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

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

Below are the detailed NCERT Class 7 Maths Chapter 3 A Peek beyond the Point question answers provided in the textbook.

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

Topics you will learn in NCERT Class 7 Maths Chapter 3 A Peek beyond the Point include:

• 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

Class 7 Maths Chapter 3 A Peek beyond the Point Solutions: Extra Question

Question:

Kushal went to a shop to buy fruits. The shopkeeper used a digital weighing machine that shows weights in kilograms and grams.

1. The weight of the apples he bought was 2.35 kg.

2. The weight of bananas was 1.7 kg.

3. The weight of the mangoes was 0.85 kg.

Now answer the following:

a) Write the weight of each fruit in words using decimal place value.
b) Convert the weight of bananas into grams. (Hint: 1 kg = 1000 g)
c) Arrange the weights of the fruits in ascending order.

Answer:

a)

2.35 kg = 2 kilograms, 3 tenths, and 5 hundredths

1.7 kg = 1 kilograms and 7 tenths

0.85 kg = 0 kg, 8 tenths, and 5 hundredths

b)

$1.7 \mathrm{~kg}=1.7 \times 1000=1700 \mathrm{~g}$

Hence, the correct answer is 1700 gm.

c)
Arranging the weights of fruits in ascending order, we get,

0.85 kg < 1.7 kg < 2.35 kg

NCERT Class 7 Maths Chapter 3 A Peek beyond the Point Notes

Careers360 has prepared these Class 7 A Peek beyond the Point Notes to make your revision smoother and faster. Additionally, these notes will help students to understand the A Peek beyond the Point NCERT solutions and solve them on their own from next time.

Addition and Subtraction of Fractions

When Denominators are the Same (Like Fractions)

• Just add or subtract the numerators.

• Keep the denominator the same.

Example:
$\frac{5}{12}+\frac{3}{12}=\frac{5+3}{12}=\frac{8}{12}=\frac{2}{3}$
$\frac{5}{12}-\frac{3}{12}=\frac{5-3}{12}=\frac{2}{12}=\frac{1}{6}$

When Denominators are Different (Unlike Fractions)

• Find the LCM of denominators.

• Convert both fractions to equivalent fractions with the same denominator.

• Add or subtract the numerators.

Example:
$\frac34+\frac29$
LCM of 4 and $9=36$
Convert: $\frac{3}{4}=\frac{27}{36},\frac{2}{9}=\frac{8}{36}$
Add: $\frac{27}{36}+\frac{8}{36}=\frac{35}{36}$

Rule for Mixed Fractions

• Convert them into improper fractions first.

• Then follow the rules of addition/subtraction.

• If needed, convert back into a mixed fraction.

Example:

$2 \frac{1}{3}+1 \frac{2}{5}$

Convert: $\frac{7}{3}+\frac{7}{5}$
LCM of 3 and $5=15$

$\frac{35}{15}+\frac{21}{15}=\frac{56}{15}=3 \frac{11}{15}$

Addition of Decimals

• Write the numbers in columns (place value form).

  • Line up the decimal points.

  • Add zeros, if necessary, to make the number of decimal places equal.

• Add as normal whole numbers from right to left.

• Put the decimal point in the answer directly below the other decimal points.

Example:
7.58 + 2.60 = 10.18

Subtraction of Decimals

• Write the numbers in columns, aligning decimal points.

• Add zeros, if required, to make the decimal places equal.

• Subtract as usual, starting from the rightmost digit.

• Place the decimal point in the answer below the other points.

Example:

7.58 - 2.60 = 4.98

NCERT Solutions for Class 7 Maths Chapter Wise

We at Careers360 compiled all the NCERT class 7 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Books and NCERT Syllabus

Students can also check the Class 7 NCERT Books and the NCERT Syllabus here:

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