Exercise 8 A
Question 1
Write these percentages as fractions and decimals :
(i) $60 \%$
Sol :
$=\dfrac{60}{100}=\dfrac{3}{5}$
=0.6
(ii) $240 \%$
Sol :
$=\dfrac{240}{100}=\dfrac{12}{5}$
=2.4
(iii) $12 \dfrac{1}{2} \%$
Sol :
$=\dfrac{25}{2}\%=\dfrac{25}{2 \times 100}$
$=\dfrac{1}{8}$
=0.125
(iv) $55 \dfrac{2}{3} \%$
Sol :
$=\dfrac{167}{3}\%$
$=\dfrac{167}{3 \times 100}=\dfrac{167}{300}$
=0.557...
Question 2
Write these numbers as percentages :
(i) $\frac{3}{4}$Sol :
Sol :
$=\dfrac{3}{4} \times 100$
=75%
(ii) $\frac{21}{40}$Sol :
Sol :
$=\dfrac{21}{40} \times 100$
$=\dfrac{105}{2}\%$
$=52\dfrac{1}{2}\%$
(iii) $0.008$
Sol :
$=\dfrac{8}{1000} \times 100$
$=\dfrac{4}{5}\%$
=0.8%
(iv) $0.375$
Sol :
$=\dfrac{375}{1000}\times 100$
$=\dfrac{75}{2}\%$
=37.5%
(v) $1.07$
Sol :
$=\dfrac{107}{100} \times 100$
=107 %
Question 3
Express the following percentage as ratios:
(i) $200 \%$
Sol :
$=\dfrac{200}{100}=\dfrac{2}{1}$
=2 : 1
(ii) $6 \dfrac{2}{3} \%$
Sol :
$=\dfrac{20}{3}\%=\dfrac{20}{3 \times 100}$
$=\dfrac{1}{15}$
=1 : 15
(iii) $75 \%$
Sol :
$=\dfrac{75}{100}=\dfrac{3}{4}$
=3 : 4
(iv) $33 \dfrac{1}{3} \%$
Sol :
$=\dfrac{100}{3}\%=\dfrac{100}{3 \times 100}$
$=\dfrac{1}{3}$
=1 : 3
Question 4
Express the following ratios as percentage:
(i) $1: 2$
Sol :
$=\dfrac{1}{2} \times 100$
=50%
(ii) $7: 10$
Sol :
$=\dfrac{7}{10} \times 100$
=70%
(iii) $21: 25$
Sol :
$=\dfrac{21}{25} \times 100$
=84%
(iv) $5: 12$
Sol :
$=\dfrac{5}{12} \times 100$
$=\dfrac{125}{3}\%=41\dfrac{2}{3}\%$
Question 5
Evaluate :
(i) $19 \%$ of $₹ 4200$
Sol :
$=\dfrac{19}{100} \times 4200$
=798 g
(ii) $75 \%$ of $440 \mathrm{~kg}$
Sol :
$=\dfrac{75}{100} \times 440$
=330 g
(iii) $30 \%$ of $20 \mathrm{~g}$
Sol :
$=\dfrac{30}{100} \times 20$
=6 g
(iv) $2.5 \%$ of $350 \mathrm{~m}$
Sol :
$=\dfrac{25}{10 \times 100} \times 350$
$=\dfrac{35}{4}$
=8.75 m
Question 6
Express the first quantity as a percentage of the second :
(i) $20 \mathrm{~g}$ of $4 \mathrm{~kg}$
Sol :
=20 g of 4000g
$=\dfrac{20}{4000} \times 100$
$=\dfrac{1}{2}\%$
=0.5%
(ii) $200 \mathrm{~mL}$ of 5 litres
Sol :
=200 ml of 5000 ml
$=\dfrac{200}{5000} \times 100$
=4%
(iii) $48 \mathrm{~cm}$ of $1 \mathrm{~m}$
Sol :
=48 cm pf 100 cm
$=\dfrac{48}{100} \times 100$
=48%
(iv) $\frac{2}{3}$ of $\frac{1}{2}$
Sol :
$=\dfrac{\frac{2}{3}}{\frac{1}{3}} \times 100$
$=\dfrac{2}{3} \times \dfrac{3}{1} \times 100$
=200 %
Question 7
(i) If $14 \%$ of a certain number is 63 , find the number.
Sol :
Let the required number be x
14% of x=63
$\dfrac{14}{100} \times x=63$
$x=\dfrac{63 \times 100}{14}$
x=450
(ii) If $12 \frac{1}{2} \%$ of a certain amount is $₹ 62.50$, find the amount.
Sol :
Let the amount be x
$12 \frac{1}{2} \%$ of x=62.50
$\dfrac{25}{2}\% \times x$=62.50
$\dfrac{25}{2\times 100} \times x$=62.50
$x=\dfrac{62.50}{100} \times \dfrac{2}{25} \times 100$
x=500
Question 8
There were 4200 people at a concert and 1400 of these were female. What percentage were male ?
Sol :
Total number of people=4200
Number of female=1400
Number of male=4200-1400=2800
Percentage of male $\dfrac{2800}{4200} \times 100$
$=\dfrac{200}{3} \%=66\dfrac{2}{3}\%$
Question 9
The cost of a theatre ticket is increased by $40 \%$ for a special movie. What is the new price if the normal price was ₹ 160.
Sol :
Normal price was ₹ 160
Increase =40% of 160$
$=\dfrac{40}{100} \times 160$
=4×16=64
New price=160+64=224
Question 10
A motorcycle originally cost $₹ 40,000$. Its cost has decreased by $22 \%$. What is its cost now ?
Sol :
Original cost of motorcycle=40000
Decrease=22% of 40000
$=\dfrac{22}{100} \times 40000$
=22×400=8800
New cost=40000-8800=31200
Question 11
The price of a garment has been reduced by $15 \%$ in a sale to $₹ 306$. Find its original price.
Sol :
Let the original cost be x
According to question,
x-15% of x=306
$x-\dfrac{15}{100}\times x=306$
$x-\dfrac{3x}{20}=306$
$\dfrac{20x-3x}{20}=306$
$\dfrac{17x}{20}=306$
$x=\dfrac{306 \times 20}{17}$
x=360
Multiple Choice Questions (MCQs)
Question 12
What percent of $\sqrt{0.0169}$ is $0.0117$ ?
(a) 0.9
(b) 0.1
(c) 0.09
(d) 9
Sol :
x% of $\sqrt{0.0169}=0.0117$
$\dfrac{x}{100} \times \sqrt{\dfrac{169}{10000}}=\dfrac{117}{10000}$
$\dfrac{x}{100} \times \dfrac{13}{100}=\dfrac{117}{10000}$
$x \times 13=\dfrac{117}{10000} \times 100 \times 100$
$x=\dfrac{117}{13}$
x=9
Question 13
What per cent is $3 \%$ of $5 \%$ ?
(a) $15 \%$
(b) $30 \%$
(c) $50 \%$
(d) $60 \%$
Sol :
Let the value be 100
x% of 5%=3%
$\dfrac{x}{100} \times \dfrac{5}{100} \times 100=\dfrac{3}{100} \times 100$
$\dfrac{x}{100} \times 5=3$
$x=\dfrac{3}{5} \times 100$
Question 14
A piece of elastic was stretched by $24 \%$ to a length of $31 \mathrm{~cm}$. Find its unstretched original length.
(a) $24 \mathrm{~cm}$
(b) $26 \mathrm{~cm}$
(c) $25 \mathrm{~cm}$
(d) $20 \mathrm{~cm}$
Sol :
Let the original length be x
According to question,
x+24% of x=31
$x+\dfrac{24}{100} \times x=31$
$\dfrac{100x+24x}{100}=31$
$\dfrac{124x}{100}=31$
$x=\dfrac{31}{124}\times 100$
x=25
High Order Thinking Skills (HOTS)
Question 15
The sum of two numbers is $\dfrac{28}{25}$ of the first number. The second number is what percent of the first?
Sol :
Let , First number be x
Then , Second number=y
$x+y=\dfrac{28}{25}$of x
$y=\dfrac{28x}{25}-x$
$y=\dfrac{28x-25x}{25}$
$y=\dfrac{3x}{25}$
$\dfrac{y}{x}=\dfrac{3}{25}$
$\dfrac{y}{x}=\dfrac{3}{25} \times 100$
$\dfrac{y}{x}=12%$
y=12% of x
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