NCERT Solutions for Class 7 Maths Chapter 3 A Peek beyond the Point

NCERT Solutions for Class 7 Maths Chapter 3: A Peek beyond the Point(Exercise)

Figure it Out

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

Solution:

(a) $\frac{3}{10}+3 \frac{4}{100}$
$=\frac{30}{100}+3+\frac{4}{100}$
$=3+\left(\frac{30}{100}+\frac{4}{100}\right)$
$=3+\frac{34}{100}=3 \frac{34}{100}$

(b) $9 \frac{5}{10} \frac{7}{100}+2 \frac{1}{10} \frac{3}{100}$
$=(9+2)+\left(\frac{5}{10}+\frac{1}{10}\right)+\left(\frac{7}{100}+\frac{3}{100}\right)$
$=11+\left(\frac{5+1}{10}\right)+\left(\frac{7+3}{100}\right)$
$=11+\frac{6}{10}+\frac{10}{100}$
$=11+\frac{6}{10}+\frac{1}{10}=11+\frac{7}{10}=11 \frac{7}{10}$

(c) $15 \frac{6}{10} \frac{4}{100}+14 \frac{3}{10} \frac{6}{100}$
$=(15+14)+\left(\frac{6}{10}+\frac{3}{10}\right)+\left(\frac{4}{100}+\frac{6}{100}\right)$
$=29+\left(\frac{6+3}{10}\right)+\left(\frac{4+6}{100}\right)$
$=29+\frac{9}{10}+\frac{10}{100}$
$=29+\frac{9}{10}+\frac{1}{10}$
$=29+\frac{10}{10}=29+1=30$

(d) $7 \frac{7}{100}-4 \frac{4}{100}$
$=(7-4)+\left(\frac{7}{100}-\frac{4}{100}\right)$
$=3+\frac{3}{100}=3 \frac{3}{100}$

(e) $8 \frac{6}{100}-5 \frac{3}{100}$
$=(8-5)+\left(\frac{6}{100}-\frac{3}{100}\right)$
$=3+\frac{3}{100}=3 \frac{3}{100}$

(f) $12 \frac{6}{10} \frac{2}{100}-\frac{9}{10} \frac{9}{100}$
$=12 \frac{62}{100}-\frac{99}{100}$
$=11+1+\frac{62}{100}-\left(\frac{99}{100}\right)$
$=11+\frac{100}{100}+\frac{62}{100}-\left(\frac{99}{100}\right)$
$=11+\frac{162}{100}-\left(\frac{99}{100}\right)$
$=11+\left(\frac{162}{100}-\frac{99}{100}\right)$
$=11+\frac{63}{100}=11 \frac{63}{100}$

Figure it Out

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 + 0.487 = 6.723

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

Figure it Out

Question 1: Convert the following fractions into decimals:

(a) $\frac{5}{100}$ (b) $\frac{16}{1000}$ (c) $\frac{12}{10}$ (d) $\frac{254}{1000}$

Solution:

(a) $\frac{5}{100}$ = 0.05

(b) $\frac{16}{1000}$ = 0.016

(c) $\frac{12}{10}$ = 1.2

(d) $\frac{254}{1000}$ = 0.254

Question 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

Solution:

(a) $0.34=\left(3 \times \frac{1}{10}\right)+\left(4 \times \frac{1}{100}\right)=\frac{3}{10}+\frac{4}{100}$

(b) $1.02=(1 \times 1)+\left(2 \times \frac{1}{100}\right)=1+\frac{2}{100}$

(c) $0.8=\left(8 \times \frac{1}{10}\right)=\frac{8}{10}$

(d) $0.362=\left(3 \times \frac{1}{10}\right)+\left(6 \times \frac{1}{100}\right)+\left(2 \times \frac{1}{1000}\right)=\frac{3}{10}+\frac{6}{100}+\frac{2}{1000}$

Question 3: What decimal number does each letter represent in the number line below?

Solution:

Total divisions between 6.4 and 6.6 = 8

Each division = $\frac{6.6-6.4}{8}$ = $\frac{0.2}{8}$ = $\frac{2}{80}$ = 0.025

Therefore,

$a=6.4+(2 \times 0.025)=6.4+0.050=6.425$

$c=6.4+(5 \times 0.025)=6.4+0.125=6.525$

$b=6.4+(6 \times 0.025)=6.4+0.150=6.550$

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) $11.10>11.01>1.101>11.011>1.01$

(b) $2.768>2.698>2.675>2.567>2.499$

(c) $4.678 \mathrm{~g}>4.666 \mathrm{~g}>4.656 \mathrm{~g}>4.600 \mathrm{~g}>4.595 \mathrm{~g}$

(d) $33.331 \mathrm{~m}>33.313 \mathrm{~m}>33.31 \mathrm{~m}>33.133 \mathrm{~m}>33.13 \mathrm{~m}$

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:

(a) Possible numbers: 14.860,40.168

30 – 14.860 = 15.14

40.168 – 30 = 10.168

Therefore, 40.168 is the closest number to 30.

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

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

Solution:

Maybe, but not always. A number with more decimal digits can be smaller.

For Example:

0.12 has 2 decimal digits.

0.111 has 3 decimal digits.

But, 0.12 > 0.111

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:

Total weight = (0.25 + 0.3 + 0.5 + 0.2 + 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:

Milk supplied in first 3 days = (3.79 + 4.2 + 4.25) L = 12.24 L

Milk supplied in 6 days = 25 L

Therefore, Milk supplied in the last three days = (25 – 12.24) L = 12.76 L

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:

Since 35.75 > 34.50,

Therefore, he has lost weight.

Change in weight = (35.75 – 34.50) kg = 1.25 kg

So, Tinku has lost 1.25 kg.

Question 10: Extend the pattern: 5.5, 6.4, 6.39, 7.29, 7.28, 6.18, 6.17, ____, _____

Solution:

Pattern alternates: +0.9, –0.01, +0.9, –0.01, –1.1, –0.01,

Therefore, the next two terms are: 5.07 (–1.1), 5.06 (–0.01).

Question 11: How many millimetres make 1 kilometre?

Solution:

1 km = 1000 m = (1000 × 1000) mm = 10,00,000 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:

45 paise = Rs 0.45

1 lakh = 1,00,000

The total insurance fee paid = (1,00,000 × 0.45) = Rs. 45,000

Question 13: Which is greater?

(a) $\frac{10}{1000}$ or $\frac{1}{10}$?

(b) One-hundredth or 90 thousandths?

(c) One-thousandth or 90 hundredths?

Solution:

(a) $\frac{10}{1000}=\frac{1}{100}=0.01$

$\frac{1}{10}=0.1$

Since $0.01<0.1$

Therefore, $\frac{10}{1000}<\frac{1}{10}$

(b) One-hundredth $=\frac{1}{100}=0.01$

90 thousandths $=\frac{90}{1000}=\frac{9}{100}=0.09$

Since $0.01<0.09$

$\therefore$ One-hundredth $<$ 90 thousandths.

(c) One thousandth $=\frac{1}{1000}=0.001$

90 hundredths $=\frac{90}{100}=\frac{9}{10}=0.9$

Since $0.001<0.9$

$\therefore$ One-thousandth < 90 hundredths.

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

$=(87 \times 1)+\left(5 \times \frac{1}{10}\right)+\left(60 \times \frac{1}{100}\right)$

$=87+\frac{5}{10}+\frac{6}{10}$

$=87+0.5+0.6=88.10$

(b) 12 tens and 12 tenths
$=(12 \times 10)+\left(12 \times \frac{1}{10}\right)$
$=120+\frac{12}{10}$
$=120+1.2=121.2$

(c) 10 tens, 10 ones, 10 tenths, and 10 hundredths
$\begin{aligned}
& =(10 \times 10)+(10 \times 1)+\left(10 \times \frac{1}{10}\right)+\left(10 \times \frac{1}{100}\right) \\
& =100+10+1+\frac{1}{10} \\
& =100+10+1+0.1 \\
& =111.1
\end{aligned}$

(d) 25 tens, 25 ones, 25 tenths, and 25 hundredths
$\begin{aligned}
& =(25 \times 10)+(25 \times 1)+\left(25 \times \frac{1}{10}\right)+\left(25 \times \frac{1}{100}\right) \\
& =250+25+\frac{25}{10}+\frac{25}{100} \\
& =250+25+2.5+0.25 \\
& =277.75
\end{aligned}$

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

Solution:

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

(b) $\frac{3}{2} =\frac{3}{2} \times \frac{10}{10}=\frac{15}{10}=1.5$

(c) $\frac{1}{4} =\frac{1}{4} \times \frac{100}{100}=\frac{25}{100}=0.25$

(d) $\frac{3}{4} =\frac{3}{4} \times \frac{100}{100}=\frac{75}{100}=0.75$

(e) $\frac{1}{5} =\frac{1}{5} \times \frac{10}{10}=\frac{2}{10}=0.2$

(f) $\frac{4}{5} =\frac{4}{5} \times \frac{10}{10}=\frac{8}{10}=0.8$

A Peek beyond the Point Class 7 NCERT Solutions: Topics

The important topics that are covered in Class 7 Maths NCERT Chapter 3, A Peek Beyond the Point, are:

• The Need for Smaller Units
• A Tenth Part
• A Hundredth Part
• Decimal Place Value
• Notation, Writing and Reading of Decimal Numbers
• Units of Measurement
• Locating and Comparing Decimals
• Addition and Subtraction of Decimals
• More on the Decimal System

NCERT Solutions for Class 7 Maths Chapter Wise

Given below are the chapter-wise NCERT solutions for class 7 mathematics, provided in one place:

NCERT Solutions for Class 7 Subject-Wise

The NCERT Solutions for Class 7, Subject-wise, can be downloaded using the links below.

• NCERT Solutions for Class 7 Maths
• NCERT Solutions for Class 7 Science

Students can also check the NCERT Books and the NCERT Syllabus here:

• NCERT Syllabus Class 7 Maths
• NCERT Books Class 7
• NCERT Syllabus Class 7