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

NCERT Solutions Class 7 Maths Chapter 3 A Peek Beyond the Point introduces students to decimal numbers in a fun and easy way. In this NCERT Solutions Class 7 Maths Chapter, students learn how to read, write, and use decimals in daily life—like when dealing with money, length, or weight. It also covers how to compare, add, and subtract decimal numbers step by step.

These NCERT Solutions are designed to help students clearly understand each concept through simple language and solved examples. With detailed explanations and practice problems, students can gain confidence and improve their calculation skills

1.0NCERT Solutions Class 7 Maths Chapter 3 A Peek Beyond the Point – Download PDF

Get your FREE PDF of NCERT Solutions Class 7 Maths Chapter 3 – A Peek Beyond the Point, created by the subject experts at ALLEN.

2.0Key Concepts in Chapter 3: A Peek Beyond the Point

1. Understanding Decimal Numbers

Decimals represent numbers that lie between whole numbers.

They are used when precision is required, such as measuring length (5.6 cm), weight (3.75 kg), or money (₹45.50).

Decimal numbers are expressed with digits to the right of the decimal point, representing parts of a whole.

2. Reading and Writing Decimals

Students learn how to read and write decimal numbers up to hundredths and thousandths.

Example: 3.07 is read as “three point zero seven”.

3. Place Value System (Decimal Version)

The place values to the right of the decimal point include:

• Tenths (1/10)
• Hundredths (1/100)
• Thousands (1/1000)
• Example: In 4.236 → 2 is tenths, 3 is hundredths, and 6 is thousandths.

4. Comparing Decimal Numbers

Students learn to compare decimal values based on place values.

Extra zeroes do not affect the value: 3.6 = 3.60

Example: 3.45 > 3.4 because 3.45 has 5 hundredths vs. 0.

5. Conversion between Fractions and Decimals

Converting decimals to fractions and vice versa strengthens understanding of their relationship.

Example: 0.25 = 25/100 = 1/4

6. Operations on Decimals

a) Addition and Subtraction

Align decimal points to perform operations correctly.

Ensure place values are matched while adding or subtracting.

b) Multiplication

Multiply as if they are whole numbers.

Place the decimal point in the result based on total decimal places in the multiplicands.

Example: 1.2 × 3.4 = 4.08

c) Division

Adjust the divisor or dividend by shifting the decimal point to make division easier.

Example: 2.5 ÷ 0.5 = 5

7. Use of Decimals in Daily Life

Understanding decimals is important for:

• Money transactions (e.g., ₹129.75)
• Measurement units (cm, kg, litre)
• Data interpretation (statistics, averages, percentages)

3.0NCERT Class 7 Maths Chapter 3 A Peek Beyond The Point: Detailed Solutions

FIGURE IT OUT-01

• Find the sums and differences: (a) $\frac{3}{10}+3\frac{4}{100}$ (b) $9\frac{5}{10}\frac{7}{100}+2\frac{1}{10}\frac{3}{100}$ (c) $15\frac{6}{10}\frac{4}{100}+14\frac{3}{10}\frac{6}{100}$ (d) $7\frac{7}{100}-4\frac{4}{100}$ (e) $8\frac{6}{100}-5\frac{3}{100}$ (f) $12\frac{6}{100}\frac{2}{100}-\frac{9}{10}\frac{9}{100}$ Sol. (a) $\frac{3}{10}+3\frac{4}{100}$ $\begin{array}{rl}& \frac{3}{10}+3\frac{4}{100}=\frac{3}{10}+3+\frac{4}{100}\\ & =3+\frac{3}{10}+\frac{4}{100}=3\frac{3}{10}\frac{4}{100}=3\frac{34}{100}\end{array}$ (b) $9\frac{5}{10}\frac{7}{100}+2\frac{1}{10}\frac{3}{100}$ $\begin{array}{rl}& 9\frac{5}{10}\frac{7}{100}+2\frac{1}{10}\frac{3}{100}\\ & =(9+2)+\left(\frac{5}{10}+\frac{1}{10}\right)+\left(\frac{7}{100}+\frac{3}{100}\right)\\ & =11+\frac{6}{10}+\frac{10}{100}\\ & =11+\frac{6}{10}+\frac{1}{10}=11\frac{7}{10}\end{array}$ (c) $15\frac{6}{10}\frac{4}{100}+14\frac{3}{10}\frac{6}{100}$ $15\frac{6}{10}\frac{4}{100}+14\frac{3}{10}\frac{6}{100}$ $=(15+14)+\left(\frac{6}{10}+\frac{3}{10}\right)+\left(\frac{4}{100}+\frac{6}{100}\right)$ $=29+\frac{9}{10}+\frac{10}{100}=29+\frac{9}{10}+\frac{1}{10}=29+\frac{10}{10}$ $=29+1=30$ (d) $7\frac{7}{100}-4\frac{4}{100}$ $7\frac{7}{100}$ $\underline{-4\frac{4}{100}}$ $=3\frac{3}{100}$ (e) $8\frac{6}{100}-5\frac{3}{100}$ $8\frac{6}{100}-5\frac{3}{100}=(8-5)+\left(\frac{6}{100}-\frac{3}{100}\right)$ $=3\frac{3}{100}$ (f) $12\frac{6}{10}\frac{2}{100}-\frac{9}{10}\frac{9}{100}$ $12\frac{6}{10}\frac{2}{100}\to 12\frac{5}{10}\frac{12}{100}\to 11\frac{15}{10}\frac{12}{100}$ $-\frac{9}{10}\frac{9}{100}-\frac{9}{10}\frac{9}{100}-\frac{9}{10}\frac{9}{100}$ $=11\frac{6}{10}\frac{3}{100}=11\frac{63}{100}$

FIGURE IT OUT-02

• 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) $6.236+0.487$
  
• 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

Sol. (a)

FIGURE IT OUT-03

• Convert the following fractions into decimals: (a) $\frac{5}{100}$ (b) $\frac{16}{1000}$ (c) $\frac{12}{10}$ (d) $\frac{254}{1000}$ Sol. (a) $\frac{5}{100}=0.05$ (b) $\frac{16}{1000}=\frac{10}{1000}+\frac{6}{1000}$ $=.01+.006=.016$ (c) $\frac{12}{10}=\frac{10}{10}+\frac{2}{10}=1+\frac{2}{10}=1.2$ (d) $\frac{254}{1000}=\frac{200}{1000}+\frac{50}{1000}+\frac{4}{1000}$ $\begin{array}{rl}& =\frac{2}{10}+\frac{5}{100}+\frac{4}{1000}\\ & =.2+.05+.004=0.254\end{array}$
• 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) $0.34=\frac{34}{100}=\frac{30}{100}+\frac{4}{100}=\frac{3}{10}+\frac{4}{100}$ (b) $1.02=\frac{102}{100}=\frac{100}{100}+\frac{2}{100}$ (c) $0.8=\frac{8}{10}$ (d) $0.362=\frac{362}{1000}=\frac{300}{1000}+\frac{60}{1000}+\frac{2}{1000}$ $=\frac{3}{10}+\frac{6}{100}+\frac{2}{1000}$
• What decimal number does each letter represent in the number line below.
  
  Sol. There are 4 divisions between 6.4 and 6.5, so each division is one-fourth part of 0.1 or $\frac{1}{10}$, i.e., $\frac{1}{40}=0.025$ 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.
• 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) $4.678\mathrm{g},4.595\mathrm{g},4.600\mathrm{g},4.656\mathrm{g},4.666\mathrm{g}$ (d) $33.13\mathrm{m},33.31\mathrm{m},33.133\mathrm{m},33.331\mathrm{m}$, 33.313 m Sol. (a) Using the decimal place value chart, we find that:

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) $2.567,2.675,2.768,2.499,2.698$

Thus, the given quantities in descending order are: $2.768>2.698>2.675>2.567>2.499$ (c) $4.678\mathrm{g},4.595\mathrm{g},4.600\mathrm{g},4.656\mathrm{g}$, 4.666 g

Thus, the given quantities in descending order are: $4.678\mathrm{g}>4.666\mathrm{g}>4.656\mathrm{g}>4.600\mathrm{g}$ $>4.595\mathrm{g}$. (d) $33.13\mathrm{m},33.31\mathrm{m},33.133\mathrm{m},33.331\mathrm{m}$, 33.313 m

Thus, the given quantitates in descending order are: $\begin{array}{rl}& 33.331\mathrm{m}>33.313\mathrm{m}>33.31\mathrm{m}>\\ & 33.133\mathrm{m}>33.13\mathrm{m}.\end{array}$

• 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 $30\to 40.168$. (b) The smallest possible decimal number between 100 and $1000\to 104.68$.
• Will a decimal number with more digits be greater than a decimal number with fewer digits? Sol. No, It is not necessary as $0.9>0.123456789$.
• 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 $=0.25\mathrm{kg}+0.3\mathrm{kg}+0.5\mathrm{kg}+0.2\mathrm{kg}+0.05\mathrm{kg}=1.3\mathrm{kg}$
• Pinto supplies $3.79\mathrm{L},4.2\mathrm{L}$, 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 $=25\mathrm{L}-(3.79\mathrm{L}+4.2\mathrm{L}+4.25\mathrm{L})$ $=25\mathrm{L}-12.24\mathrm{L}=12.76\mathrm{L}$
• 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 $35.75\mathrm{kg}>34.50\mathrm{kg}$, Tinku has lost weight. Now, the change in the weight $=35.75\mathrm{kg}-34.50\mathrm{kg}=1.25\mathrm{kg}$
• Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 8.18, 8.17, $\_\_\_\_$ , $\_\_\_\_$ . Sol. Let us analyse the given pattern: $5.5(+0.9)\to 6.4(-0.01)\to 6.39(+0.9)\to$ $7.29(-0.01)\to 7.28(+0.9)\to 8.18(-0.01)$ $\to 8.17$. 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.
• How many millimetres make 1 kilometre? Sol. We know that $1\mathrm{km}=1000\mathrm{m}$ and $1\mathrm{m}=$ 1000 mm . Therefore, $1\mathrm{km}=1000\times 1000\mathrm{mm}=$ 1000000 mm
• 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 $=45\mathrm{p}=₹0.45$. So, total insurance fee paid for 1 lakh passengers $=₹0.45\times 100000=₹45000$.
• Which is greater? (a) $\frac{10}{1000}$ or $\frac{1}{10}$ ? (b) One-hundredth or 90 thousandths? (c) One-thousandth or 90 hundredths? Sol. (a) As $\frac{10}{1000}=\frac{1}{10}$ and $\frac{1}{10}=\frac{10}{100}$ So $\frac{1}{10}>\frac{10}{1000}$ (b) As one- hundredths $=\frac{1}{100}$ and 90 thousandths = $\frac{90}{1000}=\frac{9}{100}$, so $\frac{9}{100}>\frac{1}{100}$ Or 90 thousandths > One-hundredth. (c) As one-thousandths $=\frac{1}{1000}$ and 90 hundredths $=\frac{90}{100}=\frac{900}{1000}$ So $\frac{1}{1000}<\frac{900}{1000}$
• Write the decimal forms of the quantities mentioned (an example is given): (a) 87 ones, 5 tenths and 60 hundredths $=88.10$ (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 $87\times 1+\frac{5}{10}+\frac{60}{100}$ $=87+\frac{11}{100}=88.10=88.10$ (b) 12 tens and 12 tenths $=12\times 10+12\times \frac{1}{10}$ $=120+1.2=121.2$ (c) 10 tens, 10 ones, 10 tenths, and 10 hundredths $=10\times 10+10\times 1+10\times \frac{1}{10}+10\times \frac{1}{100}$ $=100+10+1+\frac{1}{10}=111.1$ (d) 25 tens, 25 ones, 25 tenths, and 25 hundredths $=25\times 10+25\times 1+25\times \frac{1}{10}+25\times \frac{1}{100}$ $=250+25+\frac{20}{10}+\frac{5}{10}+\frac{20}{100}+\frac{5}{100}$ $=275+2+\frac{5}{10}+\frac{2}{10}+\frac{5}{100}=277.75$
• Using each digit $0-9$ not more than once, fill the boxes below so that the sum is closest to 10.5 .
  
  Sol.
  
• Write the following fractions in decimal form: (a) $\frac{1}{2}$ (b) $\frac{3}{2}$ (c) $\frac{1}{4}$ (d) $\frac{3}{4}$ (e) $\frac{1}{5}$ (f) $\frac{4}{5}$ Sol. (a) $\frac{1}{2}\times \frac{5}{5}=\frac{5}{10}=0.5$ (b) $\frac{3}{2}\times \frac{5}{5}=\frac{15}{10}=\frac{10}{10}+\frac{5}{10}=1+\frac{5}{10}=1.5$ (c) $\frac{1}{4}\times \frac{25}{25}=\frac{25}{100}=\frac{20}{100}+\frac{5}{100}$ $=\frac{2}{10}+\frac{5}{100}=0.25$ (d) $\frac{3}{4}\times \frac{25}{25}=\frac{75}{100}=\frac{70}{100}+\frac{5}{100}$ $=\frac{7}{10}+\frac{5}{100}=0.75$ (e) $\frac{1}{5}\times \frac{2}{2}=\frac{2}{10}=0.2$ (f) $\frac{4}{5}\times \frac{2}{2}=\frac{8}{10}=0.8$

4.0Key Features of NCERT Solutions Class 7 Maths Chapter 3 : A Peek Beyond the Point

• Concept-Oriented Explanations: Every fundamental concept, from the definition of tenths to complex decimal operations, is explained with clear and precise language, making it easy for students to grasp even challenging ideas.
• Step-by-Step Problem Solving: Solutions to all textbook exercises are provided with detailed, step-by-step working. This methodical approach helps students understand the logic and process behind each answer, fostering independent problem-solving skills.
• Real-Life Examples: The solutions connect decimal concepts to practical scenarios, like measuring lengths or converting units, reinforcing the real-world relevance and application of decimals.
• Adherence to NCERT Syllabus: Meticulously aligned with the latest NCERT curriculum (Ganita Prakash), our solutions ensure that students are studying the most current and relevant content for their examinations.
• Focus on Place Value: A strong emphasis is placed on understanding decimal place values, which is fundamental to performing all operations correctly and comparing decimals accurately.
• Downloadable PDF: Available in a convenient PDF format, enabling students to study offline, anytime, and anywhere, ensuring uninterrupted learning.