Maharashtra Board Class 6 Maths Chapter 5 Decimal Fractions Set 14 Solutions

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 14 Solutions Maharashtra Board

Std 6 Maths Practice Set 14 Solutions Answers

Question 1. In the table below, write the place value of each of the digits in the number 378.025.

🎯 Exam Tip: Accurately identifying the place value of each digit, especially in decimal fractions, is crucial for correctly performing calculations and understanding numerical magnitude.

Question 2. Solve :
(i) 905.5 - 27.197
(ii) 39 + 700.65
(iii) 40 + 27.7 + 2.451
Answer: Solution:
(i) 905.5 + 27.197
905.500
+ 27.197
932.697
(ii) 39 + 700.65
39.00
+ 700.65
739.65
(iii) 40 + 27.7 + 2.451
40.000
27.700
+ 2.451
70.151 In simple words: This question involves performing addition operations with decimal numbers. Aligning the decimal points and adding column by column, similar to whole number addition, yields the correct sums.

🎯 Exam Tip: When adding or subtracting decimal numbers, always ensure that the decimal points are vertically aligned. Adding zeros to the end of numbers can help maintain proper place value alignment for accurate results.

Question 3. Subtract:
(i) 85.96-2.345
(ii) 632.24 - 97.45
(iii) 200.005 - 17.186
Answer: Solution:
(i) 85.96-2.345
85.960
- 2.345
83.615
(ii) 632.24 - 97.45
632.24
- 97.45
534.79
(iii) 200.005 - 17.186
200.005
- 17.186
182.819 In simple words: This question focuses on subtracting decimal numbers. The key is to align the decimal points, add trailing zeros if necessary to equalize the number of decimal places, and then perform subtraction column by column.

🎯 Exam Tip: Proper alignment of decimal points is critical for subtraction. Ensure the number with fewer decimal places is padded with zeros to avoid errors in calculation. Double-check borrowing when subtracting.

Question 4. Avinash traveled 42 km 365 m by bus, 12 km 460 in by car and walked 640 m. How many kilometers did he travel altogether? (Write your answer in decimal fractions)
Answer: Solution:
Distance traveled in bus = 42 km 365 m
= 42 km + \( \frac{365}{1000} \) km
= 42 km + 0.365 km
42.000
+ 0.365
42.365
= 42.365 km
Distance travelled in car = 12 km 460 m
= 12 km + \( \frac{460}{1000} \) km
= 12 km + 0.460 km
12.000
+ 0.460
12.460
= 12.460 km
Distance walked = 640 m
= \( \frac{640}{1000} \) = 0.640 km
... Total distance travelled = Distance travelled in bus + Distance travelled in car + Distance walked
= 42.365 + 12.460 + 0.640
42.365
12.460
+ 0.640
55.465
= 55.465 km
.. Distance travelled altogether by Avinash is 55.465 km. In simple words: To find the total distance Avinash traveled, all distances (by bus, car, and walking) must be converted to kilometers and then added together. This involves converting meters to kilometers by dividing by 1000 and then summing the resulting decimal values.

🎯 Exam Tip: Always ensure all units are consistent before performing calculations (e.g., convert all distances to kilometers). Pay close attention to decimal point alignment during addition to avoid calculation errors.

Question 5. Ayesha bought 1.80 m of cloth for her salwaar and 2.25 for her kurta. If the cloth costs Rs 120 per metre, how much must she pay the shopkeeper?
Answer: Solution:
Total length of cloth bought = 1.80 m + 2.25 m
= 4.05 m
1.80
+2.25
4.05
Cost of 1 m of cloth = Rs 120
... Cost of 4.05 m of cloth = 4.05 x 120
= \( \frac{405}{100} \) × 120 = \( \frac{405×120}{100×1} = \frac{48600}{100} \) = Rs 486
.. Amount to be paid to the shopkeeper is Rs 486. In simple words: First, calculate the total length of cloth Ayesha bought by adding the lengths for the salwaar and kurta. Then, multiply this total length by the cost per meter to find the total amount she needs to pay.

🎯 Exam Tip: Pay attention to units and ensure all quantities are in the same unit (e.g., meters) before adding. For cost calculations, be careful with decimal multiplication, especially when converting decimals to fractions.

Question 6. Sujata bought a watermelon weighing 4.25 kg and gave 1 kg 750 g to the children in her neighbourhood. How much of it does she have left?
Answer: Solution:
Total weight of watermelon = 4.25 kg
Weight of watermelon given to children = 1 kg 750 g
= 1 kg + \( \frac{750}{1000} \) kg
= 1 kg + 0.75 kg
1.000
+0.750
1.750
= 1.75 kg
.. Weight of watermelon left = Total weight of watermelon - Weight of watermelon given to children
= 4.25 kg - 1.75 kg
4.25
- 1.75
2.50
= 2.5 kg
.. Weight of watermelon left with Sujata is 2.5 kg. In simple words: To determine the remaining weight, first convert the weight given to children from kilograms and grams into a single decimal kilogram value. Then, subtract this amount from the total weight of the watermelon Sujata initially bought.

🎯 Exam Tip: Always convert all measurements to a consistent unit (e.g., kilograms) before performing calculations. Accurately converting grams to kilograms (divide by 1000) is key to solving such problems correctly.

Question 7. Anita was driving at a speed of 85.6 km per hour. The road had a speed limit of 55 km per hour. By how much should she reduce her speed to be within the speed limit?
Answer: Solution:
Speed at which Anita is driving = 85.6 km per hr.
Speed limit = 55 km per hr.
.. Anita should reduce her speed by 85.6 km per hr - 55 km per hr.
85.6
- 55.0
30.6
= 30.6 km per hr.
.. Anita should reduce her speed by 30.6 km per hour to be within the speed limit. In simple words: To find out how much Anita needs to reduce her speed, simply subtract the speed limit from her current driving speed. This difference will be the required reduction.

🎯 Exam Tip: When dealing with speed-related problems, clearly identify the current speed and the limit. The difference between these two values indicates the necessary adjustment.

Maharashtra Board Class 6 Maths Chapter 4 Operations On Fractions Practice Set 14 Intext Questions And Activities

Question 1. Nandu went to a shop to buy a pen, notebook, eraser and a paint box. The shopkeeper told him the prices. A pen costs four and a half rupees, an eraser one and a half, a notebook six and a half and a paintbox twenty-five rupees and fifty paise. Nandu bought one of each article. Prepare his bill. If Nandu gave a 100 rupee note, how much money does he get back? (Textbook pg. no. 29)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक दुकान में नन्दू नाम के ग्राहक को खरीदारी करते हुए दर्शाता है, जहाँ एक दुकानदार उसे सामान के दाम बता रहा है। दाईं ओर, एक बिल का प्रारूप दिखाया गया है जिसमें "अक्षय वस्तु भंडार" का नाम, क्रम संख्या, विवरण, मात्रा और राशि के कॉलम हैं, जिसमें एक पेन की खरीद दर्ज है और कुल योग के लिए जगह छोड़ी गई है।
Answer: Nandu will get _ rupees back.
Solution:
100-38 = 62.00
The prepared bill is as follows:

🎯 Exam Tip: When creating bills, ensure all item prices are correctly listed and summed. For change calculation, perform simple subtraction. This problem also tests the understanding of converting halves and fifty paise into decimal rupees.

Question 2. Take a pen and notebook with you when you go to the market with your parent. Note the weight of every vegetable your mother buys. Find out the total weight of those vegetables. (Textbook pg. no. 30)
Answer: Solution:
(Students should attempt this activity on their own.) In simple words: This is a practical activity where students need to visit a market with their parents, record the weights of purchased vegetables, and then calculate the total weight.

🎯 Exam Tip: For practical activities, focus on accurate observation and recording of data. When calculating total weight, ensure consistent units of measurement (e.g., all in kg or all in grams) for correct summation.

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