Additional Questions - Chapter 1 Number Systems Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise $1.1$

Question $1 .$
When Malar woke up her temperature was $102^{\circ} \mathrm{F}$. Two hours later it was $3^{\circ}$ lower, what was her temperature then?
Solution:
Initially Malar's temperature $=102^{\circ} \mathrm{F}$
After two hours it lowered $3^{\circ} \Rightarrow-3^{\circ} \mathrm{F}$
Here present temperature $=102^{\circ}+\left(-3^{\circ}\right)=99^{\circ} \mathrm{F}$

Question $2 .$
An elevator is on the twentieth floor. It goes down 11 floors and then up 5 floors. What floor is the elevator on now?
Solution:
Present location of the elevator $=20$ th floor
If it goes down 11 floor $\Rightarrow(-11)$
$=20+(-11)=9$ th floor
If it goes up 5 floor $\Rightarrow 9+5$
$=14$ th floor
 

Question $3 .$
$16+_{+}=16 .$ The property expressed here is
Solution:
$16+0=16$.
0 is the additive identity on integers.

Exercise $1.2$
Question $1 .$
Roman civilization began in $509 \mathrm{BC}$ and ended in $476 \mathrm{AD}$. How long did Romans civilization last.
Solution:
From the start of common era no. of years upto 476
$\mathrm{AD}=476$
From $509 \mathrm{BC}$ to start of common era years $=509$
Total years Roman civilization last $=476+509=985$ years.

Question $2 .$
A submarine was situated 450 feet below sea level. If it descends 300 feet. What is its new position?
Solution:
Position of submarine $=-450 \mathrm{ft}$.
Again it descends 300 feet $\Rightarrow-300$ feet
$\therefore$ New position $=-450+(-300)=-750 \mathrm{ft}$.
$\therefore$ It was 750 feet below sea level.
 

Question $3 .$
In January the high temperature recorded was $90^{\circ} \mathrm{F}$ and the low temperature was $-2^{\circ} \mathrm{F}$. Find the difference between the high and the low temperatures?
Solution:
The high temperature recorded $=90^{\circ} \mathrm{F}$
The low temperature recorded $=-2^{\circ} \mathrm{F}$
Difference $=90^{\circ} \mathrm{F}-\left(-2^{\circ} \mathrm{F}\right)$
$=90^{\circ} \mathrm{F}+\left(\right.$ Additive inverse of $\left.-2^{\circ} \mathrm{F}\right)$
$=90^{\circ} \mathrm{F}+\left(+2^{\circ} \mathrm{F}\right)=92^{\circ} \mathrm{F}$

Exercise $1.3$
Question $1 .$
Ani is scuba diving. She descends 5 feet below sea level. She descends the same distance 4 more times. What is Anis final elevation?
Solution:
Ani descends 5 feet below sea level once she descends 4 more times
$\therefore$ She descends $(5 \times 4)+5$ feet in total $=20+5=25$ feet below sea level
 

Question $2 .$
The price of a plant reduced $₹ 6$ per week for 7 weeks. By how much did the price of the plant change over the 7 weeks?
Solution:
The price of plant reduced in a week $=₹ 6$
$\therefore$ The price reduced in 7 weeks $=7 \times 6=42$

Question $3 .$
The product of three integers is $-3$. Determine all of the possible values for the three factors?
Solution:
Product of three integers $=-3$
Possible factors are $(1 \times-1 \times 3),(-1 \times-1 \times-3),(1 \times 1 \times-3)$

Exercise $1.6$
Question 1 .
Simplify the following using suitable properties.
(a) $(-1650) \times(-2)+(-1650) \times(-98)$
(b) $(9150 \times 405)-(8150 \times 405)$
Solution:
(a) $(-1650) \times(-2)+(-1650) \times(-98)$
$=1650[(-1) \times(-2)+(-1) \times-98]=1650(2+98)$ [Distributive property]
$=1650 \times 100=1,65,000$
(b) $(9150 \times 405)-(8150 \times 405)$
$=405(9150-8150)=405 \times 1000$
$=4,05,000$

Question 2 .
Which is greater: $(9+7) \times 1000$ or $9+7 \times 1000$ ?
Solution:
$\begin{aligned}
&(9+7) \times 1000=16 \times 1000=16,000 \\
&9+7 \times 1000=9+7000=7,009 \\
&16,000>7009 \\
&\therefore(9+7) \times 1000>[9+7 \times 1000]
\end{aligned}$

Question $3 .$
Simiplify: $80 \div[240 \div(-24)]+7$
Solution:
We have
$80 \div[240 \div(-24)]+7$
$=80 \div\left[\frac{240}{-24}\right]+7$
$=80 \div(-10)+7=-\left[\frac{80}{10}\right]+7=(-8)+7=-1$