In Text Questions Try These (Text book Page No. 14,16, 18, 19, 22) - Chapter 1 - Number Systems - Term 1 - 7th Maths Guide Samacheer Kalvi Solutions

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(Try These Text book Page No. 14)
Question $1 .$
Fill in the blanks.
(i) $(-7)-(-15)=+8$
$-7-(-15)=-7+\overline{(\text { Additive inverse of-15) }}$
$=-7+15=+8$
(ii) $12-(-7)=19 \quad 12-(-7)=19$
(iii) $-4-(-5)=1$

Question $2 .$
Find the values and compare the answers.
(i) $15-12$ and $12-15$
(ii) $-21-32$ and $-32-(-21)$
Solution:
(i) $15-12=3 \& 12-15=12+(-15)=-3$
$15-12 \Rightarrow 12-15$
(ii) $-21-32=(-21)+(-32)=-53$
Also $-32-(-21)=(-32)+(+21)=-11 ;-53<-11$
$-21-32 \square(-32)-(-21)$

Question $3 .$
Is associative property true for subtraction of integers. Take any three examples and check. Solution:
Consider the numbers 1,2 and 3 . Now $(1-2)-3=-1-3=-4$
Also $1-(2-3)=1-(-1)=1+1=2$
$\therefore(1-2)-\neq 1-(2-3)$
$\therefore$ Associative property is not true for subtraction of integers.

(Try These Textbook Page No. 16)
Question $1 .$
Find the product of the following
(i) $(-20) \times(-45)=+900$ [As we know the product of two negative integers is positive, the answer is $+900$.]
(ii) $(-9) \times(-8)=72[\because$ Product of two negative integers is positive $]$
(iii) $(-30) \times 40 \times(-1)=(+1200)$ [Product of two integers with opposite sings is negative integer.
$(-30) \times 40 \times(-1)=(-1200) \times(-1)=+1200)]$
(iv) $(-50) \times 2 \times(-10)=-1000$ [Product of two integers with opposite signs is negative.
$(+50) \times 2 \times(-10)=100 \times(-10)=-1000)$ ]
 

Question $2 .$
Complete the following table by multiplying the integers in the corresponding row and column headers.

Solution:
We know that
(i) product of two positive integers is positive
(ii) product of two negative integers is
(ii) product of two negative integers is positive
(iii) product of integers with opposite sign is negative.
$\therefore$ The table will be as follows:

Question $3 .$
Which of the following is incorrect?
(i) $(-55) \times(-22) \times(-33)<0$
(ii) $(-1521) \times 2511<0$
(iii) $2512-1525<0$
(iv) $(1981) \times(+2000)<0$
Solution:
(iii) and (iv) are incorrect because $2512-1252$ is a positive integer.
Also $(+1981) \times(+2000)$ is a positive integer.

(Try These Textbook Page No. 18)
Question $1 .$
Find the product and check for equality
(i) $18 \times(-5)$ and $(-5) \times 18$
Solution:
Here $18 \times(-5)=-90$ Also $(-5) \times 18=-90$
$\therefore 18 \times(-5)=(-5) \times 18$
(ii) $31 \times(-6)$ and $(-6) \times 31$
Solution:
Here $31 \times(-6)=-186$ Also $(-6) \times 31=-186$
$\therefore 31 \times(-6)=(-6) \times 31$
(iii) $4 \times 51$ and $51 \times 4$
Solution:
Here $4 \times 51=204$ Also $51 \times 4=204$
$\therefore 4 \times 51=51 \times 4$

Question 2 .
Prove the following.
(i) $(-20) \times(13 \times 4)=[(-20) \times 13] \times 4$
Solution:

$\text { LHS }=(-20) \times(13 \times 4)=(-20) \times 52=-1040$
LHS $=(-20) \times(13 \times 4)=(-20) \times 52=-1040$ RHS $=[(-20) \times 13] \times 4=(-260) \times 4=-1040$ LHS $=$ RHS $\therefore(-20) \times(13 \times 4)=[(-20) \times 13] \times 4$ (ii) $[(-50) \times(-2)] \times(-3)=(-50) \times[(-2) \times(-3)]$ Solution: LHS $=[(-50) \times(-2)] \times(-3)=100 \times(-3)=-300$ RHS $=(-50) \times[(-2) \times(-3)]=(-50) \times 6=300$ LHS $=$ RHS $\therefore[(-50) \times(-2)] \times(-3)=(-50) \times[-2) \times(-3)]$
Solution:
(iii) $[(-4) \times(-3)] \times(-5)=(-4) \times[(-3) \times(-5)]$
Solution:
LHS $=[(-4) \times(-3)] \times(-5)=12 \times(-5)=-60$
RHS $=(-4) \times[(-3) \times(-5)]=(-4) \times 15=-60$
LHS $=$ RHS
$\therefore[(-4) \times(-3)] \times(-5)=(-4) \times[(-3) \times(-5)]$

(Try These Textbook Page No. 19)
Question $1 .$
Find the values of the following and check for equality:
(i) $(-6) \times(4+(-5))$ and $((-6) \times 4)+((-6) \times(-5))$
Solution:
$(-6) \times(4+(-5))=(-6) \times(-1)=6$.
$((-6) \times 4)+((-6) \times(-5))=(-24)+30=6$
Hence $(-6) \times(4+(-5))=((-6) \times 4)+((-6) \times(-5))$
(ii) $(-3) \times[2+(-8)]$ and $[(-3) \times 2]+[(-3) \times 8]$
Solution:
$(-3) \times[2+(-8)]=(-3) \times(-6)=18$
Also $[(-3) \times 2]+[(-3) \times 8]=(-6)+(-24)=-30$
$(-3) \times[2+(-8)] \neq[(-3) \times 2]+[(-3) \times 8]$

Question $2 .$
Prove the following.
(i) $[(-5) \times(-76)]+[(-5) \times 8]$
Solution:
$\begin{aligned}
&\text { LHS }=(-5) \times[(-76)+8]=(-5) \times(-68) \\
&=+340 \\
&\text { RHS }=[(-5) \times(-76)]+[(-5) \times 8] \\
&=+380+(-40)=+380-40 \\
&=+340 \\
&\text { LHS }=\text { RHS } \\
&\therefore(-5) \times[(-76)+8]=[(-5) \times(-76)]+[(-5) \times 8]
\end{aligned}$

(Try These Textbook Page No. 19)
Question $1 .$
Find the values of the following and check for equality:
(i) $(-6) \times(4+(-5))$ and $((-6) \times 4)+((-6) \times(-5))$
Solution:
$(-6) \times(4+(-5))=(-6) \times(-1)=6$.
$((-6) \times 4)+((-6) \times(-5))=(-24)+30=6$
Hence $(-6) \times(4+(-5))=((-6) \times 4)+((-6) \times(-5))$
(ii) $(-3) \times[2+(-8)]$ and $[(-3) \times 2]+[(-3) \times 8]$
Solution:
$(-3) \times[2+(-8)]=(-3) \times(-6)=18$
Also $[(-3) \times 2]+[(-3) \times 8]=(-6)+(-24)=-30$
$(-3) \times[2+(-8)] \neq[(-3) \times 2]+[(-3) \times 8]$

Question $2 .$
Prove the following.
(i) $[(-5) \times(-76)]+[(-5) \times 8]$
Solution:
$\begin{aligned}
&\text { LHS }=(-5) \times[(-76)+8]=(-5) \times(-68) \\
&=+340 \\
&\text { RHS }=[(-5) \times(-76)]+[(-5) \times 8] \\
&=+380+(-40)=+380-40 \\
&=+340 \\
&\text { LHS }=\text { RHS } \\
&\therefore(-5) \times[(-76)+8]=[(-5) \times(-76)]+[(-5) \times 8]
\end{aligned}$

(Try These Text book Page No. 22)
Question $1 .$
(i) $(-32) \div 4=$
(ii) $(-50) \div 50=$
(iii) $30 \div 15=$
(iv) $-200 \div 10=$
(v) $-48 \div 6=$
Solution:
(i) $-8$
(ii) $-1$
(iii) 2
(iv) $-20$
(v) $-8$