RBSE Solutions Class 6 Maths Chapter 13 Ratio and Proportion Exercise 13.3

Question 1. If the cost of 1 quintal of sugar is Rs 2700, find the cost of 1 kg sugar.
Answer: First, we need to know that 1 quintal is equal to 100 kg. The cost of 100 kg of sugar is Rs. 2700. To find the cost of 1 kg, we divide the total cost by the total kilograms. So, we divide Rs. 2700 by 100. This calculation gives us Rs. 27. Therefore, 1 kg of sugar costs Rs. 27.
In simple words: 1 quintal means 100 kg. If 100 kg sugar costs Rs. 2700, then 1 kg sugar will cost Rs. 2700 divided by 100, which is Rs. 27.
🎯 Exam Tip: Remember the common conversions, like 1 quintal = 100 kg, as these are often used in ratio and proportion problems.

Question 2. The bus fare for 200 km is Rs 150. What will be the fare for 500 km?
Answer: The bus fare for traveling 200 km is Rs 150. To find the fare for 1 km, we divide the total fare by the distance: \( \frac { 150 }{ 200 } \) Rs. Once we have the fare per km, we can multiply it by 500 to find the fare for 500 km.
\( \text{Fare for 1 km} = \frac { 150 }{ 200 } \) Rs.
\( \implies \) \( \text{Fare for 500 km} = \frac { 150 }{ 200 } \times 500 \)
\( \implies \) \( = \frac { 150 \times 5 }{ 2 } \)
\( \implies \) \( = 75 \times 5 \)
\( \implies \) \( = 375 \) Rs.
So, the bus fare for 500 km will be Rs. 375.
In simple words: First, find out how much it costs for 1 km (Rs. 150 divided by 200). Then, multiply that cost by 500 to get the total fare for 500 km, which is Rs. 375.

🎯 Exam Tip: When solving problems involving distance and fare, always calculate the cost per unit distance (like per km) first, then multiply by the new total distance.

Question 3. If the interest on Rs. 700 is Rs. 168 for a certain amount of time. What will be the interest on Rs. 1500 for the same time and same rate of interest?
Answer: We know that the interest on Rs. 700 is Rs. 168. To find the interest for Rs. 1, we can divide the interest by the principal amount: \( \frac { 168 }{ 700 } \) Rs. Since the time and rate of interest are the same, we can then multiply this per-rupee interest by Rs. 1500 to find the interest on Rs. 1500.
\( \text{Interest for Rs. 1} = \frac { 168 }{ 700 } \) Rs.
\( \implies \) \( \text{Interest for Rs. 1500} = \frac { 168 }{ 700 } \times 1500 \)
\( \implies \) \( = \frac { 168 \times 15 }{ 7 } \)
\( \implies \) \( = 24 \times 15 \)
\( \implies \) \( = 360 \) Rs.
Therefore, the interest on Rs. 1500 for the same time and rate will be Rs. 360.
In simple words: If Rs. 700 gives Rs. 168 interest, then find how much interest Rs. 1 gives. Then multiply that by Rs. 1500 to get the interest for Rs. 1500, which is Rs. 360.

🎯 Exam Tip: When comparing interest amounts for different principal values with the same time and rate, find the interest per unit of principal first to simplify calculations.

Question 5. 5 persons can save 15 litres of water if they take bath using a bucket instead of shower. Then 25 persons can save how many litres of water?
Answer: We are told that 5 people can save 15 liters of water. To find out how much water 1 person can save, we divide the total water saved by the number of people: \( \frac { 15 }{ 5 } \) liters. Since 1 person saves 3 liters, we can then multiply this by 25 to find the total water saved by 25 people.
\( \text{1 person can save} = \frac { 15 }{ 5 } = 3 \) litres of water
\( \implies \) \( \text{25 person can save} = 3 \times 25 \)
\( \implies \) \( = 75 \) litres of water
So, 25 persons can save a total of 75 liters of water.
In simple words: If 5 people save 15 litres of water, then 1 person saves 3 litres. So, 25 people will save 25 times 3 litres, which is 75 litres of water.

🎯 Exam Tip: For problems involving multiple individuals, always calculate the amount for a single person first, then scale it up for the larger group.

Question 6. A motorbike travels 120 km in 2 litres of petrol. How many litres of petrol will it require to travel 300 km?
Answer: The motorbike uses 2 litres of petrol to travel 120 km. To find out how much petrol is needed for 1 km, we divide the total petrol by the distance: \( \frac { 2 }{ 120 } \) litres. Then, to find the petrol needed for 300 km, we multiply this amount by 300. This helps us calculate the exact fuel efficiency.
\( \text{Petrol required for travelling 1 km} = \frac { 2 }{ 120 } \) litres
\( \implies \) \( \text{Petrol required for travelling 300 km} = \frac { 2 }{ 120 } \times 300 \)
\( \implies \) \( = \frac { 600 }{ 120 } \)
\( \implies \) \( = 5 \) litres
Therefore, 5 litres of petrol will be needed to travel 300 km.
In simple words: The bike uses 2 litres for 120 km. So, for 1 km, it uses \( \frac { 2 }{ 120 } \) litres. For 300 km, it will use \( \frac { 2 }{ 120 } \times 300 \), which is 5 litres.

🎯 Exam Tip: In fuel consumption problems, always determine the fuel used per unit distance (e.g., litres per km) first to easily calculate for any other distance.

Question 7. A train travels 130 km in 2 hours. How much time it required to cover 520 km with the same speed?
Answer: The train covers 130 km in 2 hours. To find the time required to travel 1 km, we divide the total time by the distance: \( \frac { 2 }{ 130 } \) hours. Since the speed is constant, we can multiply this time-per-km by 520 km to find the total time needed for that longer distance.
\( \text{Time required for travelling 1 km} = \frac { 2 }{ 130 } \) hours
\( \implies \) \( \text{Time required for travelling 520 km} = \frac { 2 }{ 130 } \times 520 \)
\( \implies \) \( = 2 \times 4 \)
\( \implies \) \( = 8 \) hours
Thus, the train will require 8 hours to cover 520 km.
In simple words: The train takes 2 hours for 130 km. So, for 1 km, it takes \( \frac { 2 }{ 130 } \) hours. For 520 km, it will take \( \frac { 2 }{ 130 } \times 520 \), which means 8 hours.

🎯 Exam Tip: For distance-time-speed problems where speed is constant, always calculate the time taken per unit distance or distance covered per unit time first.

Question 9. Geeta pays Rs. 10500 as rent for 3 months. How much does she has to pay for a whole year, (if the rent per month remains same)?
Answer: Geeta pays Rs. 10500 for 3 months of rent. To find the rent for 1 month, we divide the total rent by the number of months: \( \frac { 10500 }{ 3 } \) Rs. Since there are 12 months in a year, we multiply the monthly rent by 12 to find the yearly rent. This calculation shows the total cost over a longer period.
\( \text{Rent for 1 month} = \frac { 10500 }{ 3 } = 3500 \) Rs.
\( \implies \) \( \text{Rent for 12 months (or 1 year)} = 3500 \times 12 \)
\( \implies \) \( = 42000 \) Rs.
Therefore, Geeta has to pay Rs. 42000 for a whole year.
In simple words: Geeta pays Rs. 10500 for 3 months, so her rent for one month is Rs. 3500. For a whole year (12 months), she will pay 12 times Rs. 3500, which is Rs. 42000.

🎯 Exam Tip: When dealing with payments over periods, always find the cost per single unit (like per month) first, then scale it to the total period requested.

Question 10. Rahim made 48 runs in 8 overs and Kabir made 54 runs in 6 overs. Who made more runs per over?
Answer: To find out who made more runs per over, we need to calculate the average runs per over for both Rahim and Kabir. For Rahim, we divide his total runs (48) by the number of overs he played (8). For Kabir, we divide his total runs (54) by his overs (6). Comparing these two averages will show us who scored better.
\( \text{Rahim made in 1 over} = \frac { 48 }{ 8 } = 6 \) Runs
\( \implies \) \( \text{Kabir made in 1 over} = \frac { 54 }{ 6 } = 9 \) Runs
Since 9 runs per over is greater than 6 runs per over, Kabir made more runs per over.
In simple words: Rahim scored 6 runs in each over (48 divided by 8). Kabir scored 9 runs in each over (54 divided by 6). Kabir scored more runs per over.

🎯 Exam Tip: To compare performances that involve different total quantities, always calculate the performance per unit (e.g., runs per over) for each case.

Question 11. Co millet is Rs. 49.50.
Answer: The problem implies that 3 kg of millet costs Rs. 49.50. We need to find the cost of 7 kg millet and the quantity of millet that can be purchased for Rs. 165.
(i) Cost of 7 kg millet:
\( \text{Cost of 3 kg millet} = \text{Rs. } 49.50 \)
\( \implies \) \( \text{Cost of 1 kg millet} = \frac { 49.50 }{ 3 } = \text{Rs. } 16.50 \)
\( \implies \) \( \text{Cost of 7 kg millet} = 16.50 \times 7 = \text{Rs. } 115.50 \)
(ii) Millet purchased in Rs. 165:
\( \text{Millet purchased in Rs. } 49.50 = 3 \text{ kg} \)
\( \implies \) \( \text{Millet purchased in Rs. 1} = \frac { 3 }{ 49.50 } \text{ kg} \)
\( \implies \) \( \text{Millet purchased in Rs. 165} = \frac { 3 }{ 49.50 } \times 165 \)
\( \implies \) \( = \frac { 3 \times 165 }{ 49.50 } = \frac { 495 }{ 49.50 } \)
\( \implies \) \( = 10 \text{ kg} \)
So, 7 kg of millet will cost Rs. 115.50, and 10 kg of millet can be purchased for Rs. 165.
In simple words: If 3 kg of millet costs Rs. 49.50, then 1 kg costs Rs. 16.50. So, 7 kg will cost Rs. 115.50. Also, if Rs. 49.50 buys 3 kg, then Rs. 1 buys \( \frac { 3 }{ 49.50 } \) kg. So, Rs. 165 will buy \( \frac { 3 }{ 49.50 } \times 165 = 10 \) kg of millet.

🎯 Exam Tip: When a question has multiple parts, break it down and solve for the unit value first (cost per kg or kg per rupee) before calculating for the specific quantities requested.