# CBSE Important Questions Class 7 Maths Chapter 2

## Important Questions Class 7 Mathematics Chapter 2 – Fractions and Decimals

Mathematics is an important subject that you study in school. We need Mathematics in our daily life too. In this chapter, you will learn about decimals and fractions more elaborately. In Class 6, students have learned about fractions and decimals.

Chapter 2 of Class 7 Mathematics under the CBSE curriculum deals with the addition or multiplication of decimals and fractions. Students must practice questions from the chapter as much as possible. They may take help from other sources because the textbook exercises have limited questions.

Extramarks is a leading company in the educational sector. Our experts have made the Important Questions Class 7 Mathematics Chapter 2 to help students to solve the questions regularly. They have collated the questions from different sources such as the CBSE sample papers, CBSE past years’ question papers, the textbook exercises, NCERT exemplars and important reference books. They have solved the questions too.

Extramarks provides all the important study materials related to CBSE and NCERT. You can download the study materials after registering on our official website. We provide CBSE syllabus, CBSE sample papers, CBSE past years’ question papers, CBSE extra questions, CBSE revision notes, NCERT books, NCERT exemplars, NCERT solutions, NCERT important questions, vital formulas and many more.

## Important Questions Class 7 Mathematics Chapter 2 – With Solutions

The subject matter of Extramarks has made the question series so that students can regularly solve questions from the chapter. They have taken help from the textbook exercises, CBSE sample papers, CBSE past years’ question papers, NCERT exemplars and important reference books. They have solved the questions, and experienced professionals have further checked the answers to ensure the best quality of the content. Thus. The Important Questions Class 7 Mathematics Chapter 2 will help students to score better in exams. The important questions are-

Question 1. Ritu ate at least (3/5) part of an apple, and then the remaining apple was eaten by ritu’s brother Shyam. How many parts of the apple did Shyam eat? Who had the larger share? And By how much?

From the above question, it is given that,

The part of apple eaten by Ritu is equal to (3/5)

And the part of apple eaten by Shyam is = 1 – the part of apple eaten by Ritu is?

= 1 – (3/5)

The LCM of numbers 1 and 5 is = 5

Now, let us change these given fractions into an equivalent fraction by taking ten as the denominator number.

= [(1/1) × (5/5)] – [(3/5) × (1/1)]

= (5/5) – (3/5)

= (5 – 3)/5

= 2/5

Hence the part of apple eaten by Shyam is (2/5)

So, (3/5) is greater than (2/5); hence, Ritu ate the larger apple.

Now, the difference between the shares is = (3/5) – (2/5)

= (3 – 2)/5

= 1/5

Thus, Ritu’s share is greater than the share of Shyam by (1/5)

Question 2. Michae finished colouring a picture on (7/12) hour. On the other hand, Vaibhav finished colouring the same picture in (3/4) hour. Who worked longer colouring the picture? By what fraction was it longer?

From the above question, it is given that,

Time taken by Michae to colour the picture is = (7/12)

Time taken by Vaibhav to colour the picture is = (3/4)

The LCM of numbers 12, 4 = 12

Now, let us change these given fractions into an equivalent fractions by using 12 as the denominator number.

(7/12) = (7/12) × (1/1) = 7/12

The same method is applied to,

(3/4) = (3/4) × (3/3) = 9/12

As seen, (7/12) is less than (9/12)

Hence, (7/12) < (3/4)

Thus, Vaibhav worked for a longer time as compared.

Now, Vaibhav worked longer time by = (3/4) – (7/12)

= (9/12) – (7/12)

= (9 – 7)/12

= (2/12)

= (1/6) of an hour.

Question 3. Vidya and Pratik went on a picnic. Their mother gave them a mineral water bottle that contained 5 litres of water. Vidya consumed around 2/5 of the water, and Pratik consumed the remaining water.

(i) How many litres of water did Vidya drink?

(ii) And What fraction of the total quantity of water did Pratik drink?

(i) From the above question, it is given that,

The amount of water in the water bottle is = 5 litres.

The amount of water consumed by Vidya is = 2/5 of 5 litres.

= (2/5) × 5

= 2 litres

Hence, the total amount of water drank by Vidya is 2 litres.

(ii) From the above question, it is given that,

Amount of water present in the water bottle = 5 litres

Then,

The amount of water consumed by Pratik = (1 – water consumed by Vidya)

= (1 – (2/5))

= (5-2)/5

= 3/5

Hence the Total Amount of water consumed by Pratik = is 3/5 of 5 litres.

= (3/5) × 5

= 3 litres

So, the total amount of water consumed by Pratik is 3 litres.

Question 4. Which of the following is greater:

(i) (2/7) of (3/4) or the fraction (3/5) of (5/8)

We have seen that,

= (2/7) × (3/4) and the (3/5) × (5/8)

Hence, By the rule of Multiplication of the fraction,

The product of fraction is equal to (product of numerator)/ (product of denominator)

Then,

= (2/7) × (3/4)

= (2 × 3)/ (7 × 4)

= (1 × 3)/ (7 × 2)

= (3/14) … [i]

And,

= (3/5) × (5/8)

= (3 × 5)/ (5 × 8)

= (3 × 1)/ (1 × 8)

= (3/8) … [ii]

Now, convert the [i] and [ii] into like fractions,

LCM of 14 and 8 become 56

Now, let us change each of these given fractions into an equivalent fraction that has 56 as its denominator number.

[(3/14) × (4/4)] is equal to (12/56) [(3/8) × (7/7)] = (21/56)

Clearly, it is seen,

(12/56) is less than (21/56)

Hence,

(3/14) is less than (3/8)

(ii) (1/2) of (6/7) or the (2/3) of (3/7)

We have seen that,

= (1/2) × (6/7) and the (2/3) × (3/7)

By the rule of Multiplication of the fraction,

Product of fraction = (product of numerator) divided by (product of denominator)

Then,

= (1/2) × (6/7)

= (1 × 6)/ (2 × 7)

= (1 × 3)/ (1 × 7)

= (3/7) … [i]

And,

= (2/3) × (3/7)

= (2 × 3)/ (3 × 7)

= (2 × 1)/ (1 × 7)

= (2/7) … [ii]

By comparing [i] and [ii],

Clearly,

(3/7) > (2/7)

Question 5. Saili plants four saplings successively in a row in her garden. The distance between the two adjacent saplings is ¾ m. Now Find the distance between the first and the last planted sapling.

From the above question, it is given that,

The distance between the two adjacent saplings is = ¾ m

The number of saplings which are planted by Saili in a row is = 4

Then, the number of the gap in the saplings = ¾ × 4

= 3

Hence The total distance between the first and the last saplings becomes = three × ¾

= (9/4) m

= 2 ¼ m

Thus, the distance between the first and the last saplings is two ¼ m.

Question 6. Lipika always reads a book for one ¾ hour every day. She reads an entire book in 6 days. How many hours in total were required by her to finish the book?

From the above question, it is clearly given that,

Lipika reads her book for = one ¾ hours every day, which is equal to 7/4 hours for six days.

The number of days she took to finish the entire book = was six days

Hence, the Total number of hours required by her to fully complete the book = (7/4) × 6

= (7/2) × 3

= 21/2

= 10 ½ hours

Thus, the total number of hours required by her to complete the book is 10 ½ hours.

Question 7. A car runs around 16 km using 1 litre of petrol. How much distance will the car cover use two ¾ litres of petrol?

From the above question, it is given that,

The total distance travelled by car in 1 litre of petrol is = 16 km.

Then,

The Total quantity of petrol becomes = two ¾ litre = 11/4 litres

The total distance travelled by car in 11/4 litres of petrol is = (11/4) × 16

= 11 × 4

= 44 km

Hence, the total distance travelled by car in 11/4 litres of petrol is 44 km.

Question 8. Find the reciprocal of each of these following fractions. Also, Classify these reciprocals as proper fractions, improper fractions and whole numbers.

(i) 3/7

Reciprocal of (3/7) is (7/3) [which is ((3/7) × (7/3)) = 1]

So, it is an improper fraction.

An improper fraction is defined as a fraction in which the numerator is always greater than its denominator.

(ii) 5/8

Reciprocal of (5/8) is (8/5) [which is ((5/8) × (8/5)) = 1]

So, it is also an improper fraction.

An improper fraction is defined as a fraction in which the numerator is greater than its denominator.

(iii) 9/7

Reciprocal of (9/7) is (7/9) [which is ((9/7) × (7/9)) = 1]

Hence, it is a proper fraction.

A proper fraction is defined as that fraction in which the denominator is greater than the numerator of their fraction.

(iv) 6/5

Reciprocal of (6/5) is (5/6) [which is ((6/5) × (5/6)) = 1]

Hence, it is a proper fraction.

A proper fraction is defined as a fraction in which the denominator is greater than the numerator of any fraction.

(v) 12/7

Reciprocal of (12/7) is (7/12) [which is ((12/7) × (7/12)) = 1]

Hence, it is a proper fraction.

A proper fraction is defined as a fraction in which the denominator is greater than the numerator of the given fraction.

(vi) 1/8

Reciprocal of the fraction (1/8) is (8/1) or 8 as [∵ ((1/8) × (8/1)) = 1]

Hence, it is a whole number.

Whole numbers are the total collection of all positive integers, including the number 0.

(vii) 1/11

Reciprocal of the fraction (1/11) is (11/1) or 11 which is [∵ ((1/11) × (11/1)) = 1]

So, it is a whole number.

Whole numbers are the total collection of all positive integers, including 0.

Question 9. Find:

(i) (7/3) ÷ 2

We have,

= (7/3) × reciprocal of 2

= (7/3) × (1/2)

= (7 × 1) / (3 × 2)

= 7/6

(ii) (4/9) ÷ 5

We have,

= (4/9) × reciprocal of 5

= (4/9) × (1/5)

= (4 × 1) / (9 × 5)

= 4/45

(iii) (6/13) ÷ 7

We have,

= (6/13) × reciprocal of 7

= (6/13) × (1/7)

= (6 × 1) / (13 × 7)

= 6/91

Question 10. Which of the following is greater?

(i) 0.5 or 0.05

By comparing the whole number, we get 0 = 0

By comparing the tenths place digit, we get 5 > 0

∴ 0.5 > 0.05

(ii) 0.7 or 0.5

By comparing the whole number, 0 = 0

By comparing the tenths place digit we receive, 7 > 5

∴ 0.7 > 0.5

(iii) 7 or 0.7

By comparing these whole numbers, 7 > 0

∴ 7 > 0.7

(iv) 1.37 or 1.49

By comparing these whole numbers, 1 = 1

By comparing the tenths place digit, we get, 3 < 4

∴ 1.37 < 1.49

(v) 2.03 or 2.30

By comparing the whole number, 2 = 2

By comparing the tenths place digit, we get, 0 < 3

∴ 2.03 < 2.30

(vi) 0.8 or 0.88

By comparing these whole numbers, 0 = 0

By comparing the tenths place digit, we get, 8 = 8

Also, by comparing the hundredths place digit, 0 < 8

∴ 0.8 < 0.88

Question 11. Express as rupees as decimals:

(i) 7 paise

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 paise = ₹ (7/100)

= ₹ 0.07

(ii) 7 rupees 7 paise

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 7 rupees 7 paise = ₹ 7 + ₹ (7/100)

= ₹ 7 + ₹ 0.07

= ₹ 7.07

(iii) 77 rupees 77 paise

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 77 rupees 77 paise = ₹ 77 + ₹ (77/100)

= ₹ 77 + ₹ 0.77

= ₹ 77.77

(iv) 50 paise

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 50 paise = ₹ (50/100)

= ₹ 0.50

(v) 235 paise

We know that,

= ₹ 1 = 100 paise

= 1 paise = ₹ (1/100)

∴ 235 paise = ₹ (235/100)

= ₹ 2.35

Question 12. (i) Express 5 centimeters in meter and kilometre

We all know that,

= 1 meter = 100 centimeter

Then,

= 1 cm = (1/100) meter

= 5 cm = (5/100)

= 0.05 m

Now,

= 1 km = 1000 meter

Then,

= 1 m = (1/1000) kilometer

= 0.05 m = (0.05/1000)

= 0. 00005 kilometer

(i) Express 35 mm in cm, m and km

We know that,

= 1 cm = 10 mm

Then,

= 1 mm = (1/10) cm

= 35 mm = (35/10) cm

= 3.5 cm

And,

= 1 meter = 100 cm

Then,

= 1 cm = (1/100) m

= 3.5 cm = (3.5/100) m

= (35/1000) m

= 0.035 m

Now,

= 1 km = 1000 m

Then,

= 1 m = (1/1000) km

= 0.035 m = (0.035/1000)

= 0. 000035 km

Question 13. Express in kilogram:

(i) 200 gram

We know that,

= 1 kg = 1000 gram

Then,

= 1 g = (1/1000) kg

= 200 g = (200/1000) kilogram

= (2/10)

= 0.2 kg

(ii) 3470 gram

We know that,

= 1 kg = 1000 gram

Then,

= 1 g = (1/1000) kilogram

= 3470 g = (3470/1000) kilogram

= (3470/100)

= 3.470 kg

(ii) 4 kg 8 g

We know that,

= 1 kg = 1000 gram

Then,

= 1 g = (1/1000) kilogram

= 4 kg 8 g is equal to 4 kg + (8/1000) kilogram

= 4 kg + 0.008

= 4.008 kilogram

Question 14. Write the following in decimal numbers in their expanded form:

(i) 20.03

We have that,

20.03 = (2 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(ii) 2.03

We have that,

2.03 is equal to (2 × 1) + (0 × (1/10)) + (3 × (1/100))

(iii) 200.03

We have,

200.03 is (2 × 100) + (0 × 10) + (0 × 1) + (0 × (1/10)) + (3 × (1/100))

(iv) 2.034

We have,

2.034 gets equal to (2 × 1) + (0 × (1/10)) + (3 × (1/100)) + (4 × (1/1000)).

Question. Write the place value of 2 in the following decimal numbers:

(i) 2.56

From the above question, we can observe that,

The place value of 2 in the decimal 2.56 is ones.

(ii) 21.37

From the above question, we can observe that,

The place value of 2 in the decimal 21.37 is tens.

(iii) 10.25

From the above question, we can observe that,

The place value of 2 in the decimal 10.25 is tenths.

(iv) 9.42

From the above question, we can observe that,

The place value of 2 in the decimal 9.42 is the hundredth.

(v) 63.352

From the above question, we can observe that,

The place value of 2 in the decimal 63.352 is the thousandth.

Question 15. Shyama bought around 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala then bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?

From the above question, it is given that,

Fruits bought by Shyama is = 5 kg 300 g

= 5 kg + (300/1000) kg

= 5 kg + 0.3 kg

= 5.3 kg

Fruits bought by Sarala is = 4 kg 800 g + 4 kg 150 g

= (4 + (800/1000)) + (4 + (150/1000))

= (4 + 0.8) kg + (4 + .150) kg

= 4.8 kg + 4.150kg

= 8.950 kg

So, Sarala bought more fruits.

Question 16. How much less is 28 km as compared to 42.6 km?

Now, we have to find the difference between 42.6 km and 28 km.

42.6

-28.0

14.6

Hence, 14.6 km less is 28 km than 42.6 km.

Question 17. Find:

(i) 1.3 × 10

On multiplying the decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 1.3 × 10 = 13

(ii) 36.8 × 10

n multiplying the decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 36.8 × 10 = 368

(iii) 153.7 × 10

On multiplying the decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 153.7 × 10 = 1537

(iv) 168.07 × 10

On multiplying the decimal by 10, the decimal point is shifted to the right by one place.

We have,

= 168.07 × 10 = 1680.7

(v) 31.1 × 100

On multiplying the decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 31.1 × 100 = 3110

(vi) 156.1 × 100

On multiplying the decimal by 100, the decimal point is shifted to the right by two places.

We have,

= 156.1 × 100 = 15610

(vii) 3.62 × 100

On multiplying a decimal by the number 100, the decimal point is shifted to the right by two places.

We have,

= 3.62 × 100 = 362

(viii) 43.07 × 100

On multiplying a decimal by the number 100, the decimal point is shifted to the right by two places.

We have,

= 43.07 × 100 = 4307

Question 18. A two-wheeler covers a total distance of 55.3 km in one litre of petrol. How much total distance will it cover in 10 litres of petrol?

From the above question, it is given that,

Distance covered by the two-wheeler in 1L of petrol = 55.3 km

Now,

Distance covered by the two-wheeler in 10L of petrol is = (10 × 55.3)

= 553 km

Hence the Two-wheeler covers a distance of 10L of petrol 553 km.

Question 19. Find the following:

(a) 2.5 × 0.3

We have,

= (25/10) × (3/10)

= (75/100)

On dividing the decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 0.75

(b) 0.1 × 51.7

We have,

= (1/10) × (517/10)

= (517/100)

On dividing the decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 5.17

(c) 0.2 × 316.8

We have,

= (2/10) × (3168/10)

= (6336/100)

On dividing the decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 63.36

(d) 1.3 × 3.1

We have,

= (13/10) × (31/10)

= (403/100)

On dividing the decimal by 100, the decimal point is shifted to the left by two places.

Then,

= 4.03.

Question 20. Find:

(i) 7 ÷ 3.5

We have,

= 7 ÷ (35/10)

= 7 × (10/35)

= 1 × (10/5)

= 2

(ii) 36 ÷ 0.2

We have,

= 36 ÷ (2/10)

= 36 × (10/2)

= 18 × 10

= 180

(iii) 3.25 ÷ 0.5

We have,

= (325/100) ÷ (5/10)

= (325/100) × (10/5)

= (325 × 10)/ (100 × 5)

= (65 × 1)/ (10 × 1)

= 65/10

= 6.5

### Benefits of Solving Important Questions Class 7 Mathematics Chapter 2

Practice is very important for students. It helps them to clear their doubts and strengthen their concepts. Often, students need more than textbook exercises. They may take help from the chapter-wise question series made by our professionals. They have collected the questions from several sources and given the solutions in the Class 7 Mathematics Chapter 2 Important Questions, and there will be multiple benefits to solving these questions. These are-

• Sometimes, students need more than textbook exercises. They would need help to search for questions from different sources. The experts have made the task easier for the students. They collected questions from the CBSE sample papers, NCERT textbook exercises, NCERT exemplars and important reference books. Apart from this, they have included a few questions from CBSE past years’ question papers. Thus, students will find the vital questions in one single pdf.
• The experts have not only included the questions, but they have solved the questions too. They have given a step-step process to solve the problems. Thus, the students can follow the solutions if they cannot solve the questions. Experienced professionals have further checked the answers to ensure the best quality of the content. Thus, the Mathematics Class 7 Chapter 2 Important Questions will also help the students to solve the questions. It will boost their confidence and generate their interest in the subject matter.
• The habit of practice is very important for students. They must solve questions regularly to strengthen their concepts of the subject matter. The experts have tried to include all the important concepts in this article. Thus, the Important Questions Class 7 Mathematics Chapter 2 will help students in several ways. It will clear their doubts and strengthen their concepts. Thus, it will help them to score better in exams. Some students are scared of this subject. It is because they need help understanding the concepts clearly. This question series will help them greatly because they can follow the solutions and learn how to solve the sums.

Extramarks is a leading company that provides all the study materials related to CBSE and NCERT. After registering on our official website, you can download these materials. We provide CBSE syllabus, NCERT books, CBSE past years’ question papers, CBSE sample papers, CBSE extra questions, CBSE revision notes, NCERT exemplars, NCERT solutions, chapter-wise important questions, vital formulas and many more. Like the Important Questions Class 7 Mathematics Chapter 2, you will find important questions for other chapters of Class 7 Mathematics. Links to the study materials are given below-

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Q.1

$\begin{array}{l}\text{What is the value of the following expression?}\\ \frac{0.1}{0.01}+\frac{0.01}{0.1}\end{array}$

1.10.10

2.10.01

3.1.10

4.1.01

Marks:1

Ans Option1
Explanation

$\frac{0.1}{0.01}+\frac{0.01}{0.1}=10+\frac{1}{10}=10+0.1=10.1$

Q.2 What is the value of 10.05 × 1.05?

1. 10.5125

2. 10.5525

3. 10.5515

4. 10.5425

Marks:1

Ans Option2
Explanation

1005 × 105 = 105525

Since 10.05 has 2 digits and 1.05 has 2 digits after the decimal point, therefore, the product must have 4 digits after decimal point.

10.05 × 1.05 = 10.5525

Q.3 Bob has given half of the whole cake to his sister Maria. Maria further divided it into three parts, ate only one part and kept the rest two parts in the fridge for later. What part of the whole cake has she kept in the fridge?

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{3}{2}$

Marks:1

Ans Option2
Explanation

$\begin{array}{l}\text{Share=}\frac{\text{1}}{\text{2}}\\ \text{The part which Maria ate}=\frac{1}{2}÷3\\ \text{Remaining part}=\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}\end{array}$

Q.4

$4+\frac{3}{5}-\frac{2}{7}$

Marks:1
Ans

L.C.M of 1, 5, 7 = 35

$\begin{array}{l}\therefore \text{\hspace{0.17em}}4+\frac{3}{5}-\frac{2}{7}=\frac{140+21-10}{35}\\ \text{=}\frac{151}{35}\end{array}$

Q.5  What is the expanded form of 205.149?

Marks:1

$2×1000+0×100+5×10+1×\frac{1}{10}+4×\frac{1}{100}+9×\frac{1}{1000}$ $2×100+0×10+5×1+1×1+4×\frac{1}{10}+9×\frac{1}{100}$ $1×100+4×10+9×1+2×\frac{1}{10}+0×\frac{1}{100}+5×\frac{1}{1000}$

Ans Option4
Explanation

 Hundreds (100) Tens (10) Ones (1) Tenths $\left(\frac{1}{10}\right)$ Hundredths $\left(\frac{1}{100}\right)$ Thousandths $\left(\frac{1}{1000}\right)$ 2 0 5 1 4 9

$\begin{array}{l}\text{The expanded form of}205.149\text{is:}\\ 2×100+0×10+5×1+1×\frac{1}{10}+4×\frac{1}{100}+9×\frac{1}{1000}\end{array}$