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

3.1 The Need for Smaller Units

NCERT In-Text Questions (Pages 46-48)

In the following figure, screws are placed above a scale. Measure them and write their length in the space provided.

Solution:

Which scale helped you measure the length of the screws accurately? Why?

Solution:

The third scale allowed us to measure the screws more accurately because each unit is divided into 10 smaller, equal parts, making the measurements more precise.

Can you explain why the unit was divided into smaller parts to measure the screws?

Solution:  Yes, because the screws are too long to be measured using whole units alone. That is why the unit is divided into smaller parts to measure their length more accurately.

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:

Students should do it by themselves.

Write the measurements of the objects shown in the picture:

Solution:

Length of the eraser is 2   4/10 cm

Length of the pencil is 4

5/10 cm

Length of the chalk is 1

4/10 cm

3.2 A Tenth Part

NCERT In-Text Questions (Pages 49-52)

For the objects shown below, write their lengths in two ways and read them aloud. An example is given for the USB cable. (Note that the unit length used in each diagram is not the same.)

The length of the USB cable is 4 and 8/10 units or 48/10  units.

Solution:

The length of the little finger is 1 and  3/10  units or 

13/10 units;

The height of the geometry box is 2 and 8/10 units or 

28/10 units; and the length of a leaf is 10 and 9/10 units or 

109/10 units

Arrange these lengths in increasing order: 

(a) $$\frac{9}{10}$$

(b) 1$$\frac{7}{10}$$

(c) $$\frac{130}{10}$$

(d) 13$$\frac{1}{10}$$

(e) 10$$\frac{5}{10}$$

(f) 7$$\frac{6}{10}$$

(g) 6$$\frac{7}{10}$$

(h) $$\frac{4}{10}$$

Solution:

(a) $$\frac{9}{10}$$ → nine-tenths

(b) $$1 \frac{7}{10}=\frac{17}{10}$$ → one and seven-tenths or seventeen tenths

(c) $$\frac{130}{10}$$ → one hundred thirty-tenths

(d) $$13 \frac{1}{10}=\frac{131}{10}$$ → Thirteen and one-tenths or one hundred thirty-one tenths

(e) $$10 \frac{5}{10}=\frac{105}{10}$$ → Ten and five-tenths or one hundred five tenths

(f) $$7 \frac{6}{10}=\frac{76}{10}$$ → Seven and six-tenths or seventh-six tenths

(g) $$6 \frac{7}{10}=\frac{67}{10}$$ → Six and seven-tenths or sixty-seven tenths

(h) $$\frac{4}{10}$$ → four-tenths

These fractional units are in increasing order as:

$$\frac{4}{10}<\frac{9}{10}<1 \frac{7}{10}<6 \frac{7}{10}<7 \frac{6}{10}<10 \frac{5}{10}<\frac{130}{10}<13 \frac{1}{10}$$

Arrange the following lengths in increasing order:

$$4 \frac{1}{10}, \frac{4}{10}, \frac{41}{10}, 41 \frac{1}{10}$$

Solution:

We can see that $$4 \frac{1}{10}=\frac{41}{10}$$ and $$41 \frac{1}{10}=\frac{411}{10}$$

Therefore, the given lengths can be arranged in increasing order as:

$$\frac{4}{10}<4 \frac{1}{10}=\frac{41}{10}<41 \frac{1}{10}$$

The lengths of the body parts of a honeybee are given. Find its total length.

Head: 2$$\frac{3}{10}$$ units

Thorax: 5$$\frac{4}{10}$$ units

Abdomen: 7$$\frac{5}{10}$$ units

Solution:

The total length of a honeybee = length of the head + length of the thorax + length of the abdomen

\[ = 2\frac{3}{10}\ \text{units} \;+\; 5\frac{4}{10}\ \text{units} \;+\; 7\frac{5}{10}\ \text{units} \]

\[ = (2 + 5 + 7) \;+\; \left( \frac{3}{10} + \frac{4}{10} + \frac{5}{10} \right) \ \text{units} = 14 + \frac{12}{10}\ \text{units} \]

\[ = 14 + \frac{10}{10} + \frac{2}{10}\ \text{units} = 14 + 1 + \frac{2}{10}\ \text{units} \]

\[ = 15 + \frac{2}{10}\ \text{units} = 15\frac{2}{10}\ \text{units} \]

A Celestial Pearl Danio’s length is 2

4/10  cm, and the length of a Philippine Goby is 9/10 cm. What is the difference in their lengths?

Solution:

$$\text{Length of Celestial Pearl Danio} = 2\frac{4}{10} \text{ cm} = \frac{24}{10} \text{ cm}$$

$$\text{Length of Philippine Goby} = \frac{9}{10} \text{ cm}$$

$$\text{Difference in length} = \frac{24}{10} - \frac{9}{10}$$

$$= \frac{15}{10} \text{ cm}$$

$$= 1\frac{5}{10} \text{ cm}$$

How big are these fish compared to your figure?

Solution:

Do it yourself.

Observe the given sequences of numbers. Identify the change after each term and extend the pattern: 

$$(a)\; 4,\; 4\frac{3}{10},\; 4\frac{6}{10},\; \underline{\hspace{1cm}},\; \underline{\hspace{1cm}},\; \underline{\hspace{1cm}},\; \underline{\hspace{1cm}}$$

Solution:

\[ \text{So, further terms are} \] \[ 4\frac{6}{10} + \frac{3}{10} = 4\frac{9}{10}, \qquad 4\frac{9}{10} + \frac{3}{10} = 4\frac{12}{10} \] \[ = 4 + 1 + \frac{2}{10} = 5\frac{2}{10} \] \[ 5\frac{2}{10} + \frac{3}{10} = 5\frac{5}{10}, \qquad 5\frac{5}{10} + \frac{3}{10} = 5\frac{8}{10} \] \[ \text{So, the pattern is} \] \[ 4,\; 4\frac{3}{10},\; 4\frac{6}{10},\; 4\frac{9}{10},\; 5\frac{2}{10},\; 5\frac{5}{10},\; 5\frac{8}{10} \]

(b) \(8 \frac{2}{10}, 8 \frac{7}{10}, 9 \frac{2}{10}\), _________, _________, _________, _________

Solution:
\[ \text{Since, the sequence is} \] \[ \begin{aligned} 8\frac{2}{10} + \frac{5}{10} &= 8\frac{7}{10}, \\ 8\frac{7}{10} + \frac{5}{10} &= 8\frac{12}{10} = 8 + 1 + \frac{2}{10} = 9\frac{2}{10}. \end{aligned} \] \[ \text{So, further terms are} \] \[ \begin{aligned} 9\frac{2}{10} + \frac{5}{10} &= 9\frac{7}{10}, \\ 9\frac{7}{10} + \frac{5}{10} &= 9\frac{12}{10} = 9 + 1 + \frac{2}{10} = 10\frac{2}{10}. \end{aligned} \] \[ \text{So, the sequence is} \] \[ 8\frac{2}{10},\; 8\frac{7}{10},\; 9\frac{2}{10},\; 9\frac{7}{10},\; 10\frac{2}{10},\; 10\frac{7}{10},\; 11\frac{2}{10} \]

(c) \(7 \frac{6}{10}, 8 \frac{7}{10}\), _________, _________, _________, _________

Solution:
\[ \text{Here, } 7\frac{6}{10} + 1\frac{1}{10} = 7 + 1 + \frac{6}{10} + \frac{1}{10} = 8 + \frac{7}{10} = 8\frac{7}{10},\ \ldots \] \[ \text{So, adding } 1 \text{ and } \frac{1}{10} \text{ to each preceding term, we get the sequence as follows:} \] \[ 7\frac{6}{10},\; 8\frac{7}{10},\; 9\frac{8}{10},\; 10\frac{9}{10},\; 12,\; 13\frac{1}{10} \]

(d) \(5 \frac{7}{10}, 5 \frac{3}{10}\), _________, _________, _________, _________

Solution:
\[ \text{We observe that,} \] \[ 5\frac{7}{10} - \frac{4}{10} = 5\frac{3}{10}, \qquad 5\frac{3}{10} - \frac{4}{10} = 4 + \frac{13}{10} - \frac{4}{10} = 4\frac{9}{10},\ \ldots \] \[ \text{So, subtracting } \frac{4}{10} \text{ from each previous term, we get the sequence as follows:} \] \[ 5\frac{7}{10},\; 5\frac{3}{10},\; 4\frac{9}{10},\; 4\frac{5}{10},\; 4\frac{1}{10},\; 3\frac{7}{10} \]

(e) \(13 \frac{5}{10}, 13,12 \frac{5}{10}\), _________, _________, _________, _________

Solution:
\[ \text{Since, } 13\frac{5}{10} - \frac{5}{10} = 13, \qquad 13 - \frac{5}{10} = 12 + 1 - \frac{5}{10} \] \[ = 12 + \frac{10}{10} - \frac{5}{10} = 12\frac{5}{10} \] \[ \text{So, by subtracting } \frac{5}{10} \text{ from each previous term, we get the sequence as follows:} \] \[ 13\frac{5}{10},\; 13,\; 12\frac{5}{10},\; 12,\; 11\frac{5}{10},\; 11,\; 10\frac{5}{10} \]

(f) \(11 \frac{5}{10}, 10 \frac{4}{10}, 9 \frac{3}{10}\), _________, _________, _________, _________

Solution:
\[ \text{Since, } 11\frac{5}{10} - 1\frac{1}{10} = 10\frac{4}{10}, \qquad 10\frac{4}{10} - 1\frac{1}{10} = 9\frac{3}{10},\ \ldots \] \[ \text{So, by subtracting } 1\frac{1}{10} \text{ from each previous term, we get the sequence as:} \] \[ 11\frac{5}{10},\; 10\frac{4}{10},\; 9\frac{3}{10},\; 8\frac{2}{10},\; 7\frac{1}{10},\; 6,\; 4\frac{9}{10} \]

3.3 A Hundredth Part

NCERT In-Text Questions (Pages 54-56)

Observe the figure below. Notice the markings and the corresponding lengths written in the boxes when measured from 0. Fill in the lengths in the empty boxes.

Solution:

For the lengths shown below, write the measurements and read out the measures in words.

Solution:

In each group, identify the longest and the shortest lengths. Mark each length on the scale.

Solution:

Figure it Out (Page 58)

Question 1.

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) $$12 \frac{6}{100} \frac{2}{100}-\frac{9}{10} \frac{9}{100}$$

Solutions:
\[ (a) \] \[ \begin{aligned} \frac{3}{10} + \frac{3}{100} &= \frac{3}{10} + \frac{3}{100} \\ &= 3 + \frac{3}{10} + \frac{4}{100} \\ &= 3 + \frac{3}{10} + \frac{4}{100} \\ &= 3\frac{3}{10}\frac{4}{100} = 3\frac{34}{100} \end{aligned} \]

\[ (b) \] \[ \begin{aligned} 9\frac{5}{10} + 2\frac{1}{100} &= 9\frac{5}{10} + 2\frac{1}{100} \\ &= (9 + 2) + \left( \frac{5}{10} + \frac{1}{10} \right) + \left( \frac{7}{100} + \frac{3}{100} \right) \\ &= 11 + \frac{6}{10} + \frac{10}{100} \\ &= 11 + \frac{6}{10} + \frac{1}{10} \\ &= 11\frac{7}{10} \end{aligned} \]

\[ (c) \] \[ \begin{aligned} 15\frac{6}{10} + \frac{4}{100} + 14\frac{3}{10} + \frac{6}{100} &= (15 + 14) + \left( \frac{6}{10} + \frac{3}{10} \right) + \left( \frac{4}{100} + \frac{6}{100} \right) \\ &= 29 + \frac{9}{10} + \frac{10}{100} \\ &= 29 + \frac{9}{10} + \frac{1}{10} \\ &= 29 + 1 = 30 \end{aligned} \]

\[ (d) \] \[ \begin{aligned} 7\frac{7}{100} - \frac{4}{100} &= 7 + \left( \frac{7}{100} - \frac{4}{100} \right) \\ &= 7 + \frac{3}{100} \\ &= 7\frac{3}{100} \end{aligned} \]

\[ (e) \] \[ \begin{aligned} 8\frac{6}{100} - 5\frac{3}{100} &= (8 - 5) + \left( \frac{6}{100} - \frac{3}{100} \right) \\ &= 3 + \frac{3}{100} \\ &= 3\frac{3}{100} \end{aligned} \]

\[ (f) \] \[ \begin{aligned} 12\frac{6}{10} - \frac{2}{100} &\longrightarrow 12\frac{5}{10} - \frac{12}{100} \\ &\longrightarrow 11\frac{15}{10} - \frac{12}{100} \\ &= 11\frac{6}{10} - \frac{3}{100} \\ &= 11\frac{63}{100} \end{aligned} \]

3.4 Decimal Place Value

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 $$\frac{1}{10}$$ times moving from left to right.

(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.

Make a few more questions of this kind and answer them.

Solution:
Students must complete this by themselves.

Make a place value table similar to the one above. Write each quantity in decimal form and terms of place value, and read the number:

(a) 2 ones, 3 tenths, and 5 hundredths

(b) 1 ten and 5 tenths

(c) 4 ones and 6 hundredths

(d) 1 hundred, 1 one, and 1 hundredth

(e) $$\frac{8}{100}$$ and $$\frac{9}{10}$$

(f) $$\frac{5}{100}$$

(g) $$\frac{1}{10}$$

(h) 2$$\frac{1}{100}$$, 4$$\frac{1}{10}$$ and 7$$\frac{7}{1000}$$

Solution:

Write these quantities in decimal form:

(a) 234 hundredths

(b) 105 tenths.

Solution:

3.5 Units of Measurement

Length Conversion

NCERT In-Text Questions (Pages 65-66)

Fill in the blanks below (mm ↔ cm)

Solution:

Fill in the blanks below (cm ↔ m)

Solution:

How many mm does 1 m have?

Solution:
1000 mm = 1 m
[As, 1 m = 100 cm = 100 × 10 mm]

Weight Conversion

NCERT In-Text Questions (Pages 67-68)

Fill in the blanks below (g ↔ kg)

Rupee-Paise Conversion

NCERT In-Text Questions (Page 69)

Fill in the blanks below (rupee ↔ paise)

3.6 Locating and Comparing Decimals

NCERT In-Text Questions (Page 70)

Name all the divisions between 1 and 1.1 on the number line.

Solution:

The small divisions between 1 and 1.1 represent the decimal values 1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08, and 1.09 on the number line.

Identify and write the decimal numbers against the letters.

Solution:

The letters shown on the number line correspond to the following decimal values:

A → 5.09, B → 5.13, C → 5.20, D → 5.31

There is Zero Dilemma!

NCERT In-Text Questions (Pages 71-73)

Can you tell which of these (0.2, 0.02, and 0.002) is the smallest and which is the largest?

Solution:
From the decimal place value chart, we see that

Among the given decimals, 0.002 (which is 2 thousandths) is the smallest, and 0.2 (which is 2 tenths) is the largest.

Which of these are the same: 4.5, 4.05, 0.405, 4.050, 4.50, 4.005, 04.50?

Solution:

From the decimal place value chart, we see that

The decimals 4.5, 4.50, and 04.50 are all equal because they represent the same value—four ones and five tenths.

Likewise, 4.05 and 4.050 are equal since they both represent four ones and five hundredths.

However, 0.405 and 4.005 are not equal because they represent different values.

Identify the decimal number in the last number line in Figure (b) denoted by ‘?’

Solution:

After marking and labelling the required section of the number line, we can see that the decimal number 3.059 corresponds to the point labelled ‘?’.

Make such number lines for the decimal number:

(a) 9.876

(b) 0.407

Solution:

In the number line shown below, what decimal numbers do the boxes labelled ‘a’, ‘b’, and ‘c’ denote?

The box with ‘b’ corresponds to the decimal number 7.5; are you able to see how? There are 5 units between 5 and 10, divided into 10 equal parts. Hence, every 2 divisions make a unit, and so every division is 1/2 unit. 

What numbers do ‘a’ and ‘c’ denote?

Solution:

Each division between 5 and 10 represents
5 and 10 is ½ ​=0.5 units.

Therefore:

• The second division after 5, marked as ‘a’, corresponds to 6.
  
  

• The ninth division, marked as ‘c’, corresponds to 9.5.
  
  

So, ‘a’ = 6 and ‘c’ = 9.5.

Using similar reasoning, find out the decimal numbers in the boxes below.

Solution:

Between 8 and 8.1, there are 10 equal divisions.
So, each division represents:

\[0.110 = 0.01 \times \frac{0.1}{10} = 0.0110 = 0.01\]

Therefore:

• The first division after 8, marked ‘d’, represents 8.01.
  
  

• The fifth division, marked ‘e’, represents 8.05.
  
  

Between 4.3 and 4.8, the difference is:

4.8−4.3=0.54.8 - 4.3 = 0.54.8−4.3=0.5

There are 10 divisions, so each division represents:

\[0.510 = 0.05 \times \frac{0.5}{10} = 0.0510 = 0.05\]

Therefore:

• The first division after 4.3, marked ‘f’, represents 4.35.
  
  

• The fourth division, marked ‘g’, represents:
  
  

4.3+(4×0.05)=4.54.3 + (4 × 0.05) = 4.54.3+(4×0.05)=4.5

• The first division after 4.8, marked ‘h’, represents:
  
  

4.8+0.05=4.854.8 + 0.05 = 4.854.8+0.05=4.85

Which decimal number is greater?

(a) 1.23 or 1.32

(b) 3.81 or 13.800

(c) 1.009 or 1.090

Solution:

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

Both numbers have the same whole number part (1), but the first number has 2 tenths, while the second has 3 tenths.
Since 2 tenths is less than 3 tenths, we get:

1.23 < 1.32

(b) Using the decimal place value chart, we find that:

The first number has 3 ones, while the second number has 1 ten and 3 ones (i.e., 13).
Since 3 is less than 13, we conclude:

3.81 < 13.800

(c) Using the decimal place value chart, we find that:

Both numbers have the same whole number part (1) and the same tenths part (0).
However, the first number has 0 hundredths, while the second has 9 hundredths.
Since 0 hundredths is less than 9 hundredths, we get:

1.009 < 1.090

Closest Decimals

NCERT In-Text Questions (Page 73)

Which of the decimal numbers 0.9, 1.1, 1.01, and 1.11 is closest to 1.09?
Solution:

Which among these is closest to 4: 3.56, 3.65, 3.099?

Solution: When arranged in ascending order, the numbers are:

3.099 < 3.56 < 3.65 < 4

Among these, 3.65 is the closest to 4.

Which among these is closest to 1: 0.8, 0.69, 1.08?

Solution:

In each case below, use the digits 4, 1, 8, 2, and 5 exactly once and try to make a decimal number as close as possible to 25.

Solution:

We can form a decimal number that is closest to 25 using the digits 4, 1, 8, 2, and 5, following the given conditions, as shown below:

3.7 Addition and Subtraction of Decimals

NCERT In-Text Questions (Pages 75)

Write the detailed place value computation for 84.691 – 77.345, and its compact form.

Solution:
We can find the difference of 84.691 – 77.345 as follows:

Figure it Out (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:

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:

Decimal Sequences

NCERT In-Text Questions (Pages 75-76)

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

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:

Make your sequences and challenge your classmates to extend the pattern.

Solution:

Do it yourself.

Estimating Sums and Differences

NCERT In-Text Questions (Pages 76)

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.”
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.

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 decimal numbers are 25.936 and 8.202.

• Sum of the whole-number parts:
  25 + 8 = 33
  
  

• Sum of the decimal numbers:
  25.936 + 8.202 = 34.138
  
  

We can see that:

33 < 34.138 < 33 + 2

Now, take another example with the numbers 1.532 and 4.536.

• Sum of the whole-number parts:
  1 + 4 = 5
  
  

• Sum of the decimal numbers:
  1.532 + 4.536 = 6.068
  
  

And here again:

5 < 6.068 < 5 + 2

What about for the sum of 25.93603259 and 8.202?

Solution:

The given numbers are 25.93603259 and 8.202.

• Sum of the whole-number parts:
  25 + 8 = 33
  
  

• Sum of the decimal numbers:
  25.93603259 + 8.202 = 34.13803259
  
  

It is clear that:

33 < 34.13803259 < 33 + 2

Similarly, come up with a way to narrow down the range of whole numbers within which the difference of two decimal numbers will lie.

Solution: When we subtract two decimal numbers, the result will always be less than the difference of their whole-number parts plus 1. Similarly, the difference will always be greater than the difference of their whole-number parts minus 1.

You can verify this by trying different pairs of decimal numbers.

3.8 More on the Decimal

System Deceptive Decimal Notation

NCERT In-Text Questions (Pages 78)

Where else can we see such ‘non-decimals’ with a decimal-like notation?

Solution:
Do it yourself.

Figure it Out (Pages 78-80)

Question 1.

Convert the following fractions into decimals:

Solution:

Question 2.

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

(а) 0.34

(b) 1.02

(c) 0.8

(d) 0.362

Solution:

Question 3.

What decimal number does each letter represent in the number line below?

Solution:

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 \(6.45\).

The first division after 6.5, denoted by ‘c’, represents the number \(6.525\).

The second division after 6.5, denoted by ‘b’, represents the number \(6.55\).

Question 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

Solution: (a) By using the decimal place value chart, we observe that:

Both numbers 11.01 and 11.10 have the same whole-number part (11).
However:

• 11.01 has 0 tenths
  
  

• 11.10 has 1 tenth
  
  

Since 1 tenth is greater than 0 tenths, we get:

11.10 > 11.01

Next, consider the numbers 1.011, 1.101, and 1.01.
All three have the same whole-number part (1), but:

• 1.011 → 0 tenths
  
  

• 1.01 → 0 tenths
  
  

• 1.101 → 1 tenth
  
  

So, 1.101 is the greatest among the three.

Now compare the remaining two:

• 1.011 has 1 hundredth
  
  

• 1.01 has 0 hundredths
  
  

So, 1.011 > 1.01

Final Descending Order

11.10>11.01>1.101>1.011>1.0111.10 > 11.01 > 1.101 > 1.011 > 1.0111.10>11.01>1.101>1.011>1.01

You can similarly arrange the other values by using a decimal place value chart to compare digits from the leftmost (units) to the rightmost decimal place.

(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 g, 4.595 g, 4.600 g, 4.656 g, 4.666 g

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

Thus, the given quantities in descending order are:

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

Question 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.

Solution:

Using the digits 1, 4, 0, 8, and 6, we can form:

(a) The decimal number closest to 30 → 40.168
(b) The smallest possible decimal number between 100 and 1000 → 104.68

Question 6.
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 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.

Solution:

The total weight of all the items Mahi purchased is:

0.25 kg + 0.3 kg + 0.5 kg + 0.2 kg + 0.05 kg = 1.3 kg.

Question 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.

Solution: 

Question 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?

Solution:

Question 10.

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

Solution:

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

We see that the sequence alternates between:

• adding 0.9, and
  
  

• subtracting 0.01
  
  

Following this pattern:

8.17 (+0.9) = 9.07
9.07 (–0.01) = 9.06

Thus, the next two numbers in the sequence are:

9.07 and 9.06.

Question 11.

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 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?

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

Question 13.

Which is greater?

(a) 10/1000

 or 

1/10 ?

(b) One-hundredth or 90 thousandths?

(c) One-thousandth or 90 hundredths?

Solution: 

Question 14.

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

Solution:

(a) 87 ones, 5 tenths, and 60 hundredths = 88.10

Question 15.

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

Question 16.

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) \(\frac{4}{5}\)

Solution:

$$(a)\; \frac{1}{2} \times \frac{5}{5} = \frac{5}{10} = 0.5$$

\[(b)\; \frac{3}{2} \times \frac{5}{5} = \frac{15}{10} = 1\frac{5}{10} = 1.5\]

\[(c)\; \frac{1}{4} \times \frac{25}{25} = \frac{25}{100} = \frac{20}{100} + \frac{5}{100} = 0.25\]

\[(d)\; \frac{3}{4} \times \frac{25}{25} = \frac{75}{100} = \frac{70}{100} + \frac{5}{100} = 0.75\]

\[(e)\; \frac{1}{5} \times \frac{2}{2} = \frac{2}{10} = 0.2\]

\[(f)\; \frac{4}{5} \times \frac{2}{2} = \frac{8}{10} = 0.8\]

Understanding Decimals in Class 7 Maths Chapter 3

Mastering NCERT Solutions for Class 7 Maths Chapter 3 A Peek Beyond the Point is crucial for developing strong decimal concepts. With well-structured solutions, students learn to measure, compare, and perform calculations with decimals confidently.

The chapter explains units, tenths, and hundredths—an essential part of the syllabus for 2025-26. Practicing all exercise-based questions will boost both problem-solving skills and concept clarity for Class 7 exams and beyond.

To score higher, students should review decimal place value and attempt different question types. Regular practice and revision are key strategies for building confidence and ensuring success in mathematics.