ML AGGARWAL CLASS 8 CHAPTER 7 Percentage Exercise 7.1

    Exercise 7.1

SOLUTION 1

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(i) 

=$\frac{2}{3}$

$\Rightarrow 0.6666 \quad \rightarrow$ Decimal

while converting decimals into percentage we 

have to multiply 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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

Sol :

(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

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

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

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

x=850

SOLUTION 13

Sol :

(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

Sol :

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

Sol :

The price of article increased from rs 960 to rs 1080

original price = 960 

Increase in price = 1080 - 960 

= 120 

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

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

12.5 %

SOLUTION 16

Sol :

(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 no. of 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

Sol :

Given total candidate = 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 candidate passed in exam 

i.e 3840 + 2380= 6720

Number of candidates who failed 

800 - 6720

= 1280

SOLUTION 18

Sol :

(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.

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

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

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

(ii) percentage of students who passed 'in

both the subjects $=100-40$

$=60 \%$

(iii) Given no. of students who failed 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

Sol :

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

Sol :

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

Sol :

(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 original 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

Sol :

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

Sol :

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

Sol :

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 %