Additional Questions - Chapter 1 - Number Systems - Term 3 - 7th Maths Guide Samacheer Kalvi Solutions

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Additional Questions and Answers
Exercise 1.1
Question 1.
Match the following:

Solution:
$1-(\mathrm{v})$
2 -(iv)
$3-(i)$
4 -(iii)
$5-$ (ii)

Question $2 .$
Round $89.357$ to the nearest whole number.
Solution:
Underlining the digit to be rounded $89.357$. Since the digit next to the underlined digit 3 which is less than 5 , the underlined digit remains the same.
$\therefore$ The nearest whole number $89.357$ rounds to 89 .

Question $3 .$
Round $110.929$ to the nearest tenths place.
Solution:

Underlining the digit to be rounded $110.229$. Since the digit next to the underlined digit is 2 which is less than 5 .
$\therefore$ The underlined digit 9 remains the same. Hence the rounded number is $110.9$

Question $4 .$
Round $87.777$ upto 2 places of decimal.
Solution:
Rounding $87.777$ upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of $87.777$ gives 87.777. Since the digit after the hundredth place value is 7 which is more than 5 , we add 1 to the underlined digit. So the rounded value of $87.777$ upto 2 places of decimal is $87.78$

Exercise $1.2$
Question 1.
If Sheela bought $2.083 \mathrm{~kg}$ of grapes and $3.752 \mathrm{~kg}$ of orange. What is the total weight of fruits Solution:

Weight of grapes $=2.083 \mathrm{Kg}$
Weight of orange $=2.752 \mathrm{Kg}$
Total weight $=(2.083+2.752) \mathrm{Kg}=4.835 \mathrm{Kg}$

Question $2 .$
Kathir bought $8.72 \mathrm{~kg}$ of sugar, $7.302 \mathrm{~kg}$ of grains. His carry bag can contain only $15 \mathrm{~kg}$ of weight. What is the remaining weight of items bought?
Solution:

Question $3 .$
Use place value grid to add $7.357$ and $13.92$.
Solution:
Let as use place value grid.

Exercise $1.3$
Question 1.
Cost of $1 \mathrm{~m}$ cloth is ₹ $6.75$. Find the cost of $14.75 \mathrm{~m}$ correct to two places of decimal.
Solution:
Cost of $1 \mathrm{~m}$ cloth $=₹ 6.75$
Cost of $14.75 \mathrm{~m}$ cloth $=14.75 \times 6.75$
= ₹ $99.5625$
=₹ $99.56$

Question $2 .$
Length of a side of a square is $18.35 \mathrm{~cm}$. Find its Area. Solution:
Side of a square $=18.35 \mathrm{~cm}$
Area of a square $=($ Side $\times$ Side $)$ sq.units
$=18.35 \times 18.35 \mathrm{~cm}^{2}$
$=336.7225 \mathrm{~cm}^{2}$

Exercise $1.4$
Question 1.
A wire of length $363.987 \mathrm{~m}$ is cut into 30 pieces. What is the length of each piece?
Solution:
Length of the wire $=363.987 \mathrm{~m}$
i.e Total length of 30 pieces $=\frac{363987}{1000} \mathrm{~m}$
$\therefore$ Length of I piece
$=12132.9 \times \frac{1}{1000}$
Length of 1 piece of wire $=12.1329 \mathrm{~m}$

Question $2 .$
A cake of $50 \mathrm{~kg}$ needs $23.4 \mathrm{~kg}$ sugar. Find the weight of cake made by $1 \mathrm{~kg}$ of sugar.
Solution:
Weight of cake made using $23.4 \mathrm{~kg}$ sugar $=50 \mathrm{~kg}$
Weight of cake made using $1 \mathrm{~kg}$ sugar $=\frac{50}{23.4}$
$=\frac{50}{23.4} \times \frac{10}{10}=\frac{500}{234}=2.1367 \mathrm{Kg}$
$=2.14 \mathrm{Kg}$
Weight of cake made using $1 \mathrm{~kg}$ sugar $=2.14 \mathrm{Kg}$
 

Question $3 .$
A pack of 20 pencils cost ₹ 94.4. What is the cost of each pencil?
Solution:
Cost of 20 pencils $=₹ 94.4$

$\begin{aligned}
\text { Cost of 1 pencil } &=\frac{94.4}{20}=\frac{944 / 10}{20 / 1}=\frac{944}{10} \times \frac{1}{20}=\frac{944}{20} \times \frac{1}{10} \\
&=\frac{47.2}{10}=4.72
\end{aligned}$