In text Questions Try These (Text Book Page No. 1, 3, 7,11) - Chapter 1 Number Systems Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Text Questions :  Chapter 1 Number Systems Term 1 Class 7th std Maths Guide Samacheer Kalvi Solutions
(Try These Textbook Page No. 1)
Question $1 .$
Write the following integers in ascending order: $-5,0,2,4,-6,10,-10$
Solution:
Plotting the points on the number line, we get

The numbers are placed in an increasing order from left to right.
$\therefore$ Ascending order: $-10<-6<-5<0<2<4<10$

Question $2 .$
If the integers $-15,12,-17,5,-1,-5,6$ are marked on the number line then the integer on the extreme left is
Solution:
The least number will be on the extreme left.
$\therefore-17$ will be on the extreme left.

Question $3 .$
Complete the following pattern:
$50, \ldots 30,20,_{-}, 0,-10, \ldots,-40$,
Solution:
The difference between the consecutive number is 10 .
$50,40,30,20,10,0,-10,-20,-30,-40,-50,-60$
 

Question $4 .$
Compare the given numbers and write "<", ">" or in the boxes.

Solution:

Question $5 .$
Write the given integers in descending order, $-27,19,0,12,-4,-22,47,3,-9,-35$.
Solution:
Separating positive and the negative integers, we get $-27,-4,-22,-9,-35$
Arranging the numbers in descending order $-4>-9>-22>-27>-35$
The positive numbers are $19,12,47,3$
Arranging in descending order, we get $47>19>12>3$
0 stands in the middle.
$\therefore$ Descending order: $47>19>12>3>0>-4>-9>-22>-27>-35$

(Try This Text Book Page No. 3)

Question $1 .$
Find the value of the following using the number line activity.
(i) $(-4)+(+3)$
(ii) $(-4)+(-3)$
(iii) $(+4)+(-3)$
Solution:
(i) $(-4)+(+3)$
To find the sum of $(-4)$ and $(+3)$, we start at zero facing positive direction continuing in the same direction and move 4 units backward to represent $(-4)$.
Since the operation is addition we maintain the same direction and move three units forward to represent $(+3)$
We land at $-1$
(ii) $(-4)+(-3)$
From zero move 4 steps backward to represent $(-4)$
From the same direction again move 3 units backward to represent $(-3)$
We land at $-7$ So $(-4)+(-3)=-7$
(iii) $(+4)+(-3)$
We start at zero facing positive direction and move 4 steps forward to represent $(+4)$ Since the operation is addition we maintain the same direction and move three units backward to represent $(-3)$. We land at $+1$.
$\text { So }(+4)+(-3)=+1$

(Try These Textbook Page No. 7)
Question $1 .$
Fill in the blanks:
(i) $20+(-11)=-(11)+20[\because$ Addition is commutative $]$
(ii) $(-5)+(-8)=(-8)+(-5)$. [\because Addition is commutative]
(iii) $(-3)+12=\underline{12}+(-3)$ [ $\because$ Addition is commutative]

Question $2 .$
Say True or False.
(i) $(-11)+(-8)=(-8)+(-11)$
(ii) $-7+2=2+(-7)$
(iii) $(-33)+8=8+(-33)$
Solution:
(i) True, because addition is commutative for intergers
(ii) True, by commutative property on intergers
(iii) True, by commutative property on intergers

Question $3 .$
Verify the following.
(i) $[(-2)+(-9)]+6=(-2)+[(-9)+6]$
(ii) $[7+(-8)]+(-5)=7+[(-8)+(-5)]$
(iii) $[(-11)+5]+(-14)=(-11)+[5+(-14)]$
(iv) $(-5)+[(-32)+(-2)]=[(-5)+(-32)+(-2)]$
Solution:
(i) $[(-2)+(-9)]+6=(-2)+[(-9)+6]$
$[(-2)+(-9)]+6=(-11)+6=-5$
Also $(-2)+[(-9)+6]=(-2)+(-3)=-5$
Both the cases the sum is $-5 .$
$\therefore-[(-2)+(-9)]+6=(-2)+[(-9)+6]$
(ii) $[7+(-8)]+(-5)=7+[(-8)+(-5)]$
Here $[7+(-8)]+(-5)=(-1)+(-5)=-6$
Also $7+[(-8)+(-5)]=7+(-13)=7-13=-6$
In both the cases the sum is $-6$.
$\therefore[7+(-8)]+(-5)=7+[(-8)+(-5)]$
(iii) $[(-11)+5]+(-14)=(-11)+[5+(-14)]$
Here $[(-11)+5]+(-14)=(-6)+(-14)=(-20)$
$(-11)+[5+(-14)]=(-11)+(-9)=(-20)$
In both the cases the sum is $-20$.
$\therefore[(-11)+5]+(-14)=(-11)+[5+(-14)]$
(iv) $(-5)+[(-32)+(-2)]=[(-5)+(-32)]+(-2)$
$(-5)+[(-32)+(-2)]=(-5)+(-34)=-39$
Also $[(-5)+(-32)]+(-2)=(-37)+(-2)=-39$
In both the cases the sum is $-39$.
$\therefore(-5)+[(-32)+(-2)]=[(-5)+(-32)]+(-2)$

(Try These Text book Page No. 11)
Question $1 .$
Do the following by using number line.
(i) $(-4)-(+3)$
Solution:
We start at zero facing positive direction move 4 units backward to represent (-4). Then turn towards negative side and move 3 units forward.
We reach $-7$.
$\therefore(-4)-(+3)=-7$.

(ii) $(-4)-(-3)$
Solution:
We start at zero facing positive direction. Move 4 units backward to represent $-4$. Then turn towards the negative side and move 3 units backwards.
We reach at-1.
$\therefore(-4)-(-3)=-1$

Question $2 .$
Find the values and compare the answers.
(i) $(-6)-(-2)$ and $(-6)+2$
Solution:
$(-6)-(-2)=-6+($ Additive inverse of-2 $)$
$=-6+(+2)=-4$
Also $(-6)+2=-4$
$\therefore(-6)-(-2)=(-6)+2$
(ii) $35-(-7)$ and $35+7$
Solution:
$35-(-7)=35+($ Additive inverse of $-7)=35+(+7)=42$
Also $35+7=42 ; 35-(-7)=35+7$
(iii) $26-(+10)$ and $26+(-10)$
Solution:
$26-(+10)=26+($ Additive inverse of $+10)=26+(-10)=16$
Also $26+(-10)=16 ; 26-(+10)=26+(-10)$

Question $3 .$
Put the suitable symbol $<,>$ or $=$ in the boxes.

Solution:
(i) $-10-8=-18 \&-10+8=-2$
(ii) $(-20)+10=-10 \&(-20)-(-10)=-10$
(iii) $-70-50=(-70)+(-50)=-20$
(iv) $100-(+100)=0 \& 100-(-100)=100+(+100)=200$
(v) $-50-30=-50+(-30)=-80$ Also $-100+20=-80$