Exercise 1.3 - Chapter 1 Number Systems Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $1.3$
Question $1 .$
Find the product of the following
(i) $0.5 \times 3$
(ii) $3.75 \times 6$
(iii) $50.2 \times 4$
(iv) $0.03 \times 9$
(v) $453.03 \times 7$
(vi) $4 \times 0.7$
Solution:
(i) $0.5 \times 3$
$5 \times 3=15$
$\therefore 0.5 \times 3=1.5$
(ii) $3.75 \times 6$

$\begin{aligned}
&375 \times 6=2250 \\
&3.75 \times 6=22.50
\end{aligned}$

(iii) $50.2 \times 4$
$502 \times 4=2008$
$50.2 \times 4=200.8$
(iv) $0.03 \times 9$
$3 \times 9=27$
$0.03 \times 9=0.27$
(v) $453.03 \times 7$
$45303 \times 7=317121$
$453.03 \times 7=3171.21$
(vi) $4 \times 0.7$
$4 \times 7=28$
$4 \times 0.7=2.8$

Question $2 .$
Find the area of the parallelogram whose base is $6.8 \mathrm{~cm}$ and height is $3.5 \mathrm{~cm}$.
Solution:
Base of the parallelogram $b=6.8 \mathrm{~cm}$
Height of the parallelogram $\mathrm{h}=3.5 \mathrm{~cm}$
Area of the parallelogram $\mathrm{A}=\mathrm{b} \times \mathrm{h}$ sq.units $=6.8 \times 3.5 \mathrm{~cm}^{2}$
Area of the parallelogram $=23.80 \mathrm{~cm}^{2}$

Question $3 .$
Find the area of the rectangle whose length is $23.7 \mathrm{~cm}$ and breadth is $15.2 \mathrm{~cm}$.
Solution:

Length of the rectangle $l=23.7 \mathrm{~cm}$
Breadth of the rectangle $\mathrm{b}=15.2 \mathrm{~cm}$
Area of the rectangle $\mathrm{A}=1 \times \mathrm{b}$ sq.units
$=23.7 \times 15.2 \mathrm{~cm}^{2}$
Area of the rectangle $=360.24 \mathrm{~cm}^{2}$

Question $4 .$
Multiply the following
1. $2.57 \times 10$
2. $0.51 \times 10$
3. $125.367 \times 100$
4. $34.51 \times 100$
$5.62 .735 \times 100$
6. $0.7 \times 10$
7. $0.03 \times 100$
8. $0.4 \times 1000$
Solution:
1. $2.57 \times 10=25.7$
2. $0.51 \times 10=5.1$
3. $125.367 \times 100=12536.7$
4. $34.51 \times 100=3451$
5. $62.735 \times 100=6273.5$
6. $0.7 \times 10=7.0$
7. $0.03 \times 100=3$
8. $0.4 \times 1000=400$

Question $5 .$
A wheel of a baby cycle covers $49.7 \mathrm{~cm}$ in one rotation. Find the distance covered in 10 rotations.
Solution:
Length covered in 1 rotation $=49.7 \mathrm{~cm}$
Length covered in 10 rotations $=49.7 \times 10 \mathrm{~cm}=497 \mathrm{~cm}$

Question $6 .$
A picture chart costs $₹ 1.50$. Radha wants to buy 20 charts to make an album. How much does she have to pay?
Solution:
Cost of 1 chart $=₹ 1.50$
Cost of 20 charts $=₹ 1.50 \times 20=₹ 30.00$
Cost of 20 charts $=₹ 30$

Question 7.
Find the product of the following.
(i) $3.6 \times 0.3$
(ii) $52.3 \times 0.1$
(iii) $537.4 \times 0.2$
(iv) $0.6 \times 0.06$
(v) $62.2 \times 0.23$
(vi) $1.02 \times 0.05$
(vii) $10.05 \times 1.05$
(viii) $101.01 \times 0.01$
(ix) $100.01 \times 1.1$
Solution:
(i) $3.6 \times 0.3$

$\begin{aligned}
&36 \times 3=108 \\
&3.6 \times 0.3=1.08
\end{aligned}$
(ii) $52.3 \times 0.1$
$\begin{aligned}
&\text { (ii) } 52.3 \times 0.1 \\
&523 \times 1=523 \\
&52.3 \times 0.1=5.23
\end{aligned}$
(iii) $537.4 \times 0.2$

$\begin{aligned}
&5374 \times 2=10748 \\
&537.4 \times 0.2=107.48
\end{aligned}$
(iv) $0.6 \times 0.06$
$\begin{aligned}
&6 \times 6=36 \\
&0.6 \times 0.06=0.036
\end{aligned}$
(v) $62.2 \times 0.23$

$\begin{aligned}
&622 \times 23=14306 \\
&62.2 \times 0.23=14.306
\end{aligned}$
(vi) $1.02 \times 0.05$
$\begin{aligned}
&102 \times 5=510 \\
&1.02 \times 0.05=0.0510
\end{aligned}$
(vii) $10.05 \times 1.05$

$\begin{aligned}
&1005 \times 105=105525 \\
&10.05 \times 1.05=10.5525
\end{aligned}$
$\begin{aligned}
&\text { (viii) } 101.01 \times 0.01 \\
&10101 \times 1=10101 \\
&101.01 \times 0.01=1.0101
\end{aligned}$
$\begin{aligned}
&\text { (ix) } 100.01 \times 1.1 \\
&1001 \times 11=110011 \\
&100.01 \times 1.1=110.011
\end{aligned}$

Objective Type Questions
Question 1.
$1.07 \times 0.1$
(i) $1.070$
(ii) $0.107$
(iii) $10.70$
(iv) $11.07$
Answer:
(ii) $0.107$

Question $2 .$
$2.08 \times 10=$
(i) $20.8$
(ii) $208.0$
(iii) $0.208$
(iv) $280.0$
Answer:
(i) $20.8$
Hint:
$\begin{aligned}
&208 \times 10=2080 \\
&2.08 \times 10=20.80=20.8
\end{aligned}$

Question $3 .$
A frog jumps $5.3 \mathrm{~cm}$ in one jump. The distance travelled by the frog in 10 jumps is
(i) $0.53 \mathrm{~cm}$
(ii) $530 \mathrm{~cm}$
(iii) $53.0 \mathrm{~cm}$
(iv) $53.5 \mathrm{~cm}$

Answer:
(iii) $53.0 \mathrm{~cm}$
Hint:
$\begin{aligned}
&53 \times 10=530 \\
&5.3 \times 10=53.0
\end{aligned}$