NCERT Solutions for Class 6 Maths Chapter 8: Decimals - Exercise 8.6

Exercise 8.6

1. Subtract:

a) \[{\text{Rs}}{\text{. 18}}{\text{.25}}\] from \[{\text{Rs}}{\text{. 20}}{\text{.75}}\]

Ans: Subtract values of the same places to get the difference.

\[ - \begin{array}{*{20}{c}} 1&8&.&5&0 \\  0&6&.&7&9 \\  \hline 1&1&.&7&1\\  \hline \end{array}\]

So, the obtained value is \[11.76\].

b) \[{\text{202}}{\text{.54}}\;{\text{m}}\] from \[{\text{250 m}}\]

  Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 2&5&0&.&0&0 \\  2&0&2&.&5&4 \\  \hline 0&4&7&.&4&6\\  \hline \end{array}\]

So, obtained difference is \[47.46\;{\text{m}}\]

c) \[{\text{Rs}}{\text{. 8}}{\text{.40}}\] from \[{\text{Rs}}{\text{. 5}}{\text{.36}}\]

Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 8&.&4&0\\  5&.&3&6\\  \hline 3&.&0&4\\ \hline \end{array}\]

So, obtained difference is \[{\text{Rs}}{\text{.}}\;3.04\]

d) \[{\text{2}}{\text{.051 km}}\] from \[{\text{5}}{\text{.206 km}}\]

Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 5&.&2&0&6 \\  2&.&0&5&1 \\  \hline 3&.&1&5&5 \\ \hline \end{array}\]

So, obtained difference is \[3.155\;{\text{km}}\]

e) \[{\text{1}}{\text{.314 kg}}\] from \[{\text{2}}{\text{.107 kg}}\]

  Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 2&.&1&0&7 \\  1&.&3&1&4 \\  \hline 1&.&7&9&3 \\ \hline \end{array}\]

So, obtained difference is \[{\text{1}}{\text{.793 kg}}\]

2. Find the value of:

a) \[9.756 - 6.28\]

Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 9&.&7&5&6 \\  6&.&2&8&{} \\  \hline 3&.&4&7&6 \\ \hline \end{array}\]

So, obtained value is \[3.476\]

b) \[21.05--15.27{\text{ }}\]

Ans: Subtract values of the same places to get the difference.

\[ - \begin{array}{*{20}{c}} 2&1&.&0&5 \\ 1&5&.&2&7 \\  \hline 0&5&.&7&8 \\ \hline \end{array}\]

So, obtained value is \[5.78\]

c) \[18.50 - 6.79\]

Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 1&8&.&5&0 \\  {}&6&.&7&9 \\ \hline 1&1&.&7&1 \\ \hline \end{array}\]

So, the obtained value is \[11.76\].

d) \[11.600 - 9.847\]

Ans: Subtract values of the same places to get the difference. 

\[ - \begin{array}{*{20}{c}} 1&1&.&6&0&0 \\  {}&9&.&8&4&7 \\  \hline 0&1&.&7&5&3\\ \hline \end{array}\]

So, the obtained value is \[1.753\].

3. Raju bought a book of ${\text{Rs}}{\text{.}}\;35.65$. He gave ${\text{Rs}}{\text{.}}\;50$ to the shopkeeper. How much money did he get back from the shopkeeper? 

Ans: The amount of money Raju will get back from the shopkeeper is the difference between the amount he gave to the shopkeeper and the price of the book.

So, amount he will get back $ = {\text{Rs}}{\text{.}}\;50 - {\text{Rs}}{\text{.}}\;35.65 = {\text{Rs}}{\text{.}}\;14.35$.

So, Raju got back ${\text{Rs}}{\text{.}}\;14.35$ from the shopkeeper.

4. Rani had \[{\text{Rs}}{\text{.}}\;18.50\]. She bought one ice-cream for \[{\text{Rs}}{\text{.}}\;11.75\]. How much money does she have now?

Ans: The amount of money Rani is having now is the difference of the amount she had to the price of ice cream.

So, amount Rani will have now $ = {\text{Rs}}{\text{.}}\;18.50 - {\text{Rs}}{\text{.}}\;11.75 = {\text{Rs}}{\text{.}}\;6.75$.

So, Rani has ${\text{Rs}}{\text{.}}\;6.75$ with her.

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?

Ans: The amount of cloth Tina is left with now is the difference in length of cloth she had to the length of cloth used in the curtain.

Convert cm into m by dividing values in cm to 100.

Length of cloth Tina had $ = 20\;{\text{m}}\;5\;{\text{cm}} = 20 + \dfrac{5}{{100}}{\text{m  =  20}}{\text{.05 m}}$

Length of cloth used in curtain$ = 4\;{\text{m}}\;50\;{\text{cm}} = 4 + \dfrac{{50}}{{100}}{\text{m  =  4}}{\text{.5 m}}$

So, amount of cloth Tina is left with her now \[ = 20.05\;{\text{m}} - 4.5\;{\text{m}} = 15.55\;{\text{m}}\].

So, Tina is left with\[15.55\;{\text{m}}\]of cloth.

6. Namita travels 20 km 50 m every 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?

Ans: The amount of distance travelled by Namita by Auto is the difference of distance she travels daily to the distance travelled by bus.

Convert m into km by dividing values in m to 1000.

Distance Travelled by Namita every day \[ = 20\;{\text{km}}\;50\;{\text{m}} = 20 + \dfrac{{50}}{{1000}}{\text{km  =  20}}{\text{.05 km}}\]

Distance Travelled by Namita by Bus \[ = 10\;{\text{km}}\;200\;{\text{m}} = 10 + \dfrac{{200}}{{1000}}{\text{km  =  10}}{\text{.2 km}}\]

So, distance travelled by her by auto \[ = 20.05\;{\text{km}} - 10.2\;{\text{km}} = 9.850\;{\text{km}}\].

So, the distance travelled by Namita by auto is \[9.850\;{\text{km}}\].

7. Aakash bought vegetables weighing 10 kg. Out of this 3 kg 500 g in onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?

Ans: The weight of potatoes bought by Akash is the difference of the total weight of vegetables to the weight of onions and tomatoes.

Convert g into kg by dividing values in g to 1000.

Weight of onions \[ = \;3\;{\text{kg}}\;500\;{\text{g}} = 3 + \dfrac{{500}}{{1000}}{\text{kg  =  3}}{\text{.5 kg}}\]

Weight of Sugar \[ = \;2\;{\text{kg}}\;75\;{\text{g}} = 2 + \dfrac{{75}}{{1000}}{\text{kg  =  2}}{\text{.75 kg}}\]

So, weight of potatoes bought by Akash \[ = 10\;{\text{kg}} - 3.5\;{\text{kg}} - 2.075\;{\text{kg}} = 4.425\;{\text{kg}}\].

So, the weight of potatoes bought by Akash is \[4.425\;{\text{kg}}\].

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.6

Opting for the NCERT solutions for Ex 8.6 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 8.6 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 8 Exercise 8.6 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 6 Maths Chapter 8 Exercise 8.6, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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