ML AGGARWAL CLASS 8 CHAPTER 7 Percentage Exercise 7.1

    Exercise 7.1

SOLUTION 1

(i) $356 \%$

$\Rightarrow \frac{356}{10025}=\frac{89}{25}$

(ii) $2 \frac{1}{2}-1$

=$\frac{5}{2} \%$

=$\frac{8}{2 \times 100}=\frac{1}{40}$

(ii)

 = $\begin{array}{ll} & 16 \frac{2}{3} \\ \Rightarrow & \frac{50}{3} \% \\ & \frac{50}{3 \times 100}=\frac{1}{6}\end{array}$

SOLUTION 2

(i) $\frac{3}{2}$

= $\frac{3}{2} \times 100 \%=150 \%$

(ii) $\frac{9}{20}$

= $\left(\frac{9}{20} \times 100\right) \%=36 \%$

(iii) $1 \frac{1}{4}$

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

SOLUTION 3

(i) $\frac{3}{4}$

$\Rightarrow 0.75$, Decimal.

when we converting decimals into percentages

then $\Rightarrow \quad 0.75 \times 100$

$\Rightarrow \quad 75 \%$

(ii) 

 $\begin{aligned} & \frac{5}{8} \\ \Rightarrow & 0.62 \\ \Rightarrow & 0.625 \times 100 \\ \rightarrow & 62.5 \% \end{aligned}$

(iii) 

=$\frac{3}{16}$

=$0.1875$

=$0.1875 \times 100$

=$18.75 \%$

SOLUTION 4

(i) 

=$\frac{2}{3}$

$\Rightarrow 0.6666 \quad \rightarrow$ Decimal

while converting decimals into percentage we 

have to multply with 100

$\Rightarrow \quad 0.6666 \times 100$

$\Rightarrow \quad 66.66 \%$

(ii)

$\begin{aligned} & \frac{5}{6} \\ &=0.83333 \\ \Rightarrow & 0.8333 \times 100 \\ \Rightarrow & 83.33 \% \end{aligned}$

(iii) 

$\begin{aligned} & \frac{4}{7} \\ \rightarrow & 0.5714 \\ &=0.5714 \times 100 \\ \Rightarrow & 57.14 \% \end{aligned}$

SOLUTION 5

(i)

 $\begin{aligned} & 17: 20 \\ \Rightarrow & \frac{17}{20} \\ \Rightarrow & \frac{17}{20} \times 10^{5} 0 \\ \Rightarrow & 17 \times 5 \\ \Rightarrow & 85 \% \end{aligned}$

(ii) $\frac{13}{18}: 18$

=$\frac{13}{18}$

=$\frac{13}{18} \times 100$

=$72.22 \%$

(iii) $93: 80$

=$\frac{93}{88} \times 100$

=$116.25 \%$

SOLUTION 6

(i) $20 \%$

=$\frac{20}{100}$

=$\quad \frac{1}{5}$

=0.2

(ii)

⇒$2.1 .$

=$\quad \frac{2}{100150}$

=$\frac{1}{50}$

= 0.02

(iii)

$3 \frac{1}{4} \%$

=$\frac{7}{4}$

=$\frac{7}{4 \times 100}$

=0.0175

SOLUTION 7

(i) $27 \%$ of 750

=$\frac{27}{100} \times 50$

=$\frac{27}{2}$

=$₹ 13.5$

(ii) $6 \frac{1}{4} \%$ of $25 \mathrm{~kg}$

⇒$\frac{25}{4} \%$ of $25 \mathrm{~kg}$

⇒$\frac{\frac{25}{4}}{100} \times 25$

⇒$\frac{25}{4 \times 100} \times 25$

⇒$\frac{25}{16} \quad$ (or) $1 \frac{9}{16} \mathrm{~kg}$

⇒$1.56 \mathrm{~kg}$

SOLUTION 8

(i)

$\begin{aligned} 300 \mathrm{~g} \text { of } 2 \mathrm{~kg} & \\ 2 \mathrm{~kg}=(2 \times 1000) \mathrm{g} &=2000 \mathrm{~g} \\ \text { Required percentage } &=\left(\frac{300}{2000} \times 100\right) \% \\ &=15 \% \end{aligned}$

(ii) $₹ 7 .50$ of 26

$₹6=(6 \times 100)$ paise

Required percentage $=\frac{750 \times 100}{600}$

$=125 \%$

SOLUTION 9

(i) 

$\begin{aligned} 50 \mathrm{~kg} \text { is } 65 \mathrm{~kg} \text {  } \\ \text { Required percentage } &=\frac{65}{50} \times 100 \mathrm{} \\ &=130 \mathrm{~kg} . \end{aligned}$

(ii) $\begin{aligned} ₹ 9 \text { is } ₹ 4 & \\ \text { Required percentage } &=\frac{4}{9} \times 100 \\ &=\frac{400}{9} . \\ &=44 \frac{4}{9} \% \end{aligned}

SOLUTION 10

(i) $16 \frac{2}{3} \%$ of number is 25

let the required number be $x$ '

According to given condition, $16 \frac{2}{3} \%$ of $x$ is 25

$\therefore \quad \frac{16 \frac{2}{3}}{100} \times x=25$

$\frac{50}{3 \times 100} \times x=25$

x=\frac{3 \times 100 \times 25}{50}$

$x=150$

(ii) let the number be 'x '

Given condition is 13.25\% of $x$ is 159

$\begin{array}{l}x=\frac{159 \times 100}{13.25} \\ x =1200\end{array}$

SOLUTION 11

(i) New number $=\left(1+\frac{30}{100}\right) \times 60$

$\because\left(1+\frac{x}{100}\right)$ -f originel number $]$

=$\left(\frac{100+30}{100}\right) \times 60$

=$\frac{130}{100} \times 60$

=78

(ii)

$\begin{aligned} \text { Naw number } &=\left(1-\frac{x}{100}\right) \text { of original } \\ &=\left(1-\frac{10}{100}\right) \times 750 \\ &=\left(\frac{100-10}{100}\right) \times 750 \\ &=\frac{90}{100} \times 750 \\ &=675 \end{aligned}$

SOLUTION 12

(i) $\quad\left(1+\frac{x}{100}\right)$ of original $=$ New number

$\left(1+\frac{15}{100}\right) \times x=299$

$\frac{115}{100} \times x=299$

$\left(\frac{100+15}{100}\right) \times x=209$

$x=\frac{299 \times 100}{115}$

x =260

(ii) $\left(1-\frac{18}{100}\right) \times x=697$

let number be $x$ 1

$\therefore \quad\left(\frac{100-18}{100}\right) \times x=697$

$\frac{82}{100} \times x=697$

$x=\frac{697 \times 100}{82}$

x=850

SOLUTION 13

(i) let the monthly salary is 'x'

Mr. khana spent 83 %  and saved 1870 

savings = 100 - 83 

= 17 % 

17 % of x = 1870 

$\begin{aligned} 17 \% \text { of } x &=1870 \\ \frac{17}{100} \times x &=1870 \\ x &=\frac{1870 \times 100}{17} \\ &=110 \times 100 \\ x &=\{11,000\end{aligned}$

∴ monthly salary is rs 11000

SOLUTION 14

let the total strength of school is 'x' 

given 38 % of the students are girls 

so = 100 - 38 

= 62 %  is boys 

Number of boys $=1023$

i.e. $62 \%$ of $x=1023$

$\frac{62}{100} \times x=1023$

$x=\frac{1023 \times 100}{62}$

=1650

SOLUTION 15

The price of article increased from rs 960 to rs 1080

original price = 960 

Increase in price = 1080 - 960 

= 120 

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

$\frac{120}{960} \times 100$

$12.5 \%$

SOLUTION 16

(i) Given , the total no of eligible voters = 1 lakh 

= 1,00,000

loser polled = 42% 

winner polled = 100- 42

 

= 58 %

Loser lost by 14400 votes.

$\therefore$ winner- loser $=14,400$

$58 /-42 \%=14,400$

$16 \%=14,400$

let the total noof voter polled

$\begin{aligned} 16 \% \text { of } &=14,400 \\ \frac{16}{100} \times x &=14,400 \\ x &=\frac{14,400 \times 100}{16} \\ x &=90,000 \end{aligned}$

∴ The percentage of voters did not vote 

= 1,00,000 - 90,000

= 10,000

i.e 10 % 

SOLUTION 17

Given total candidated = 8000

60% were boys 

i. e $\frac{60}{100} \times 8000=4800$

girls = 8000 - 4800

= 3200 

passed candidates was

$\begin{aligned}80 \% \text { of boys } &=\frac{80}{100} \times 4800 \\90 \% \text { of } \dot{9}+girls &=3840 \\&=\frac{90}{100} \times 3200 \\&=2880 .\end{aligned}$

Total no of candidated passed in exam 

i.e 3840 + 2380= 6720

Number of candidates who failed 

800 - 6720

= 1280

SOLUTION 18

(i) Given, $\frac{1}{4}$ of students failed in both in

English and maths ie, $25 \%$

$35 \%$ students failed in maths

$30 \%$ students failed in English.

$\begin{aligned} \therefore \text { percentage of students who failod in } \\ \text { only maths } &=35 \%-25 \% \\ &=10 \% \end{aligned}$

$\begin{aligned} \text { percentage of students who failed in } \\ \text { only english } &=30 \% 25 \% \\ &=5 \% \end{aligned}$

$\begin{aligned} \therefore \text { percentate of students who farled in } \\ \text { any of the subjects } &=25+i 0+5 \\ &=40 \% \end{aligned}$

(ii) percentage of students who passed 'in

both the subjects $=100-40$

$=60 \%$

(iii) Given noof students who frailed only in

$\begin{aligned} \text { english }=25 \quad \Rightarrow \quad 5 \%=25 \\ \therefore \text { Total noof students }=100 \% &=\frac{100}{5} \times 25 \\ &=500 \% \end{aligned}$

SOLUTION 19

Let the price of the article be 'x' 

The price of article increased by 16% 

So $\left(1+\frac{16}{100}\right) \times x=1479$

$\left(\frac{100+16}{100}\right) \times x=1479$

$\frac{116}{100} \times x=1479$

$x=\frac{1479 \times 100}{116}$

$x=1275$

$\therefore$ Original price of article is $\{12751$

SOLUTION 20

Let the prathibha weight is 'x' kg 

prathiba weight reduced by 15%

So   $\left(1-\frac{15}{100}\right) \times x=59.5$

  $\left(\frac{100-15}{100}\right) \times x=59.5$

  $\frac{85}{100} \times x=59.5$

  $x=\frac{59.5 \times 100}{85}$

  $x=70$

SOLUTION 21

(i) As per given condition,

shop reduces all its prices by 15\%

 the original price is rs 40

$\therefore$ Let the cost of an article is

$\begin{aligned} x &=40-15 \% \text { of } 40 \\ &=40-\frac{15}{100} \times 40 \\ &=40-6 \\=& 34\end{aligned}$

(ii) Let the orginal price be 'x'.

The article sold at $2.20 .40$

$\therefore x-15 \% o x=20.40$

$x-\frac{15}{100} x x=20.40$

$\frac{100 x-15 x}{100}=20.40$

$\frac{85 x}{100}=20 \% 0$

$x=\frac{20.40}{85} \times 100$

$x=24 .$

Original price of article is $2.24$

SOLUTION 22

The original price is rs 200 

increases by 10 % 

⇒ $200+10 \%$ of 200

⇒$200+\frac{10}{100} \times 200$

⇒$200+20$

⇒220

decreases by 10 % 

⇒ $220-10 \%$ of 220

⇒$220=\frac{10}{104} \times 220$

⇒$220=22$

⇒198

original price is 200

final price is 198 

No, the final price is not same as original one .

SOLUTION 23

let 'x' be the number of parrots initially chandini have 20% of the parrots away and 5%  of them died 

No. of parrots remaining $\begin{aligned} \text { now } &=\left[1-\left(\frac{20}{100}+\frac{5}{100}\right)\right] \times(x) \\ &=0.75 x \end{aligned}$

Now,

        45% of the remaining parrots were sold 

⇒ 55%  of remaining parrots were with chandini 

Therefore ,

No of parrots chandini is having finally = $=\left(\frac{55}{100}\right) \times 0.75 x$ -------①

But

given no of parrots chandini is having = 33 -------②

From ①&②

33 is equal to $\left(\frac{55}{100}\right){0.75 x}$

⇒ $33=\left(\frac{55}{100}\right) \times(0.75) \times x$

⇒ x =80

∴ chandini had purchased 80 parrots 

SOLUTION 24

Let ''x be the maximum marks

 'y' be the minimum pass marks 

A candidate gets 36 % in examination and fails by 24 marks 

⇒ $(0.36) \times(x)=y-24$ ------①

Another candidate gets 43% in an examination and gets 18 marks more than that of pass marks

⇒ $(0.43) \times(x)=y+18$--------②

⇒ solving equation (1) and (2), we get

=0.36 x+24=0.43 x-18

⇒$\frac{0.7 x}{10}=42$

⇒ x=600 maximum marks 

Substituting x = 600 in equation 1 

⇒ $(0.3 t) \times 600=y-24$

⇒ y= 240

percentage of pass marks = $\frac{y}{x} \times 100 \%$

⇒  $\frac{240}{600} \times 100$

% of pass marks = 40 %

    Exercise 7.2

Question 1

(i) C.P 400  S.P = 468

 $\begin{aligned} \text { S.P }>\text { c.p , profit } &=\text { s.p-C.p } \\ &=468-400 \\ &=68 \end{aligned}$

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

$=\left(\frac{68}{400} \times 100\right) \%$

$=17 .$

(ii) C.P = 13600  , S.P = 12104

As C.p $>$ S. P $\begin{aligned} \text { Loss } &=c.p-s \cdot p\\ &=13,600-12,10 y \\ &=1496 \end{aligned}$

loss percentage  

$\begin{aligned} &=\left(\frac{\text { Loss }}{c . p} \times 100\right) \% \\ &=\left(\frac{1496}{13600} \times 100\right) \% \\ &=11 \% \end{aligned}$ 

Question 2

Given $S \cdot P=1636.25$, gain $=96.25$

As $Gain=S \cdot p-C p$

⇒ $\begin{aligned} C \cdot P=S \cdot p-G a i n &=1636.25-96.25 \\ &=1540 . \end{aligned}$

⇒ Gain percentage

 $\begin{aligned} &=\left(\frac{G a i n}{c \cdot p} \times 100\right) \% \\ &=\left(\frac{96 \cdot 25}{1540} \times 100\right) \% \\ &=6.25 \% \end{aligned}$

Question 3

Given 

         S.P = 770  LOSS = 110

As loss = c.p - s.p 

c.p = loss + s.p 

= 110 + 700

⇒ 880

Loss percentage 

$\begin{aligned} &=\left(\frac{\text { Loss }}{c . p} \times 100\right) \% \\ &=\left(\frac{110}{880} \times 100\right) \% \\ &=12.5 \% \end{aligned}$

Question 4

c. $p$ of 1 dozen eges $=9.60$

c. $p$ of 25 doten eggs $=(25 \times 9.6)=240$

Total no of eqrs $=25 \times 12=300$ eggs

Out of 300 eqgs, 30 eggs were broken

So the remaining no of eggs were 300 - 30 = 270 eggs 

Given  

       S.P of each egg = Rs 1 

S. of 270 eggs $=270 \times 1=270$

As $S \cdot p>c. p$, he always gets profit on given

So the gain percentage = $\left(\frac{gain }{c \cdot p} \times 100\right) \%$

Gain = S.P - C.P 

⇒ 270 - 240 = 30 

Gain percentage = $\left(\frac{30}{240} \times 100\right) \%$

$=12.5 \%$

Question 5

C.p of an article          

                    $\begin{aligned} &=20,000+1400 \text { (repairs) } \\ &=21,400 \end{aligned}$

profit percentage $=20 \%$

$\frac{\text { profit }}{\text { c.p }} \times 100=20$

Profit $x  \times100=20 \times 21,400$

$(s \cdot p-(\cdot p)=20 \times 214$

S. $p-21,400=4280$

$\begin{array}{l}S p=21,400+4280 \\ s p=25680\end{array}$

Selling price of an article = 25,680

Question 6

C.p of bicycles includes

i) 200 bicycles at 1200. pes bicyde $=(200 \times 1200)=240000$

ii) 301- pes birycle on transportation $=200 \times 30=6000$

iii) $40001-$ on advertising $=4000$

Total cost price of bicycles $=2,40,000+6000+4000$

S.p of 200 bicycles $=200 \times 1350$

As $\mathrm{S} \cdot \mathrm{P}>\mathrm{C} \mathrm{p}$, there is always a gain

So $g a i n=s. p-c .p$

$\begin{array}{l}=2,70,000-2,50,000 \\=20,000\end{array}$

$=\frac{Gain}{c . p} \times 100$

$=\frac{20,000}{2,50000} \times 100$

$=8 \%$

Question 7

Let S.P be ₹ x ,

then c.p 90 % of x 

$=\frac{9}{10} x$

$\begin{aligned} \text { Profit } &=s \cdot p-c \cdot p \\ &=x-\frac{9 x}{10} \\ &=\frac{x}{10} \end{aligned}$

$\begin{aligned} \text { profit percentage }=&\left(\frac{\text { Profit }}{\text { c. } p} \times 100\right) \% \\ &=\left(\frac{(x / 10)}{(9 \times 10)} \times 100\right) \% \\ &=\frac{100}{9} \% \\ &=11.11 \% \end{aligned}$

Question 8

(i) c.p of 4 note boots $= ₹35$

then c.s of 1 note book $=\frac{35}{4}=8.75$ 

Sp of 5 note books $=58$

then $s \cdot p$ of 1 notebook $=\frac{58}{5}=11.6 \%$

As. $S \cdot p>C$. $p$, there is always a gain

Gain = s.p - c.p = 11.6 - 8.75

= 2.85

Gain percentage 

$\begin{aligned} &=\frac{G a i n}{C \cdot p} \times 100 \\ &=\frac{2.85}{8.25} \times 100 \\ &=32.57 \% \end{aligned}$

(ii) Number of notebooks to be sold = $\frac{\text { total profit }}{\text { profit on one notebooks }}$

$=\frac{171}{2.85}$

=60 

Question 9

Cost price of 3 bananas = Rs 1 

The c .p of 1 banana = Rs $\frac{1}{3}=0.33$

S.p of 4 bananas = Rs 1

Then S.P of 1 banana = Rs $\frac{1}{4}=0.25$

As C.p $>$ s.p, These is always a loss.

$\begin{aligned} \text { Loss } &=C \cdot p-s \cdot p \\ &=\frac{1}{3}-\frac{1}{4} \end{aligned}$

$=\frac{1}{12}$

$\begin{aligned} \text { Loss percentage } &=\frac{\text { Loss }}{c . p} \times 100 \% \\ &=\frac{(1 / 12)}{(1 / 3)} \times 100 \% \\ &=\frac{100}{4} \% \\ &=25 \% \end{aligned}$

Question 10

Given S.P of 5 pens = C.p of 7 pens

let cost price of one pen be x then 

C.p of  7 pens = Rs 7x

It is given 

S.P of 5 pens = c.p of 7 pens 

S.p of 5 pens = rs 7x 

S.p of one pen = $\frac{7 x}{5}$

As $S \cdot P>C.$ p $\quad$ there is a profit

$\begin{aligned} \text { profit } &=S\cdot p-c \cdot p \\ &=\frac{7 x}{5}-x=\frac{2 x}{5} \end{aligned}$

∴ $\begin{aligned} \text { Profit percentage } &=\frac{\text { profit }}{C \cdot p} \times 100 \% \\ &=\frac{(2 x / 5)}{x} \times 100 \% \\ &=\frac{200}{5} \% \\ &=40 \% \end{aligned}$

Question 11

i) $C \cdot P=2360$, profit $=8 \%$

As. Profit percentage $=\frac{\text { profit }}{C. p} \times 100$

$\begin{aligned} \text { profit } \times 100 &=8 \times 2360 \\ \text { profit } &=\frac{8 \times 2360}{100} \\ S.P-C \cdot p &=188.8 \\ S \cdot p &=2360+188 \cdot 8 \\ S \cdot P &=2548.8 \end{aligned}$

(ii) $C \cdot P=380 \quad ;$ loss $=7.5 \%$

Loss percentage $=\frac{\text { loss }}{\text { C.P }} \times 100$

Loss $=\frac{7.5 \times 380}{100}$

$\begin{aligned} c\cdot p-s .p &=28.5 \\ s \cdot p &=380-28.5 \\ s \cdot p &=351.5 \end{aligned}$

Question 12

c. $p$ of dozen eqrs $=E 18$

then c.p of 1 egg=$\frac{18}{12}=21.5$

⇒Profit $=50 \%$

$\frac{\text { S. } P-C . P}{C. P} \times 100=50$

$s \cdot p-c \cdot p=9$

$S \cdot p=18+9$

$S \cdot p=27$

S.p of 1 egg  =$\frac{27}{12}=22.25$

Question 13

Let the no of wrist watches are x 

cost of x wrist watches Rs 60,000

C.P one third of wrist watches will worth 20,000

i) As we know $\frac{1}{3} \mathrm{rd}$ are sold at a 30% profit

$\begin{aligned} S \cdot p &=\left[1+\frac{P}{100}\right] \text { of } C. P \\ &=\left[1+\frac{30}{100}\right] \times 20,000 \\ &=\frac{130}{100} \times 20,000 \\ &=26,000 \end{aligned}$

(ii) $s \cdot p=\left[1+\frac{p}{100}\right]$ of $c.p$ $\left[\because\right.$ As $\frac{1}{3}$ rd are sold

at $20 \%$ gian $]$

$=\left[1+\frac{20}{100}\right] \times 20,000$

$=\frac{120}{100} \times 20,000$

$=24,000$

(iii) $S \cdot P=\left(1-\frac{l}{100}\right)$ of c.p [Remaining are sold at

a $5 \%$. Less $]$

$=\left[1-\frac{5}{100}\right] \times 20,000$

$=\frac{95}{100} \times 20,000$

$=19,000$

Total cost price = ₹ 60,000

$\begin{aligned} \text { Selling price } &=[26,000+24,000+19,000] \\ &=69000 \end{aligned}$

As $S P>C P$ these is always a profit

$\begin{aligned}G a i n=S \cdot p-C \cdot p &=69,000-60,000 \\&=9,000\end{aligned}$

Gain percentage

$=\begin{aligned} &=\frac{G a i n}{c \cdot p} \times 100 \% \\ &=\frac{9,000}{69,000} \times 100 \% \\ &=15 \% \end{aligned}$

Question 14

C.p of a Laptop $= 40,000$

C.P of a mobile phene =  24,000

Total c.p of whole Transaction $=40,000+24,000$

=64,000

As shopkeeper made a profit of $8 \%$ on laptop

$\text { So. } \begin{aligned}s \cdot p &=\left(1+\frac{p}{100}\right) \text { of } c \cdot p \\&=\left[1+\frac{8}{100}\right] \times 40,000 \\&=\frac{108}{100} \times 40,000 =43,200\end{aligned}$

Also, he made a loss of $12 \%$ on mobile phone

$\begin{aligned}S \cdot p &=\left[1-\frac{l}{100}\right] \text { of } C \cdot P \\&=\left[1-\frac{12}{100}\right] \times 24,000 \\&=\frac{88}{100} \times 24,000 \\&=21,120\end{aligned}$

Total S.p on whOle Transaltion $=43,200+21,120$

=64,320 .

As, $S \cdot P>C\cdot p$ there is always a gain

$\begin{aligned}\text { Gain }=S \cdot p-c \cdot p &=64,320-64,000 \\&=320 \\\text { Gain percentage } &=\frac{G a i n}{C i p} \times 100\end{aligned}$

$=0.5 \%$

Question 15

C.p of 40 chairs $=(40 \times 175)=7,000$

Desired gain on whole deal $=10 \%$

s.p of all chairs $=\left[1+\frac{10}{100}\right] \times 7,000$

$=\frac{110}{100} \times 7,000$

$=7,700$

One-Fourt of all articles $=\frac{1}{4} \times 40=10$

C.P of 10 articles $=10 \times 175=1750$

As these articles are sold at a loss of $8 \%$

s.p of these articles $=\left[1-\frac{8}{100}\right]$ of 1750

$=\frac{92}{100} \times 1750$

$=1610$

Selling price of remaining i .e 30 chairs = 7700- 1610 = 6090

∴ S.P of each of the remaining chairs = $\frac{6090}{30}$

 =₹203

Question 16

S.P of two electronic gadgets = Rs 44,000 (each)

for first gadget 

S.p = rs 44,000  , profit = 10% , C.P = ?

$\begin{aligned} 44,000 &=\left(1+\frac{10}{100}\right) \text { of }(. P) \\ C \cdot P &=Rs\left(44,000 \times \frac{100}{110}\right)=Rs 40,000 \end{aligned}$

For second gadget:

$\begin{aligned} S \cdot p=44,000 &, \text { loss }=12 \%, C \cdot p=9 \\ \text { S. } \rho &=\left[1-\frac{1}{100}\right] \text { of } C \cdot P \\ 44,000 &=\left[1-\frac{12}{100}\right] \text { of } C \cdot P \\ C \cdot P &=\left\{\left[44,000 \times \frac{100}{88}\right]=\{50,000 .\right.\end{aligned}$

Then, Tofal cost price $=40,000+50,000=90,000$

Total selling $\rho_{i}(e=44,000+44,000=88,000$

Loss $=c.p-s \cdot p=90,000-88,000=2,000$

Loss percentage

 $\begin{aligned} &=\frac{\text { Loss }}{C.{P}} \times 100 \\ &=\frac{2,000}{99,000} \times 10 \% \\ &=\frac{20}{9} \\ &=2.22 \% \end{aligned}$

Question 17

Manufacturing price of a $T$.V set $=\{12,000$

Shopkeeper Sold to a dealer at a profit of $20 \%$

Now s.p of de T.V $\operatorname{set}=\left[1+\frac{20}{100}\right]$ of $C p$

$\begin{aligned} S \cdot p &=\frac{120}{100} \times 12,000 \\ &=1.4,000 \end{aligned}$

Dealer sold to a customer at $12.5 \%$ profit

Now Dealer's s.p will become cost price

So, New selling price to Customer

$\begin{array}{l}=\left[1+\frac{12.5}{100}\right] \text { of } C P \\ =\frac{112.5}{100} \times 14,000\end{array}$

= 16,200

So the customs bas to pay rs 16200 for T .V SET 

Question 18

(i) S. $p=450$, hoss $=10 \%$

$\begin{aligned} \% \text { Loss } &=\frac{\text { Loss }}{c \cdot p} \times 100 \\ 10 &=\left[1-\frac{s \cdot p}{c \cdot p}\right] \times 100 \\ 1-\frac{s \cdot p}{c \cdot p} &=\frac{1}{10} \end{aligned}$

$\frac{S \cdot p}{C .P}=1-\frac{1}{10}=\frac{9}{10}$

$C \cdot p=\frac{450 \times 10}{9}$

$C \cdot P=₹ 500$

(ii) $\begin{aligned} S p=690 &, \text { profit }=15 \% \\ S \cdot p &=\left[1+\frac{p}{100}\right] \text { of }C\cdot p\\ 690 &=\left[1+\frac{15}{100}\right] \times C . p \\ C \cdot \rho &=\frac{690 \times 100}{115} \\ C \cdot &=Rs 600 \end{aligned}$

Question 19

If S.P = 3920 , gain = 12 %

$S \cdot p=\left[1+\frac{p}{100}\right]$ of $c \cdot p$

$3920=\left[1+\frac{12}{100}\right] \times C . p$

$c \cdot p=\frac{3920 \times 100}{112}$

C.P = 3,500

Now S.p= 4375

As s.p $>$ c.p $\quad$ Gain $=S \cdot P-C \cdot P$

$=4,375-3,500$

$=875$

Gain percentage

$\begin{aligned} &=\left[\frac{G a i n}{c \cdot p} \times 100\right] \% \\ &=\left[\frac{875}{3,500} \times 100\right] \% \\ &=25 \% \end{aligned}$

Question 20

$S \cdot P=1334$, Loss $=8 \%$

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

$1334=\left[1-\frac{8}{100}\right] \times C . p$

$\begin{aligned} C \cdot p &=\frac{1334 \times 100}{92} \\c\cdot p&= 1450 \end{aligned}$

Given profit $=12 \frac{1}{2} \%=12.5 \%$

Now $S \cdot p=\left[1+\frac{p}{100}\right] \times(c.p)$

$=\left[1+\frac{12.5}{100}\right] \times 1450$

$S \cdot p=\frac{112.5 \times 1450}{100}$

$S \cdot p=21631.25$

Question 21

$\begin{aligned} S \cdot p=252 \cdot & \text { Gain }=5 \% \\ S \cdot P=&\left[1+\frac{P}{100}\right] \times C \cdot \rho \\ C \cdot P &=\frac{s.p \times 100}{100+P} \\ &=\frac{252 \times 100}{100+5} \\ & C \cdot p=\frac{25200}{105}=240 \end{aligned}$

$S \cdot p=? \quad$ if $\quad g a i n=35 \%$

$S \cdot p=\left[1+\frac{p}{100}\right]$ of $c \cdot p$

$=\left[1+\frac{35}{100}\right] \times 240$

$S \cdot 1=\frac{135 \times 240}{100}$

$S \cdot p=2324$

Question 22

Let the selling price of a bag be ₹ x

profit = 12 % 

$x=\left(1+\frac{12}{100}\right) d f \cdot p$

$\begin{aligned} x=\frac{112}{100} & \text { of }(\cdot p\\ c \cdot p &=\frac{100 x}{112} \end{aligned}$

To make $18 \%$ profit

$S \cdot p=\left[1+\frac{18}{100}\right] \text { of } c \cdot p$

$=\frac{118}{100} \times \frac{100 x}{112}=\frac{59 x}{56}$

According to given in formation, $\frac{59 x}{56}=x+39$

$\frac{59 x-x}{56}-39$

$\frac{3 x}{56}=3913$

$x=56 \times 13$

$x=728=s.p$

$\begin{aligned} \text { Cost price of } b a g &=\frac{100 \times x}{112} \\ &=\frac{100 \times 728}{112} \\ \text { Cost price of bag } &=₹650 \end{aligned}$

Question 23

Let the S.p of Sweater be $x$, loos $=5 \%$

$\begin{array}{l}x=\left[1-\frac{5}{160}\right] \text { of } c \cdot p \\ \quad C \cdot p=\frac{100 x}{95}\end{array}$

To make $15 \%$ profit

$\begin{aligned} \text { S. } p=\left[1+\frac{15}{100}\right] \text { of } C \cdot p &=\left[1+\frac{15}{100}\right] \times \frac{100 x}{95} \\ &=\frac{115}{100} \times \frac{100 x}{95}=\frac{23 x}{19} \end{aligned}$

According to given information, $\frac{23 x}{19}=x+260$

$\begin{array}{c}\frac{4 x}{19}=260 \\ x=65 \times 19 \\ x=1235\end{array}$

$\therefore$ Selling price of . Sweater $=1,235$

Question 24

Let the selling price be $x$ " Loss $=8 \%$

$\begin{array}{c}x=\left[1-\frac{8}{100}\right] \text { of } c \cdot p \\ c \cdot p=\frac{100 x}{92}\end{array}$

To make a profit of 12 % 

$\begin{aligned} S \cdot P=\left[1+\frac{12}{100}\right] \ of c\cdot p&=\left[\frac{112}{100} \times \frac{100 x}{92}\right] \\ &=\frac{28 x}{23} . \end{aligned}$

According to given information, $\frac{28 x}{23}=x+150$

$\frac{5 x}{23}=150^{30}$

Selling price, $x=23 \times 30=₹690$

    Exercise 7.3

Question 1

(i)

Marked price $=E 575$, discount $=12 \%$

Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$

$\begin{array}{l}12=\frac{\text { Discont }}{575} \times 100 \\ \text { Discount }=\frac{12 \times 575}{100}\end{array}$

Discount =₹69

Selling price = Marked price - Discounnt $=575-69$

=₹506

(ii)  Printed price $= 12,750$, discount $=8 \frac{1}{3} \%$

Discount percentage $=\left(\frac{\text { Discount }}{\text { Marked price }} \times 100\right) \%$

$\frac{25}{3}=\frac{D \text { discount }}{12,750} \times 100$

Discount $=\frac{25}{3} \times \frac{12,750}{1004}=1062.5$

Selling price $=$ Marked price or printed price - discount

$=12,750-1062.5$

$=11,687.5$

Question 2

(i) Marked price $=\{780$, selling price $=2721.5$

Selling price = Marked price - Discount

Discount $=$ Marked price - selling price

$=780-721.5$

$=58.5$

Discount percentage $=\frac{\text { Discount }}{\text { Marked price }} \times 100 \%$

$=\frac{58.5}{780} \times 100 \%$

$=7.5 \%$

(ii) Advertised price (or)

Marked price $=₹28,500$

selling price $=₹24510$.

Selling price = Marked rice - Discount

$\begin{aligned} \text { Dis count } &=\text { Marked price - selling price } \\ &=28,500-24,510 \\ &=3990 \\ \text { Discount. Percentage } &=\frac{\text { Discount }}{\text { marked price }} \times 100 \end{aligned}$

$=\frac{3990}{28500} \times 100 \%$

$=14 \%$

Question 3

Marked price $=₹ 30$. (each)

Discount \% $=\frac{\text { Discount }}{\text { Marked price }} \times 100$

Discount $=\frac{15 \times 30}{100}=4.5$

Discount on one note book $=4.5$

Discount on dozen note books $=12 \times 4.5=54$

Marked price on doren hote books $=12 \times 30=360$

Selling price $=$ Marked price - Ducount

$=360-54$

$=₹ 306$

Question 4

Selling price = Rs 728     discount = 9 % 

$\begin{aligned} s \cdot p &=\left(1-\frac{d}{100}\right) \text { of } M \cdot p \\ 7 28 &=\left(1-\frac{9}{100}\right) \times M \cdot \rho \\ M \cdot P &=\frac{728 \times 100}{91}=₹ 800\end{aligned}$

Question 5

Marked price $=₹ 800 \quad$ Discount $=20 \%$

i) Selling price $=?$

$\begin{aligned} \text { Discount } & \%=\frac{\text { Discant }}{M \cdot \rho} \times 100 \\ &(O R) \\ S \cdot P &=\left[1-\frac{d}{100}\right] \text { of } M \cdot p \\=&\left[1-\frac{20}{100}\right] \times 800 \\ &=\frac{80}{100} \times 800 \\ S \cdot p &=₹640\end{aligned}$

(ii)  profit = 25 %

$S \cdot p=\left[1+\frac{p}{100}\right]$ of $(\cdot p)$

$640=\left[1+\frac{25}{100}\right] \times(p)$

$C \cdot p=\frac{640 \times 100}{125}$

C . P= ₹512

Question 6

Marked price $= 2,250 \quad$ Dis cant $=12 \%$ profit $=10 \%$

(i) $\begin{aligned} s \cdot p &=\left[1-\frac{d}{100}\right] \text { of } M \cdot p \\ &=\left[1-\frac{12}{100}\right] \times 2250 \\ &=\frac{88}{100} \times 2250 \end{aligned}$

=₹ 1980

(ii) $S \cdot P=\left[1+\frac{P}{100}\right]$ of $C \cdot p$

$1980=\left[1+\frac{10}{100}\right] \times C .P$

$\begin{aligned} C.p&=\frac{1980 \times 100}{110} \\ &=₹1800 \end{aligned}$

Question 7

Cost price = Rs 650   Discount = 20%   profit = 20 % 

(i) $\begin{aligned} S \cdot p &=\left[1+\frac{p}{100}\right] \text { of } C \cdot p \\ &=\left[1+\frac{20}{100}\right] \times 650 \\ &=\frac{120}{109} \times 65 \\ S \cdot P &=780 \end{aligned}$

(ii) $S \cdot P=\left[1-\frac{d}{100}\right]$ of $M P$

$780=\left[1-\frac{20}{100}\right] \times M \cdot P$

$M \cdot P=\frac{780 \times 100}{80}$

M. $P=₹975$

Question 8

Cost price = 1200    profit = 80 %    Discount = 15%  

(i) $\begin{aligned} \text { Marked price } &=1200+(80 \%+(p)\\ &=1200+\left(\frac{80}{100} \times 1200\right) \\ &=1200+960 \\ &=2160\end{aligned}$

$\left[1-\frac{d}{100}\right] \times M \cdot P$

$\left[1-\frac{15}{100}\right] \times 2160$

$\frac{85}{100} \times 2160$

$=1836$

(iii)

 $\begin{aligned} \text { profil percentage } &=\left(\frac{S \cdot P-C .P}{c .p} \times 100\right) \% \\ &=\left(\frac{1836-1200}{1200} \times 10\right) \% \\ &=\frac{636}{1200} \times 100 \\ &=53 \% \end{aligned}$

Question 9

Cost price of an article = ₹ 1600

(i) since the cost price is 20 %  below the marked price 

$C \cdot P=M \cdot p-20 \%$ of $r$

$1600=M \cdot P-\frac{20}{100} \times M \cdot P$

$\begin{aligned} 1600 &=\left[1-\frac{20}{100}\right] \times \mathrm{M} \cdot P \\ M \cdot P &=\frac{1600 \times 100}{80} \\ M \cdot P &= 2000 \end{aligned}$

(ii) Discount $=16 \%$

$\begin{aligned} S \cdot P=&\left[1-\frac{d}{100}\right] \times \mathrm{M} \cdot \mathrm{p} \\=&\left[1-\frac{16}{100}\right] \times 2000 . \\ &=\frac{8 y}{100} \times 2000 \\ S \cdot p &=1680 \end{aligned}$

(iii) 

$\begin{aligned} \text { profit percentage } &=\left[\frac{s .p}{C{.p}} \times 100\right] \% \\ &=\left[\frac{1680-1100}{1600} \times 100\right] \% \\ &=\frac{80}{1600} \times 100 \% \\ &=5 \% \end{aligned}$

Question 10

(i)

Discount $=20 \% \quad$,profit =20%   ,selling price $= 360$.

$S \cdot p=\left[1-\frac{d}{100}\right]$ of $M \cdot P$

$360=\left[1-\frac{20}{100}\right] \times M \cdot 1$

$M \cdot P=\frac{360 \times 100}{80}$

$M \cdot P=450$

(ii) $s \cdot p=\left[1+\frac{p}{100}\right]$ of $c \cdot p$

$360=\left[1+\frac{20}{1 \cdot 0}\right] \times(\cdot p)$

$C \cdot p=\frac{360 \times 100}{120}$

$C \cdot p=₹ 300$

Question 11

Marked price of a refrigerator $=28,600$.

The selling price of a refrigerator is

$=\left[1-\frac{10}{100}\right]\left[1-\frac{5}{100}\right]$ of $\mathrm{M} \cdot \mathrm{p}$

$=\frac{90}{100} \times \frac{95}{10^{\circ}} \times 28600$

$= 24,453$

Question 12

Let the market price be 'x'

first dealer 

$\begin{aligned} \text { S. } P &=\left[1-\frac{15}{100}\right]\left[1-\frac{5}{100}\right] \text { of } \mathrm{M} \cdot \mathrm{P} \\ &=\frac{85}{100} \times \frac{95}{100} \times x \\ &=\frac{17 \times 19 \times x}{20 \times 20}=0.8075 \mathrm{x} \end{aligned}$

Second dealer ;

$\begin{aligned} s \cdot p &=\left[1-\frac{20}{100}\right] \text { of } m \cdot p \\ &=\frac{80}{100} \times x \\ &=\frac{4 x}{5}=0.8 x \end{aligned}$

As the second dealer offer price is less compared to first order

So the second dealer is best offer

Question 13

Let the marked price of an article be' $x^{\prime}$.

and a single discount of $d \%$ be equivalent to

two given successive discounts of $30 \%$ and $10 \%$, then

$\left(1-\frac{d}{100}\right)$ of $\bar{₹} x=\left(1-\frac{36}{100}\right)\left(1-\frac{10}{100}\right) \overline{₹} x$

$1-\frac{d}{100}=\frac{78}{100} \times \frac{90}{100}$

$1-\frac{d}{100}=\frac{63}{100}$

$\frac{d}{100}=\frac{37}{100}$

$d=37$

hence a discount of 37 % is equivalent to two given successsive discount 

    Exercise 7.4

Question 1

(i) Cost price of a tawel = Rs 50 

Sales tax = 5 % of 50 = Rs $\frac{5}{100} \times 50$

= Rs 2.5

$\begin{aligned} \therefore \text { Buying price } &=\text { cost price }+\text { sales Tax } \\ &=50+2.5=\  52.5 \end{aligned}$

(ii) Cost price of flour = Rs 15 per kg 

 Then cost price of 5kg flour = 5 x 15 = Rs 75 

sales Tax = 5 % of 75 = 

$\frac{5}{100} \times 75$

Rs $3.75$

Buying  price = cost price + Sales tax 

= 75 + 3 .75 = Rs 78 .75

Question 2

(i) Let the original price of $T \cdot V$ be $x$.

$\therefore V A T=8 \%$ of $x=₹\left(\frac{8}{100} \times  x\right)= ₹\frac{2 x}{25} .$

Price including $V A I=₹\left(x+\frac{2 x}{25}\right)= \frac{27 x}{25}$

$\begin{aligned} \therefore \frac{27 x}{25} &=13,500 \\ x &=\frac{13,500 \times 25}{27} \\ &=₹12,500 \end{aligned}$

(ii) Now 

$\begin{aligned} \frac{27 x}{25} &=180 \\ x &=\frac{180 \times 25}{27} \\ & x=₹166.67 \end{aligned}$

Question 3

let the original pice of $A$ c be $\left\{x^{\circ}\right.$

$\therefore V A T=₹8 \%$ of $x=₹\frac{2 x}{25}$

price including $V A T=₹\left(x+\frac{2 \pi}{25}\right)=₹\frac{27 x}{25}$

$\begin{aligned} \therefore \quad \frac{27 x}{25} &=34,992 \\ x &=\frac{34,992 \times 25}{27} \\ x &= ₹32,400 . \end{aligned}$

Question 4

Price including $V A 1=Rs  1296$

Original price of shirt $=Rs 1200$

Let VAT be ' $x$ "

price including VAT = $x \%$ of original price + original price

$1296=\left(\frac{1}{100} \times 1200+1200\right)$

$12 x=1296-1200$

$12 x=96$

$x=18 \%$

$\therefore \quad V A T=8 \%$

Question 5

Price of purse including $8 \% V A I=523.8$

Let the original price be Rs x

$\therefore$ pripe including $V A T=$ original price $+8 \%$ of orsinal

$523.8=x+8 \% d x$

$523.8=x+\frac{8 x}{100}=x+\frac{2 x}{25}$

$\frac{27 x}{25}=5238$

x=Rs 485

Now VAI increased by $10 \%$

New selling price 

$\begin{aligned} &=485+10 \% \text { of } 485 \\ &=485+\frac{10}{100} \times 485 . \\ &=485+48.5 \end{aligned}$

= Rs 533.5

ஃ  Now selling price of purse = Rs 533.5

Question 6

Marked price $=\sum 4800$

Rate of discount $=10 \%$

$\text { Dis count }=\frac{10}{100} \times 4,800= 480$

S.P of wall hanging $=$ M.P $-$ Discount

$=4,800-480$

= Rs 4,320 

Now VAT $8 \%$ of $4320=₹ \frac{8}{100} \times 4320=345.6$

Bill amount $=4,320+345 \cdot 6=₹ 4665.6$

Hence, the Customer has to pay $4665.6$ in Cash to puschace

Question 7

Let the reduced price of washing machine be Rs x 

VAT = 9 % of x = Rs $\left(\frac{9}{100} \times x\right)= \frac{91}{100}$

Amount paid by Amit = x $+\frac{9 x}{100}$ = $\frac{109 x}{100}$

As Amit has 10,900 to purchase it,

$\therefore \frac{1091}{100}=10,900 \Rightarrow x=\frac{10,9 \times \times 100}{109}=10,000$

$\therefore$ The reduced price of washing machine $= 10,000$.

So, amount reduced $=10,900-10,000$

$=2900$.

Hence, the amount reduced by shopkeeper is  Rs 900