NCERT Solutions for Class 7 Maths Chapter 4 - Another Peek Beyond the Point

4.1 A Quick Recap of Decimals, 4.2 Decimal Multiplication

Figure It Out (Pages 73-74)

Question 1.
Recall that a tenth is 0.1, a hundredth is 0.01, and so on. Find the following products in tenths, hundredths, and so on:
(a) 6 × 4 tenths = 24 tenths
(b) 7 × 0.3
(c) 9 × 5 hundredths

Answer:
(a) 6 × 4 tenths = 24 tenths

(b) 0.3 = 3 tenths

Therefore,
7 × 0.3 = 7 × 3 tenths
= 21 tenths

(c) 9 × 5 hundredths
= 45 hundredths

Question 2.
Find the products:
(a) 27.34 × 6
(b) 4.23 × 3.7
(c) 0.432 × 0.23

Answer:

$\begin{gathered}(a)\ \text{Here, }27.34\times6=\dfrac{2734}{100}\times6=\dfrac{16404}{100}\\[6pt]=\underset{\text{(2 decimals)}}{164.04}\\[12pt](b)\ \text{Here }\underset{\text{(2 decimals)}}{4.23}\times\underset{\text{(1 decimal)}}{3.7}=\dfrac{423\times37}{100\times10}\\[6pt]=\dfrac{15651}{1000}=\underset{\text{(3 decimals)}}{15.651}\\[12pt](c)\ \text{We have }\underset{\text{(3 decimals)}}{0.432}\times\underset{\text{(2 decimals)}}{0.23}=\dfrac{9936}{1000\times100}\\[6pt]=\underset{\text{(5 decimals)}}{0.09936}\end{gathered}$

Question 3.
Thejus needs 1.65 m of cloth for a shirt. How many metres of cloth are needed for 3 shirts?

Answer:
Cloth needed for 1 shirt = 1.65 m

Cloth needed for 3 shirts
= 1.65 × 3

= (165/100)× 3

= 495 / 100

= 4.95 m

Therefore, 4.95 m of cloth is needed for 3 shirts. 

Question 4.
Meenu bought 4 notebooks and 3 erasers. The cost of each book was ₹ 15.50, and each eraser was ₹ 2.75. How much did she spend in all?

Answer:
Here, the cost of 1 notebook = ₹ 15.50

Cost of 4 notebooks
= 4 × ₹15.50
= (4/1550)× 100
= 6200/100
= ₹62

Cost of 3 erasers
= 3 × ₹2.75
= (3 × 275)/100
= 825/100
= ₹8.25

Total amount spent
= ₹62 + ₹8.25
= ₹70.25

Therefore, Meenu spent ₹70.25 in all. 

Question 5.
The thickness of a rupee coin is 1.45 mm. What is the total height of the cylinder formed by stacking 36 rupee coins one on top of the other? Write the answer in centimetres.

Answer:
Thickness of 1 coin = 1.45 mm

Thickness of 36 coins
= 36 × 1.45
= (36×145)/100
= 5220/100
= 52.2 mm

Since 10 mm = 1 cm,

52.2 mm = 52.2 ÷ 10
= 5.22 cm

Therefore, the cylinder's total height is 5.22 cm. 

Question 6.
The price of 1 kg of oranges is ₹ 56.50. What is the price of 2.250 kg of oranges? Can we write 56.50 as 56.5 and 2.250 as 2.25 and multiply? Will we get the same product? Why?

Answer:
Price of 1 kg of oranges = ₹56.50

Price of 2.250 kg
= 56.50 × 2.250
= (5650×2250)/ 100×1000
= 12712500/100000
= ₹127.125

Yes, we can write 56.50 as 56.5 and 2.250 as 2.25.

56.5 × 2.25 = 127.125

We get the same product because trailing zeros in a decimal do not change its value.

Therefore, 56.50 = 56.5 and 2.250 = 2.25, so the product remains the same. 

Question 6.
Dwarakanath buys notebooks at a wholesale price of ₹23.6 per piece and sells each notebook at ₹30/-. How much profit does he make if he sells 50 books in a week?

Answer:
Profit on one notebook
= Selling price − Cost price
= ₹30 − ₹23.6
= ₹6.4

Profit on 50 notebooks
= ₹6.4 × 50
= ₹320

Therefore, Dwarakanath makes a profit of ₹320. 

Question 7.
Given that 18 × 12 = 216, find the products:
(a) 18 × 1.2
(b) 18 × 0.12
(c) 1.8 × 1.2
(d) 0.18 × 0.12
(e) 0.018 × 0.012
(f) 1.8 × 12
In which of the cases above is the product less than 1?

Answer:
Given: 18 × 12 = 216 ...(i)

(a) 18 × 1.2
= 18 × 12/10
= 216/10 [Using (i)]
= 21.6 (1 decimal place)

(b) 18 × 0.12
= 18 × 12/100
= 216/100 [Using (i)]
= 2.16 (2 decimal places)

(c) 1.8 × 1.2
= (18/10) × (12/10)
= 216/100 [Using (i)]
= 2.16 (2 decimal places)

(d) 0.18 × 0.12
= (18/100) × (12/100)
= 216/10000 [Using (i)]
= 0.0216 (4 decimal places)

(e) 0.018 × 0.012
= (18/1000) × (12/1000)
= 216/1000000 [Using (i)]
= 0.000216 (6 decimal places)

(f) 1.8 × 12
= (18/10) × 12
= 216/10 [Using (i)]
= 21.6 (1 decimal place)

When you multiply two positive numbers less than 1, their product is also less than 1.

Therefore, the product is less than 1 in (d) and (e).

Question 8.
In which of the following multiplications is the product less than 1? Can you find the answer without actually doing the multiplications?
(a) 7 × 0.6
(b) 0.7 × 0.6
(c) 0.7 × 6
(d) 0.07 × 0.06

Answer:
We can identify the products less than 1 by looking at the size of the numbers.

(a) 7 × 0.6 → Greater than 1

(b) 0.7 × 0.6 → Less than 1

(c) 0.7 × 6 → Greater than 1

(d) 0.07 × 0.06 → Less than 1

Therefore, the product is less than 1 in (b) and (d). 

Question 9.
Multiply the following numbers by 10, 100, and 1000 to complete the table.

Answer:

4.3 Decimal Division

Figure It Out (Page 83)

Question 1.
Find the quotient by converting the denominator to 1, 10, 100, or 1000, and verify the solution using long division (division by place value).
(a) 18/5
(b) 415/4
(c) 1217/2
(d) 4827/8

Answer:
The fraction bars are missing in the extracted question. Based on the accompanying solution, these represent 18/5, 415/4, 1217/2 and 4827/8. 

(a) 18 ÷ 5

To convert the denominator 5 into 10, multiply both the numerator and denominator by 2.

18/5 = (18×2)/ 5×2= 36/10
= 3.6

Verification: 18 ÷ 5

Dividing 1 ten and 8 ones into 5 equal parts.1 < 5

It means we need to regroup 1 ten as 10 ones,

i.e., 10 + 8 = 18 ones

18 ones ÷ 5

3 ones remain.

To divide 3 ones into 5 equal parts.

Regroup the 3 ones as 30 Tenths. (Place a decimal while regrouping ones into tenths).

30 Tenths ÷ 5 = 6

Then, 18 ÷ 5 = 3.6

$\begin{array}{r@{\;}l@{\;}l}&\overset{\text{}}{\phantom{03.6}}&\\[-4pt]5\ \big)&18&\big(03.6\\[-2pt]&\underline{-15}&\\&30&\\&\underline{-30}&\\&0&\end{array}$

Therefore, 18 ÷ 5 = 3.6.

(b) 415 ÷ 4

415/4 = (415×25)/4×25= 10375/100
= 103.75

$\begin{array}{r@{}l}4&\overline{)\,415}\,(\,103.75\\&\phantom{)\,}\underline{-4\phantom{00}}\\&\phantom{)\,4}15\\&\phantom{)\,}\underline{-12}\\&\phantom{)\,41}30\\&\phantom{)\,4}\underline{-28}\\&\phantom{)\,41}20\\&\phantom{)\,41}\underline{20}\\&\phantom{)\,41}\ \times\end{array}$

Therefore, 415 ÷ 4 = 103.75.

(c) 1217 ÷ 2

1217/2 = 1217×5/2×5= 6085/10
= 608.5

$\begin{array}{r@{}l}&\quad\ \ \\2&\overline{)\,1217}\,(\,608.5\\&\phantom{)\,}\underline{-12\phantom{00}}\\&\phantom{)\,}\ 017\\&\phantom{)\,0}\underline{-16}\\&\phantom{)\,01}10\\&\phantom{)\,0}\underline{-10}\\&\phantom{)\,011}0\end{array}$

Therefore, 1217 ÷ 2 = 608.5.

(d) 4827 ÷ 8

4827/8 = 4827×125/8×125= 603375/1000
= 603.375

$\begin{array}{r@{}l}8&\overline{)\,4827}\,(\,0603.375\\&\phantom{)\,}\underline{-48\phantom{00}}\\&\phantom{)\,}\ 027\\&\phantom{)\,0}\underline{-24}\\&\phantom{)\,02}30\\&\phantom{)\,02}\underline{-24}\\&\phantom{)\,02}60\\&\phantom{)\,02}\underline{-56}\\&\phantom{)\,02}40\\&\phantom{)\,02}\underline{-40}\\&\phantom{)\,02}\ \times\end{array}$

Therefore, 4827 ÷ 8 = 603.375.

Question 2.
Choose the correct answer:
(a) 1526/4 = _____
(i) 38.15
(ii) 380.15
(iii) 381.5
(iv) 381.05
(b) 3567/8 = _____
(i) 4458.75
(ii) 44.5875
(iii) 445.875
(iv) 4458.75

Answer:
The fraction bars are missing again in the extracted question. From the given options and reference answer:

(a) 1526/4

$\begin{array}{r@{}l}4&\overline{)\,1526}\,(\,381.5\\&\phantom{)\,}\underline{-12\phantom{00}}\\&\phantom{)\,}\ 32\\&\phantom{)\,}\underline{-32}\\&\phantom{)\,0}\ 06\\&\phantom{)\,0}\underline{-4}\\&\phantom{)\,0}\ 20\\&\phantom{)\,0}\underline{-20}\\&\phantom{)\,00}\ 0\end{array}$

Hence, option (iii) is correct.

(b) Given 3567/8

By using the Long Division Method:

$\begin{array}{r@{}l}8&\overline{)\,3567}\,(\,445.875\\&\phantom{)\,}\underline{-32\phantom{00}}\\&\phantom{)\,}\ 36\\&\phantom{)\,}\underline{-32}\\&\phantom{)\,0}\ 47\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 70\\&\phantom{)\,00}\underline{-64}\\&\phantom{)\,00}\ 60\\&\phantom{)\,00}\underline{-56}\\&\phantom{)\,00}\ 40\\&\phantom{)\,00}\underline{-40}\\&\phantom{)\,000}\ 0\end{array}$

Hence, option (iii) is correct.

Question 3.
What is the quotient?
(a) 132 ÷ 4 = _____
(b) 13.2 ÷ 4 = _____
(c) 1.32 ÷ 4 = _____
(d) 0.132 ÷ 4 = _____

Answer:
(a) 132 ÷ 4 = 33

$\begin{array}{r@{}l}&\quad\ \ \\4&\overline{)\,132}\,(\,33\\&\phantom{)\,}\underline{-12\phantom{0}}\\&\phantom{)\,}\ 12\\&\phantom{)\,}\underline{-12}\\&\phantom{)\,0}\ 0\end{array}$

(b) 13.2 ÷ 4 = 3.3

$\begin{array}{r@{}l}&\quad\ \ \\40&\overline{)\,132}\,(\,3.3\\&\phantom{)\,}\underline{-120\phantom{0}}\\&\phantom{)\,}\ 120\\&\phantom{)\,}\underline{-120}\\&\phantom{)\,00}\ 0\end{array}$

(c) 1.32 ÷ 4 = 0.33

$\begin{array}{r@{}l}&\quad\ \ \\400&\overline{)\,1320}\,(\,0.33\\&\phantom{)\,}\underline{-1200\phantom{0}}\\&\phantom{)\,}\ 1200\\&\phantom{)\,}\underline{-1200}\\&\phantom{)\,000}\ 0\end{array}$

(d) 0.132 ÷ 4 = 0.033

$\begin{array}{r@{}l}&\quad\ \ \\4000&\overline{)\,132000}\,(\,0.033\\&\phantom{)\,}\underline{-12000\phantom{0}}\\&\phantom{)\,}\ 12000\\&\phantom{)\,}\underline{-12000}\\&\phantom{)\,0000}\ 0\end{array}$

Question 4.
What is the quotient?
(a) 126 ÷ 8 = _____
(b) 12.6 ÷ 8 = _____
(c) 1.26 ÷ 8 = _____
(d) 0.126 ÷ 8 = _____
(e) 0.0126 ÷ 8 = _____

Answer:
(a) 126 ÷ 8 = 15.75

$\begin{array}{r@{}l}&\quad\ \ \\8&\overline{)\,126}\,(\,15.75\\&\phantom{)\,}\underline{-8\phantom{0}}\\&\phantom{)\,}\ 46\\&\phantom{)\,}\underline{-40}\\&\phantom{)\,0}\ 60\\&\phantom{)\,0}\underline{-56}\\&\phantom{)\,0}\ 40\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 0\end{array}$

(b) 12.6 ÷ 8 = 1.575

$\begin{array}{r@{}l}&\quad\ \ \\8&\overline{)\,12.6}\,(\,1.575\\&\phantom{)\,}\underline{-8\phantom{0}}\\&\phantom{)\,}\ 46\\&\phantom{)\,}\underline{-40}\\&\phantom{)\,0}\ 60\\&\phantom{)\,0}\underline{-56}\\&\phantom{)\,0}\ 40\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 0\end{array}$

(c) 1.26 ÷ 8 = 0.1575

$\begin{array}{r@{}l}&\quad\ \ \\8&\overline{)\,1.26}\,(\,0.1575\\&\phantom{)\,}\underline{-0\phantom{00}}\\&\phantom{)\,}\ 12\\&\phantom{)\,}\underline{-8}\\&\phantom{)\,0}\ 46\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 60\\&\phantom{)\,00}\underline{-56}\\&\phantom{)\,00}\ 40\\&\phantom{)\,00}\underline{40}\\&\phantom{)\,000}\ 0\end{array}$

(d) 0.126 ÷ 8 = 0.01575

$\begin{array}{r@{}l}&\quad\ \ \\8&\overline{)\,0.126}\,(\,0.01575\\&\phantom{)\,}\underline{-0\phantom{00}}\\&\phantom{)\,}\ 12\\&\phantom{)\,}\underline{-8}\\&\phantom{)\,0}\ 46\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 60\\&\phantom{)\,00}\underline{-56}\\&\phantom{)\,00}\ 40\\&\phantom{)\,00}\underline{-40}\\&\phantom{)\,000}\ 0\end{array}$

(e) 0.0126 ÷ 8 = 0.001575 

$\begin{array}{r@{}l}8&\overline{)\,0.0126}\,(\,0.001575\\&\phantom{)\,}\underline{\ 0\phantom{000}}\\&\phantom{)\,}\ 00\\&\phantom{)\,}\underline{\ 00}\\&\phantom{)\,00}\ 12\\&\phantom{)\,00}\underline{-8}\\&\phantom{)\,000}\ 46\\&\phantom{)\,000}\underline{-40}\\&\phantom{)\,0000}\ 60\\&\phantom{)\,0000}\underline{-56}\\&\phantom{)\,0000}\ 40\\&\phantom{)\,0000}\underline{-40}\\&\phantom{)\,00000}\ 0\end{array}$

Figure It Out (Pages 86-87)

Question 1.
Express the following fractions in decimal form:
(a) 2/5
(b) 13/4
(c) 4/50
(d) 5/8

Answer:
The fraction bars are missing in the extracted text. The accompanying solutions show the fractions as 2/5, 13/4, 4/50 and 5/8. 

(a) 2/5

Multiply both the Nr and Dr by 2.

2/5×2/2=4/10

Now, put the decimal: 4/10 = 0.4

Hence, 2/5 in decimal form is 0.4.

(b) 13/4

Multiply both the Nr and Dr by 25.

13/4×25/25=325/100

Now, put a decimal

325/100 = 3.25

Hence, 13/4 in decimal form is 3.25.

(c) 4/50

Multiply both the Nr and Dr by 2.

4/50×2/2=8/100

Now, put the decimal

8/100 = 0.08

Hence, 4/50 in decimal form is 0.08.

(d) 5/8

Multiply both the Nr and Dr by 125.

5/8×125/125=625/1000

Now, put the decimal

6251000 = 0.625

Hence, 58 in decimal form is 0.625.

Question 2.
Find the quotients:
(a) 24.86 ÷ 1.2
(b) 5.728 ÷ 1.52

Answer:
(a) 24.86 ÷ 1.2
=24.86/1.2
=2486×10/12×100
=2486/120
$\begin{array}{r@{}l}120&\overline{)\,2486}\,(\,20.7166\\&\phantom{)\,}\underline{-240\phantom{00}}\\&\phantom{)\,}\ 860\\&\phantom{)\,}\underline{-840}\\&\phantom{)\,0}\ 200\\&\phantom{)\,0}\underline{-120}\\&\phantom{)\,0}\ 800\\&\phantom{)\,0}\underline{-720}\\&\phantom{)\,00}\ 80\end{array}$

∴ Quotient = 20.7166…

(b) 5.728 ÷ 1.52

Question 3.
Evaluate the following using the information 156 × 12 = 1872.
(a) 15.6 × 1.2 = _____
(b) 187.2 ÷ 1.2 = _____
(c) 18.72 ÷ 15.6 = _____
(d) 0.156 × 0.12 = _____

Answer:
Given:

156 × 12 = 1872

⇒ 156 = 1872/12 …….(ii)

⇒ 12 = 1872/156 …….(iii)

(a) Now converting division into a fraction

15.6 × 1.2 = 156×12/10×10=1872/100 = 18.72 [using (i)]

(b) Converting division into a fraction: 187.2 ÷ 1.2

187.2 ÷ 1.2 = 187.2/1.2=1872/12 = 156 [using (ii)]

(c) Converting division into a fraction: 18.72 ÷ 15.6, we get

18.72 ÷ 15.6 = 18.72/15.6=1872×10/156×100 = 12/10 = 1.2 [Using (iii)]

(d) Here 0.156 × 0.12 = 156×12/1000×100 [Using (i)]

= 1872/1000×100

= 0.01872 

Question 4.
Evaluate the following:
(a) 25 ÷ _____ = 0.025
(b) 25 ÷ _____ = 250
(c) 25 ÷ _____ = 2.5
(d) 25 ÷ 10 = 25 × _____
(e) 25 ÷ 0.10 = 25 × _____
(f) 25 ÷ 0.01 = 25 × _____

Answer:

$\begin{array}{l}(a)\ \text{Let } 25÷ x=0.025\\\qquad\dfrac{25}{0.025}=x\ \Rightarrow\ x=\dfrac{25\times1000}{25}=1000\\(b)\ 25÷ x=250\\\qquad\Rightarrow x=\dfrac{25}{250}=\dfrac{1}{10}=0.1\\(c)\ 25÷ x=2.5\\\qquad\Rightarrow x=\dfrac{25}{2.5}=\dfrac{25\times10}{25}=1\times10=10\\(d)\ \text{Let } 25÷10=25\times x\\\qquad\dfrac{25}{10}=25\times x\ \Rightarrow\ x=\dfrac{25}{10\times25}=\dfrac{1}{10}=0.1\\(e)\ \text{Let } 25÷0.10=25\times x\\\qquad\Rightarrow\dfrac{25}{0.10\times25}=x\ \Rightarrow\ x=\dfrac{1}{0.10}=\dfrac{100}{10}=10\\(f)\ \text{Let } 25÷0.01=25\times x\\\qquad x=\dfrac{25}{0.01\times25}=x\ \Rightarrow\ x=\dfrac{1}{0.01}=\dfrac{100}{1}=100.\end{array}$

Question 5.
Find the quotients:
(a) 2.46 ÷ 1.5 = _____
(b) 2.46 ÷ 0.15 = _____
(c) 2.46 ÷ 0.015 = _____
Is the quotient obtained in 24.6 ÷ 1.5 the same as the quotient obtained in 2.46 ÷ 0.15?

Answer:
(a) Converting 2.46 ÷ 1.5 into a fraction

$\begin{array}{l}\dfrac{2.46}{1.5}=\dfrac{246\times10}{15\times100}=\dfrac{246}{150}\\150\overline{)\,246}\,(\,1.64\\\phantom{150)\,}\underline{-150\phantom{0}}\\\phantom{150)\,}\ 960\\\phantom{150)\,}\underline{-900}\\\phantom{150)\,0}\ 600\\\phantom{150)\,0}\underline{-600}\\\phantom{150)\,00}\ 0\end{array}$

Quotient = 1.64

(b) Converting 2.46 ÷ 0.15 into a fraction, we get

2.46/0.15=246×100/15×100=246/15

$\begin{array}{r@{}l}15&\overline{)\,246}\,(\,16.4\\&\phantom{)\,}\underline{-15\phantom{0}}\\&\phantom{)\,}\ 96\\&\phantom{)\,}\underline{-90}\\&\phantom{)\,0}\ 60\\&\phantom{)\,0}\underline{-60}\\&\phantom{)\,00}\ 0\end{array}$

Quotient = 16.4

(c)Converting 2.46 ÷ 0.015 into a fraction, we get 2.46/0.015=246×1000/15×100=2460/15

$\begin{array}{r@{}l}15&\overline{)\,246}\,(\,16.4\\&\phantom{)\,}\underline{-15\phantom{0}}\\&\phantom{)\,}\ 96\\&\phantom{)\,}\underline{-90}\\&\phantom{)\,0}\ 60\\&\phantom{)\,0}\underline{-60}\\&\phantom{)\,00}\ 0\end{array}$

Also,

24.6 ÷ 1.5 = 16.4

and

2.46 ÷ 0.15 = 16.4

Therefore, yes, both quotients are the same. 

Question 6.
A 4 m long wooden block has to be cut into 5pieces of equal length. What is the length of each piece?

Answer:
Total length of the wooden block = 4 m

Number of equal pieces = 5

Length of each piece
= 4 ÷ 5
= 0.8 m

$\begin{array}{r@{}l}\text{Now }5&\overline{)\,4}\,(\,0.8\\&\phantom{)\,}\underline{-0\phantom{0}}\\&\phantom{)\,}\ 40\\&\phantom{)\,}\underline{-40}\\&\phantom{)\,0}\ 0\end{array}$

Therefore, each piece is 0.8 m long. 

Question 7.
If the perimeter of a regular polygon with 12 sides is 208.8 cm, what is the length of its side?

Answer:
Perimeter = 208.8 cm

Number of sides = 12

Length of each side
= 208.8 ÷ 12
= 17.4 cm

$\begin{array}{r@{}l}12&\overline{)\,208.8}\,(\,17.4\\&\phantom{)\,}\underline{-12\phantom{0.0}}\\&\phantom{)\,}\ 88\\&\phantom{)\,}\underline{-84}\\&\phantom{)\,0}\ 48\\&\phantom{)\,0}\underline{-48}\\&\phantom{)\,00}\ 0\end{array}$

Therefore, each side of the polygon is 17.4 cm long. 

Question 8.
3 litres of watermelon juice is shared equally among 8 friends. How much watermelon juice will each get? Express the quantity of juice in millilitres.

Answer:
Total juice = 3 litres = 3000 ml

Number of friends = 8

Juice received by each friend = ⅜ litres

= ⅜ ×1000ml 

= 3000 ÷ 8
= 375 ml

$\begin{array}{r@{}l}8&\overline{)\,3000}\,(\,375\\&\phantom{)\,}\underline{-24\phantom{00}}\\&\phantom{)\,}\ -60\\&\phantom{)\,}\underline{\ \ 56}\\&\phantom{)\,0}\ 40\\&\phantom{)\,0}\underline{-40}\\&\phantom{)\,00}\ 0\end{array}$

Therefore, each friend will get 375 ml of watermelon juice. 

Question 9.
A car covers 234.45 km using 12.6 litres of petrol. What distance is travelled per litre?

Answer:
Total distance = 234.45 km

Petrol used = 12.6 litres

$\begin{gathered}\text{Distance per litre}=\dfrac{\text{Total distance}}{\text{Total petrol}}\\[2pt]=\dfrac{234.45}{12.6}\\[1pt]=\dfrac{23445\times10}{126\times100}\\[1pt]=\dfrac{23445}{1260}=18.607\ \text{km}\\[]\text{Now}\\[4pt]1260\big)\overline{23445}\big(18.607\\[-2pt]\begin{array}{r}\underline{-1260}\\10845\\\underline{-10080}\\7650\\\underline{-7560}\\9000\\\underline{-8820}\\180\end{array}\end{gathered}$

Therefore, the car travels approximately 18.607 km per litre. 

Question 10.
13.5 kg of flour (aata) was distributed equally among 15 students. How much flour did each student receive?

Answer:
Total flour = 13.5 kg

Number of students = 15

Flour received by each student
= 13.5 ÷ 15
= 0.9 kg

Now
$\begin{array}{r@{}l}15&\overline{)\,13.5}\,(\,.9\\&\phantom{)\,}\underline{-13.5}\\&\phantom{)\,00}\ 0\end{array}$

Therefore, each student received 0.9 kg of flour. 

4.4 Look Before You Leap!

Figure It Out (Pages 93-95)

Question 1.
A 210-gram packet of peanut chikki costs ₹70.5, while a 110-gram packet of potato chips costs ₹33.25. Which is cheaper?

Answer:
Cost of peanut chikki per gram
= ₹70.5 ÷ 210
≈ ₹0.3357

Cost of potato chips per gram
= ₹33.25 ÷ 110
≈ ₹0.3023

Since ₹0.3023 < ₹0.3357, potato chips are cheaper per gram. 

Question 2.
Write the decimal number at the arrow mark:

Answer:
(i) The interval from 3.1 to 3.2 is divided into 10 equal parts.

Value of each division
= 0.1 ÷ 10
= 0.01

The arrow is at the sixth mark after 3.1.

3.1 + (6 × 0.01)
= 3.1 + 0.06
= 3.16

(ii) The difference between 2.15 and 2.17 is 0.02.

Value of each division
= 0.02 ÷ 10
= 0.002

The arrow is at the sixth mark after 2.15.

2.15 + (6 × 0.002)
= 2.15 + 0.012
= 2.162 

Question 3.
Shyamala bought 3 kg of bananas at ₹ 30/- per kg. She counted 35 bananas in all. She sells each banana for ₹ 5/-. How much profit does she make selling all the bananas?

Answer:
Cost of 3 kg of bananas
= 3 × ₹30
= ₹90

Amount received by selling 35 bananas
= 35 × ₹5
= ₹175

Profit
= ₹175 − ₹90
= ₹85

Therefore, Shyamala makes a profit of ₹85. 

Question 4.
A teacher placed textbooks that are 2.5 cm thick on a bookshelf. The teacher wanted to place 80 textbooks on the shelf. The bookshelf is 160 cm long. How many books could be placed on the shelf? Was there any space left? If yes, how much?

Answer:
Thickness of one textbook = 2.5 cm

Length of bookshelf = 160 cm

Number of books that can be placed
= 160 ÷ 2.5
= 64

Space occupied by 64 books
= 64 × 2.5
= 160 cm

Therefore, 64 books can be placed on the shelf, leaving no space. 

Question 5.
Fill in the following blanks appropriately:

Answer:
Here, (i) 1 km = 1000 m

5.5 km = 5.5 × 1000 = 5500 m

(ii) 1 m = 100 cm

35 cm = 35/100 = 0.35 m

(iii) 1 cm = 10 mm

14.5 cm = 14.5 × 10 = 145 mm

(iv) 1 kg = 1000 g

68 g = 68/1000 = 0068 kg

(v) 1 m = 1000 mm

9.02 m = 9.02 × 1000 = 9020 mm

(vi) 1 l = 1000 ml

125.5 ml = 125.5/1000 = 0.1255 l

Question 6.
Sridharacharya set the following problem in his book, Patiganita. “6(¼) is divided by 2(½), and 60(¼) is divided by 3(½). Tell the quotients separately.” Can you try solving it by converting the fractions to decimals?

Answer:

$\begin{array}{l}\text{Given }6\dfrac{1}{4}=6+\dfrac{1}{4}=6+0.25=6.25\\\text{and }\ 2\dfrac{1}{2}=2+\dfrac{1}{2}=2+0.5=2.5\\\qquad 60\dfrac{1}{4}=60+\dfrac{1}{4}=60+0.25=60.25\\\qquad 3\dfrac{1}{2}=3+0.5=3.5\\\text{Now }6\dfrac{1}{4}\text{ is divided by }2\dfrac{1}{2}.\\\Rightarrow\ \dfrac{6.25}{2.5}=\dfrac{625\times10}{25\times100}=\dfrac{625}{250}\\250\overline{)\,625}\,(\,2.5\quad\therefore\ \text{Quotient}=2.5\\\phantom{250)\,}\underline{-500\phantom{0}}\\\phantom{250)\,}\ 1250\\\phantom{250)\,}\underline{-1250}\\\phantom{250)\,0}\ 0\\\text{Also }60\dfrac{1}{4}\text{ is divided by }3\dfrac{1}{2}.\\\Rightarrow\ \dfrac{60.25}{3.5}=\dfrac{6025\times10}{35\times100}=\dfrac{6025}{350}\\\therefore\ 350\overline{)\,6025}\,(\,17.21\\\phantom{\therefore\ 350)\,}\underline{-350\phantom{0}}\\\phantom{\therefore\ 350)\,}\ 2525\\\phantom{\therefore\ 350)\,}\underline{-2450}\\\phantom{\therefore\ 350)\,0}\ 750\\\phantom{\therefore\ 350)\,0}\underline{-700}\\\phantom{\therefore\ 350)\,00}\ 500\\\phantom{\therefore\ 350)\,00}\underline{-350}\\\phantom{\therefore\ 350)\,00}\ 150\end{array}$

Quotient = 17.21

Question 7.
Fill the boxes in at least 2 different ways:

Answer:
One possible set is:

(a) 1.2 × 2 = 2.4
and
0.4 × 6 = 2.4

(b) 2.9 × 5 = 14.5
and
14.5 × 1 = 14.5

Question 8.
Find the following quotients, given that 756 ÷ 36 = 21:
(a) 75.6 ÷ 3.6
(b) 7.56 ÷ 0.36
(c) 756 ÷ 0.36
(d) 75.6 ÷ 360
(e) 7560 ÷ 3.6
(f) 7.56 ÷ 0.36

Answer:

$\begin{array}{l}(a)\ \text{Given }756÷36=21\hfill\ldots(i)\\\qquad\text{Here }\dfrac{75.6}{3.6}=\dfrac{756\times10}{36\times10}=\dfrac{756}{36}=21\qquad[\text{Using }(i)]\\(b)\ \text{Here }\dfrac{7.56}{0.36}=\dfrac{756\times100}{36\times100}=\dfrac{756}{36}=21\qquad[\text{Using }(i)]\\(c)\ \text{Here }\dfrac{756}{0.36}=\dfrac{756\times100}{36}=21\times100=2100\qquad[\text{Using }(i)]\\(d)\ \text{Here }\dfrac{75.6}{360}=\dfrac{756}{360\times10}=\dfrac{756}{360\times10}\\\qquad=\dfrac{21}{100}=0.21\qquad[\text{Using }(i)]\\(e)\ \text{Here }\dfrac{7560}{3.6}=\dfrac{7560}{36}\times10=\dfrac{756}{36}\times10\times10\\\qquad=21\times10\times10=2100\qquad[\text{Using }(i)]\\(f)\ \text{Here }\dfrac{7.56}{0.36}=\dfrac{756\times100}{36\times100}=\dfrac{756}{36}=21\qquad[\text{Using }(i)]\end{array}$

Question 9.
Find the missing cells if each cell represents a ÷ b:

Answer:
The table itself is image-based, so keep the original table on the page.

Using the relationship shown in the reference:

1517 ÷ 37 = 41

Therefore,

151.7 ÷ 3.7 = 41

15.17 ÷ 0.37 = 41

1.517 ÷ 37 = 0.041

Question 10.
Using the digits 2, 4, 5, 8, and 0, fill the boxes ◻◻. ◻ × ◻.◻ to get the:
(a) maximum product
(b) minimum product
(c) product greater than 150
(d) product nearest to 100
(e) product nearest to 5

Answer:

$\begin{array}{l}(a)\ \boxed{4}\,\boxed{2}\,.\,\boxed{0}\times\boxed{8}\,.\,\boxed{5}=357\\(b)\ \boxed{4}\,\boxed{5}\,.\,\boxed{8}\times\boxed{0}\,.\,\boxed{2}=9.16\\(c)\ \boxed{8}\,\boxed{5}\,.\,\boxed{4}\times\boxed{2}\,.\,\boxed{0}=170.8\\(d)\ \boxed{2}\,\boxed{0}\,.\,\boxed{5}\times\boxed{4}\,.\,\boxed{8}=98.4\\(e)\ \boxed{4}\,\boxed{5}\,.\,\boxed{8}\times\boxed{0}\,.\,\boxed{2}=9.16\end{array}$

Question 11.
Sort the following expressions in increasing order:
(a) 245.05 × 0.942368
(b) 245.05 × 7.9682
(c) 245.05 ÷ 7.9682
(d) 245.05 ÷ 0.942368
(e) 245.05
(f) 7.9682

Answer:
Let A = 245.05, B = 0.942368, C = 7.9682

We note that B < 1 and C > 1

(a) Now A × B = 245.05 × 0.942368 < 245.05

(∴ Multiplying a number by a value less than 1 results in a smaller number)

(b) A × C = 245.05 × 7.9682 > 245.05

(Multiplying a number by a value greater than 1 results in a larger number)

(c) A ÷ C = 2.4505 ÷ 7.9682 < 245.05

(Dividing a number by a value greater than 1 results in a smaller number)

(d) A ÷ B = 245.05 ÷ 0.942368 > 245.05

(Dividing a number by a value greater than 1 results in a smaller number)

(e) Now, the expression is less than 245.05

∴ 0.942368 is closer to 1 than 7.9682 is to 1.

0.942368 will result in a value closer to 245.05 than dividing by 7.9682.

∴ (c) < (a)

Again, 0.942368 is closer to 1 than 7.9682 is to 1.

∴ Dividing by 0.942368 yields a value closer to 245.05 than multiplying by 7.9682.

∴ (d) < (b)

Also, (f) 7.9682 is much smaller than 245.05, so it will be the smallest value.

(f) ∴ 7.9682 < (e) 245.05 ÷ 7.9682

Combining all, we get (f), (c), (a), (e),(d), (b).

Revise Another Peek Beyond the Point with Class 7 Maths Solutions

The NCERT Solutions for Class 7 Maths Chapter 4 help students understand decimal multiplication, division and place value through clear, step-by-step answers.

• Strengthen decimal concepts: Practise multiplication, division and decimal place values.

• Improve calculation accuracy: Learn to place decimal points correctly and avoid common mistakes.

• Solve practical problems: Apply decimals to questions involving money, measurements and quantities.

• Check your answers: Compare your working with the solved textbook questions.

• Revise before exams: Use the NCERT Solutions for Class 7 Maths Chapter 4 PDF to revisit difficult questions and practise independently.

Important Study Material Links for Chapter 4: Another Peek Beyond the Point

NCERT Solutions for Class 7 Maths Ganita Prakash I and II (New Syllabus)

Important Related Links for NCERT Class 7 Maths