NCERT Solutions for Class 7 Maths Chapter 2 (EX 2.5)
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Exercise:  Exercise  2.5 
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Access NCERT Solutions for Class 7 Maths Chapter 2  Fractions and Decimals
Exercise  2.5
1. Which is greater:
(i) $ \text{0}\text{.5}\,\text{or}\,\text{0}\text{.05}$
Ans: Finding the greater term, $ 0.5>0.05$ .
(ii) $ \text{0}\text{.7}\,\text{or}\,\text{0}\text{.5}$
Ans: Finding the greater term, $ 0.7>0.5$ .
(iii) $ \text{7 or 0}\text{.7}$
Ans: Finding the greater term, $ 7>0.7$ .
(iv) $ \text{1}\text{.37 or 1}\text{.49}$
Ans: Finding the greater term, $ 1.37<1.49$ .
(v) $ \text{2}\text{.03 or 2}\text{.30}$
Ans: Finding the greater term, $ 2.03<2.30$ .
(vi) $ \text{0}\text{.8 or 0}\text{.88}$
Ans: Finding the greater term, $ 0.8<0.88$ .
2. Express as rupees using decimals:
(i) $ \text{7}\,\text{paise}$
Ans: Expressing the term as rupees,
$ \text{7}\,\text{paise=Re}\text{.}\dfrac{\text{7}}{\text{100}}\text{=Re}\text{.0}\text{.07}$
(ii) $ \text{7 rupees 7 paise}$
Ans: Expressing the term as rupees,$ \text{7}\,\text{rupees}\,\,\text{7}\,\,\text{paise=Rs}\text{.7+Re}\text{.}\dfrac{\text{7}}{\text{100}}\text{=Rs}\text{.7+Rs}\text{.0}\text{.07=Rs}\text{.7}\text{.07}$
(iii) $ \text{77 rupees 77 paise}$
Ans: Expressing the term as rupees,$ \text{77}\,\text{rupees}\,\,\text{77}\,\,\text{paise=Rs}\text{.77+Re}\text{.}\dfrac{\text{77}}{\text{100}}\text{=Rs}\text{.77+Rs}\text{.0}\text{.77=Rs}\text{.77}\text{.77}$
(iv) $ \text{50 paise}$
Ans: Expressing the term as rupees,
$\text{50}\,\text{paise=Re}\text{.}\dfrac{\text{50}}{\text{100}}\text{=Re}\text{.0}\text{.50}$
(v) $ \text{235 paise}$
Ans: Expressing the term as rupees,
$ \text{235}\,\text{paise=Re}\text{.}\dfrac{\text{235}}{\text{100}}\text{=Rs}\text{.2}\text{.35}$
3. Find
(i) Express $ \text{5}$ cm in metre and kilometer.
Ans: Expressing $ 5$ cm in meter and kilometer,
$ \because \,\,\text{100}\,\text{cm=1}\,\text{meter}$
$ \therefore \,\,\text{1}\,\text{cm=}\dfrac{\text{1}}{\text{100}}\,\text{meter}\Rightarrow \text{5}\,\text{cm=}\dfrac{\text{5}}{\text{100}}\text{=0}\text{.05}\,\text{meter}$
And, $ \because \,\,\text{1000}\,\text{meters=1}\,\text{kilometers}$
$ \therefore \,\,\text{1}\,\text{meter=}\dfrac{\text{1}}{\text{1000}}\,\text{kilometers}$ $ \Rightarrow \text{0}\text{.05}\,\text{meter=}\dfrac{\text{0}\text{.05}}{\text{1000}}\text{=0}\text{.00005}\,\,\text{kilometer}$
(ii) Express $ \text{35}$ mm in cm, m and km.
Ans: Expressing $ 35$ mm in cm, m and km.
$ \because \,\,\text{10}\,\text{mm=1}\,\text{cm}$
$ \therefore \,\,\text{1}\,\text{mm=}\dfrac{\text{1}}{\text{10}}\,\text{cm}\Rightarrow \text{35}\,\text{mm=}\dfrac{\text{35}}{\text{10}}\text{=3}\text{.5}\,\text{cm}$
And, $ \because \,\,\text{100}\,\text{cm=1}\,\text{meter}$
$ \therefore \,\,\,\text{1}\,\text{cm=}\dfrac{\text{1}}{\text{100}}\,\text{meter}\Rightarrow \text{3}\text{.5}\,\text{cm=}\dfrac{\text{3}\text{.5}}{\text{100}}\text{=0}\text{.035}\,\text{meter}$
Also, $ \because \,\,\text{1000}\,\text{meters=1}\,\text{kilometers}$
$ \therefore \,\,\text{1}\,\text{meter=}\dfrac{\text{1}}{\text{1000}}\,\text{kilometer}$ $ \Rightarrow \,\,\text{0}\text{.035}\,\text{meter=}\dfrac{\text{0}\text{.035}}{\text{1000}}\text{=0}\text{.000035}\,\text{kilometer}$
4. Express in kg:
(i) $ \text{200}\,\text{g}$
Ans: Converting from grams to kilograms,
$ \text{200}\,\text{g=}\left( \text{200 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1000}} \right)\,\text{kg=0}\text{.2}\,\text{kg}$
(ii) $ \text{3470}\,\text{g}$
Ans: Converting from grams to kilograms,
$ 3470\,\text{g=}\left( \text{3470 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1000}} \right)\,\text{kg=3}\text{.470}\,\text{kg}$
(iii) $ \text{4}\,\text{kg}\,\text{8}\,\text{g}$
Ans: Converting from grams to kilograms,$ \text{4}\,\text{kg}\,\text{8}\,\text{g=4}\,\text{kg}\,\text{+}\left( \text{8 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1000}} \right)\,\text{kg=4}\,\text{kg+0}\text{.008}\,\text{kg=4}\text{.008}\,\text{kg}$
5. Write the following decimal numbers in the expanded form:
(i) $ \text{20}\text{.03}$
Ans: Converting the decimal number in expanded form,
$ 20.03=2\times 10+0\times 1+0\times \dfrac{1}{10}+3\times \dfrac{1}{100}$
(ii) $ \text{2}\text{.03}$
Ans: Converting the decimal number in expanded form, $ 2.03=2\times 1+0\times \dfrac{1}{10}+3\times \dfrac{1}{100}$
(iii) $ \text{200}\text{.03}$
Ans: Converting the decimal number in expanded form, $ 200.03=2\times 100+0\times 10+0\times 1+0\times \dfrac{1}{10}+3\times \dfrac{1}{100}$
(iv) $ \text{2}\text{.034}$
Ans: Converting the decimal number in expanded form, $ 2.034=2\times 1+0\times \dfrac{1}{10}+3\times \dfrac{1}{100}+4\times \dfrac{1}{1000}$
6. Write the place value of $ \text{2}$ in the following decimal numbers:
(i) $ \text{2}\text{.56}$
Ans: The place value of $ 2$ in $ 2.56$ $ =2\times 1=2\,$ ones
(ii) $ \text{21}\text{.37}$
Ans: The place value of $ 2$ in $ 21.37=2\times 10=2$ tens
(iii) $ \text{10}\text{.25}$
Ans: The place value of $ 2$ in $ 10.25=2\times \dfrac{1}{10}=2$ tenths
(iv) $ \text{9}\text{.42}$
Ans: The place value of $ 2$ in $ 9.42=2\times \dfrac{1}{100}=2$ hundredth
(v) $ \text{63}\text{.352}$
Ans: The place value of $ 2$ in $ 63.352=2\times \dfrac{1}{1000}=2$ thousandth.
7. Dinesh went from place$ \text{ }\!\!~\!\!\text{ A}$ to place $ \text{B}$ and from there to place$ \text{C}$ .
$ \text{A}$ is $ \text{7}\text{.5}$ km from $ \text{B}$ and $ \text{B}$ is $ \text{12}\text{.7}$ km from$ \text{C}$ . Ayub went from place $ \text{A}$ to place $ \text{D}$ and from there to place$ \text{C}$ . $ \text{D}$ is $ \text{9}\text{.3}$ km from $ \text{A}$ and $ \text{C}$ is $ \text{11}\text{.8}$ km from$ \text{D}$ . Who travelled more and by how much?
Ans: Given: The distance travelled by Dinesh when he went from place $ \text{A}$ to place $ \text{B = 7}\text{.5 km}$ and from place$ \text{B to C = 12}\text{.7 km}$
According to the figure,
The total distance covered by Dinesh $ \text{= AB + BC }$
Substituting the values,
$ \text{=7}\text{.5 + 12}\text{.7 = 20}\text{.2 km}$
The total distance covered by Ayub $ \text{= AD + DC }$
Substituting the values,
$ \text{=9}\text{.3 + 11}\text{.8 = 21}\text{.1 km}$
Comparing the total distance covered by Ayub and Dinesh,
$ \text{21}\text{.1 km 20}\text{.2 km}$
Hence, Ayub covered $ \text{21}\text{.1  20}\text{.2 = 0}\text{.9 km = 900m}$ more distance as compared to Dinesh.
8. Shyam bought $ \text{5 kg 300 g}$ apples and $ \text{3 kg 250 g}$ mangoes. Sarala bought $ \text{4 kg 800 g}$ oranges and $ \text{4 kg 150 g}$ bananas. Who bought more fruits?
Ans: Given:
The total weight of fruits bought by Shyam$ \text{ = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g}$
And the total weight of fruits bought by Sarala$ \text{= 4 kg 800 g + 4 kg 150 g = 8 kg 950 g}$
Comparing the quantity of fruits bought by Shyam and Sarala, $ \text{8}\,\text{kg}\,\text{550}\,\text{g8}\,\text{kg}\,\text{950}\,\text{g}$
Observe that quantity of fruits bought by Sarala is greater.
Hence, Sarala bought more fruits than Shyam.
9. How much less is $ \text{28 km}$ then $ \text{42}\text{.6 km}$ ?
Ans: Given: The two distances are $ \text{42}\text{.6}\,\text{km}\,\text{and}\,\text{28}\,\text{km}$
Finding the difference of $ \text{42}\text{.6}\,\text{km}\,\text{and}\,\text{28}\,\text{km}$ ,
$ \text{42}\text{.628}\text{.0=14}\text{.6}\,\text{km}$
Hence, $ \text{14}\text{.6}\,\text{km}$ less is $ \text{28}\,\text{km}$ then $ \text{42}\text{.6}\,\text{km}$ .
NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5
Opting for the NCERT solutions for Ex 2.5 Class 7 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 2.5 Class 7 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 inhouse subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite wellversed 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 7 Maths Chapter 2 Exercise 2.5 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.
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