GSEB Class 6 Maths Solutions Chapter 8 Decimals Exercise 8.6

Gujarat Board Textbook Solutions Class 6 Maths Chapter 8 Decimals Ex 8.6

Question 1. Subtract:
(a) 18.25 from 20.75
(b) 202.54 m from 250 m
(c) Rs 5.36 from Rs 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Answer:
(a) To find the difference when we take \( 18.25 \) from \( 20.75 \), we arrange the numbers and calculate:
      \( 20.75 \)
   \( -18.25 \)
   \( \rule{3em}{0.4pt} \)
      \( 02.50 \)
Therefore, the result of \( 20.75 - 18.25 \) is \( 2.50 \).

(b) To get \( 202.54 \text{ m} \) from \( 250 \text{ m} \), we first change \( 250 \text{ m} \) into \( 250.00 \text{ m} \) and then set up the subtraction:
      \( 250.00 \)
   \( -202.54 \)
   \( \rule{3em}{0.4pt} \)
      \( 047.46 \)
So, \( 250 \text{ m} - 202.54 \text{ m} \) gives us \( 47.46 \text{ m} \).

(c) To calculate Rs \( 5.36 \) taken from Rs \( 8.40 \), we carry out the operation:
      \( 8.40 \)
   \( -5.36 \)
   \( \rule{3em}{0.4pt} \)
      \( 3.04 \)
Hence, Rs \( 8.40 \) minus Rs \( 5.36 \) equals Rs \( 3.04 \).

(d) To find \( 2.051 \text{ km} \) less than \( 5.206 \text{ km} \), we do the math:
      \( 5.206 \)
   \( -2.051 \)
   \( \rule{3em}{0.4pt} \)
      \( 3.155 \)
As a result, \( 5.206 \text{ km} - 2.051 \text{ km} \) comes out to \( 3.155 \text{ km} \).

(e) To determine \( 0.314 \text{ kg} \) from \( 2.107 \text{ kg} \), we complete the subtraction:
      \( 2.107 \)
   \( -0.314 \)
   \( \rule{3em}{0.4pt} \)
      \( 1.793 \)
Therefore, \( 2.107 \text{ kg} \) minus \( 0.314 \text{ kg} \) leaves \( 1.793 \text{ kg} \).
In simple words: When subtracting decimals, always line up the decimal points and then subtract as usual, borrowing from the left if needed.

Exam Tip: For decimal subtraction, it is crucial to align the decimal points vertically before beginning the operation. If one number has fewer decimal places, you can add zeros to the end to make them match, ensuring correct place value calculation.

Question 2. Find the value of:
(a) 9.756 - 6.28
(b) 21.05 - 15.27
(c) 18.5 - 6.79
(d) 11.6 - 9.847
Answer:
(a) To find the value of \( 9.756 - 6.28 \), we first make the number of decimal places equal by writing \( 6.28 \) as \( 6.280 \). Then, we perform the subtraction:
      \( 9.756 \)
   \( -6.280 \)
   \( \rule{3em}{0.4pt} \)
      \( 3.476 \)
So, \( 9.756 - 6.28 = 3.476 \).

(b) To determine the value of \( 21.05 - 15.27 \), we arrange the numbers and subtract:
      \( 21.05 \)
   \( -15.27 \)
   \( \rule{3em}{0.4pt} \)
      \( 05.78 \)
Thus, \( 21.05 - 15.27 = 5.78 \).

(c) To calculate \( 18.5 - 6.79 \), we first write \( 18.5 \) as \( 18.50 \). Then, we proceed with the subtraction:
      \( 18.50 \)
   \( -6.79 \)
   \( \rule{3em}{0.4pt} \)
      \( 11.71 \)
Therefore, \( 18.5 - 6.79 = 11.71 \).

(d) To find the value of \( 11.6 - 9.847 \), we first make the decimal places equal by writing \( 11.6 \) as \( 11.600 \). Next, we perform the subtraction:
      \( 11.600 \)
   \( -9.847 \)
   \( \rule{3em}{0.4pt} \)
      \( 01.753 \)
Hence, \( 11.6 - 9.847 = 1.753 \).
In simple words: When finding the difference between decimals, always add zeros to the shorter number so both have the same number of digits after the decimal point before you start subtracting.

Exam Tip: When subtracting decimals, it's a good practice to write the numbers vertically, aligning the decimal points. Add placeholder zeros to the right of the decimal point for numbers with fewer decimal places to avoid errors.

Question 3. Raju bought a book for Rs 35.65. He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Answer: Raju gave Rs \( 50 \) to the shopkeeper. The book's cost was Rs \( 35.65 \). To find the money Raju got back, we subtract the cost of the book from the amount he paid. We write Rs \( 50 \) as Rs \( 50.00 \) to align the decimal places for subtraction:
      \( 50.00 \)
   \( -35.65 \)
   \( \rule{3em}{0.4pt} \)
      \( 14.35 \)
Thus, Raju received Rs \( 14.35 \) back from the shopkeeper.
In simple words: Raju paid with Rs 50. The book cost Rs 35.65. To see how much money he got back, we take the book's price away from the money he gave.

Exam Tip: In word problems involving money, clearly identify the initial amount and the amount spent. Ensure units (like Rupees) are consistent and decimal points are aligned for calculations.

Question 4. Rani had 18.50. She bought one ice-cream for Rs 11.75. How much money does she have now?
Answer: Rani initially had Rs \( 18.50 \). She purchased an ice-cream for Rs \( 11.75 \). To determine how much money she has remaining, we subtract the cost of the ice-cream from her initial amount:
      \( 18.50 \)
   \( -11.75 \)
   \( \rule{3em}{0.4pt} \)
      \( 06.75 \)
Therefore, Rani now possesses Rs \( 6.75 \).
In simple words: Rani had Rs 18.50. She spent Rs 11.75 on ice cream. To find out what she has left, we take away the money she spent from her starting money.

Exam Tip: For problems about spending money, remember that 'remaining money' means you need to subtract the amount spent from the original amount. Always ensure currency symbols are used correctly.

Question 5. Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Answer: The total cloth Tina had was \( 20 \text{ m } 5 \text{ cm} \). She used \( 4 \text{ m } 50 \text{ cm} \) to create a curtain. To discover the remaining cloth, we must subtract the used length from the total length. First, convert both lengths into meters:
We know that \( 1 \text{ cm} = \frac{1}{100} \text{ m} \).
So, \( 20 \text{ m } 5 \text{ cm} = (20 + \frac{5}{100}) \text{ m} = (20 + 0.05) \text{ m} = 20.05 \text{ m} \).
And \( 4 \text{ m } 50 \text{ cm} = (4 + \frac{50}{100}) \text{ m} = (4 + 0.50) \text{ m} = 4.50 \text{ m} \).
Now, we subtract the used length from the total length:
      \( 20.05 \)
   \( -04.50 \)
   \( \rule{3em}{0.4pt} \)
      \( 15.55 \)
Thus, the length of cloth left with Tina is \( 15.55 \text{ m} \).
In simple words: Tina started with some cloth and cut a part of it. To find out how much is still there, change all the lengths to meters, then take away the part she cut from the total.

Exam Tip: When dealing with measurements like meters and centimeters, always convert them to a single unit (usually the larger one, like meters or kilometers) before performing calculations. This helps in avoiding common errors with mixed units.

Question 6. A person travels a total distance of 20 km 50 m in a day. Out of this, she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Answer: The total distance to be covered by the person is \( 20 \text{ km } 50 \text{ m} \). She travels \( 10 \text{ km } 200 \text{ m} \) by bus. To calculate the remaining distance covered by auto, we subtract the bus distance from the total distance. First, convert both distances into kilometers:
We know that \( 1 \text{ m} = \frac{1}{1000} \text{ km} \).
So, \( 20 \text{ km } 50 \text{ m} = (20 + \frac{50}{1000}) \text{ km} = (20 + 0.050) \text{ km} = 20.050 \text{ km} \).
And \( 10 \text{ km } 200 \text{ m} = (10 + \frac{200}{1000}) \text{ km} = (10 + 0.200) \text{ km} = 10.200 \text{ km} \).
Now, we subtract the distance travelled by bus from the total distance:
      \( 20.050 \)
   \( -10.200 \)
   \( \rule{3em}{0.4pt} \)
      \( 09.850 \)
Thus, the distance travelled by auto is \( 9.850 \text{ km} \).
In simple words: First, change all distances to kilometers. Then, take away the distance travelled by bus from the total distance to find out how far was travelled by auto.

Exam Tip: For problems involving distances in mixed units (km and m), convert all measurements to a single unit, typically kilometers, by dividing meters by 1000. This ensures accurate decimal operations.

Question 7. Aakash bought vegetable weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Answer: Aakash purchased a total of \( 10 \text{ kg} \) of vegetables. The weight of onions is \( 3 \text{ kg } 500 \text{ g} \), and the weight of tomatoes is \( 2 \text{ kg } 75 \text{ g} \). To find the weight of potatoes, we first determine the combined weight of onions and tomatoes, and then subtract this total from the overall weight of vegetables. Convert all weights to kilograms:
We know that \( 1 \text{ g} = \frac{1}{1000} \text{ kg} \).
Weight of onions = \( 3 \text{ kg } 500 \text{ g} = (3 + \frac{500}{1000}) \text{ kg} = 3.500 \text{ kg} \).
Weight of tomatoes = \( 2 \text{ kg } 75 \text{ g} = (2 + \frac{75}{1000}) \text{ kg} = 2.075 \text{ kg} \).
Total weight of onions and tomatoes = \( 3.500 \text{ kg} + 2.075 \text{ kg} \):
      \( 3.500 \)
   \( +2.075 \)
   \( \rule{3em}{0.4pt} \)
      \( 5.575 \)
So, the combined weight of onions and tomatoes is \( 5.575 \text{ kg} \).
Now, subtract this from the total weight of vegetables \( (10 \text{ kg}) \). We write \( 10 \text{ kg} \) as \( 10.000 \text{ kg} \) for calculation:
      \( 10.000 \)
   \( -5.575 \)
   \( \rule{3em}{0.4pt} \)
      \( 04.425 \)
Therefore, the weight of potatoes is \( 4.425 \text{ kg} \).
In simple words: First, convert all weights to kilograms. Then, add the weights of onions and tomatoes together. Finally, take that total away from the full weight of all vegetables to find the weight of the potatoes.

Exam Tip: When a problem involves multiple items and a total, it's often best to first sum the known parts (converting units if necessary) and then subtract that sum from the total to find the unknown part.