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
No comments:
Post a Comment