Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 16 Solutions

Question 1. If, 317 x 45 = 14265, then 3.17 × 4.5 = ?
Answer:
Solution:
3.17 x 4.5
= 14.265
In simple words: The product of the numbers 3.17 and 4.5 is found by first multiplying the whole number parts (317 and 45) and then placing the decimal point based on the total number of decimal places in the original numbers.

🎯 Exam Tip: When multiplying decimals, count the total number of decimal places in the factors to accurately position the decimal point in the final product.

Question 2. If, 503 × 217 = 109151, then 5.03 × 2.17 = ?
Answer:
Solution:
5.03 x 2.17
= 10.9151
In simple words: To find the product of 5.03 and 2.17, multiply 503 by 217 and then adjust the decimal point according to the total number of decimal places in both factors.

🎯 Exam Tip: Understand that multiplying numbers with decimals involves the same multiplication process as whole numbers, with an additional step of correctly placing the decimal point based on the sum of decimal places in the original numbers.

Question 3.
(i) 2.7 x 1.4
(ii) 6.17 x 3.9
(iii) 0.57 x 2
(iv) 5.04 × 0.7
Answer:
Solution:
(i) 2.7 x 1.4
\( = \frac{27}{10} \times \frac{14}{10} \)
\( = \frac{27 \times 14}{10 \times 10} = \frac{378}{100} \)
= 3.75

(ii) 6.17 x 3.9
\( = \frac{617}{100} \times \frac{39}{10} \)
\( = \frac{617 \times 39}{100 \times 10} = \frac{24063}{1000} \)
= 24.063

(iii) 0.57 x 2
\( = \frac{57}{100} \times \frac{2}{1} \)
\( = \frac{57 \times 2}{100 \times 1} = \frac{114}{100} \)
= 1.14

(iv) 5.04 × 0.7
\( = \frac{504}{100} \times \frac{7}{10} \)
\( = \frac{504 \times 7}{100 \times 10} = \frac{3528}{1000} \)
= 3.528
In simple words: Each part of this question involves multiplying decimal numbers by converting them into fractions, performing the multiplication, and then converting the result back to a decimal.

🎯 Exam Tip: Remember that the number of zeros in the denominator of the fractional representation of a decimal determines the number of decimal places in the resulting decimal number.

Question 4. Virendra bought 18 bags of rice, each bag weighing 5.250 kg. How much rice did he buy altogether? If the rice costs Rs 42 per kg, how much did he pay for it?
Answer:
Solution:
Weight of one bag of rice = 5.250 kg
Number of bags of rice = 18
.. Total Weight = 18 × 5.250
\( = \frac{18}{1} \times \frac{525}{100} = \frac{18 \times 525}{1 \times 100} = \frac{9450}{100} = 94.5 \text{ kg} \)
Cost of 1 kg of rice = Rs 42
.. Cost of 94.5 kg of rice = 42 × 94.5
\( = \frac{42}{1} \times \frac{945}{10} = \frac{42 \times 945}{1 \times 10} = \frac{39690}{10} = 3969 \)
.. Total rice bought by Virendra is 94.5 kg, and the amount paid for it is Rs 3969.
In simple words: First, calculate the total weight of rice by multiplying the number of bags by the weight of each bag. Then, calculate the total cost by multiplying the total weight of rice by the cost per kilogram.

🎯 Exam Tip: Clearly state all given information and show each step of the calculation, ensuring correct decimal multiplication and unit consistency, especially when dealing with money and weight.

Question 5. Vedika has 23.5 metres of cloth. She used it to make 5 curtains of equal size. If each curtain required 4 metres 25 cm to make, how much cloth is left over?
Answer:
Solution:
We know, that 1 m = 100 cm
Cloth required to make 1 curtain = 4 m 25 cm
\( = 4 \text{ m} + \frac{25}{100} \text{ m} \)
= 4 m + 0.25 m
= 4.25 m
.. Cloth required to make 5 curtains = 5 × 4.25
\( = \frac{5}{1} \times \frac{425}{100} = \frac{5 \times 425}{1 \times 100} = \frac{2125}{100} \)
= 21.25 m
Cloth remaining with Vedika = Total cloth with Vedika – Cloth used
= 23.5 m - 21.25 m
= 2.25 m
.. The length of cloth remaining with Vedika is 2.25 m.
In simple words: First, convert the cloth required for one curtain from meters and centimeters to meters only. Then, calculate the total cloth needed for five curtains. Finally, subtract the used cloth from the total cloth Vedika had to find the remaining cloth.

🎯 Exam Tip: Remember to convert all measurements to a common unit (e.g., meters) before performing calculations. Pay close attention to decimal subtraction for accuracy in finding the remaining quantity.