000 grams.
   • Ratio = $500:2000$.
4. Simplify the Ratio: We can divide both sides by 500.
   • $500÷500=1$
   • $2000÷500=4$
5. So, the answer is $1:4$. Mastering these conversions helps with ssc gd ratio and proportion questions pdf practice.

Comparing Ratios: Key to SSC GD Ratio and Proportion Questions

Sometimes you need to know which ratio is bigger or smaller. This is called comparing ratios.

How to Compare Ratios

To compare ratios, we turn them into fractions and find a common denominator (the bottom number).

• If the bottom numbers are the same, we just look at the top numbers to see which fraction is bigger.
• This method is very helpful when solving complex ssc gd ratio and proportion questions.

Solved Examples: Comparing Ratios

Example 1: Comparing Two Ratios

Which ratio is larger: $2:3$ or $3:4$?

Solution:

1. Turn Ratios into Fractions:
   • $2:3$ becomes $\frac{2}{3}$
   • $3:4$ becomes $\frac{3}{4}$
2. Find the Common Denominator: The LCM of 3 and 4 is 12.
3. Change the First Fraction: To make the bottom 12, we multiply 3 by 4. We must multiply the top by 4.
   • $\frac{2}{3}=\frac{2\times 4}{3\times 4}=\frac{8}{12}$
4. Change the Second Fraction: To make the bottom 12, we multiply 4 by 3. We must multiply the top by 3.
   • $\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12}$
5. Compare: We compare $\frac{8}{12}$ and $\frac{9}{12}$. Since $9>8$, the second ratio is larger.
6. So, the answer is $3:4$ is the larger ratio.

Example 2: Ordering Ratios

Arrange the ratios $1:2$, $2:5$, and $3:4$ in ascending order (smallest to largest).

Solution:

1. Write as Fractions: $\frac{1}{2}$, $\frac{2}{5}$, $\frac{3}{4}$.
2. Find the Common Denominator: The LCM of 2, 5, and 4 is 20.
3. Convert All Fractions to Denominator 20:
   • $\frac{1}{2}=\frac{1\times 10}{2\times 10}=\frac{10}{20}$
   • $\frac{2}{5}=\frac{2\times 4}{5\times 4}=\frac{8}{20}$
   • $\frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20}$
4. Order the Numerators: The order is 8, 10, 15.
5. Write the Ratios in Order: $2:5$, $1:2$, $3:4$.
6. So, the answer is $2:5,1:2,3:4$.

Example 3: Quick Comparison (Decimal Method)

Which is smaller: $1:5$ or $1:6$?

Solution:

1. Turn Ratios into Decimals: This is a fast way to check for ssc gd ratio and proportion questions.
   • $1:5=\frac{1}{5}=0.20$
   • $1:6=\frac{1}{6}\approx 0.166$
2. Compare Decimals: Since $0.166$ is smaller than $0.20$.
3. Identify the Smaller Ratio: $1:6$ is smaller.
4. So, the answer is $1:6$.

Example 4: Finding the Missing Ratio

If $A:B$ is smaller than $5:6$, which of the following could be $A:B$? ($4:5$ or $6:7$)

Solution:

1. Convert $5:6$ to Decimal: $5÷6\approx 0.833$.
2. Check Option 1 ($4:5$): $4÷5=0.80$.
3. Check Option 2 ($6:7$): $6÷7\approx 0.857$.
4. Compare: We need a ratio smaller than $0.833$. $0.80$ is smaller, but $0.857$ is larger.
5. So, the answer is $4:5$.

What is Proportion?

Proportion means that two ratios are equal to each other. It is the foundation of many ssc gd ratio and proportion questions pdf problems.

• If $A:B$ is the same as $C:D$, we say they are in proportion.
• We write this as $A:B::C:D$. (The $::$ means “is in proportion to”).
• This means $\frac{A}{B}=\frac{C}{D}$.

The Rule of Proportion

In any proportion, the product of the Extremes (the outside numbers) is equal to the product of the Means (the inside numbers).

• In $A:B::C:D$:
  • Extremes are A and D.
  • Means are B and C.
• Formula: $A\times D=B\times C$.

Solved Examples: Proportion

Example 1: Checking for Proportion

Are the numbers 2, 4, 6, and 12 in proportion?

Solution:

1. Set up the Ratios: We check if $2:4$ is equal to $6:12$.
2. Identify Means and Extremes:
   • Extremes: 2 and 12.
   • Means: 4 and 6.
3. Apply the Rule: Multiply the Extremes and the Means.
   • Product of Extremes: $2\times 12=24$.
   • Product of Means: $4\times 6=24$.
4. Compare: Since $24=24$, the numbers are in proportion.
5. So, the answer is Yes, they are in proportion.

Example 2: Finding the Fourth Proportional

Find the fourth proportional to 3, 9, and 4.

Solution:

1. Set up the Proportion: Let the missing number be $x$.
   • $3:9::4:x$
2. Apply the Means and Extremes Rule:
   • Extremes: $3\times x$
   • Means: $9\times 4$
   • $3\times x=9\times 4$
3. Calculate the Product of Means: $9\times 4=36$.
   • $3x=36$
4. Solve for x: Divide 36 by 3.
   • $x=\frac{36}{3}$
   • $x=12$
5. So, the answer is 12. This is a standard type of ssc gd ratio and proportion problem.

Example 3: Finding the Mean Proportional

Find the mean proportional between 4 and 9.

Solution:

1. Understand Mean Proportional: The mean proportional is the middle term when the two middle terms are the same.
   • $4:x::x:9$
2. Apply the Rule:
   • Product of Extremes: $4\times 9=36$.
   • Product of Means: $x\times x=x^2$.
   • $x^2=36$
3. Solve for x: We need the number that, when multiplied by itself, gives 36.
   • $x=\sqrt{36}$
   • $x=6$.
4. So, the answer is 6.

Example 4: Finding the Third Proportional

Find the third proportional to 5 and 10.

Solution:

1. Set up the Proportion: The third proportional means the second number is repeated as the mean.
   • $5:10::10:x$
2. Apply the Rule:
   • Product of Extremes: $5\times x$
   • Product of Means: $10\times 10$
   • $5x=100$
3. Solve for x:
   • $x=\frac{100}{5}$
   • $x=20$.
4. So, the answer is 20.

Direct Proportion in SSC GD Ratio and Proportion

Direct proportion means that if one quantity increases, the other quantity increases too. If one quantity decreases, the other decreases too.

• Example: If you buy more pens, you pay more money.
• The ratio between the two quantities always stays the same (it is constant).
• Formula: $\frac{A_1}{B_1}=\frac{A_2}{B_2}$

Solved Examples: Direct Proportion

Example 1: Simple Cost Calculation

If 5 pencils cost ₹25, how much will 8 pencils cost?

Solution:

1. Identify the Relationship: This is Direct Proportion (more pencils = more cost).
2. Set up the Ratio: Let $x$ be the cost of 8 pencils.
   • $\frac{{\text{Pencils}}_1}{{\text{Cost}}_1}=\frac{{\text{Pencils}}_2}{{\text{Cost}}_2}$
   • $\frac{5}{25}=\frac{8}{x}$
3. Cross-Multiply (Means and Extremes):
   • $5\times x=25\times 8$
   • $5x=200$
4. Solve for x:
   • $x=\frac{200}{5}$
   • $x=40$.
5. So, the answer is ₹40. Practice these ssc gd ratio and proportion questions often.

Example 2: Distance and Time

A car travels 150 km in 3 hours. How far can it travel in 5 hours at the same speed?

Solution:

1. Identify the Relationship: Direct Proportion (more time = more distance).
2. Set up the Ratio: Let $x$ be the distance traveled in 5 hours.
   • $\frac{{\text{Distance}}_1}{{\text{Time}}_1}=\frac{{\text{Distance}}_2}{{\text{Time}}_2}$
   • $\frac{150}{3}=\frac{x}{5}$
3. Cross-Multiply:
   • $150\times 5=3\times x$
   • $750=3x$
4. Solve for x:
   • $x=\frac{750}{3}$
   • $x=250$.
5. So, the answer is 250 km.

Example 3: Finding the Constant Ratio

If $x$ and $y$ are in direct proportion, and $x=10$ when $y=5$, find $y$ when $x=18$.

Solution:

1. Find the Constant Ratio (k): In direct proportion, $k=\frac{x}{y}$.
   • $k=\frac{10}{5}=2$.
2. Use the Constant for the New Values: The ratio must still be 2.
   • $\frac{x_2}{y_2}=2$
   • $\frac{18}{y}=2$
3. Solve for y:
   • $18=2\times y$
   • $y=\frac{18}{2}$
   • $y=9$.
4. So, the answer is 9.

Example 4: Unitary Method for Direct Proportion

If 12 kg of rice costs ₹480, what is the cost of 7 kg of rice?

Solution:

1. Find the Cost of One Unit (1 kg):
   • Cost of 1 kg = $\frac{\text{Total Cost}}{\text{Total Weight}}$
   • Cost of 1 kg = $\frac{480}{12}=40$. (₹40 per kg).
2. Calculate the New Cost:
   • Cost of 7 kg = $7\times (\text{Cost of 1 kg})$
   • Cost of 7 kg = $7\times 40=280$.
3. So, the answer is ₹280. This is a fast way to solve ratio and proportion ssc gd questions pdf problems.

Inverse Proportion and SSC GD Ratio and Proportion Questions PDF

Inverse proportion means that if one quantity increases, the other quantity decreases.

• Example: If you hire more workers (increase), the time taken to finish the job decreases.
• The product of the two quantities always stays the same (it is constant).
• Formula: $A_1\times B_1=A_2\times B_2$

Solved Examples: Inverse Proportion

Example 1: Men and Work

4 workers can build a wall in 12 days. How many days will it take 6 workers to build the same wall?

Solution:

1. Identify the Relationship: Inverse Proportion (more workers = less time).
2. Set up the Product Rule: Let $x$ be the number of days for 6 workers.
   • ${\text{Workers}}_1\times {\text{Days}}_1={\text{Workers}}_2\times {\text{Days}}_2$
   • $4\times 12=6\times x$
3. Calculate the Product:
   • $48=6x$
4. Solve for x:
   • $x=\frac{48}{6}$
   • $x=8$.
5. So, the answer is 8 days. This is a classic example of ssc gd ratio and proportion questions pdf.

Example 2: Speed and Time

A car traveling at 60 km/h takes 2 hours to reach a city. How long will it take if the car travels at 40 km/h?

Solution:

1. Identify the Relationship: Inverse Proportion (slower speed = more time).
2. Set up the Product Rule: Let $x$ be the new time.
   • ${\text{Speed}}_1\times {\text{Time}}_1={\text{Speed}}_2\times {\text{Time}}_2$
   • $60\times 2=40\times x$
3. Calculate the Product:
   • $120=40x$
4. Solve for x:
   • $x=\frac{120}{40}$
   • $x=3$.
5. So, the answer is 3 hours.

Example 3: Food Supply

A fort has enough food for 100 soldiers for 30 days. If 50 more soldiers join the fort, how long will the food last?

Solution:

1. Calculate Total Soldiers: $100+50=150$ soldiers.
2. Identify the Relationship: Inverse Proportion (more soldiers = fewer days the food lasts).
3. Set up the Product Rule: Let $x$ be the new number of days.
   • ${\text{Soldiers}}_1\times {\text{Days}}_1={\text{Soldiers}}_2\times {\text{Days}}_2$
   • $100\times 30=150\times x$
4. Calculate the Product:
   • $3000=150x$
5. Solve for x:
   • $x=\frac{3000}{150}$
   • $x=20$.
6. So, the answer is 20 days.

Example 4: Taps and Tanks

4 taps can fill a tank in 10 hours. How many taps are needed to fill the same tank in 8 hours?

Solution:

1. Identify the Relationship: Inverse Proportion (less time = more taps needed).
2. Set up the Product Rule: Let $x$ be the number of taps needed.
   • ${\text{Taps}}_1\times {\text{Time}}_1={\text{Taps}}_2\times {\text{Time}}_2$
   • $4\times 10=x\times 8$
3. Calculate the Product:
   • $40=8x$
4. Solve for x:
   • $x=\frac{40}{8}$
   • $x=5$.
5. So, the answer is 5 taps.

Solving Word Problems: Ratio and Proportion SSC GD Questions

Word problems combine all the concepts we have learned. We must read carefully to decide if the problem is about simple ratio, direct proportion, or inverse proportion. These are common ratio and proportion ssc gd questions.

Solved Examples: Word Problems

Example 1: Dividing Money in a Ratio

Divide ₹1200 between A and B in the ratio $5:7$. How much does B get?

Solution:

1. Find Total Parts: $5+7=12$ total parts.
2. Find the Value of One Part:
   • Value of 1 part = $\frac{\text{Total Money}}{\text{Total Parts}}$
   • Value of 1 part = $\frac{1200}{12}=100$.
3. Calculate B’s Share: B gets 7 parts.
   • B’s share = $7\times 100=700$.
4. So, the answer is B gets ₹700. This is a basic ssc gd ratio and proportion distribution problem.

Example 2: Ratio Change Problem

The ratio of two numbers is $3:4$. If 5 is added to each number, the new ratio becomes $4:5$. Find the original numbers.

Solution:

1. Represent the Numbers: Let the original numbers be $3x$ and $4x$.
2. Set up the New Ratio Equation: When 5 is added to both, the ratio is $4:5$.
   • $\frac{3x+5}{4x+5}=\frac{4}{5}$
3. Cross-Multiply:
   • $5\times (3x+5)=4\times (4x+5)$
   • $15x+25=16x+20$
4. Solve for x: Move the smaller $x$ to the right and the smaller number to the left.
   • $25-20=16x-15x$
   • $5=x$
5. Find the Original Numbers:
   • First number: $3x=3\times 5=15$.
   • Second number: $4x=4\times 5=20$.
6. So, the answer is 15 and 20.

Example 3: Combined Proportion (Chain Rule)

10 men working 6 hours a day can complete a job in 15 days. How many days will 5 men working 10 hours a day take to complete the same job?

Solution:

1. Identify the Relationship: This is a mix of inverse proportions (more men/hours = less days).
2. Use the Formula: $M_1\times H_1\times D_1=M_2\times H_2\times D_2$
   • $M$ = Men, $H$ = Hours, $D$ = Days.
3. Put in the Numbers: Let $x$ be the unknown number of days.
   • $10\times 6\times 15=5\times 10\times x$
4. Calculate Both Sides:
   • $900=50x$
5. Solve for x:
   • $x=\frac{900}{50}$
   • $x=18$.
6. So, the answer is 18 days. These complex ssc gd ratio and proportion questions pdf problems require careful setup.

Example 4: Mixture Problem

A mixture of milk and water is 60 liters. The ratio of milk to water is $2:1$. How much more water must be added to make the ratio $1:2$?

Solution:

1. Find Initial Quantities: Total parts = $2+1=3$.
   • Value of 1 part = $\frac{60}{3}=20$ liters.
   • Milk: $2\times 20=40$ liters.
   • Water: $1\times 20=20$ liters.
2. Set up the New Ratio: Milk (40L) stays the same. We add $x$ liters of water. The new ratio is $1:2$.
   • $\frac{\text{Milk}}{\text{Water + Added Water}}=\frac{1}{2}$
   • $\frac{40}{20+x}=\frac{1}{2}$
3. Cross-Multiply:
   • $40\times 2=1\times (20+x)$
   • $80=20+x$
4. Solve for x:
   • $x=80-20$
   • $x=60$.
5. So, the answer is 60 liters of water must be added.