NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3 PDF

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3  A Peek Beyond the Point provide clear and step-by-step explanations based on the latest CBSE syllabus. 

These Class 7 Maths Ganita Prakash Chapter 3 A Peek Beyond the Point solutions are designed to strengthen students’ basic understanding of decimals through practical examples, mental math strategies, and detailed question answers. Ideal for revision and exam preparation, these solutions ensure conceptual clarity and boost confidence in handling decimal-based problems.

Class 7 Maths Ganita Prakash Chapter 3 A Peek Beyond the Point

Chapter 3  A Peek Beyond the Point from the Class 7 Maths Ganita Prakash textbook introduces students to the world of decimal numbers. This chapter helps learners move beyond whole numbers and understand the concept of numbers after the decimal point, such as tenths, hundredths, and thousandths.

Class 7 Maths Ganita Prakash Chapter 3 Explanations cover key ideas such as place value in decimals, converting decimals to fractions, and solving problems using mental maths and estimation. This chapter encourages students to explore how decimals are used in everyday situations, making learning both practical and engaging. With activities, visual aids, and simple examples, it lays a strong foundation for understanding decimal operations in higher classes.

Class 7 A Peek Beyond the Point Question Answers

Below are the solutions to all the questions from Class 7 Maths Ganita Prakash Chapter 3  A Peek Beyond the Point. Class 7 Maths Ganita Prakash Chapter 3 Solutions include detailed explanations for decimal place values, comparison of decimal numbers, operations on decimals, and real-life word problems. These solutions aim to build a strong conceptual base and improve students' problem-solving skills.

 NCERT In-Text Questions (Pages 46-48)

Q.1. Which scale helped you measure the length of the screws accurately? Why?
Solution:
The third scale helped us to measure the length of the screws accurately because each unit length has been further divided into 10 equal parts.

Q.2. Can you explain why the unit was divided into smaller parts to measure the screws?
Solution:
Yes, as the screws are so long that cannot be measured in exact unit length. Therefore, the unit was divided into smaller parts.

Q.3. Measure the following objects using a scale and write their measurements in centimetres (as shown earlier for the lengths of the screws): pen, sharpener, and any other object of your choice.
Solution:
Using a scale, the measurements of the objects are as follows:

• Pen – 14.5 cm

• Sharpener – 3.8 cm

• Eraser (chosen object) – 4.2 cm

Note: The measurements may slightly vary depending on the size and brand of the objects.

NCERT In-Text Questions (Pages 49-52)

Figure it Out (Page 58)

3.4 Decimal Place Value

Q.1. NCERT In-Text Questions (Pages 61-64)

We can ask similar questions about fractional parts:
(a) How many thousandths make one unit?
(b) How many thousandths make one tenth?
(c) How many thousandths make one hundredth?
(d) How many tenths make one ten?
(e) How many hundredths make one ten?
Solution:
We know that the value of a place in the decimal place value chart becomes ten times at every step moving from right to left, and it becomes 1 10 times moving from left to right. Therefore,

(a) 1000 thousandths make one unit.
(b) 100 thousandths make one tenth.
(c) 10 thousandths make one hundredth.
(d) 100 tenths make one ten.
(e) 1000 hundredths make one ten.

NCERT In-Text Questions (Page 73)

Q.1. Which of the decimal numbers 0.9, 1.1, 1.01, and 1.11 is closest to 1.09?
Solution:
Arranging the decimal numbers in ascending order, we have 0.9 < 1.01 < 1.09 < 1.1 < 1.11
Among the neighbours of 1.09, 1.01 is  8 100  away from 1.09, whereas 1.1 is  1 100  away from 1.09. Therefore, 1.1 is closest to 1.09.

Q.2. Which among these is closest to 4: 3.56, 3.65, 3.099?
Solution:
Arranging the decimal numbers in ascending order, we have 3.099 < 3.56 < 3.65 < 4
3.65 is closest to 4 among the given decimal numbers.

NCERT In-Text Questions (Pages 75)

Question 1.
Find the sums.
(a) 5.3 + 2.6
(b) 18 + 8.8
(c) 2.15 + 5.26
(d) 9.01 + 9.10
(e) 29.19 + 9.91
(f) 0.934 + 0.6
(g) 0.75 + 0.03
(h) 6.236 + 0.487
Solution:

(a) 5.3 + 2.6

= 7.9

(b) 18 + 8.8

= 26.8

(c) 2.15 + 5.26

= 7.41

(d) 9.01 + 9.10

= 18.11

(e) 29.19 + 9.91

= 39.10

(f) 0.934 + 0.6

= 1.534

(g) 0.75 + 0.03

= 0.78

(h) 6.236 + 0

= 6.236

Question 2.
Find the differences.
(a) 5.6 – 2.3
(b) 18 – 8.8
(c) 10.4 – 4.5
(d) 17 – 16.198
(e) 17 – 0.05
(f) 34.505 – 18.1
(g) 9.9 – 9.09
(h) 6.236 – 0.487
Solution:
(a) 5.6 – 2.3

= 3.3

(b) 18 – 8.8

= 9.2

(c) 10.4 – 4.5

= 5.9

(d) 17 – 16.198

= 0.802

(e) 17 – 0.05

= 16.95

(f) 34.505 – 18.1

= 16.405

(g) 9.9 – 9.09

= 0.81

(h) 6.236 – 0.487

= 5.749

Decimal Sequences

NCERT In-Text Questions (Pages 75-76)

Q.1. Observe this sequence of decimal numbers and identify the change after each term.
4.4, 4.8. 5.2, 5.6, 6.0,….
We can see that 0.4 is being added to a term to get the next term.
Continue this sequence and write the next 3 terms.
Solution:
4.4, 4.8. 5.2, 5.6, 6.0, 6.4, 6.8, 7.2

Q.2. Similarly, identify the change and write the next 3 terms for each sequence given below. Try to do this computation mentally.
(a) 4.4, 4.45, 4.5,..…
(b) 25.75, 26.25, 26.75,……
(c) 10.56, 10.67, 10.78,….…
(d) 13.5, 16, 18.5,….…
(e) 8.5, 9.4, 10.3,……
(f) 5, 4.95, 4.90,……
(g) 12.45, 11.95, 11.45,……
(h) 36.5, 33, 29.5,……
Solution:
(a) 4.4, 4.45, 4.5, …

Change: +0.05
Next 3 terms: 4.55, 4.6, 4.65

(b) 25.75, 26.25, 26.75, …

Change: +0.5
Next 3 terms: 27.25, 27.75, 28.25

(c) 10.56, 10.67, 10.78, …

Change: +0.11
Next 3 terms: 10.89, 11.00, 11.11

(d) 13.5, 16, 18.5, …

Change: +2.5
Next 3 terms: 21, 23.5, 26

(e) 8.5, 9.4, 10.3, …

Change: +0.9
Next 3 terms: 11.2, 12.1, 13

(f) 5, 4.95, 4.90, …

Change: −0.05
Next 3 terms: 4.85, 4.80, 4.75

(g) 12.45, 11.95, 11.45, …

Change: −0.5
Next 3 terms: 10.95, 10.45, 9.95

(h) 36.5, 33, 29.5, …

Change: −3.5
Next 3 terms: 26, 22.5, 19

Estimating Sums and Differences

NCERT In-Text Questions (Pages 76)

Q.1. Sonu has observed sums and differences of decimal numbers and says, “If we add two decimal numbers, then the sum will always be greater than the sum of their whole number parts. Also, the sum will always be less than 2 more than the sum of their whole number parts.”

Solution: 
Let us use an example to understand what his claim means:
If the two numbers to be added are 25.936 and 8.202, the claim is that their sum will be greater than 25+ 8 (whole number parts) and will be less than 25 + 1 + 8 + 1.

Q.2. What do you think about this claim? Verily if this is true for these numbers. Will it work for any 2 decimal numbers?
Solution:
The given numbers are 25.936 and 8.202.
Sum of whole number parts = 25 + 8 = 33
Sum of given decimal numbers = 25.936 + 8.202 = 34.138
Clearly, 33 < 34.138 < (33 + 2)
Let another two decimal numbers 1.532 and 4.536
Sum of whole number part = 1 + 4 = 5
Sum of given decimal numbers = 1.532 + 4.536 = 6.068
Clearly, 5 < 6.068 < 5 + 2
So, the claim is true for the sum of any two decimal numbers.

What about for the sum of 25.93603259 and 8.202?
Solution:
The given numbers are 25.93603259 and 8.202.
Sum of whole number parts = 25 + 8 = 33
Sum of given decimal numbers = 25.93603259 + 8.202 = 34.13803259
Clearly, 33 < 34.13803259 < (33 + 2)

NCERT In-Text Questions (Pages 78)

Question 1.
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.
Solution:
Using the digits 1, 4, 0, 8, and 6, we can make:
(a) The decimal number closest to 30 → 40.168.
(b) The smallest possible decimal number between 100 and 1000 → 104.68

Question 2.
Will a decimal number with more digits be greater than a decimal number with fewer digits?
Solution:
No. It is not necessary as 0.9 > 0.123456789.

Question 3.
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.
Solution:
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.

Question 4.
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.
Solution:
The total quantity of milk supplied to the dairy in the last three days = Total 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

Question 5.
Tinku weighed 35.75 kg in January and 34.50 kg in February. Has he gained or lost weight? How much is the change?
Solution:
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.

Question 6.
Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, ____, _____
Solution:
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.

Question 7.
How many millimetres make 1 kilometre?
Solution:
We know that 1 km = 1000 m and 1 m = 1000 mm
Therefore, 1 km = 1000 × 1000 mm= 1000000 mm

Question 8.
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?
Solution:
The insurance fee paid for 1 passenger = 45 p = ₹ 0.45
So, total insurance fee paid for 1 lakh passengers = ₹ 0.45 × 100000 = ₹ 45000

NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3 PDF Download

Students looking to strengthen their understanding of decimals can now access the NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3  A Peek Beyond the Point in a simple and easy-to-follow format. These solutions are available in PDF form, making it convenient for students to download and study offline at any time. 

Class 7 Maths Ganita Prakash Chapter 3 Solutions PDF includes all question answers and detailed explanations, helping learners practice effectively and build confidence in decimal concepts. 

Benefits of Using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3

Here are the benefits of using NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3 A Peek Beyond the Point:

1. The solutions explain each decimal-related concept in a step-by-step manner, helping students grasp the topic with ease.

2. With mental calculation tips and patterns, students become faster and more accurate in solving decimal problems.

3. These solutions are based on the latest Class 7 Maths Ganita Prakash curriculum, ensuring relevant and exam-focused learning.

4. Regular practice of questions improves logical thinking and analytical skills in decimal operations.

5. Students can easily download the solutions in PDF form, making it convenient to study anytime, anywhere.