NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Ex 8.5) Exercise 8.5

Exercise 8.5

1. Find the sum in each of the following:

1. $\mathbf{0.007 + 8.5 + 30.08}$

Ans: Add values of the same places to get the sum.

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\  {}&{}&0&.&0&0&7 \\  {}&{}&8&.&5&{}&{} \\ {}&3&0&.&0&8&{} \\ \hline {}&3&8&.&5&8&7\\ \hline \end{array}$

$\therefore 0.007 + 8.5 + 30.08 = 35.587$

2. $\mathbf{15 + 0.632 + 13.8}$

Ans: Add values of the same places to get the sum. 

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\ {}&1&5&.&0&0&0 \\ {}&{}&{}&.&6&3&2 \\ {}&1&3&.&8&{}&{} \\\hline{}&2&9&.&4&3&2\\ \hline \end{array}$

$\therefore 15 + 0.632 + 13.8= 29.432$

3. $\mathbf{27.076 + 0.55 + 0.08}$

Ans: Add values of the same places to get the sum. 

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\ {}&2&7&.&0&7&6 \\ {}&{}&0&.&5&5&{} \\ {}&{}&0&.&0&8&{} \\ \hline {}&2&7&.&7&0&6\\ \hline \end{array}$ 

$27.076 + 0.55 + 0.08= 27.706$

4. $\mathbf{25.65 + 9.005 + 3.7}$

Ans: Add values of the same places to get the sum. 

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\ {}&2&5&.&6&5&{} \\  {}&{}&9&.&0&0&5 \\ {}&{}&3&.&7&{}&{} \\ \hline {}&3&8&.&3&5&5\\ \hline \end{array}$

$\therefore 25.65 + 9.005 + 3.7= 38.355$

5. $\mathbf{0.75 + 10.425 + 2}$

Ans: Add values of the same places to get the sum. 

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\ {}&{}&0&.&7&5&{} \\  {}&1&0&.&4&2&5 \\ {}&{}&2&.&{}&{}&{} \\ \hline {}&1&3&.&1&7&5\\ \hline \end{array}$ 

$\therefore 0.75 + 10.425 + 2= 13.175$

6. $\mathbf{280.69 + 25.2 + 38}$

Ans: Add values of the same places to get the sum. 

$ + \begin{array}{*{20}{c}} H&T&O&.&{Tenth}&{Hund.}&{Thou.} \\ 2&8&0&.&6&9&{} \\ {}&2&5&.&2&{}&{} \\ {}&3&8&.&{}&{}&{} \\ \hline 3&4&3&.&8&9&{}\\ \hline \end{array}$ 

$\therefore 280.69 + 25.2 + 38= 343.89$

2. Rashid spent \[{\text{Rs}}{\text{. }}{\mathbf{35}}.{\mathbf{75}}\] for Maths book and \[{\text{Rs}}{\text{. }}{\mathbf{32}}.{\mathbf{60}}\] for Science book. Find the total amount spent by Rashid.

Ans: Add up the sums of price of two books to get the total amount spent.

Total Amount\[ = {\text{Rs}}{\text{. }}{\mathbf{35}}.{\mathbf{75}} + {\text{Rs}}{\text{. }}{\mathbf{32}}.{\mathbf{60}} = {\text{Rs}}{\text{.}}\;68.35\]

So, the total amount spent by Rashid is \[{\text{Rs}}{\text{.}}\;68.35\].

3. Radhika’s mother has her \[{\text{Rs}}{\text{. }}{\mathbf{10}}.{\mathbf{50}}\] and her father gave her \[{\text{Rs}}{\text{. }}{\mathbf{15}}.{\mathbf{80}}\] . Find the total amount given to Radhika by the parents. 

Ans: Add up the amount given by mother and her father to Radhika to get the total amount given to Radhika by the parents.

Total Amount\[ = {\text{Rs}}{\text{. }}{\mathbf{10}}.{\mathbf{50}} + {\text{Rs}}{\text{. }}{\mathbf{15}}.{\mathbf{80}} = {\text{Rs}}{\text{.}}\;26.30\]

So, the total amount given to Radhika by the parents is \[{\text{Rs}}{\text{.}}\;26.30\].

4. Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trousers. Find the total length of cloth bought by her.

Ans: Add up the length of cloth for shirt and length of cloth for trouser to get the total length of cloth brought by her.

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

Length of cloth for shirt$ = 3\;{\text{m}}\;20\;{\text{cm}} = 3 + \dfrac{{20}}{{100}}{\text{m  =  3}}{\text{.2 m}}$

Length of cloth for trouser$ = 2\;{\text{m}}\;5\;{\text{cm}} = 2 + \dfrac{5}{{100}}{\text{m  =  2}}{\text{.05 m}}$

Total length of cloth\[ = {\text{3}}{\text{.2 m}} + 2.05{\text{ m}} = 5.{\text{25 m}}\]

So, total length of cloth brought by Nasreen is \[5.{\text{25 m}}\]

5. Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?

Ans: Add up the distance walked in morning and distance walked in evening to get the total distance walked by him.

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

Distance Travelled by Naresh in Morning\[ = 2\;{\text{km}}\;35\;{\text{m}} = 2 + \dfrac{{35}}{{1000}}{\text{km  =  2}}{\text{.035 km}}\]

Distance Travelled by Naresh in Evening\[ = 1\;{\text{km}}\;7\;{\text{m}} = 1 + \dfrac{7}{{1000}}{\text{km  =  1}}{\text{.007 km}}\]

Total distance walked by him\[ = {\text{2}}{\text{.035 km}} + {\text{1}}{\text{.007 km}} = 3.{\text{042 km}}\]

So, total distance walked by Naresh is \[3.{\text{042 km}}\]

6. Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m by foot in order to reach her school. How far is her school from her residence?

Ans: Add up the distance travelled by bus, car and foot in morning and distance walked in evening to get the total distance walked by him.

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

Distance Travelled by Sunita by bus\[ = \;15\;{\text{km}}\;268\;{\text{m}} = 15 + \dfrac{{268}}{{1000}}{\text{km  =  15}}{\text{.268 km}}\]

Distance Travelled by Sunita by car\[ = \;7\;{\text{km}}\;7\;{\text{m}} = 7 + \dfrac{7}{{1000}}{\text{km  =  7}}{\text{.007 km}}\]

Distance Travelled by Sunita by foot\[ = \;500\;{\text{m}} = \dfrac{{500}}{{1000}}{\text{km  =  0}}{\text{.5 km}}\]

Total Distance Travelled by Sunita\[ = {\text{15}}{\text{.268 km}} + {\text{7}}{\text{.007 km}} + {\text{0}}{\text{.5 km  =  22}}{\text{.775 km}}\]

So, total distance walked by Sunita is \[{\text{22}}{\text{.775 km}}\]

7. Ravi purchases 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850 g flour. Find the total weight of his purchases.

Ans: Add up the weight of rice, weight of sugar and weight of flour to get the total weight purchased by Ravi.

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

Weight of Rice\[ = \;5\;{\text{kg}}\;400\;{\text{g}} = 5 + \dfrac{{400}}{{1000}}{\text{kg  =  5}}{\text{.4 kg}}\]

Weight of Sugar\[ = \;2\;{\text{kg}}\;20\;{\text{g}} = 2 + \dfrac{{20}}{{1000}}{\text{kg  =  2}}{\text{.02 kg}}\]

Weight of Flour\[ = \;10\;{\text{kg}}\;850\;{\text{g}} = 10 + \dfrac{{850}}{{1000}}{\text{kg  =  10}}{\text{.85 kg}}\]

Total weight purchased by Ravi\[ = {\text{5}}{\text{.4 kg}} + 2.{\text{02 kg}} + 10.{\text{85 kg  =  18}}{\text{.270 kg}}\]

So, total weight purchased by Ravi is \[{\text{18}}{\text{.270 kg}}\]

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Exercise 8.5

Opting for the NCERT solutions for Ex 8.5 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 8.5 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 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 8.5 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 8.5, 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. 

Do not delay any more. Download the NCERT solutions for Class 6 Maths Chapter 8 Exercise 8.5 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.