ML AGGARWAL CLASS 7 Chapter 9 Linear Equations and Inequalities Excercise 9.1 ~ LearnsCBSE

Excercise: 9.1

Question 1

(i)

$\begin{aligned} 2(3-2 x) &=13 \\ 6-4 x &=13 \\-4 x &=13-6 \\-4 x &=7 \\ x=-7 / 4 \end{aligned}$

ii) $\begin{aligned} \frac{3}{5} y-2=& 7 / 10 \\ \frac{3}{5} y &=7 / 10+2 \\ \frac{3}{5} y &=\frac{27}{10} \\ y &=\frac{27}{10} \times \frac{5}{3}=9 / 2 \end{aligned}$

= $9 / 2$

Question 2

i) $\frac{x}{2}=5+x / 3$

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

$\frac{3 x-2 x}{6}=5$

$x=5 \times 6$

$x=30$

ii) $2(x-3 / 2)=11$

$2 x-2\times 3 / 2=11$

$2 x-3=11$

$2 x=3+11$

$x=14 /2$

$x=7$

Question 3

i) $7(x-2)=2(2 x-4)$

$7 x-14=4 x-8$

$7 x-4 x=14-8$

$3 x=6$

$x= 6 / 2$

x = 2

ii)$21-3(x-7)=x+20$

$21-3 x+21=x+20$

$4 x=42-20$

$4 x=\frac{22}{11}$

x = 22\4

x = 11\2

Question 4

i) $3 x-\frac{1}{3}=2(x-1 / 2)+5$

$\frac{9 x-1}{3}=\frac{2(2 x-1)}{2}+5$

$\frac{9 x-1}{3}=2 x-1+5$

$\frac{9 x-1}{3}=2 x+4$

$9 x-1=6 x+12$

$9 x-6 x=12+1$

$3 x=13$

$x=13 / 3$

ii) $\frac{2 m}{3}-\frac{m}{5}=7$

$\frac{(2m) \times 5-3 m}{15}=7$

$\frac{10 m-3 m}{15}=7$

$\frac{7 m}{15}=7$

$m=7 \times \frac{15}{7}$

$m=15$

Question 5

i) $\frac{x+1}{5}-\frac{x-7}{2}=1$

$\frac{2(x+1)-5(x-7)}{10}=1$

$2 x+2-5 x+35=10$

$-3 x=10-35-2$

$-3 x=-27$

$x=\frac{-27}{-3}$

$x=9$

ii)

$\begin{aligned} \frac{3 p-2}{7}-\frac{p-2}{4}=2 & \\ \frac{4(3 p-2)-7(p-2)}{7 \times 4}=2 & \\ \frac{12 p-8-7 p+14}{28} &=2 \\ 5 p+6 &=56 \\ 5 p &=56-6 \\ 5 p &=50 \\ p &=50 / 5 \end{aligned}$

P = 10

Question 6

i) $\frac{1}{2}(x+5)-\frac{1}{3}(x-2)=4$

$\frac{3(x+5)-2(x-2)}{3 \times 2}=4$

$\frac{x+2 x+4=4 \times 6}{x+19}=24$

$x+19=24$

$x=24-19$

$x=5$

ii) $\frac{2 x-3}{1}-\frac{x-5}{2}=\frac{x}{6}$

$\frac{2(2 x-3)-6(x-5)}{6 \times 2}=\frac{x}{6}$

$4 x-6-6 x+30=$ $\frac{x}{6} \times {12}$

$-2 x+24 .=2 x$

$4 x=24$

$x=24 / 4$

x = 6

Question 7

i) $\frac{x-4}{7}-\frac{x+4}{5}=\frac{x+3}{7}$

$\frac{5(x-4)-7(x+4)}{7 \times 5}=\frac{x+3}{7}$

$5 x-20-7 x-28=\left(\frac{x+3}{7}\right) 35$

$\begin{aligned}-2 x-48 &=5 x+15 \\-48-15 &=5 x+2 x \\ 7 x &=-63 \\ x &=\frac{-63}{7} \\ x &=-9 \end{aligned}$

ii) $\frac{x-1}{5}+\frac{x-2}{2}=\frac{x}{3}+1$

$\frac{2(x-1)+5(x-2)}{5 \times 2}=\frac{x+3}{3}$

$\frac{2 x-2+5 x-10}{10}=\frac{x+3}{3}$

$\frac{7 x-12}{10}=\frac{x+3}{3}$

$3(7 x-12)=10(x+3)$

$21 x-36=10 x+30$

$11 x=66$

$x=\frac{66}{11}$

$x=6$

Question 8

$\begin{aligned} y+1.2 y &=4.4 \\ 2.2 y &=4.4 \\ y &=\frac{4.4}{2.2} \\ y &=2 \end{aligned}$

ii) $15 \%$ of $x=21$

$\frac{15}{100} \times x=21$

$x=\frac{21 \times 100}{15}$

x = 140

Question 9

i) $2 p+20 \%$ of $(2 p-1)=7$

$2 p+\frac{20}{100}(2 p-1)=7$

$2 P+\frac{1}{5}(2 p-1)=7$

$5 \times 2 P+2 P-1=35$

$10 P+2 P=1+35$

$\begin{aligned} 12 p &=36 \\ p &=36 / 12 \\ p=3 \end{aligned}$

ii) $3(2 x-1)+25 \%$ of $x=97$

$3(2 x-1)+\frac{95}{100} \times x=97$

$6 x-3+\frac{1}{4} \times x=97$

$\frac{4(6 x-3)+x}{4}=97$

$24 x-12+x=97 \times 4$

$25 x=388+12$

$25 x=400$

$x=\frac{400}{25}$

x= 16

Question 10

$x^{4}-3 x^{3}-p x-5=2.3 .$

Given $x=-2$

$(-2)^{4}-3(-2)^{3}-p(-2)-5=23$

$16-3(-8)+2 p-5=23$

$16+24+2 p-5=23$

$2 P=23-35$

$2 p=-12$

p = -12\2

p = -6

Excercise: 9.2

Question 1

Let the required number = x

5 times the number = 5x

7 added to 5 times the number = 5x + 7

According to the problem

$5 x+7=57$

$5 x=50$

$x=50 / 5$

$x=10$

Question 2

Let the required number = x

$\frac{1}{4}^{\text {th }}$ of the number is 3 more than 7

$\frac{1}{4} x=7+3$

$\frac{1}{4} x=10$

$\quad x=40$

Question 3

Let required number = x

A number is greater than 15 and it is Less than 51 then

x - 15 = 51 -x

2x = 66

x = 33

Question 4

Let required number = x

$\frac{1}{2}$ is subtracted from a number = $x-1 / 2$

Multiplied by 4 = $4(x-1 / 2)$

Given result = 5

$4(x-1 / 2)=5$

$24\times\left(\frac{2 x-1}{y}\right)=5$

$4 x-2=5$

$4 x=5+2$

$4 x=7$

$x=7 / 4$

Question 5

Let the required number = x , 80- x

The greater number exceeds twice the smaller by 11 is

$x=2(80-x)+11$

$x=160-2 x+11$

$3 x=171$

$x=171 / 3$

x=57 And other number is 80 - 57 = 23

Question 6

Three consecutire odd natureal number are

$2 x+1, \quad 2 x+3, \quad 2 x+5$

Given sum is = 87

$2 x+2 x+3+2 x+5=87$

$6 x=87-9$

$6 x=78$

$x=78 / 6$

x = 13

Required number are 27, 29 , 31

Question 7

Let number of boys = x

Number of girls = $\frac{2}{5} x$.

Total no. of students = 35

$x+\frac{2}{5} x=35$'

$5 x+2 x=35 \times 5$

$7 x=35 \times 5$''

$x=\frac{35 \times 5}{7}$

$x=25$

Number of girls in the class is = $\frac{2}{5} \times 25$ = 10

Question 8

Let number of chairs = x

A house wife purchased certain number of chairs and two tables = 2800

$(250 \times x)+(2 \times 400)=2800$

$(250 \times x)+(800)=2800$

$250 \times x=2800-800$

$x=\frac{2000}{250}$

$x=8$

Number of chairs purchased = 8

Question 9

Let Aparna's monthly salary= x

Then over time payment = x - 16560

According to the problem

$\begin{aligned} x+x-16560 &=27840 \\ & 2 x=44,400 \\ &=\frac{44,400}{2} . \\ x=& 22,200 \end{aligned}$

Aparna monthly salary= 22,200

Question 10

Let 5 rupee coins = x

2 rupees coins = 80 -x

According to the problem total amount is 232 rupees

$\begin{aligned} 5 x+2(80-x) &=232 \\ 5 x+160-2 x &=232 \\ 3 x &=232-160 \\ 3 x &=72 \\ x=& 72 / 3 \\ x &=24 \end{aligned}$

Number of 5 rupees coins = 24

Question 11

Let purse contains 10 rupees notes = x

50 rupees notes is one less = x - 1

According to the problem total amount in purse is = 550

$10 x+50(x-1)=550$

$10 x+50 x-50=550$

$60 x=600$

$x=600 / 60$

$x=10$

Number of 50 rupees note = 10-1 = 9

Question 12

Let present age = x

After 12 years = x + 12

3 Times as old was 4 years ago

$x+12=3(x-4)$

$x+12=3 x-12$

$2 x=24$

$x=24 / 2$

$x=12$

Present age is 12

Question 13

Given sides of isosceles triangle are

$3 x-1, \quad 2 x+2,2 x$

Two equal sides are

$3 x-1=2 x+2$

$3 x-2 x=2+1$

$x=3$

perimeter of triangle $=3 x-1+2 x+2+2 x$

$=7 x+1$

$=7(3)+1$

$=22$

Question 14

Let breadth = x

Length of the rectanangular = 3x - 6

According to the problem perimeter is = 148

$\begin{aligned} 2(x+3 x-6) &=148 \\ 2(4 x-6) &=148 \\ 8 x-12 &=148 \\ 8 x &=160 \\ x &=80 \end{aligned}$

Length 54 , breadth = 20

Question 15

Difference of each angle = 20 '

Let the angles be = x and x + 20

Sum of the complementary angle is 90'

$x+x+20=90$

$2 x=90-20$

$2 x=70$

$x=35$

One angles , 35' other angle is (35+ 20) = 5

Excercise: 9.3

Question 1

i) x <-2

Replacement sent {$-5,-3,-1,0,1,3,4$}

Solutin set { -5, -3}

ii) x>1