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

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

i)

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

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

Replacement set $\{-5,-3,-1,0,1,3,4\}$

Solution set $\{3,4\}$

iii) $x \geqslant-1$

Replacement set $\{-5,-3,-1,0,1,3,4\}$

Solution set $\{-1,0,1,3,4\}$

iv) $-5<x<3$

Replacement set $\{-5,-3,-1,0,1,3,4\}$

Solution set $\{-3,-1,0,1\}$

v) $-3 \leq x<4$

Replacement set $\{-5,-3,-1,0,1,3,4\}$

Solution set $\{-3,-1,0,1,3\}$

vi) $0 \leq x<7$

Replacement set $\{-5,-3,-1,0,1,3,4\}$

solution set $\{0,1,3,4\}$

Question 2

(i) $\begin{aligned} & x \leq 3 \\ & \text { Solution set }=\{1,2,3\} \end{aligned}$

(Diagram should be added)

ii) $x<4$

Solution set $=\{0,1,2,3\}$

(Diagram should be added)

iii) $-2 \leq x<4$

Solution set : $\{-2,-1,0,1,2,3\}$

(Diagram should be added)

iv) $-3 \leqslant x<2$

Solution set $=\{-3,-2,-1,0,1\}$

(Diagram should be added)

Question 3

(i)

 $\begin{aligned} 4-x>-2 & \\-4+4-x &>-2+(-4) \\-x &>-6 \\ x &<6 \end{aligned}$ Add both side $(-4)$

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ML AGGARWAL CLASS 7 Chapter 8 Algebraic Expressions Excercise =8.1

 Excercise =8.1

Question 1 

(i) $3 x+6$

(ii) $\quad 13-5 x$

(ii) $\quad x^{2}+y^{2}$

(iv) $\quad 3 p q+7$

(v) $\quad x^{2}-3 x$

ii) $\quad mn-(m+n)$

Question 2

A taxi charges = 9\km

fixed charges = 80

Taxo fixed for x km is 

9x + 50

Question 3

(i) $\quad 5 a-3 b+c$

(ii. $\quad x^{2}-5 x+6 .$

(iii) $x^{2} y+x y-x y^{2}$

Question 4

(i) $3,-7 x$

ii. $2,-5 a, \frac{3}{2} b$

iii, $3 x^{5}, 4 y^{3},-7 x y^{2}, 3$

Question 5

i) -4x + 5y

Term - 4x , 5y

Factor -4x  5y

ii) $\quad x y+2 x^{2} y^{2}$

Term $: x y \quad, \quad 2 x^{2} y^{2}$

'

factor: $x, y \quad 2, x, x, y, y$

iii) $1.2 a b-2.4 b+3.6 a$

Term $: 1.2 a b \quad-2.4 b \quad 3.6 a$

farror : $1.2, a, b \quad-2.4, b \quad 3.6, a$

Question 6

i) $8 x+3 y^{2}$

(to be added)

ii)$y-y^{3}$

(to be added)

iii) $5 x y^{2}+7 x^{2} y$

(to be added)

iv) $-a b+2 b^{2}-3 a^{2}$

(to be added)

Question 7

i) $-7$

ii) $-2$

iii) 6

iv) $2 / 3$

Question 8

i) $-4 b$

ii) $5 y_{2}$

iii) $-1$

iv) $-3 x y$

Question 9

i) $-y^{2} z^{3}$

ii) $72^{3}$

(iii) $-7 y z^{2}$

iv) $-x y z$

Question 10

i) Non - constant term = -7x

Numerical coefficients = -7

ii) $1+2 x-3 x^{2}$

Non-constant term $2 x,-3 x^{2}$

Numerical coefficients $=2,-3$

iii) 1.2a + 0.8 b 

Non- constant term = 1.2a , 0.8b

Numerical coefficient = 1.2, 0.8

Question 11

(i) $13 y^{2}-8 x y$

$-8 x y$

coefficient of $x=-8 y$

ii) 7x - $x y^{2}$

$7 x,-x y^{2}$

Coefficient of x = 7 $-y^{2}$

(iii)

 $\begin{aligned} 5 &-7 x y z+4 x^{2} y \\ &-7 x y z, 4 x^{2} y \end{aligned}$

Coefficient of x = -7xyz , 4xy

Question 12

i)

 $\begin{aligned} 8-x y^{2} \\-& x y^{2} \end{aligned}$

Coefficient  of $y^{2}=-x$

ii)

 $\begin{aligned} 5 y^{2} &+7 x-3 x y^{2} \\ & 5 y^{2},-3 x y^{2} \end{aligned}$

Coefficient  of $y^{2}=5,-3 x$

iii) 

$\begin{aligned} 2 x^{2} y &-15 x y^{2}+7 y^{2} \\-& 15 x y^{2}, 7 y^{2} \end{aligned}$

Coefficient  of $y^{2}=-15 x^{2}, 7$

Question 13

i) 4y-72 binomial

ii) $-5 x y^{2}$ monomial

iii) $x+y-x y \quad$ trinomial

iv) $a b^{2}-5 b-3 a$ trinomial

v) $4 p^{2} q-5 p q^{2} \quad$ binomial

(vi) 2017 monomial

(vii) $1+x+x^{2}$ trinamial

(viii) $5 x^{2}-7+3 x+4$ trinomial

Question 14

i) $-7 x, 5 / 2^{x}$ Like term 

ii) $-29 x,-29 y$ unlike term 

iii) $2 \mathrm{xy}, 2 \mathrm{xy} 2$ unlike term 

iv) $4 m^{2} p, 4 m p^{2}$ unlike term

v) $\quad 12 x z, 12 x^{2} z^{2}$ unlike term

vi) $-59 q, 79 p$ like term

Question 15

i) $x^{2} y,-2 x^{2} y$

ii) $3 a^{2} b,-6 x^{2} b, 2 a b c$, 4abc

iii) $10 p q,-7 p q, 78 q p .$

$7P,2405 P .$

$8 q,-100 q$

$-p^{2} q^{2}, 12 q^{2} p^{2}$

$-23,41$

$-5 p^{2}, 701 p^{2}$

$13 p^{2} q, q p^{2}$

Question 16

ii, 8

ii, 1

iii, 0

iv, 2

Question 17

$\begin{array}{ll}\text { (i) } & 3 \\ \text {(ii)} & 4 \\ \text ({iii)} 5\end{array}$

Question 18

i) True 

ii) False 

iii) False 

iv) False 

v) True 

vi) False 

vii) False 

 Excercise =8.2

Question 1

i)

 $\begin{aligned} 7 x,-3 x & \\ 7 x-3 x &=4 x \end{aligned}$

ii) $6 x,-11 x$

$6 x-11 x=-5 x$

iii) $5 x^{2},-9 x^{2}$

$5 x^{2}-9 x^{2}=-4 x^{2}$

iv) $3 a b^{2},-5 a b^{2}$

$3 a b^{2}-5 a b^{2}=-4 a b^{2}$

v) $\frac{1}{2} p q,-1 / 3 p q$

$=\frac{1}{2} p q-\frac{1}{3} p q$

$=\frac{3 p q-2 p q}{3 \times 2}$

$=\frac{p q}{6}$

vi)$5 x^{3} y,-\frac{2}{3} x^{3} y$

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

$=\frac{15 x^{3} y-2 x^{3} y}{3}$

$=\frac{13 x^{3} y}{3}$

Question 2

i)

 $\begin{aligned} 3 x-5 x, 7 x & \\ 3 x-5 x+7 x=5 x \end{aligned}$

ii) 7xy, $2 x y$, $-8 x y$

=$27 x y+2 x y-8 x y$

=xy

iii) $-2 a b c, 3 a b c, a b c$

$-2 a b c+3 a b c+a b c=2 a b c$

iv) $3 m n,-5 m n, 8 m n,-4 m n$

$3 m n-5 m n+8 m n-4 m n=2 m n$

 

v) $2 x^{3}, 3 x^{3},-4 x^{3},-5 x^{3}$

$2 x^{3}+3 x^{3}-4 x^{3}-5 x^{3}$

$5 x^{3}-9 x^{3}=-4 x^{3}$

Question 3

i) $8 b-32$

ii) $8 m^{2}-11 m+10$

iii) $7 z^{3}+12 z^{2}-202$

iv) $8 x^{2} y+$ 8xy $^{2}-4 x^{2}-7 y^{2}$

v) p-q

vi) $\quad a+a b$

vii) $ \quad 4 y^{2}-3 y$

Question 4

i) $5 x y,-7 x y, 3 x^{2}$

=$5 x y-7 x y+3 x^{2}$

=$3 x^{2}-2 x y$

ii) $4 x^{2} y_{2}-3 x y^{2},-8 x y^{2}, 5 x^{2} y$

=$4 x^{2} y-3 x y^{2}-5 x y^{2}+5 x^{2} y$

=$9 x^{2} y-8 x y^{2}$

iii) $-7 m n+5,12 m n+12,8 m n-8,-2 m n-3$

=$-7 m n+5+18 m n+12+8 m n-8-2 m n-3$

=$11 m n-4$

iv) $a+b-3, \quad b-a+3, a-b+3$

=$a+b-3+b-a+3+a-b+3$

=$\quad a+b+3$

v)  $14 x+10 y-12 x y-13, \quad 18-7 x-10 y+8 x y, 4 x y$

=$14 x+10 y-12 x y-13+18-7 x-10 y+8 x y+4 x y$

=$7 x+5$

vi) $5 m-7 n, \quad 3 n-4 m+2,2 m-3 m n-5$

=$5 m-7 n+3 n-4 m+2+2 m-3 m n-5$

=$3 m-4 n-3 m n-3$

vii)$7 a^{2}-5 a+2, \quad 3 a^{2}-7, \quad 2 a+a, \quad 1+2 a-5 a^{2}$

=$7 a^{2}-5 a+2+3 a^{2}-7+2 a+9+1+2 a-5 a^{2}$

=$5 a^{2}-a+5$

Question 5

i) $2 x^{2}+3 y^{2}-5 x y+5 x^{2}-y^{2}+6 x y-3 x^{2}$

=$x^{2}(2+3)+y^{2}(3-1)+x y(-5+6)$

=$4 x^{2}+2 y^{2}+x y$

ii) $3 x y^{2}-5 x^{2} y+7 x y-8 y^{2} x-4 x y+6 x^{2} y$

=$x y^{2}(3-8)+x^{2} y(-5+6)+x y(7-4)$

=$-5 x y^{2}+6 x^{2} y+3 x y$

iii) $5 x^{4}-7 x^{2}+8 x-1+3 x^{3}-9 x^{2}+7-3 x^{4}+11 x-2+8 x^{2}$

=$x^{4}(5-3)+x^{2}(-7-9+8)+x(8+11)+3 x^{3}-1+7-2$

=$2 x^{4}-8 x^{2}+19 x+3 x^{3}+4$

=$2 x^{4}+3 x^{3}-8 x^{2}+19 x+4$

Question 6

i) $y^{2}-\left(-5 y^{2}\right)$

$y^{2}+5 y^{2}=6 y^{2}$

ii) $-2 x y-(-7 x y)$

$-2 x y+7 x y$

$5 x y$

iii) $b(5-a)-a(b-5)$

$5 b-a b-a b+5 a$

$5 a+5 b-2 a b$

iv) $4 m^{2}-3 m n+8-\left(-m^{2}+5 m n\right)$

$4 m^{2}-3 m n+8+m^{2}-5 m n$

$5 m^{2}-8 m n+8$

v)$3 a b-2 a^{2}-2 b^{2}-\left(5 a^{2}-7 a b+5 b^{2}\right)$

$3 a b-2 a^{2}-2 b^{2}-5 a^{2}+7 a b-5 b^{2}$

$10 a b-7 a^{2}-7 b^{2}$  

vi) $5 p^{2}+3 q^{2}-p q-\left(4 p q-5 q^{2}-3 p^{2}\right)$

$5 p^{2}+3 q^{2}-p q-4 p q+5 q^{2}+3 p^{2}$

$8 p^{2}+8 q^{2}-5 p q$

vii) $7 x^{2}-8 x y+3 y^{2}-5-\left(7 x y+5 x^{2}-7 y^{2}+3\right)$

$7 x^{2}-8 x y+3 y^{2}-5-7 x y-5 x^{2}+7 y^{2}-3$

$x^{2}(7-5)+x y(-8-7)+y^{2}(3+7)-8$

$2 x^{2}-15 x y+10 y^{2}-8$

viii) $\left(x^{4}-3 x^{3}-2 x^{2}+3\right)-\left(2 x^{4}-7 x^{2}+5 x+3\right)$

$x^{4}-3 x^{3}-2 x^{2}+3-2 x^{4}+7 x^{2}-5 x-3$

$x^{4}(1-2)-3 x^{3}+x^{2}(-2+7)-5 x$

$-x^{4}-3 x^{3}+5 x^{2}-5 x$

Question 7

Sum of ( 10p - r ) and 5P+ 2q is 

$=10 p- r+5 p+2 q$

=$15 p+2 q-r$

$P-2 q+r$ Subtract from $15 p+2 q-r$

=$(15 p+2 q-r)-(p-2 q+r)$

=$15 p+2 q-r-p+2 q-r$

=$p(15-1)+q(2+2)+r(-1-1)$

=$14 p+4 q-2 r$

Question 8

Sum of $4+3 x$ and $5-4 x+2 x^{2}$

$2 x^{2}-4 x+5+4+3 x$

$2 x^{2}-x+9$

Sum of $3 x^{2}-5 x$ and $-x^{2}+2 x+5$

=$3 x^{2}-5 x+\left(-x^{2}+2 x+5\right)$

=$2 x^{2}-3 x+5$

$2 x^{2}-x+9$ is substracted from $2 x^{2}-3 x+5$

=$2 x^{2}-3 x+5-\left(2 x^{2}-x+9\right)$

=$2 x^{2}-3 x+5-2 x^{2}+x-9$

=$2x-4$

Question 9

Sum is $x^{2}+y^{2}+5 x y$

Subtract $x^{2}-y^{2}+2 x y$ from $x^{2}+y^{2}+5 x y$

$x^{2}+y^{2}+5 x y$

$x^{2}-y^{2}+2 x y$

$-$  $+$   $-$

---------------------------------

          $2 y^{2}+3 x y$

Question 10

$-7 m n+2 m^{2}+3 n^{2}$

$2 m n+m^{2}+n^{2}$

$-$    $-$          $-$

------------------------------------

$-9 m n+m^{2}+2 n^{2}$

Question 11

The required

$y^{4}-12 y^{2}+y+14-\left(17 y^{3}+34 y^{2}-51 y+68\right)$

$y^{4}-17 y^{3}+y^{2}(-12-34)+y(1+51)+14-68$

$y^{4}-17 y^{3}-46 y^{2}+52 y-54$

Question 12

The required

$93 p^{2}-55 p+4-\left(13 p^{3}-5 p^{2}+17 p-90\right)$

$93 p^{2}-55 p+4-13 p^{3}+5 p^{2}-17 p+90$

$-13 p^{3}+98 p^{2}-72 p+94$

Question 13

The required expressions

$3 x^{2}-4 y^{2}+5 x y+20$

$-x^{2}-y^{2}+6 x y+20$

$+$        $+$       $-$       $-$

-------------------------------------

$4 x^{2}-2 y^{2}-x y$

Question 14

Sum of $2 y^{2}+3 y z,-y^{2}-y z-z^{2}, \quad y z+2 z^{2}$ is

$2 y^{2}+3 y z-y^{2}-y z-z^{2}+y z+2 z^{2}$

$y^{2}(2-1) 7 y z(3-1+1)+z^{2}(-1+2)$

$y^{2}+3 y z+z^{2}$

 Excercise =8.3

Question 1

i. $3 m-5$

Given $m=2$

$\quad 3(2)-5=6-5=1$

ii) $9-5 m$

$m=2$

$9-5(2)=9-10$

$=-1$

iii, $3 m^{2}-2 m-7$

$m =2$

$3(2)^{2}-2(2)-7$

$3 \times 4-2 \times 2-7$

$12-4-7$

= 1 Ans

iv) $\frac{5}{2} m-4$

$m=2$

=$\frac{5}{2} \times 2-4$

$5-4$ = 1

Question 2

i)

 $\begin{aligned} 4 p+7 & \\ p &=-2 \\ & 4(-2)+7=-8+7=-1 \end{aligned}$

ii) $-38^{2}+4 p+7$

$-3(-2)^{2}+4(-2)+7$

$-3 \times 4-8+7$

$-12-8+7$

$-13$

iii) $-2 p^{3}-3 p^{2}+4 p+7$

=$-2 x(-2)^{3}-3(-9)^{2}+4(-2)+7$

=$-2 x-8-3 \times  4+4 x-2+7$

=$16-12-8+7$

=3

Question 3

(i) $a^{2}+b^{2}$

$a=2, b=2$

$(2)^{2}+(2)^{2}$

$4+4=8$

ii) $a^{2}+a b+b^{2}$

$a=2 \quad b=2$

$(2)^{2}+2 x a+(2)^{2}$

=$4+4+4$

$=12$

iii) $a^{2}-b^{2}$

=$(2)^{2}-(2)^{2}$

=$4-4$

=0

Question 4

i) $2 a^{2}+b^{2}+1$

$a=0 \quad b=-1$

=$2 \times(0)^{2}+(-1)^{2}+1$

=$0+1+1$

=2

ii) $a^{2}+a b+2$

=$(0)^{2}+0 \times -1+2$

=0

iii) $2 a^{2} b+2 a b^{2}+a b$

=$2(0)^{2}(-1)+2(0)(-1)^{2}+0(-1)$

= 0

Question 5

Given $p=-10$

The value of $p^{2}-2 p-100$

=$(-10)^{2}-2(-10)-100$

=$100+20-100$

=20

Question 6

Given = 10

The value of $z^{3}-3 z+30$

=$(10)^{3}-3(10)+30$

=$1000-30+30$

=1000

Question 7

Given x = 2

The value of $x+7+4(x-5)$ is

$x+7+4 x-20$

$5 x-13$

$5 \times 2-13$

$10-13=-3$

ii) Given $x=2$

The value of $3(x+2)+5 x-7$

=$3 x+6+5 x-7$

=$8 x-1$

=$8(+2)-1$

=$16-1$

=15

iii) Given x = 2

The value of $6 x+5(x-2)$

=$6 x+5 x-10$

=$11 x-10$

=$11 \times 2-10$

=$22-10$

=12

iv) Given x = 2

The value of $4(2 x-1)+3 x+11$

=$8 x-4+3 x+11$

=$11 x+7$

=$11 \times 2+7$

=$22+7$

= 29

Question 8

(i) Given a=-1, b=-2

=$2 a-2 b-4-5+a$

=$2(-1)-2(-2)-4-5+(-1)$

=$-2+4-4-5-1$

=$-8$

ii) Given $a=-1, \quad b=-2$

The value of $2\left(a^{2}+a b\right)+3-a b$

=$2 a^{2}+2 a b+3-a b$

=$2 a^{2}+a b+3$

=$2(-1)^{2}+(-1)(-2)+3$

=$2+2+3$

=7

 Excercise =8.4

Question 1

i) $2 n+1$

Number of shapes

1

2

3

No of Line segments

3

5

7

n shape of letter are formed then algrabraic equation is 2n + 1

ii) Number of shapes

       1

       2

       3

No, of line segments 

      5

      8

     11

Algrebraic equation is 3n + 2

Question 2

i) Number of shapes

       1

       2

       3

   $\vdots$

No, of line segments

     4

     7

    10

   $\vdots$

Alegebraic expression is 3nm

ii) Number of shapes 

     1

     2

     3

No. of line segments

    6

    11

     16

Algebraic expression is 5n + 1

iii) Number of shapes

         1

         2

         3

No. of line segments 

         7

         12

          17

Algebraic expression is 5n+ 2

Question 3

i) $2 n+1$

$n=5 \quad 2(5)+1=11$

$n=10 \quad 2(10)+1=21$

$n=100 \quad 2(100)+1=201$

ii) $3 n+1$

$n=5 \quad 3(5)+1=16$

$n=10 \quad 3(10)+1=31$

$n=100 \quad 3(100)+1=301$

iii) $3 n+2$

$n=5 \quad 3(5)+2=17$

$n=10 \quad 3(10)+2=32$

$n=100 \quad 3(100)+2=302$

iv) $n=5(5)+1=26$

$n=10 \quad 5(10)+1=51$

$n=100 \quad 5(100)+1=501$

v) $5 n+2$

$n=5 \quad 5(5)+2=27$

$n=10 \quad 5(10)+2=52$

$n=100 \quad 5(100)+2=502$

vi) $4 n+3$

$n=5 \quad 4(5)+3=23$

$n=10 \quad 4(10)+3=43$

$n=100 \quad 4(100)+3=403$

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ML AGGARWAL CLASS 7 Chapter 7 Percentage and Its Applications EXERCISE:7.1

 EXERCISE:7.1

Question 1

i) $25 \%=\frac{25}{100}=\frac{1}{4}$

ii) $150 \%=\frac{150}{100}=\frac{3}{2}$.

iii) $7 \frac{1}{2} \%=\frac{15}{2} \%=\frac{15 / 2}{100}=\frac{15}{2 \times 100}=\frac{3}{40}$

iv) $33 \frac{1}{3} \%=\frac{100}{3} \%=\frac{100 / 3}{100}=\frac{100}{3 \times 100}=\frac{1}{3} .$

Question 2

i) $\frac{1}{8}=\left(\frac{1}{8} \times 100\right) \%=12.5 \%$

ii) $\frac{5}{4}=\left(\frac{5}{4} \times 100\right) \%=125 \%$

iii) $\frac{9}{16}=\left(\frac{9}{16} \times 100\right) \%=56 \frac{1}{4} \%$

iv) $\frac{3}{7}=\left(\frac{3}{2} \times 100\right) \%=42 \frac{6}{7} \%$

v) $\frac{11}{15}=\left(\frac{11}{15} \times 100\right) \%=73 \frac{1}{3} \%$

vi) $\frac{13}{8}=\frac{11}{8}=\left(\frac{11}{8} \times 100\right) \%=137 \frac{1}{2} \%$

Question 3

i) Given 6 students out of 40 students in a class are absent 

percentage of students are absent $=\frac{6}{40} \times 100 \%$

=15%

ii) Given, Antony secured 384 marks out of 500 maiks.

$\therefore$ percentage of marks secured $=\left(\frac{384}{500} \times 100\right) \%$

$=76.8 \%$

iii) Given , A shop has 500 shirts 

Out of 500 shirts , 15 are defective 

Now, percentage of shirts are defective = $\left(\frac{15}{500} \times 100\right) \%$

=3%

iv) Given,

Vani has 20 gold bangels 

Also she has 10 silver bangles 

Now, Total number of bangles = 20+ 10 = 30 bangles 

∴ percentage of gold bangles = ($\left.\frac{20}{30} \times 100\right) \%$

=66.67 %

$\begin{aligned} \text { percentage of Silver bangles } &=\left(\frac{10}{30} \times 100\right) \% \\ &=33.34 \% \end{aligned}$

v) Total number of voters = 120

Out of 120 , 90 of them voted 

Then out of 120, votes did not vote = 120-90=30

ஃPercentage of voters did not vote = $\frac{30}{120} \times 100 \%$

=$25 \%$

Question 4

i) Shaded part $=\frac{3}{4}$

percentage of Shaded part $=\left(\frac{3}{4} \times 100\right) \%$

$=75 \%$

ii) Shaded part $=\frac{2}{6}=\frac{1}{3}$

percentage of Shaded part $=\left(\frac{1}{3} \times 100\right) \%$

$=33.34 \%$

iii) Shaded part $=\frac{5}{8}$

percentage of shaded part $=\left(\frac{5}{8} \times 100\right) \%=62.5 \%$

Question 5

i) $14 \%=\frac{14}{100}=\frac{7}{50} .$

ii) $1 \frac{3}{4} \%=\frac{7}{4} \%=\frac{7}{4 \times 100}=\frac{7}{400} .$

iii) $33 \frac{1}{3} \%=\frac{100}{3} \%=\frac{100}{3 \times 100}=\frac{1}{3} .$

iv) $37.5 \%=\frac{37.5}{100}=\frac{375}{1000}=\frac{3}{8} .$

Question 6

i) $\frac{5}{4}=\left(\frac{5}{4} \times 100\right) \%=125 \%$

ii) $\frac{1}{1}=\left(\frac{1}{1} \times 100\right) \%=100 \%$

iii) $\frac{2}{3}=\left(\frac{2}{3} \times 100\right) \%=66.67 \%$

iv) $\frac{9}{16}=\left(\frac{9}{16} \times 100\right) \%=56.25 \%$

Question 7

Given alloy consists of 7 parts of zinc and 33 parts of copper 

Total alloy contains = 33+ 7 = 40 

ஃ percentage of copper in alloy = $\left(\frac{33}{40} \times 100\right) \%$

=$82.5 \%$

Question 8

Given , Calcium, carbon and sand in the ratio 12:3:10

Sum of ratio = 12+ 3+ 10 = 25

Percentage of carbon in the chalk = ($\frac{3}{25} \times 100$) % =12%

Question 9

Given Total money = Rs 2500

It is divided among Ravi, Raju and Roy

Out of total money , ravi gets two parts 

Raju gets three parts 

Roy gets five parts 

Total no. of parts = 2+ 3+ 5= 10

Ravi get money = Rs $\frac{2}{10} \times 2500$

=Rs 500

$\begin{aligned} \text { Raju gets money } &=3 \frac{3}{10} \times 2500 \\ &= 750 rs. \\ \text { Roy gets money } &= \frac{5}{10} \times 2500 \\ &= 1250 rs . \end{aligned}$

Percentage of Ravi get money $\begin{aligned} &=\left(\frac{500}{2500} \times 100\right) \% \\ &=20 \% \end{aligned}$

$\begin{aligned} \text { percentage of Raju gets money } &=\left(\frac{750}{2500} \times 100\right) \% \\ &=30 \% \end{aligned}$

$\begin{aligned} \text { Percentage of Roy geT money } &=\left(\frac{1250}{2500} \times 100\right) \% \\ &=50 \% \end{aligned}$

Question 10

i) $28 \%=\frac{28}{100}=0.28 .$

ii) $3 \%=\frac{3}{100}=0.03$

iii) $0.44 \%=\frac{0.44}{100}=0.0044$

iv) $37 \frac{1}{2} \%=\frac{75}{2} \%=\frac{75}{2 \times 100}=0.335$

Question 11

i) $0.65=\frac{65}{100}=\left(\frac{65}{100} \times 100\right) \%=65 \%$ 

ii) $0.90=\frac{90}{100}=\left(\frac{90}{100} \times 100\right) \%=90 \%$

iii) $2.1=\frac{21}{10}=\left(\frac{21}{10} \times 100\right) \%=210 \%$

iv) $0.02=\frac{2}{100}=\left(\frac{2}{100} \times 100\right) \%=2 \%$

Question 12

i) Given percentage of students in a class are girls = 42% 

Actual percentage of students in a class will be 100%

Percentage of students in a class are boys = (100-42)% =58%

ii) A basket have full of apples , oranges and mangoes 

Percentage of apples = 50 %

Percentage of oranges = 30%

Percentage of total oranges , apples and mangoes = 100 % 

ஃNow percentage of mangoes = [$100-(50+30)$]%

$=[100-80] %$

$=20 %$

 EXERCISE:7.2

Question 1

i) $15 %$ of $250=\frac{15}{100} \times 250=\frac{3}{20} \times 250=375$

ii) $25\% $ of 120 litres $=\frac{25}{100} \times 120=\frac{1}{4} \times 120=30$.

iii) $1 \%$ of 1 hour $=\frac{1}{100} \times 3600 \mathrm{sec}=36$ seconds.

iv) $75 \%$ of $\mathrm{Kg}=\frac{75}{100} \times 1000 \mathrm{~g}=\frac{3}{4} \times 1000 \mathrm{gr} \mathrm{ms}=750 \mathrm{~g}$

V) $120 \%$ of $\ 250=\frac{120}{100} \times 250$=Rs 300

vi) $0.6 \%$ of $2 \mathrm{Km}=\frac{0.6}{100} \times 2000 \mathrm{~m}=12 \mathrm{~m}$

Question 2

Given , 8% children of a class like getting wet = 25 

Now , children like getting wet=$\frac{8}{100} \times 25$

=$\frac{2}{25} \times 25$

=2

Question 3

Given, 

Out of 20 in the fridge, Vasundara ate = 3 ice creams 

Percentage of icecreams , she ate = $\frac{3}{20} \times 100 \%$

=15 %

Question 4

i) Required percentage $=\left(\frac{20}{50} \times 100\right) \%=\frac{200}{5} \%=40 \%$

ii) Required percentage $=\left(\frac{60}{40} \times 100\right) \%=\frac{300}{2} \%=150 \%$

iii) Required percentage $=\left(\frac{90cm}{4 \cdot 5 m} \times 100\right) \%=\left(\frac{90}{4.5 \times 100} \times 100\right) \%$

$=\left(\frac{90}{450} \times 100\right) \%$

$=\frac{100}{5} \%$

$=20 \%$

iv) $5.6 \mathrm{~kg}=5.6 \times 1000 \mathrm{~g}=5600 \mathrm{~g}$

$\begin{aligned} \text { Required Percentage }=\left(\frac{3509}{56008} \times 100\right) \% &=\frac{350}{56} \% \\ &=6.25 \% \end{aligned}$

Question 5

i) 12 of $80=\left(\frac{12}{80} \times 100\right) \%=\frac{120}{8} \%=15 \%$

ii) $\quad$ 4 rupees $=4 \times 100$ paide $=400$ paise

25 paise of 400 paise = $\left(\frac{25}{400} \times 100\right) \%=\frac{25}{4} \%=6.25 \%$

iii) $2 \mathrm{~kg}=2 \times 1000 \mathrm{~g}=2000 \mathrm{~g}$

$300 \mathrm{~g}$ of $200 \mathrm{~g}=\left(\frac{300}{2000} \times 100\right) \%=\frac{30}{2} \%=15 \%$

Question 6

Percentage increase $=\left(\frac{\text { intrease in value }}{\text { Original value }} \times 100\right) \%$

A school team won 4 games last years, and this year the team won 6 games. 

Increase in the games won = 6-4=2

ஃ percentage increase = $\begin{aligned} &( \left.\frac{2}{4} \times 100\right) \% \\ & \frac{100}{2} \\=& 50 \% \end{aligned}$

Question 7

Original price = Rs 80

Decrease in price = Rs 80 - Rs 60

 

= Rs 20

Percentage Decrease $=\left[\frac{D ecr e a s e \text { in value }}{\text { original value }} \times 100\right] \%$

$=\left[\frac{20}{80} \times 100\right] \%$

$=\frac{100}{4} \%$

$=25 \%$

Question 8

 In Childhood , petrol price was = Rs 1 per litre 

Now the price of petrol was = Rs 65 per litre

Increase in the Value of price = Rs 65 - Rs 1

= Rs 64

ஃ Percentage increase = $\left(\frac{64}{1} \times 100\right) \%$

$6400 \%$

Question 9

Last years, the cost of basmati rice= Rs 40 kg 

Also , percentage increase = 20 %

∴This price , this year will be increased by 

= $\frac{20}{100} \times 40$

=8 a $\mathrm{kg}$

∴ $\begin{aligned} \therefore \text { The price of Bamati rice, tis year } &=40+8 \\ &= ₹ 48 \text { kg. } \end{aligned}$

Question 10

Number of students took exam = 300

Percentage failed = 28%

Number of students failed $=\frac{28}{100} \times 300$

=84 

∴ Now , the number of students passed= 300- 84

=216

Question 11

In a constituency, number of voters = 15,000

Percentage of voters who voted = 60% 

∴ Number of votes who voted = $\frac{60}{100} \times 15000$

=9000

Question 12

Length of a flag pole painted green = 20%

Painted yellow = 455

Remaining painted red = 100-(20+ 45)

=100-65

= 35%

Total length of pole = 18m

Length of pole painted red =$\frac{35}{100} \times 18m$

=6.3m

Question 13

A chalk contains , calcium = 10% 

Carbon = 3% 

Oxygen = 12%

and the remaining is sand = 100- (10+3+12)

= 100-25

= 75%

Amount of carbon in $2 \frac{1}{2}$ kg chalk $=\frac{3}{100} \times \frac{5}{2} \times 1000 \mathrm{~g}$ = 75g

Amount of Calcium in $2 \frac{1}{2} \mathrm{~kg} \ chalk \mathrm{}=\frac{10}{100} \times \frac{5}{2} \times 1000 \mathrm{~g}$ = 250g

$\begin{aligned} \text { Amount of } \text { Sand } &=\frac{75}{100} \times \frac{5}{2} \mathrm{~kg} \\ &=1.875 \mathrm{~kg} \end{aligned}$

Question 14

i) $25 \%$ of $x$ is $9 \Rightarrow \frac{25}{100} \times x=9$

$\frac{x}{4}=9$

$x=4 \times 9$

$x=36$

ii) $75 \%$ of $x$ is $15 \Rightarrow \frac{75}{100} \times x=15$

$\frac{3 x}{4}=15$

$x=\frac{15 \times 4}{3}$

$x=20$

iii) $12 \%$ of it is Rs 1080

$\begin{aligned} \Rightarrow \frac{12}{100} \times x &=1080 \\ x &=\frac{1080 \times 100}{12} \\ x &=9000 . \end{aligned}$

iv) $8 \%$ of it is 40 litr $\Rightarrow \frac{8}{100} \times x=40$

$x=\frac{40 \times 100}{8}$

$x=500$

Question 15

Mohini Saved salary =Rs 400

percentage Saved $=10 \%$ of total salary

i.e $\frac{10}{100} \times x=400$

$x=\frac{400 \times 100}{10}$

$x=4000$

$\therefore$ Salary =Rs 4000

Question 16

Number of good apples in basket = 42

percentage of the apples in a basket go bad = 16%

Remaining , percentage of apples go good = 100- 16= 84%

Let Total no of apples be x

i.e 84% of x = 42

 $\frac{84}{100} \times x=42$

$x=\frac{42 \times 100}{84}$

$x=50$

∴ Total number of apples = 50

Question 17

Varun got secured marks = 251 marks 

and got failed by 19 marks 

If he gets passed, then he will get = 251+19= 270 marks

percentage of marks to get pass = 45%

Let maximum marks be 'x'

i.e 45% of x = 270

$\begin{aligned} \frac{45}{100} \times x &=270 \\ x &=\frac{270 \times 100}{45} \\ x &=600 \end{aligned}$

∴ Maximum marks $=600$

Question 18

In a rainy day , percentage of students 

present in a school = 94% 

Then percentage of students absent = 100- 94% 

= 6%

Also given , number of students absent on that 

day = 174 

Let total strength of school be x 

i,e 

$\begin{aligned} 6 \% \text { of } x &=134 . \\ \frac{6}{100} \times x &=174 \\ x &=\frac{174 \times 100}{6} \\ x &=2900 . \end{aligned}$

Total strength of school = 2900

Question 19

Percentage of population in a town are men = 40%

Those are women = 39%

Then percentage of population are children = 100-(39+40)

= 100-79

 Number of children - 12,600

Let the total population be 'x '

i.e 21% of x = 12,600

$\begin{aligned} \frac{21}{100} \times x &=12,600 \\ x &=\frac{12,600 \times 100}{21} \\ x&=60,000 \end{aligned}$

∴ Now the number of men = 40% of total 

$=\frac{40}{100} \times 60,000$

$=24,000$

Question 20

Price of watch is increased by 15% 

Increase in price is Rs 90

percentage increase = $\frac{\text { Increase in value }}{\text { original value }} \times 100$

∴ i.e $15=\frac{90}{\text { original value }} \times 100$

$\begin{aligned} \therefore \text { original price } &=\frac{90 \times 100}{15} \\ \text { original price } &=2600 \end{aligned}$

Question 21

i)Let the original number be x 

Increase in the number = 30% of x $=\frac{30}{100} \times x=\frac{3 x}{10}$

$\therefore$ New number $=x+\frac{3 x}{10}$

According to given Condition, $x+\frac{3 x}{10}=39$

$\begin{aligned} 10 x+3 x=39 \times 10 \Rightarrow 13 x &=390 \\ x &=\frac{390}{13}=30 \end{aligned}$

Hence , the original number is 30

ii) Let the original number be x 

Decrease in number = 8% of x = $\frac{8}{100} \times x=\frac{2 x}{25}$

∴New number = $x-\frac{2 x}{25}$

Accoding to given information, $x-\frac{2 x}{25}=506$.

$25 x-2 x=506 \times 25$

$23 x=506 \times 25$

$x=\frac{506 \times 25}{23}$

$x=550$

Hence, the original number is 550

Question 22

Percentage reduced = 7%

Let the original number be x 

Decreased in number = 7% of x =$\frac{7}{100} \times x=\frac{7 x}{100}$

∴ New number = $x-\frac{7 x}{100}=\frac{93 x}{100}$.

According to given , $\frac{93 x}{100}=165$

$x=\frac{465 \times 100}{93}$

$x=500$

∴ Original price = Rs 500

 EXERCISE:7.3

Question 1

Cost price =Rs 60, selling price =₹ 874

Profit = Selling price - cost price 

= 8 74- 760

= Rs 114

profit percentage = $\left(\frac{p r of i t}{c \cdot p} \times 100\right) \%$

$=\left(\frac{114}{760} \times 100\right) \%$

$=15 \%$

Question 2

Cost price =Rs 2500 ; selling price = Rs 2300

Loss $=$ cost price - selling price

= Rs 2500- Rs 2300

= Rs 200

Loss percent = ($\frac{\text { Loss }}{\text { C.P }} \times 100$)

$\begin{aligned}=&\left(\frac{200}{2500} \times 100\right) \cdot \% \\=8 \% . \end{aligned}$

Question 3

i) Cost price = Rs 250 ; Selling price = Rs 325

As S.p >C.P , Profit = S.P - C.p 

=325-250

=Rs 75 

profit percent $=\left(\frac{\text { profit }}{c \cdot p} \times 100\right) \%$

$=\left(\frac{75}{250} \times 100\right) \% .$

$=30 \%$

ii) Cost price = Rs 250 , Selling price = Rs 150

As C.P >S.P , Loss = C.P -S.P

=250- 150

= Rs 100

Loss percent $\left.=\frac{\text { Loss }}{C \cdot \rho} \times 100\right) \%$.

$=\left(\frac{100}{250} \times 100\right) \%$

$=40 \%$

Question 4

1st offer :

Cost price = Rs 4800

Profit = $13 \frac{1}{3} \%$ of Cost price = $\frac{40}{3} \times \frac{1}{100} \times 4800$

=640

Selling price = 480+ 640 i.e Cost price +profit = Rs 25440

2nd offer: 

Cost price = Rs 3640

Loss = 15 % of cost price = $=\frac{15}{100} \times 3640$

=Rs 546

$\begin{aligned} \text { Selling price } &=\text { cost price - Loss } \\ &=3640-546 \\ &= 3094 rs. \end{aligned}$

Selling price of 1st and 2nd offer = 5440+ 3094

= Rs 8440

As S.P > C.P , He always get gain 

i.e Gain = S.P - C.P

= 8534- 8440

= Rs 94

Question 5

cost price of 24 Tables $=24 \times 450$

= Rs 10,800

$\begin{aligned} \text { Selling price of } 16 \text { of them } &=16 \times 600 \\ &=9600 \end{aligned}$

Remaining i.e 24-16 =8 were sold 

i.e now s.p of 8 tables = 8 $\times $ 400

= 3200

∴ Total selling price = 9600+ 3200

= 1,2800

As S.P > C.P there is always a gain 

Gain = S.p - C.P 

12,800- 10,800

= Rs 2000

Question 6

Selling price = Rs 810 ;$ profit =Rs 60 

$\begin{aligned} \text { As } \text { profit } &=S \cdot p-c \cdot p \\ c \cdot p &=s \cdot p-\text { profit } \\ &=810-60 \end{aligned}$

Cost price =Rs 750

$\begin{aligned} \text { profit percent } &=\left(\frac{\text { profit }}{c \cdot p} \times 100\right) \% \\ &=\left(\frac{60}{750} \times 100\right) \% \\ &=8 \% \end{aligned}$

Question 7

Selling price -= Rs 3906; Loss = Rs 294

$\begin{aligned} \text { Loss }=&C \cdot p-s \cdot p\\ c \cdot p &=\text { Loss }+S \cdot p \\ &=294+3906 \\ &=4,200 . \end{aligned}$

Loss percent = $\left(\frac{\operatorname{loss} }{C.p} \times 100\right) \%$

$=\left(\frac{294}{4,200} \times 100\right) \%$

$=7 \%$

Question 8

C .P=Rs120, Loss percent =10%

Loss pescent $=\frac{\text { Loss }}{\text { c.p }} \times 100$

$\begin{aligned} \text { Loss } &=\frac{\text { Loss percent } c.p}{100} \\ &=\frac{10 \times 120}{100}=₹12\end{aligned}$

$\begin{aligned} \text { Loss }=& C \cdot p-s \cdot p \\ S \cdot p=&c\cdot p-\text { Loss }\\ &=120-12 \\ &=\ ₹108 \end{aligned}$

Question 9

cost price $=₹ 10,000 ;$ profit $=20 \%$

profit $\%=\frac{\text { Profit }}{C \cdot p} \times 100$

$20=\frac{\text { profit }}{10,000} \times 100$

profit $=\frac{20 \times 10,000}{10}$

profit $=20,00$

As profit = S.P - C.P

S.P =Profit + C.P 

= 2000+ 10,000

S.P = Rs 12,000

∴ Selling price = Rs 12,000

Question 10

selling price =Rs 300; profit $=20 \%$

profit percentage $=\left(\frac{\text { profit }}{\text { c.p }} \times 100\right) \%$.

$=\left(\frac{S \cdot p-C\cdot p}{C \cdot p} \times 100\right) \%$

$=\left[\frac{s \cdot p}{c \cdot p}-1\right] \times 100$

$20=\left(\frac{300}{c \cdot p}-1\right) \times 100$

$\frac{300}{c \cdot p}-1=\frac{20}{100}$

$\cdot \frac{300}{c \cdot p}-1=\frac{1}{5}$

$\frac{300}{C \cdot p}=1+\frac{1}{5}=\frac{6}{5}$

$\frac{300}{C \cdot p}=\frac{6}{5}$

$C \cdot P=\frac{300 \times 5}{6}$

$c \cdot$ P=Rs 250

$\therefore$ cost price =Rs 250 

Question 11

Selling price = Rs 320 ; Loss percent = 20 %

Loss percent = $\frac{\text { Loss }}{\text { c.p }} \times 100$

$=\frac{C p-s . p}{c \cdot p} \times 100$

$20=\left(1-\frac{320}{c \cdot p}\right) \times 100$

$1-\frac{320}{c \cdot p}=\frac{20}{100}$

$1-\frac{320}{c \cdot p}=\frac{1}{5}$

$\frac{320}{c \cdot p}=1 \frac{1}{5}$

$\frac{320}{6 \cdot p}=\frac{5-1}{5}$

$\frac{320}{c \cdot p}=\frac{4}{5}$

$c \cdot p=\frac{320 \times 5}{4}$

$c \cdot$ p=Rs 400

∴ Cost price = Rs 400

Question 12

selling price = Rs 522 ; profite 16%

profit $\%=\left(\frac{\text { profit }}{c \cdot p} \times 100\right)$

$=\frac{S \cdot p-c \cdot p}{C.p} \times 100$

$=\left(\frac{s \cdot p}{c \cdot p}-1\right) \times 100$

$16=\left(\frac{522}{c . p}-1\right) \times 100$

$\frac{522}{c.p}-1=\frac{16}{10}$

$\frac{522}{c \cdot p}-1=\frac{4}{25}$

$\frac{522}{c \cdot p}=1+\frac{4}{25}$

$\frac{522}{c \cdot p}=\frac{29}{25}$

$C \cdot p=\frac{522 \times 25}{29}$

$C \cdot$ p=Rs 750

$\therefore$ cost price =Rs 70

Question 13

selling price $=57360$; Loss $\%=8 \%$

$\begin{aligned} \text { Loss pescent } &=\left(\frac{\text { Loss }}{\text { c.p }} \times 100\right) \% \\ & \left.=\frac{C \cdot p-s \cdot p}{C \cdot p} \times 100\right) \end{aligned}$

$=\left(1-\frac{s \cdot p}{c \cdot p}\right) \times 100$

$8=\left(1-\frac{\partial 360}{c \cdot p}\right) \times 1$

$\frac{8}{100}=1-\frac{7360}{c. p}$

$\frac{2}{25}=1=\frac{7360}{c .p}$

$\frac{7360}{c p}=1-\frac{2}{25}$

$\frac{7360}{c \cdot p}=\frac{23}{25}$

$c.{p}=\frac{7360 \times 25}{23}$

$c \cdot$ p=Rs 8,000

cost price =Rs 8000

Question 14

Selling price= Rs 3168 ; Loss = $12 \%$

Loss percentage $=\frac{\text { Loss }}{C \cdot p} \times 100$

$=\left[1-\frac{S \cdot p}{C \cdot p}\right] \times 100$

$12=\left[1-\frac{3168}{c.p}\right] \times 100$

$1-\frac{3168}{6.8}=\frac{12}{100}$

$1-\frac{3168}{c\cdot p}=\frac{3}{25}$

$\frac{3168}{6 . p}=1-\frac{3}{25}$

$\frac{3168}{c. p}=\frac{22}{25}$

$c \cdot p=\frac{3168 \times 25}{23}$

C.p =Rs 3600

Given selling price = 3870

As S.P > C.P he gains 

So gain = S.P - C.P = 3870-3600

= 270

Gain percentage = ($\frac{\text { Gain }}{\text { c.p }} \times 100$)%

$=\left(\frac{270}{3600} \times 100\right) \%$

=75 %

Question 15

selling price =Rs4550, Loss =9 %

Loss percent $=\left[1-\frac{S \cdot p}{C.p}\right] \times 100$

$9=\left[1-\frac{4550}{c \cdot p}\right] \times 100$

$1-\frac{4550}{c \cdot p}=\frac{9}{100}$

$\frac{4550}{c \cdot p}=1-\frac{9}{100}$

$\frac{4550}{c \cdot p}=\frac{91}{100}$

$c \cdot p=\frac{4550 \times 100}{91}$

$C \cdot$ p=Rs 5000

As given selling price = 4825

As C.P > S.P, so he lose 

Loss = C.P - S.P

= 5000- 4825

Loss = Rs 175

Loss percent =  ($\frac{\text { Loss }}{c. p} \times 100$)

$=\left(\frac{175}{5000} \times 100\right)$

$=3.5 \%$

 EXERCISE:7.4

Question 1

Simple Interest $=\frac{\text { principal } \times \text { Rate } \times \text { Time }}{100}$

i.e $I=\frac{P \times R \times T}{100}$.

i) $\quad p=350 \quad ; \quad R=11 \% \quad T=2$ years

$I=\frac{350 \times 11 \times 2}{100}$

I=Rs 77

Total amount $=S \cdot I+P$

$=77+350$

=Rs 427

ii) $p=20,000 \quad T=4 \frac{1}{2}=\frac{9}{2}$ years $; R=8 \frac{1}{2}=\frac{17}{2} \%$

$\begin{aligned} I=& \frac{20,00 \times \frac{17}{2} \times \frac{9}{2}}{160} \\ &=\frac{20000 \times 17 \times 9}{4 \times 100} \\ &=₹ 7650 \end{aligned}$

∴ Amount = principal + I

=20,000+ 7,650

= 27,650

iii)  $p=₹ 648 ; R=16 \frac{2}{3}=\frac{50}{3} ; T=8$ months = $\frac{8}{12}$ years

$I=\frac{648 \times \frac{50}{3} \times \frac{8}{12}}{100}$

$I=\frac{648 \times 50 \times 8}{36 \times 100}$

I=Rs 73

$\begin{aligned} \text { Amount } &=S \cdot 1+P \\ &=73+648 \\ &=₹ 721 \end{aligned}$

Question 2

i) $S \cdot I=200, \quad p=2,500, \quad R=4 \%$

$I=\frac{P \times R \times I}{100}$

Time, $T=\frac{100 \times \underline{1}}{P \times R} .$

$T=\frac{100 \times 200}{2,500 \times 4}$

$T=2$ years

ii) $S \cdot I=2730, \quad P=12,000, R=6 \frac{1}{2}=\frac{13}{2}$

$\begin{aligned} T=& \frac{100 \times I}{P \times R} \\=& \frac{100 \times 2730}{12,000 \times 13 / 2} \\ &=\frac{100 \times 2730 \times 2}{12000 \times 13} \end{aligned}$

$T=\frac{7}{2}$ years $=3 \frac{1}{2}$ years

Question 3

i) $P=1560, \quad I=585, \quad T=3$ yeass

$I=\frac{p \times R \times T}{100}$

Rate of interest R= $\frac{100 \times I}{P \times T}$

$R=\frac{100 \times 585}{1560 \times 3}=(1.25 \times 100) \%$

$R=\frac{25}{2} \%=12 \frac{1}{2} \%$

ii) $I=325, \quad p=1625, \quad T=2 \frac{1}{2}=\frac{5}{2}$ years

$\begin{aligned} R &=\frac{100 \times I}{P \times J} \\ &=\frac{100 \times 325}{1625 \times 5 / 2} \\ &=\frac{100 \times 325 \times 2}{1625 \times 5} \\ &=8 \% \end{aligned}$

Question 4

i) $R=16 \% ; T=2 \frac{1}{2}$ years $=\frac{5}{2}$ years, $I=3840$

$I=\frac{P R I}{100}$

$P=\frac{100 \times I}{R \times T}$

$P=\frac{100 \times 3840}{16 \times 5 / 2}$

$P=\frac{100 \times 3840 \times 2}{16 \times 5}$

P=Rs 9600

∴ Principal = Rs 9600

ii) $R=7 \frac{1}{2}=\frac{15}{2} \% \quad ; \quad T=2$ years 4 months $\quad I=2730$

$=\left(2+\frac{4}{12}\right)$ year

$=\left(2+\frac{1}{3}\right)$ years

$=\frac{7}{3}$ years

$\begin{aligned} P &=\frac{100 \times I}{R \times T} \\ &=\frac{100 \times 2730}{\frac{15}{2} \times \frac{7}{3}} \\ &=\frac{100 \times 2730 \times 6}{15 \times 7} \end{aligned}$

Principal p = Rs 15,600

Question 5

i) Amount = Rs 1320; Principal = Rs 1200

S.I = A-P = 1320 - 1200

S.I = 120

$I=\frac{P \times R \times I}{100}$

$R=\frac{100 \times I}{P \times T}$

$R=\frac{100 \times 120}{1200 \times 2}$

$R=5 \%$ per annum

ii) Amount=Rs 400; principal =Rs 300

$S: I=A-P=400-300$

I=Rs 100 

$R=\frac{100 \times 100}{300 \times 2}$

$R=\frac{50}{3}=16 \frac{2}{3} \%$ pes annam.

Question 6

i) $A=1950, P=1250, R=16 \%$

$I=A-P=1950-1250= 700$

$I=\frac{P \times R \times T}{100}$

$T=\frac{100 \times I}{P \times R}$

$T=\frac{100 \times 700}{1250 \times 16}$

$\quad$ Time, $T=\frac{7}{2}$ years

ii) $A=8447.50, P=6540 ; R=12 \frac{1}{2}=\frac{25}{2} .$

$I=A-P=8447.5-6540$

$I=19075$

Time, $\begin{aligned} T &=\frac{100 \times 1907.5}{6540 \times 25 / 2} \\ T &=\frac{100 \times 19075 \times 2}{6540 \times 25} \end{aligned}$

Time, $T=\frac{7}{3}$ years

Question 7

$R=4 \% \quad A=16,240, \quad P=14,000$

$\begin{aligned} I=& A-P=16,240-14,000 \\ & I=2,240 . \\ \text { Time } &=\frac{100 \times I}{P \times R} \\ \text { Time } &=\frac{100 \times 2,240}{14,000 \times 4} \\ &=4 \text { years } \end{aligned}$

Question 8

$T=6$ years, Given Amount invested trebled

So A 3 $\times$ Principal

A = 3p

$\begin{aligned} I=& A-\rho=3 p-p \\ & I=2 p \\ I=& \frac{P \times R \times f}{100} \\ R &=\frac{100 \times 1}{p \times 1} \\ & R=\frac{100 \times 20}{P \times 6} \\ & R=\frac{100}{3}=33 \frac{1}{3} \% \text { pes annm } \end{aligned}$

Question 9

i) $A=4,500 ; R=20 \% T=5$ years

$I=A-P$

$I=4,500-P$

Also $I=\frac{p \times R \times {j}}{100}$

$4,500-P=\frac{P \times 20 \times 5}{100}$

$4,500-P=P$

$\begin{aligned} p+p=& 4,500 \\ 2 p=4,500 \\ \text { principal, } P=2250 \end{aligned}$

ii) $A=2420, R=4, \quad T=2 \frac{1}{2}$ yeass $=\frac{5}{2}$ years

$I=A-P$

$I=2420-P$

Also $I \cdot \frac{P \times R \times 1}{100}$

$2420-p=\frac{p \times 4 \times 5}{2 \times 100}$

$2420-R=\frac{1}{10}$

$P+\frac{p}{10}=2420$

$\frac{11P}{10}=2420$

$P=\frac{2420 \times 10}{11}$

$P=2200$

$\therefore$ principal, $p=2,200$

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