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NCERT Solutions for Class 7 Maths Ganita Prakash Chapter 3 A...

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

Author at PW

February 27, 2026

Mathematics becomes a lot more interesting when we start exploring what happens between whole numbers. Chapter 3 of the current Ganita Prakash syllabus is where decimals and rational numbers are first taught to students. It can be quite tricky for most students since they are first introduced to numbers that are not whole. This is why it is so important for students to have good class 7 maths chapter 3 solutions.

With the help of these NCERT solutions, students can learn how numbers behave on both sides of the decimal point. This chapter specifically deals with the transition from the integer part to the fractional part. This provides a great foundation for more complex concepts such as algebra. With the help of the class 7 maths chapter 3 summary, students can learn how to use the number line and basic math operations to confidently express and calculate numbers.

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Chapter 3 Overview A Peek Beyond the Point

The title “A Peek Beyond the Point” refers to the decimal point, which opens the door to smaller units. While students may already know about positive decimals, this chapter also introduces negative decimals. Here, your class 7 integers notes come in handy since the rules for positive and negative numbers are the same.

Students are encouraged to think about these numbers in real-life situations, such as the measurement of a pencil or the decrease in temperature below zero. Class 7 maths chapter 3solutions are designed in such a way that it becomes easy to understand the real-life connections.

Key topics covered in Chapter 3 include:

Decimal Place Value: Understanding tenths, hundredths, and thousandths.
The Number Line: Learning to plot both positive and negative decimals.
Symmetry: Using integers to understand how -0.5 and 0.5 are opposites.
Comparison: Figuring out which decimal is larger using place values.

Decimal Place Value Chart Class 7 Maths

To solve problems accurately, you must know how digits shift to the right of the ones place. Students often search for a Decimal Place Value Chart Class 7 Maths to identify the value of a digit "beyond the point" during their revision.

Number	Tens (10)	Ones (1)	Tenths (1/10)	Hundredths (1/100)
15.25	1	5	2	5
0.08	0	0	0	8
7.3	0	7	3	0

The "point" separates the whole number part from the fractional part. Each step to the right makes the value ten times smaller. This chart is the first step toward finding the right class 7 maths chapter 3 solutions.

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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

= 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

Read More: NCERT Solutions for Class 7 Maths Chapter 1
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

= 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

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

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Benefits of Using Class 7 Ganita Prakash Chapter 3 Solutions

Selecting quality study material can make a huge difference in your grades. Here are the reasons why our class 7 maths chapter 3 solutions are so important to you:

These solutions are strictly written according to the latest edition because they incorporate the discovery-based learning approach.
Using mental calculation hints and patterns, students can quickly and correctly solve decimal questions.
Our solutions are written according to the latest Class 7 Maths Ganita Prakash syllabus, so you can be assured that the study material is up-to-date and exam-oriented.
We explain how negative decimals are a representation of positive decimals, which is an important topic in class 7 integers notes.
Practicing questions on decimals will help develop your logical and analytical skills.
You can easily download our solutions in PDF format, which is very convenient for studying anywhere and anytime.