NCERT Solutions

Chapter 3 A Peek Beyond The Point

Exercise 3.3

NCERT Solutions Class 7 Maths Chapter 3 A Peek Beyond The Point Exercise 3.3

Exercise 3.3 covers how we use decimals in our daily lives. For example, you will learn how to add and subtract decimal values, and understand how a decimal in the real world works.

The NCERT Solutions is in accordance with the latest NCERT syllabus and will help you clarify every question in a step by step manner. This exercise is important for your exams and regular practice helps you build your confidence in similar Maths problems. They are also suitable for effective revision during your exams.

1.0Download NCERT Solutions Class 7 Maths Chapter 3 A Peek Beyond The Point Exercise 3.3: Free PDF

Exercise 3.3 includes real-life decimal problems related to money. These NCERT Solutions for Class 7 Maths Chapter 3 are explained clearly in a step by step manner. Click the “Download PDF” link to download the free PDF of the solutions:

NCERT Solutions Class 7 Maths Chapter 3 Exercise 3.3

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2.0Key Concepts in Exercise 3.3 of Class 7 Maths Chapter 3

This exercise focuses on using decimals in real-life situations. Key ideas include:

Solving problems involving money with decimals
Addition and subtraction of decimal values
Understanding word problems using decimal amounts
Building accuracy in decimal-based calculations

3.0NCERT Class 7 Maths Chapter 3: Other Exercises

NCERT Solutions Class 7 Maths Chapter 3: Exercise 3.1

NCERT Solutions Class 7 Maths Chapter 3: Exercise 3.2

NCERT Solutions Class 7 Maths Chapter 3: Exercise 3.3

4.0NCERT Class 7 Maths Chapter 3 Exercise 3.3: Detailed Solutions

1. Convert the following fractions into decimals:

(a) 5/100

(b) 16/1000

(c) 12/10

(d) 254/1000

Sol. (a) 5/100 = 0.05

(b) 16/1000 = 10/1000 + 6/1000 = .01 + .006 = .016

(c) 12/10 = 10/10 + 2/10 = 1 + 2/10 = 1.2

(d) 254/1000 = 200/1000 + 50/1000 + 4/1000

= 2/10 + 5/100 + 4/1000

= .2 + .05 + .004 = 0.254

2. Convert the following decimals into a sum of tenths, hundredths and thousandths:

(a) 0.34

(b) 1.02

(c) 0.8

(d) 0.362

Sol. (a) 0.34 = 34/100 = 30/100 + 4/100 = 3/10 + 4/100

(b) 1.02 = 102/100 = 100/100 + 2/100

(c) 0.8 = 8/10

(d) 0.362 = 362/1000 = 300/1000 + 60/1000 + 2/1000 = 3/10 + 6/100 + 2/1000

3. What decimal number does each letter represent in the number line below.

NCERT Solutions class 7 excericse 3.1

Sol. There are 4 divisions between 6.4 and 6.5, so each division is one-fourth part of 0.1 or 1/10, i.e., 1/40=0.025 unit.

Therefore, the second division after 6.4, denoted by ' a ', represents the number 6.45, while the first division after 6.5 , denoted by 'c', represents the number 6.525, and the second division after 6.5, denoted by 'b', represents the number 6.55.

4. Arrange the following quantities in descending order:

(a) 11.01, 1.011, 1.101, 11.10, 1.01

(b) 2.567,2.675,2.768,2.499,2.698

(c) 4.678 g,4.595 g,4.600 g,4.656 g,4.666 g

(d) 33.13 m,33.31 m,33.133 m,33.331 m, 33.313 m

Sol. (a) Using the decimal place value chart, we find that:

Tens	Ones	Decimal point	Tenths	Hundredths	Thousandths
1	1	.	0	1
	1	⋅	0	1	1
	1	.	1	0	1
1	1	.	1	0
	1	.	0	1

Here, the two numbers 11.01 and 11.10 have 11 whole-number parts, but the first number has 0 tenths, whereas the second number has 1 tenth. Therefore, 11.10 > 11.01. The three numbers 1.011, 1.101, and 1.01 have 1 whole number part, but the

first and third numbers have 0 tenths, whereas the second number has 1 tenth. Therefore, 1.101 is the greatest among the three. Now, comparing the remaining two numbers, we get 1.011>1.01.

Thus, the numbers in descending order are:

11.10>11.01>1.101>1.011>1.01.

(b) 2.567,2.675,2.768,2.499,2.698

Tens	Ones	Decimal point	Tenths	Hundre dths	Thousa ndths
	2		5	6	7
	2	.	6	7	5
	2	-	7	6	8
	2	⋅	4	9	9
	2	⋅	6	9	8

Thus, the given quantities in descending order are:

2.768>2.698>2.675>2.567>2.499

(c) 4.678 g,4.595 g,4.600 g,4.656 g, 4.666 g

Tens	Ones	Decimal point	Tenths	Hundre dths	Thousa ndths
	4	.	6	7	8
	4	⋅	5	9	5
	4	-	6	0	0
	4	.	6	5	6
	4	-	6	6	6

Thus, the given quantities in descending order are:

4.678 g>4.666 g>4.656 g>4.600 g

>4.595 g.

(d) 33.13 m,33.31 m,33.133 m,33.331 m, 33.313 m

Tens	Ones	Decimal point	Tenths	Hundre dths	Thousa ndths
3	3	.	1	3
3	3	-	3	1
3	3	-	1	3	3
3	3	-	3	3	1
3	3	-	3	1	3

Thus, the given quantitates in descending order are:

33.331 m>33.313 m>33.31 m>33.133 m>33.13 m.

5. Using the digits 1,4,0,8 and 6 make:

(a) the decimal number closest to 30

(b) the smallest possible decimal number between 100 and 1000.

Sol. Using the digits 1,4,0,8 and 6 , we can make:

(a) The decimal number closet to 30→40.168.

(b) The smallest possible decimal number between 100 and 1000→104.68.

6. Will a decimal number with more digits be greater than a decimal number with fewer digits?

Sol. No, It is not necessary as 0.9>0.123456789.

7. Mahi purchases 0.25 kg of beans, 0.3 kg of carrots, 0.5 kg of potatoes, 0.2 kg of capsicums, and 0.05 kg of ginger. Calculate the total weight of the items she bought.

Sol. The total weight of the items Mahi bought =0.25 kg+0.3 kg+0.5 kg+0.2 kg+0.05 kg=1.3 kg

8. Pinto supplies 3.79 L,4.2 L, and 4.25 L of milk to a milk dairy in the first three days. In 6 days, he supplies 25 litres of milk. Find the total quantity of milk supplied to the dairy in the last three days.

Sol. The Total quantity of milk supplied to the dairy in the last three days = Toal milk supplied in the 6 days-

Total milk supplied in the first 3 days

=25 L−(3.79 L+4.2 L+4.25 L)

=25 L−12.24 L=12.76 L

9. Tinku weighed 35.75 kg in January and 34.50 kg in February. Has he gained or lost weight? How much is the change ?

Sol. Since 35.75 kg>34.50 kg, Tinku has lost weight.

Now, the change in the weight =35.75 kg−34.50 kg=1.25 kg

10. Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, ____ , ____ .

Sol. Let us analyse the given pattern:

5.5(+0.9)→6.4(−0.01)→6.39(+0.9)→

7.29(−0.01)→7.28(+0.9)→8.18(−0.01)

→8.17.

So, the sequence follows an increasing trend of 0.9 and then a decreasing trend of 0.01 alternatively.

Thus, the next two numbers are 9.07 and 9.06.

11. How many millimetres make 1 kilometre?

Sol. We know that 1 km=1000 m and 1 m= 1000 mm .Therefore, 1 km=1000×1000 mm= 1000000 mm

12. Indian Railways offers optional travel insurance for passengers who book e-tickets. It costs 45 paise per passenger. If 1 lakh people opt for insurance in a day, what is the total insurance fee paid?

Sol. The insurance fee paid for 1 passenger =45p=₹0.45.

So, total insurance fee paid for 1 lakh passengers =₹0.45×100000=₹45000.

13. Which is greater?

(a) 10/1000 or 1/10?

(b) One-hundredth or 90 thousandths?

(c) One-thousandth or 90 hundredths?

Sol.

(a) Compare 10/1000 and 1/10.

10/1000 simplifies to 1/100.

1/10 = 10/100.

Since 10/100 > 1/100,

1/10 is greater than 10/1000.

(b) Compare One-hundredth (1/100) and 90 thousandths (90/1000).

90/1000 simplifies to 9/100.

Since 9/100 > 1/100,

90 thousandths is greater than One-hundredth.

(c) Compare One-thousandth (1/1000) and 90 hundredths (90/100).

90/100 = 900/1000.

Since 900/1000 > 1/1000,

90 hundredths is greater than One-thousandth.

14. Write the decimal forms of the quantities mentioned:

(a) 87 ones, 5 tenths and 60 hundredths

(b) 12 tens and 12 tenths

(c) 10 tens, 10 ones, 10 tenths, and 10 hundredths

(d) 25 tens, 25 ones, 25 tenths, and 25 hundredths

Sol.

(a) 87 ones, 5 tenths, and 60 hundredths

= (87 * 1) + (5 * 1/10) + (60 * 1/100)

= 87 + 0.5 + 0.60

= 88.1

(b) 12 tens and 12 tenths

= (12 * 10) + (12 * 1/10)

= 120 + 1.2

= 121.2

(c) 10 tens, 10 ones, 10 tenths, and 10 hundredths

= (10 * 10) + (10 * 1) + (10 * 1/10) + (10 * 1/100)

= 100 + 10 + 1 + 0.10

= 111.1

(d) 25 tens, 25 ones, 25 tenths, and 25 hundredths

= (25 * 10) + (25 * 1) + (25 * 1/10) + (25 * 1/100)

= 250 + 25 + 2.5 + 0.25

= 277.75

15. Using each digit 0−9 not more than once, fill the boxes below so that the sum is closest to 10.5.

Sample Questions on ncert solutions

Sol.

NCERT Solutions clas 7 maths

16. Write the following fractions in decimal form:

(a) 1/2

(b) 3/2

(c) 1/4

(d) 3/4

(e) 1/5

(f) 4/5

Sol.

(a) 1/2

Multiply numerator and denominator by 5:

1/2 * 5/5 = 5/10

= 0.5

(b) 3/2

Multiply numerator and denominator by 5:

3/2 * 5/5 = 15/10

= 10/10 + 5/10

= 1 + 0.5

= 1.5

(c) 1/4

Multiply numerator and denominator by 25:

1/4 * 25/25 = 25/100

= 0.25

(d) 3/4

Multiply numerator and denominator by 25:

3/4 * 25/25 = 75/100

= 0.75

(e) 1/5

Multiply numerator and denominator by 2:

1/5 * 2/2 = 2/10

= 0.2

(f) 4/5

Multiply numerator and denominator by 2:

4/5 * 2/2 = 8/10

= 0.8

5.0Key Features and Benefits of Class 7 Maths Chapter 3 Exercise 3.3

The exercise to the updated NCERT syllabus with realistic situations involving decimals.
It also teaches how decimals are practically worked within our daily life situations, such as money and shopping.
The pdf includes step-by-step solutions in a simplified clear language to ease learning.
Supports in the preparation for the CBSE exam and other competitive exam like the olympiads by developing word problem skills.
Regular practice also improves the thinking and problem solving skills

NCERT Class 6 Maths Ch. 3 A Peek Beyond the Point Other Exercises:-

Exercise 3.3

NCERT Solutions for Class 7 Maths Other Chapters:-

Chapter 1: Large Numbers Around Us

Chapter 2: Arithmetic Expressions

Chapter 3: A Peek Beyond the Point

Chapter 4: Expressions Using Letter-Numbers

Chapter 5: Parallel and Intersecting Lines

Chapter 6: Number Play

Chapter 7: A Tale of Three Intersecting Lines

Chapter 8: Working with Fractions

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