ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.2

Exercise 1.2

Question 1

Sol :
(i) $\quad 2 \frac{2}{3}=\frac{2 \times 3+2}{3}=\frac{8}{3}$

=$2 \frac{2}{3}-\left(-\frac{3}{7}\right)$

=$\frac{8}{3}-\left(-\frac{3}{7}\right)$

=$\frac{8}{3}+\frac{3}{7}$

=$\frac{(8 \times 7)+(3 \times 3)}{21} \quad$  LCM of $3,7=21$

=$\frac{56+9}{21}$

=$\frac{65}{21}$

$\frac{-4}{9}-\left(3 \frac{5}{8}\right)$

$\frac{-4}{9}-\left(\frac{29}{8}\right)$

$-\frac{4}{9}-\frac{29}{8}$

$\frac{-4 \times 8-(29 \times 9)}{72} \quad$ LCM of $9,18=72$

$-\frac{293}{72}$


(ii) $-3 \frac{1}{5}-\left(-4 \frac{7}{9}\right)$

$-\frac{16}{5}-\left(\frac{-43}{9}\right)$

$\frac{-16}{5}+\frac{43}{9}$

$\frac{(-16 \times 9)+(43 \times 5)}{45}$ LCM OF 9,15=45

$\frac{-144+215}{45}$

$\frac{71}{45}$


Question 3

Sol :
Let the Unknown number as x.

 $\begin{aligned}-\frac{5}{11}+x &=\frac{-7}{8} \\ x &=\frac{-7}{8}-\left(-\frac{5}{11}\right) \\ &=\frac{-7}{8}+\frac{5}{11} \end{aligned}$    

$=\frac{(-7 \times 11)+(8 \times 5)}{88} \quad$ LCM Of $11,8=88$

$=\frac{-77+45}{88}$

$x=\frac{-32}{88}$

$x=\frac{-4}{11}$


(ii)   let the unknown number as x.  

$\begin{aligned}-\frac{2}{7}+x &=\frac{3}{5} \\ x &=\frac{3}{5}-\left(-\frac{2}{7}\right) \\ x &=\frac{3}{5}+\frac{2}{7} \\ &=\frac{(3 \times 7)+(2 \times 5)}{35} \\ &=\frac{21+10}{35} \\ x &=\frac{31}{35} \end{aligned}$


Question 4

Sol :
Let the Unknown number be x

$-4 \frac{3}{5}-x=-3 \frac{1}{2}$

$-\frac{23}{5}-x=-\frac{7}{2}$

$\begin{aligned}-\frac{23}{5} &=-\frac{7}{2}+x \\ x &=-\frac{23}{5}+\frac{7}{2} \\ &=\frac{(-23 \times 2)+(7 \times 5)}{10} \end{aligned}$      LCM OF 5,2=10 

⇒$\frac{-46+35}{10}$

x⇒$\frac{-11}{10}$


Question 5

Sol :

$\left[\frac{-5}{7}+\left(\frac{-8}{3}\right)\right]-\left[\frac{5}{2}+\left(\frac{-11}{12}\right)\right]$

$\left[\frac{(-5 \times 3)+(-8 \times 7)}{21}\right]-\left[\frac{5 \times 6+(-11 \times 1)}{12}\right]$

$\frac{-284-133}{84}-\frac{417}{84}=\frac{-139}{28}=-\frac{139}{28} \|$


Question 6

Sol :

$\quad x=\frac{4}{9} ; \quad y=\frac{-7}{12}$

Consider

x-y =  $\frac{4}{9}-\left(-\frac{7}{12}\right)$

$=\frac{4}{9}+\frac{7}{12}$

⇒$\frac{(4 \times 4)+(7 \times 3)}{36}$               (∴LCM OF 9,12= 36)

⇒$\frac{16+21}{36}$

⇒$x-y=\frac{37}{36}$

Consider     

y-x = $\frac{-7}{12}-\left(\frac{4}{9}\right)$

    

⇒$\frac{-7}{12}-\frac{4}{9}$

⇒$\frac{(-7 \times 3)-(4 \times 4)}{36}$         LCM of  9, 12=36

⇒$\frac{-21-16}{36}$

y-x ⇒$\frac{-37}{36}$

ஃx-y ≠ y-x


Question 7

Sol :

$x=\frac{4}{9} ;  \quad y=\frac{-7}{12} ; \quad z=\frac{-2}{3}$

Consider

$x-(y-z)=\frac{4}{9}-\left(\frac{-7}{12}-\left(\frac{-2}{3}\right)\right)$

$=\frac{4}{9}-\left(\frac{-7}{12}+\frac{2}{3}\right)$

⇒$\frac{4}{9}-\left(\frac{(-7 \times 1)+(2 \times 4)}{12}\right)$

⇒$\frac{4}{9}-\left(\frac{-7+8}{12}\right)$

⇒$\frac{4}{9}-\frac{1}{12}$

⇒$\frac{(4 \times 4)-(1 \times 3)}{36}$

=$\frac{16-3}{36}$

$x-(y-z)$  =$\frac{13}{36}$ 

       Consider       ( x-y)-z  =$\left[\frac{4}{9}-\left(\frac{-7}{12}\right)\right]-\left(\frac{-2}{3}\right)$

$=\left[\frac{4}{9}+\frac{7}{12}\right]+\frac{2}{3}$

$=\left[\frac{(4 \times 4)+(-7 \times 3)}{36}\right]+\frac{2}{3}$

$=\frac{16+21}{36}+\frac{2}{3}$

$=\frac{37}{36}+\frac{2}{3}$

$=\frac{(37 \times 1)+(2 \times 12)}{36}$

⇒$\frac{37+24}{36}$

$(x-y)-z=\frac{61}{36}$

∴ $x-(y-z) \neq(x-y)-z$


Question 8

Sol :

(i) ⇒$\frac{2}{3}-\frac{4}{5}$

$\frac{(2 \times 5)-(4 \times 3)}{15}$     LCM = 3,5 = 15

⇒$\frac{10-12}{15}$

⇒$\frac{-2}{5}$   It is a Rational Number

So. given statement is False


(ii)true

$\frac{-5}{7}+\frac{5}{7}=0$                              

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more