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Page No 2:

Question 1:

Show the following numbers on a number line. Draw a separate number line for each example.

(1) 32, 52, -32

(2) 75, -25, -45

(3) -58, 118

(4) 1310, -1710

Answer:


1 32,52,-32 can be represented on the number line as follows.

 
2 75,-25,-45 can be represented on the number line as follows.

 
3 -58,118 can be represented on the number line as follows.

 
4 1310,-1710 can be represented on the number line as follows.
 

Page No 2:

Question 2:

Observe the number line and answer the questions.

(1) Which number is indicated by point B?
(2) Which point indicates the number 134?
(3) State whether the statement, 'the point D denotes the number 52, is true of false.

Answer:

(1) We observe that each unit on the number line is divided into 4 equal parts.
      Now, B is the tenth point on the left of 0.
      So, B indicates -104 on the number line .
2 134=74=7×14
      So, the seventh point on the right of 0 is C that indicates 134 on the number line.
(3) The point D is the tenth point on the right of 0. So, D indicates 104 on the number line.
      Now, 104 = 52
      So, D denotes 52 on the number line. Hence, the given statement is true.



Page No 3:

Question 1:

Compare the following numbers.
(1) −7, −2

(2) 0, 95

(3) 87, 0

(4) 54, 14

(5) 4029, 14129

(6)  −1720, 1320

(7) 1512716

(8) 258, 94

(9) 1215, 35

(10) 711, 34

Answer:

(1) We know that, 7 > 2

∴ −7 < −2.

(2) We know that, a negative number is always less than 0.

∴ 0>-95.

(3) We know that, a positive number is always greater than 0.

∴ 87>0

(4) We know that, −5 < 1.

∴ -54<14.

(5) We know that, 40 < 141.

∴ 4029<14129.

(6) We know that, −17 < −13.

-1720<-1320.

7 1512=15×412×4=6048; 716=7×316×3=2148
Now, 6048>2148
∴ 1512>716.

(8) Let us first compare 258 and 94.
258=25×18×1=258; 94=9×24×2=188
Now, 258>188
∴ 258>94
∴ -258<-94.

9 1215=12×115×1=1215; 35=3×35×3=915
Now, 1215>915
∴ 1215>35.

(10) Let us first compare 711 and 34.
711=7×411×4=2844; 34=3×114×11=3344
Now, 2844<3344
∴ 711<34
∴ -711>-34.



Page No 4:

Question 1:

Write the following rational numbers in decimal form.
(1) 937

(2) 1842

(3) 914

(4) 1035

(5) −1113

Answer:

(1) The given number is 937.

∴ 937 = 0.243243.... = 0.243
The decimal form of 937 is 0.243.

(2) The given number is 1842.

∴ 1842 = 0.428571428571.... = 0.428571
The decimal form of 1842 is 0.428571.

(3) The given number is 914.

∴ 914 = 0.6428571428571.... = 0.6428571
The decimal form of 914 is 0.6428571.

(4) The given number is -1035.

∴ 1035 = 20.6
The decimal form of -1035 is −20.6.

(5) The given number is -1113.

∴ 1113 = 0.846153846153.... = 0.846153
The decimal form of -1113 is -0.846153.



Page No 5:

Question 1:

The number 2 is shown on a number line. Steps are given to show 3 on the number line using 2. Fill in the boxes properly and complete the activity.

Activity:

∙ The point Q on the number line shows the number ........
∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
∙ Right angled ∆ ORQ is obtained by drawing seg OR.
l(OQ) = 2, l(QR) = 1

∴ by Pythagoras theorem,
[l(OR)]2 = [l(OQ)]2 + [l(QR)]2
  0   2 +   0   2 =   0    +    0  
 
=    0  

l(OR) =    0  
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line 3.

 

Answer:


∙ The point Q on the number line shows the number 2.
∙ A line perpendicular to the number line is drawn through the point Q. Point R is at unit distance from Q on the line.
∙ Right angled ∆ ORQ is obtained by drawing seg OR.
∙ l(OQ) = 2l(QR) = 1
∴ by Pythagoras theorem,
[l(OR)]2 = [l(OQ)]2 + [l(QR)]2
               = 22+12=2+1 
               = 3
∴ l(OR) = 3
Draw an arc with centre O and radius OR. Mark the point of intersection of the line and the arc as C. The point C shows the number line 3.



Page No 6:

Question 2:

Show the number 5 on the number line.

Answer:


Draw a number line as shown in the figure. Let the point O represent 0 and point Q represent 2. Draw a perpendicular QR at Q on the number line such that QR = 1 unit. Join OR. Now, ∆OQR is a right angled triangle. 
By Pythagoras theorem, we have
OR2 = OQ2 + QR2
= (2)+ (1)2
= 4 + 1
= 5
∴ OR = 5
Taking O as the centre and radius OR = 5, draw an arc cutting the number line at C.
Clearly, OC = OR = 5.
Hence, C represents 5 on the number line.

Page No 6:

Question 3:

Show the number 7 on the number line.

Answer:

 
Draw a number line as shown in the figure and mark the points O, A and B on it such that OA = AB = 1 unit. The point O represents 0 and B represents 2. At B, draw CB perpendicular on the number line such that BC = 1 unit. Join OC. Now, ∆OBC is a right angled triangle.
In ∆OBC, by Pythagoras theorem
(OC)2 = (OB)2 + (BC)2
= (2)2 + (1)2
= 4 + 1
= 5
∴ OC = 5
Taking O as centre and radius OC = 5, draw an arc cutting the number line at D.
Clearly, OC = OD = 5
At D, draw ED perpendicular on the number line such that ED = 1 unit. Join OE. Now, ∆ODE is a right angled triangle.
In ∆ODE, by Pythagoras theorem
(OE)2 = (OD)2 + (DE)2
= (5)2 + (1)2
= 5 + 1
= 6
∴ OE = 6
Taking O as centre and radius OE = 6, draw an arc cutting the number line at F.
Clearly, OE = OF = 6
At F, draw GF perpendicular on the number line such that GF = 1 unit. Join OG. Now, ∆OFG is a right angled triangle.
In ∆OFG, by Pythagoras theorem
(OG)2 = (OF)2 + (FG)2
= (6)2 + (1)2
= 6 + 1
= 7
∴ OG = 7
Taking O as centre and radius OG = 7, draw an arc cutting the number line at H.
Clearly, OG = OH = 7
Hence, H represents 7 on the number line.



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