Exercise 1.1
Question 1
(i) Ascending order is -18,-9,-4,0,3,8,12
(ii) Desending order is 12, 8, 3, 0, -4, -9, -18
(iii)From left -17,-14,-11,-3,-5,-3,1,4,7,9,11
Question 2
B is -6
D is -4
H is 0
J is 2
M is 5
O is 7
Question 3
Ascending order is -5,-4,0,4,7
(Daigram should be added)
Question 4
Rohit scores in five successive rounds were 15, -3, -7 , 12,8
Total score at end = Sum of all scores
=15+(-3)+(-7)+12+8
=15-3-3+20=25
∴ Rohit total score at end = 25
Question 5
Ruchi deposited on monday = Rs 4370
She withdrew on tuesday = Rs 2875
On tuesday , the money left = 4370 - 2875
=Rs 1495
Next day i.e on wednesday
She deposited = Rs 1550
Total money in bank = 1550+1495
=Rs 3045
∴ Balance on thusday =Balance on wednesday
=Rs 3045
Question 6
(Daigram should be added)
Final position from p is 28 - 37= -9
Question 7
Let the unfilled boxes be 1st row 2nd Box be p
3rd Box be Q
2nd row 3rd Box be R
3rd row Sequentially x, y , z
Given information all the sum of integers in each row colume and each digram is -6
i.e $\begin{aligned}-1+3+1 &=-6 \\ 2+x &=-6 \\ x &=-6-2 \\ x &=-8 \end{aligned}$
= - 8 + (-2)+Q=-6
= - 8 - 2+Q= -6
= -10 + Q= -6
Q= - 6+10 = 4
$\begin{aligned}-1+p+4 &=-6 \\ p+3 &=-6 \\ p &=-6-3 \\ &P= -9 \end{aligned}$
$\begin{aligned}-8+5+z &=-6 \\-3+z &=-6 \\ z &=-6+3 \\ z &=-3 \end{aligned}$
$\begin{array}{|c|c|c|}\hline-1 & P & Q \\\hline 3 & -2 & R \\\hline x & y & z \\\hline\end{array} \Rightarrow \begin{array}{|c|c|c|}\hline-1 & -9 & 4 \\\hline 3 & -2 & 7 \\\hline-8 & 5 & -3 \\\hline\end{array}$
Question 8
(i) $|-13|-|9|$
We know $|-a|=a$
$=13-9=4$
(ii) $|13-5|-|-9|$
$=|8|-|-9|$
$=8-9$
$=-1$
(iii) $|3 5-21|-| 8-3 \mid$
$=|14|-|5|$
$=14-5$
$=9$
Question 9
Ascending order is -102,-51,-39,-6,-5,0,7,35
Question 10
Desending order is 4208 , 139, 0 ,-31, -97,-203
Question 11
(i) True ,it lies left to 0
(ii) False, it will have predecessor and successor
(iii) True
(iv)True
Question 12
(i) $(-11)+(-7) ;(-11)-(-7)$
$=-11-7 \quad ;(-11+7)$
$=-18 \quad ;-4$
As $-18<-4$
(ii)
$\begin{aligned} 23-41+11 & ; 23-41-11 \\=34-41 & ; 23-52 \\=-7 & ;-29 \\ \therefore-7 &>-29 \end{aligned}$
(iii) $40-(-39)+(-5)$
$=40+39-5$
$=79-5$
$=74$
=$40+(-39)-(-5)$
=$40-39+5$
=$1+5$
=6
∴ 74>6
(iv)
$\begin{aligned}(-3)+13-(15) &: 25-(-2)+(-33) \\-3+13-15 & ; 25+2-33 \\ 13-18 & ; 27-33 \\-5 & ;-6 . \\-5 &>-6 \end{aligned}$
Exercise 1.2
Question 1
(i) $-5,2$ i.e $-5+2=-3$
$-6,3$ i.e $-6+3=-3$
(ii) $-2,3$ i.e $-2-3=-5$ (vice vessa not possible)
(iii)8,4 i.e 8-4=4 (vice vessa not possible)
Question 2
(i) $\begin{aligned}-8, &-13 \\ \text { i.e } &-8-(-13) \\ &=-8+13 \\ &=5 \end{aligned}$
(ii) $\begin{aligned}-12,4 & \\ \text { ie } &-12+4 \\ &=-8 \end{aligned}$
(iii) $-1,2$
i.e $-1-2=-3$
Question 3
$-6,-8$
1.e $[-6-(-8)]$
$\quad=-6+8=2$
Question 4
A scored -30 ,20,0
Total Team A scored = -30+20 +0 = -10
Team B Scored $20,0,-30$
Total team B Scored = $20+0+(-30)=-10$
Scored by team A and B are same = -10 , Yes
Question 5
$\begin{aligned} S u m &=-72+237+84+72+(-184)+(-37) \\ &=-72+393-184-37 \\ &=393-293 \\ &=100 \end{aligned}$
Exercise 1.3
Question 1
(i) $7 \times(-35)=-245$
(ii) $(-13) \times(-15)=195$
(iii) $(-12) \times(-11) \times(-10)=(-12) \times(110)=-1320$
(iv) $(-13) \times 0 \times(-24)=0 .$
(v) $(-1) \times(-2) \cdot(-3) \times(+4)=-24$
(vi) $(-3) \times(-6) \times(-2) \times(-1)=18 \times 2=36$
Question 2
(i) $37 \times[6+(-3)]$
If a,b,c are integers , then $a \times(b+c)$ = $a \times b+a \times c$
$=37 \times 6+37 \times(-3)$
∴ Both are equal , Verified
(ii) $(-21) \times[(-6)+(-4)]=(-21) \times(-6)+(-21) \times(-4)$
If a ,b , c are integers, then $a \times(b+c)=a \times b+a \times c$
=$(-21) \times(-6)+(-21) \times(-4)$
∴ Verified
Question 3
(i)
$\begin{aligned} 8 \times 53 \times(-125)=& 8 \times(-125) \times 53 \\=&-1000 \times 53 \\=&-53,000 \end{aligned}$
(ii)
$\begin{aligned}(-8) \times(-2) \times 3 \times(-5)=&(-8) \times(-5) \times(-2) \times 3 \\ &=40 \times 6 \\ &=240 . \end{aligned}$
(iii)
$\begin{aligned}(-6) \times 2 \times(-8) \times 5=&(-6) \times(-8) \times 2 \times 5 \\ &=48 \times 10 \\ &=480 . \end{aligned}$
(iv)
$\begin{aligned} 15 \times(-25) \times(-4) \times(-10) &=15 \times(-10) \times(-25) \times(-4) \\ &=-150 \times 100 \\ &=-15,000 \end{aligned}$
(v) $26 \times(-48)+(-48) \times(-36)$
$=(-48)[26+(-36)]$
$=(-48)[26-36]$
$=(-48) \times(-10)$
$=480 .$
(vi) $724 \times(-56)+(-724) \times 44$
$=724 \times(-56)+724 \times(-44)$
$=724 \times[-56+(-44)]$
$=724 \times[-56-44]$
$=724 \times(-100)$
$=-72400$
(vii)
$\begin{aligned}(-47) \times 102=&(-43) \times(100+2) \\=&-47 \times 100+(-47) \times 2 \\=&-4700+(-94) \\=&-4700-94 \\ &=-4794 . \end{aligned}$
(viii) $(-39) \times(-97)$
$(-39) \times(-100+3)$
$(-39) \times(-100)+(-39) \times 3$
$=3900+(-117)$
$=3900-117$
$=3783$
Question 4
(i)
$\begin{aligned}(-4) \times 1 &=44 \\ x &=\frac{44}{-4}=\frac{44 x-1}{-4 x-1} \\ & x=\frac{-44}{4}=-11 \end{aligned}$
(ii)
$\begin{aligned} 7 \times 1 . &=-42 \\ x &=\frac{-42}{7} \\ x &=-6 . \end{aligned}$
(iii)
$\begin{aligned} x \times(-13) &=143 \\ x=& \frac{143}{-13} \times \frac{-1}{-1} \\ x &=\frac{-143}{13} \\ & x=-11 \end{aligned}$
(iv) (-5) $\times$__ =0
Any number Multiplied with '0' We get '0
So 0 is answer
i.e -5$times$ 0= 0
Question 5
Fro every one hour the temperature lowered at a rate of $5^{\circ} \mathrm{C}$
For 8 hours it will be lowered by $5 \times 8=40^{\circ} \mathrm{C}$
Room temperature after free process= 32 - 40
$=-8^{\circ} \mathrm{C}$
Question 6
Total number of question = 10
Marks given for correct answer = 5 and
Marks given for incorrect answer = -2
(i) Rohit gets 4 correct and 5 incorrect answer
Rohit score = 4 $\times$ 5+6 $\times(-2)=20-12=8$
(ii) Seema gets 5 correct and 5 incorrect answer
Seema's score = $5 \times 5+5 \times(-2)=25-10=15$
(iii) As ritu attempted 7 questions and only 2 question are
correct , so number of incorrect question are 7-2=5
Ritu's score = $2 \times 5+5 \times(-2)=10-10=0$
Question 7
(i) Let pair of integers be x, y
Product is - 15 i.e $x y=-15$
$y=\frac{-15}{x}$ ___(1)
Difference = 8
x-y = 8
From (1) $x-\left(\frac{-15}{x}\right)=8$
$\frac{x^{2}+15}{x}=8$
$x^{2}+15=8 x$
$x^{2}-8 x+15=0$
$x^{2}-5 x-3 x+15=0$
$x(x-5)-3(x-5)=0$
$(x-3)(x-5)=0$
$x=3, x=9$
$y=\frac{-15}{3} ; \quad y=-\frac{15}{5}$
$y=-5 ; \quad y=-3$
The pair of integers are -3, -5 , 5, -3
Alternative method for (ii)
Product is - 36
Difference is 15
We know
$12 \times(-3)=-36 ;(-12) \times 3=-36$
$9 \times(-4)=-36 \quad ;(-9) \times 4=-36$
Thus , we have pair of integers $12,-3 ;-12,3 ; 9,-4 ;-9,4$
Such that product of each pair is -36
But Difference of $9,-4=9-(-4)=9+4=13 \mathrm{~cm}$
Difference of $-9,4=-9-4=-13$
Hence the required pair are $12,-3$ or $-12,3$.
Exercise 1.4
Question 1
(i) $(-36) \div(-9)$
$\begin{aligned}=+(36 \div 9) &=+4 \\ &=4 \end{aligned}$
(ii) $150 \div(-25)$
$=-(150+25)$
$=-6$
(iii) $(-270) \div 27$
$=-(270+27)$
=-10
(iv) $(-59)+(59)$
$=-(59 \div 59)$
=-1
(v) $0 \div(-17)$
$=0$
(vi) $(-784) \div(-56)$
$=+(784 \div 56)$
$=+14$
$=14$
Question 2
(i) $13 \div[(-2)+1]$
$=13 \div[-2+1]$
$=13 \div[-1]$
$=-[13 \div 1]$
=-13
(ii) $(-47) \div[(-45)+(-2)]$
$=(-47) \div[-45-2]$
$=(-47) \div(-47)$
$=+[47 \div 47]$
=1
(iii) $[(-6)+5] \div[(-2)+1]$
$=[-6+5] \div[-2+1]$
$=(-1) \div(-1)$
=1
(iv) $[(-48) \div(-6)] \div(-2)$
$=[+(48 \div 6)] \div(-2)$
$=[8] \div(-2)$
$=-[8+2]$
=-4
Question 3
$(a+b) \div c$
$=[(-225) \div 15] \div(-3)$
$=[-(225 \div 15)] \div(-3)$
$=(-15) \div(-3)$
$=+(15+3)$
$=5$
$a \div(b+c)$
$=(-225) \div(15 \div(-3))$
$=(-225) \div(-(15 \div 3))$
$=(-225) \div(-5)$
$=+(225 \div 5)$
=45
$5 \neq 45$
Verified
Question 4
(i)
$\begin{aligned} & a \div(b+c) \\=&(-10) \div(1+1) \\=&(-10) \div(2) \\=&-(10 \div 2) \\=&-5 \end{aligned}$
$(a \div b)+(a \div c)$
$=(-10+1)+(-10 \div 1)$
$\equiv-10+(-10)$
$=-10-10$
$=-20$
$-5 \neq-20$
$\therefore$ verified
(ii)
$\begin{aligned} & a \div(b+c) \\=& 12\div(1+(-2)) \\=& 12 \div(1-2) \\=& 12+(-1) \\=&-(12 \div 1) \quad ; \\=&-12 \end{aligned}$
$(a \div b)+(a \div c)$
$=(12 \div 1)+(12 \div(-2))$
$=12+[-(12 \div 2)]$
=$12+[-6]$
=$12-6$
=6
$-12 \neq 6$
verified.
Question 5
We know $a \div a=1 \quad-a \div a=-1$
i) $239 \div 239=1$
ii) $(-85) \div 85=-1$
iii) $(-213) \div(-213)=1$
iv) $(-43) \div(-1)=43$
v) $(-84) \div(-21)=4$
vi) $(-66) \div 22=-3$
Question 6
$a \div b=-3$
$\frac{a}{b}=-3 \Rightarrow a=-3 b$
$b=1 ; a=-3 \times 1=-3$
$b=2 ; a=-3 \times 2=-6$
$b=3 ; a=-3 \times 3=-9$
$b=4: a=-3 \times 4=-12$
$b=5 ; a=-3 \times 5=-15$
∴ Five p;airs are $(-3,1)(-6,2)(-9,3)(-12, 4)(-15,5)$
Question 7
(i) Marks Scored for 14 Correct answer = $14 \times 3=42$
Sachin's Score = 24
∴ Marks obtained for incorrect answer $=-42+24$
=-18
∴ The number of incorrect answer = $(-18) \div(-2)$
=9
(ii) Marks Scored for 9 correct answers $=9 \times 3$
$=27$
Nalini's Score = -7
Marks obtained for incorrect answers = $-7-27$
$-34$
Marks given for one incorrect answer = -2
∴ The number of incorrect answer = $(-34) \div(-2)$
$=3 4+2$
=17
Question 8
Elevator descends at a rate of $6 \mathrm{~m} / \mathrm{min}$.
Descend starts from 10m above ground level
It has to reach the shaft 350m below ground level
Total 360 m it has to reach
$\begin{aligned} \text { Time } &=\frac{\text { Distance }}{\text { speed }} \\ &=\frac{360}{6} \\ &=60 \mathrm{~min} \end{aligned}$
$\therefore$ Time $=1$ honr
$\therefore$ It will take 1hour to reach the shaft
Exercise 1.5
Question 1
$7-8 \div(-2)+3 \times(-4)$
According to BODMAS rule
$=7-(-4)+3 \times(-4)$
$=7+4+(-12)$
$=11-12$
$=-1$
Question 2
$9-\{7-24 \div(8+6 \times 2-16)\}$
$=9-\{-24 \div(8+12-16)\}$
$=9-\{7-24 \div(4)\}$
$=9-\{7-6\}$
$=9-1$
$=8$
Question 3
$-11-[-6-\{3-5(8 \div 4-1)\}]$
$=-11-[-6-\{3-5(2-1)\}]$
$=-11-[-6-\{3-5(1)\}]$
$=-11-[-6-\{3-5\}]$
$=-11-[-6-(-2)]$
$=-11-[-6+4]$
$=-11-(-2)$
$=-11+2$
$=-9$
Question 4
$(-3) \times(-12) \div(-4)+3 \times 6$
$=(-3) \times 3+3 \times 6$
$=(-3) \times 3+18$
$=-9+18$
$=9$
Question 5
$\begin{aligned} & 14 \div(3(2-3+4))-9(5-3) \\=& 14 \div(6-9+12)-45+27 \\=& 14 \div(9)-18 \\=& \frac{14}{9}-18 \end{aligned}$
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