Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 17 Solutions

Question 1. Carry out the following divisions.
(i) 4.8÷2
(ii) 17.5÷5
(iii) 20.6÷2
(iv) 32.5÷25
Answer:
(i) \( 4.8 \div 2 \)
\( = \frac{48}{10} \div \frac{2}{1} \)
\( = \frac{48}{10} \times \frac{1}{2} \)
\( = \frac{48}{20} \)
\( = \frac{48 \times 5}{20 \times 5} \)
\( = \frac{240}{100} \)
\( = 2.4 \)
(ii) \( 17.5 \div 5 \)
\( = \frac{175}{10} \div \frac{5}{1} \)
\( = \frac{175}{10} \times \frac{1}{5} \)
\( = \frac{175}{50} \)
\( = \frac{175 \times 2}{50 \times 2} \)
\( = \frac{350}{100} \)
\( = 3.5 \)
(iii) \( 20.6 \div 2 \)
\( = \frac{206}{10} \div \frac{2}{1} \)
\( = \frac{206}{10} \times \frac{1}{2} \)
\( = \frac{206}{20} \)
\( = \frac{206 \times 5}{20 \times 5} \)
\( = \frac{1030}{100} \)
\( = 10.3 \)
(iv) \( 32.5 \div 25 \)
\( = \frac{325}{10} \div \frac{25}{1} \)
\( = \frac{325}{10} \times \frac{1}{25} \)
\( = \frac{325}{250} \)
\( = \frac{325 \times 4}{250 \times 4} \)
\( = \frac{1300}{1000} \)
\( = 1.3 \) In simple words: To divide a decimal number by a whole number, convert the decimal to a fraction, then multiply by the reciprocal of the divisor. Simplify the fraction to get the final decimal answer.

🎯 Exam Tip: Remember to convert decimal numbers to fractions before performing division, especially when the divisor is a whole number, to simplify calculations and avoid errors in decimal point placement.

Question 2. A road is 4 km 800 m long. If trees are planted on both its sides at intervals of 9.6 m, how many trees were planted?
Answer: Length of road = 4 km 800 m
\( = 4 \times 1000 \text{ m} + 800 \text{ m} \)
\( = 4000 \text{ m} + 800 \text{ m} \)
\( = 4800 \text{ m} \)
Number of trees on one side \( = 4800 \div 9.6 \)
\( = \frac{4800}{1} \div \frac{96}{10} \)
\( = \frac{4800}{1} \times \frac{10}{96} \)
\( = \frac{48000}{96} \)
\( = 500 \)
\( \therefore \) Number of trees on both sides \( = 2 \times \) number of trees on one side
\( = 2 \times 500 = 1000 \)
If the trees are planted at the beginning of the road, then
Total number of trees \( = 1000 + 2 = 1002 \)
\( \therefore \) Total number of trees planted is 1000 or 1002. In simple words: First, convert the road's length to meters. Then divide the total length by the interval between trees to find trees on one side. Multiply by two for both sides and add two more if trees are planted at the very beginning of the road.

🎯 Exam Tip: Always ensure consistent units (e.g., meters) before performing calculations. Remember to consider both sides of the road and the initial tree placement for accuracy in tree planting problems.

Question 3. Pradnya exercises regularly by walking along a circular path on a field. If she walks a distance of 3.825 km in 9 rounds of the path, how much does she walk in one round?
Answer: Total distance walked in 9 rounds = 3.825 km
\( \therefore \) Distance walked in 1 round \( = 3.825 \div 9 \)
\( = \frac{3825}{1000} \div \frac{9}{1} \)
\( = \frac{3825}{1000} \times \frac{1}{9} \)
\( = \frac{425}{1000} \)
\( = 0.425 \text{ km} \)
\( \therefore \) Total distance walked in 1 round is 0.425 km. In simple words: To find the distance walked in one round, simply divide the total distance covered by the number of rounds.

🎯 Exam Tip: Direct division is key here. Ensure correct decimal placement in the quotient, as it's a common area for errors in such problems.

Question 4. A pharmaceutical manufacturer bought 0.25 quintal of hirada, a medicinal plant, for Rs 9500. What is the cost per quintal of hirada? (1 quintal = 100 kg)
Answer: Cost of 0.25 quintal of hirada = Rs 9500
\( \therefore \) Cost of 1 quintal of hirada \( = 9500 \div 0.25 \)
\( = 9500 \div \frac{25}{100} \)
\( = 9500 \times \frac{100}{25} \)
\( = \frac{950000}{25} \)
\( = \frac{950000 \times 4}{25 \times 4} \)
\( = \frac{3800000}{100} \)
\( = \text{Rs } 38,000 \)
\( \therefore \) Cost per quintal of hirada is Rs 38,000. In simple words: To find the cost per quintal, divide the total cost by the number of quintals purchased.

🎯 Exam Tip: When dividing by a decimal, convert the divisor to a whole number by multiplying both numerator and denominator by a power of 10. Pay close attention to currency units (Rs).

Maharashtra Board Class 6 Maths Chapter 4 Operations On Fractions Practice Set 17 Intext Questions And Activities

Question 1. Maths is fun! (Textbook pg. no. 34)
1. Consider any three-digit number (say 527).
2. Multiply the number by 7. Then multiply the product obtained by 13, and this product by 11.
3. The found product is 5,27,527.
Take two or three other numbers. Do the same multiplication and find out how it is done.
Answer: Solution:
\( 7 \times 13 \times 11 = 1001 \)
\( \therefore 527 \times 1001 = 527 \times (1000 + 1) \)
\( = (527 \times 1000) + (527 \times 1) \)
\( = 527000 + 527 = 527527 \)
Thus, when any three-digit number is multiplied with 1001, the product obtained is a six-digit number in which the original three-digit number is written back to back twice.
(Students may consider any other three-digit numbers and verify the property.) In simple words: Multiplying any three-digit number by 1001 (which is \( 7 \times 13 \times 11 \)) results in a six-digit number where the original three-digit number appears twice consecutively.

🎯 Exam Tip: Recognize patterns in multiplication, especially with numbers like 1001. This property demonstrates a useful shortcut for certain types of number problems.