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Ex-7.9, Decimals, Class 6, Maths RD Sharma Solutions | RD Sharma Solutions for Class 6 Mathematics PDF Download

Question 1

Subtract

  1. 46.23

-37.5

= 8.73

  1. 128.40

-53.05

= 75.35

  1. 45.03

-27.80

= 17.23

  1. 23.930

-5.946

17.984

 

Question 2

Subtract:

  1. 9.756

-6.280

3.476

  1. 21.05

-15.27

5.78

  1. 18.50

-6.79

= 11.71

  1. 48.10

-0.37

47.73

  1. 108.032

-86.800

=21.232

  1. 91.001

-72.900

= 18.101

  1. 100.0

-26.32

= 73.68

 

Question 3

The sum of two numbers is 100. If one of them is 78.01. Find the other one ?

Ans. 3

One number is 78.01

Suppose the other number is x

The sum of these numbers is 100

Therefore , 78.01 + x = 100

= x = 100 -78.01

=x = 21.99

The other number is 21.99

 

Question 4

Waheeda’s school is at a distance 5 km 350 m from her house. She travels 1 km 70 m on foot and the rest she travels by bus. How much distance does she travel by bus?

Ans. 4

Distance travelled on foot = 1 km 70 m

Suppose the distance travelled by bus = x km

Total distance of school from residence = 5km 350 m = 5.350 km

So, 1.070 +x = 5.350

= x= 5.350 – 1.070

= x = 4.280 km

Therefore, distance travelled by bus is 4.280 km

 

Question 5

Raju bought a book of Rs.35.65. he gave Rs.50.35 to the shopkeeper . How much money did he pay back to the shopkeeper?

Price of the book = Rs. 365

Amount given to the shopkeeper = Rs50

Therefore, balance returned by the shopkeeper = Rs. ( 50-35.65)

= Rs . 14.35

 

Question 6

Raju bought a water melon weighing 5 kg 200g. Out of this she gave 2 kg 750 g to her neighbor. What is the weight of the watermelon left with ruby?

Ans. 6

Weight of the watermelon when bought = 5 kg 200g = 5.200 kg

Weight of the watermelon given to the neighbor = 2 kg 750 g = 2.750 kg

Therefore, weight of the watermelon left with ruby = weight of the watermelon when bought – weight of the watermelon when given to the neighbor

= (5.200- 2.750) kg

= 2.450 kg

So, the weight of the water melon left with ruby = 2.450 kg

 

Question 7

Victor drove 89.05 km on Saturday and 73.9 km on Sunday. How many kilometers did he drove more on Saturday?

Ans. 7

Distance travelled on Saturday = 89.050 km

Distance travelled on Sunday = 73.9 km

Subtracting the distance travelled on Sunday from the distance travelled on Saturday = 89.050-73.9 km

= 15.15 km

Therefore , victor drove 15.15km more on Saturday .

 

Question 8

Raju bought a book of Rs.35.65. he gave Rs.50.35 to the shopkeeper. How much money did he pay back to the shopkeeper?

Ans. 8

Price of the book = Rs. 365

Amount given to the shopkeeper = Rs50

Therefore, balance returned by the shopkeeper = Rs. ( 50-35.65)

= Rs . 14.35

 

Question 9

Gopal travelled 125.5 km by bus , 14.25 km by pony and the rest of the distance to kedarnath on foot. If he covered a total distance of 150 km , how much did he travel on foot ?

Ans. 9

Distance travelled by bus = 125.5 km

Distance travelled by pony = 14.25 km

Suppose the distance travelled on foot = x

Total distance = 150 km = distance travelled by bus + distance travelled on pony + distance travelled travelled on foot

= 150 = 125.5 + 14.25 + x

= 150 = 139.75 + x

= x = 10.25 km

Therefore, distance travelled on foot= 10.25 km

 

Question 10

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?

Ans. 10

Length of cloth originally = 20 m 5 cm = 2.05 m

Length of cloth cut for curtain = 4m 50 cm = 4.50 m

Therefore, length of cloth left with tina = length of cloth originally – length of cloth cut for curtain

= 20.05 – 4.50 m

= 15.55 m

Length of the cloth left with tina = 15.55 m

 

Question 11

Vineeta bought a book of Rs. 18.9, a pen of Rs. 8.50 and some papers for Rs. 5.05. She gave fifty rupee to the shopkeeper. How much balance did she get back?

Ans. 11

Price of the book = Rs. 18.90

Price of the pen = Rs 8.50

Price of the paper = Rs. 5.05

Total price of three items = Rs( 18.90+8.50+5.05)

= Rs 32.45

Total amount given to the shopkeeper = Rs 50

Balance received = Rs (50-32.45)

= Rs. 17.55

 

Question 12

Tanuj walked 8.62 km on Monday, 7.05 km on Tuesday, and some distance on Wednesday. If he walked 21.01 km in three days, how much distance did he walk on Wednesday?

Ans. 12

Distance travelled on Monday = 8.62 km

Distance travelled on Tuesday = 7.05 km

Suppose the distance travelled on Wednesday = x km

Total distance covered =

21.01 = 8.62+7.05 +x

= x= 21.01 – 15.67

= x= 5.34

Therefore, Tanuj walked 5.34 km on Wednesday .

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FAQs on Ex-7.9, Decimals, Class 6, Maths RD Sharma Solutions - RD Sharma Solutions for Class 6 Mathematics

1. What are the applications of decimals in real life?
Ans. Decimals are used in various real-life applications such as money, measurements, and calculations involving precise quantities. For example, when we go shopping, prices are often displayed with decimals, and we use decimals to calculate the total amount. In science, decimals are used to represent precise measurements such as temperature, weight, and length. Additionally, decimals are used in financial transactions, scientific research, and many other practical situations.
2. How can decimals be converted into fractions?
Ans. Decimals can be converted into fractions by following a simple procedure. To convert a decimal into a fraction, we count the number of decimal places and write the decimal as the numerator of the fraction. The denominator is determined by the place value of the rightmost digit of the decimal. For example, to convert 0.75 into a fraction, we count two decimal places and write 75 as the numerator. The denominator is determined by the place value of the rightmost digit, which is 100. Therefore, 0.75 is equivalent to 75/100, which can be simplified further if needed.
3. How can decimals be compared?
Ans. Decimals can be compared by comparing the digits in each place value. Start from the leftmost place value and compare the digits one by one. If the digits are equal, move to the next place value and continue the comparison. If the digits are not equal, the decimal with the larger digit in that place value is greater. If all the digits are equal up to a certain place value, but one decimal has additional digits, it is considered greater. For example, to compare 0.75 and 0.8, we start by comparing the tenths place (5 and 8). Since 8 is greater than 5, 0.8 is greater than 0.75.
4. How can decimals be added and subtracted?
Ans. Decimals can be added and subtracted by aligning the decimal points and performing the arithmetic operation as usual. Start from the rightmost digit and add or subtract the digits in each place value, carrying over or borrowing when necessary. If there are missing digits in one decimal, consider them as zeros. For example, to add 0.75 and 0.4, we align the decimal points and add the digits in each place value (5+0.4 and 7+0). The sum is 1.15. Similarly, subtraction can be performed by aligning the decimal points and subtracting the digits in each place value.
5. How can decimals be multiplied and divided?
Ans. Decimals can be multiplied and divided using the standard multiplication and division algorithms. To multiply decimals, ignore the decimal point and multiply the numbers as if they were whole numbers. Count the total number of decimal places in both numbers and place the decimal point in the product accordingly. For example, to multiply 0.75 and 0.4, we multiply 75 and 4, which gives 300. Since there are two decimal places in total, the product is 0.30. Division of decimals is performed similarly, but the decimal point is placed in the quotient based on the total number of decimal places in the dividend and divisor.
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