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#### Question 1:

Convert each of the following fraction into a percentage:

(i) $\frac{47}{100}$
(ii)  $\frac{9}{20}$
(iii) $\frac{3}{8}$
(iv) $\frac{8}{125}$
(v) $\frac{19}{500}$
(vi) $\frac{4}{15}$
(vii) $\frac{2}{3}$
(viii) $1\frac{3}{5}$

We have the following:

(i)

(ii)

(iii)

(iv) $\frac{8}{125}=\left(\frac{8}{125}×100\right)%=\left(\frac{8×4}{5}\right)%=\left(\frac{32}{5}\right)%=6.4%$

(v) $\frac{19}{500}=\left(\frac{19}{500}×100\right)%=\left(\frac{19}{5}\right)%=3.8%$

(vi) $\frac{4}{15}=\left(\frac{4}{15}×100\right)%=\left(\frac{4×20}{3}\right)%=\left(\frac{80}{3}\right)%=26\frac{2}{3}%$

(vii) $\frac{2}{3}=\left(\frac{2}{3}×100\right)%=\left(\frac{200}{3}\right)%=66\frac{2}{3}%$

(viii) $1\frac{3}{5}=\frac{8}{5}=\left(\frac{8}{5}×100\right)%=\left(8×20\right)%=160%$

#### Question 2:

Convert each of the following into a fraction:

(i) 32%
(ii) $6\frac{1}{4}%$
(iii) $26\frac{2}{3}%$
(iv) 120%
(v) 6.25%
(vi) 0.8%
(vii) 0.06%
(viii) 22.75%

We have the following:

(i) $32%=\left(\frac{32}{100}\right)=\frac{8}{25}$

(ii) $6\frac{1}{4}%=\left(\frac{25}{4}\right)%=\left(\frac{25}{4}×\frac{1}{100}\right)=\frac{1}{16}$

(iii) $26\frac{2}{3}%=\left(\frac{80}{3}\right)%=\left(\frac{80}{3}×\frac{1}{100}\right)=\left(\frac{4×1}{3×5}\right)=\frac{4}{15}$

(iv) $120%=\left(\frac{120}{100}\right)=\frac{6}{5}=1\frac{1}{5}$

(v) $6.25%=\left(\frac{6.25}{100}\right)=\left(\frac{625}{100×100}\right)=\left(\frac{25}{400}\right)=\frac{1}{16}$

(vi) $0.8%=\left(\frac{0.8}{100}\right)=\left(\frac{8}{10×100}\right)=\left(\frac{8}{1000}\right)=\frac{1}{125}$

(vii) $0.06%=\left(\frac{0.06}{100}\right)=\left(\frac{6}{100×100}\right)=\left(\frac{6}{10000}\right)=\frac{3}{5000}$

(viii)  $22.75%=\left(\frac{22.75}{100}\right)=\left(\frac{2275}{100×100}\right)=\frac{91}{400}$

#### Question 3:

Express each of the following as a ratio:

(i) 43%
(ii) 36%
(iii) 7.5%
(iv) 125%

We have:

(i)

(ii)

(iii)

(iv)

#### Question 4:

Convert each of the following ratios into a percentage:

(i) 37 : 100
(ii) 16 : 25
(iii) 3 : 5
(iv) 5 : 4

We have the following:

(i) 37 : 100 = $\frac{37}{100}=\left(\frac{37}{100}×100\right)%=37%$

(ii) 16 : 25 = $\frac{16}{25}=\left(\frac{16}{25}×100\right)%=\left(16×4\right)%=64%$

(iii) 3 : 5 =$\frac{3}{5}=\left(\frac{3}{5}×100\right)%=\left(3×20\right)%=60%$

(iv) 5 : 4 = $\frac{5}{4}=\left(\frac{5}{4}×100\right)%=\left(5×25\right)%=125%$

#### Question 5:

Convert each of the following into decimal form:

(i) 45%
(ii) 127%
(iii) 3.6%
(iv) 0.23%

We have the following:

(i) 45% = $\left(\frac{45}{100}\right)=0.45$

(ii) 127% = $\left(\frac{127}{100}\right)=1.27$

(iii) 3.6% =$\left(\frac{3.6}{100}\right)=\left(\frac{36}{10×100}\right)=\frac{36}{1000}=0.036$

(iv) 0.23% =$\left(\frac{0.23}{100}\right)=\left(\frac{23}{100×100}\right)=\frac{23}{10000}=0.0023$

#### Question 6:

Convert each of the following decimals into a percentage:

(i) 0.6
(ii) 0.42
(iii) 0.07
(iv) 0.005

We have:

(i) 0.6 = (0.6 $×$ 100)% = 60%
(ii) 0.42 = (0.42 $×$ 100)% = 42%
(iii) 0.07 = (0.07 $×$ 100)% = 7%
(iv) 0.005 = (0.005 $×$ 100)% = 0.5%

#### Question 7:

Find:

(i) 32% of 425
(ii)
(iii) 6.5% of 400
(iv) 136% of 70
(v) 2.8% of 35
(vi) 0.6% of 45

We have:
(i) 32% of 425 = $\left(\frac{32}{100}×425\right)=\left(\frac{32×17}{4}\right)=\left(8×17\right)=136$

(ii) $16\frac{2}{3}%$ of 16 = $\frac{50}{3}%$ of 16 = $\left(\frac{50}{3×100}×16\right)=\left(\frac{1}{6}×16\right)=\frac{8}{3}=2\frac{2}{3}$

(iii) 6.5% of 400 = $\left(\frac{6.5}{100}×400\right)=\left(\frac{65}{10×100}×400\right)=\left(\frac{65×4}{10}\right)=\frac{260}{10}=26$

(iv) 136% of 70 = $\left(\frac{136}{100}×70\right)=\left(\frac{136×7}{10}\right)=\left(\frac{952}{10}\right)=95.2$

(v) 2.8% of 35 = $\left(\frac{2.8}{100}×35\right)=\left(\frac{28}{10×100}×35\right)=\left(\frac{14×7}{100}\right)=\frac{98}{100}=0.98$

(vi) 0.6% of 45  = $\left(\frac{0.6}{100}×45\right)=\left(\frac{6}{10×100}×45\right)=\left(\frac{3×45}{5×100}\right)=\left(\frac{3×9}{100}\right)=\frac{27}{100}=0.27$

#### Question 8:

Find:

(i) 25% of Rs 76
(ii) 20% of Rs 132
(iii) 7.5% of 600 m
(iv)
(v) 8.5% of 5 kg
(vi) 20% of 12 litres

We have the following:

(i) 25% of Rs 76 = Rs $\left(76×\frac{25}{100}\right)$ = Rs $\left(76×\frac{1}{4}\right)$ = Rs 19

(ii) 20% of Rs 132 = Rs $\left(132×\frac{20}{100}\right)$= Rs $\left(132×\frac{1}{5}\right)$ = Rs 26.4

(iii) 7.5% of 600 m =

(iv) $3\frac{1}{3}%$ of 90 km = $\frac{10}{3}%$of 90 km =

(v) 8.5% of 5 kg =          [âˆµ1 kg = 1000 g]

(vi) 20% of 12 L =

#### Question 9:

Find the number whose 13% is 65.

Let x be the required number.

Then, 13% of x = 65
$\left(\frac{13}{100}×x\right)=65$
x = $\left(65×\frac{100}{13}\right)=500$
Hence, the required number is 500.

#### Question 10:

Find the number whose $6\frac{1}{4}%$ is 2.

Let x be the required number.
Then, $6\frac{1}{4}%$ of x = 2
$\left(6\frac{1}{4}%×x\right)=2$
$\left(\frac{25}{400}×x\right)=2$
⇒  x = $\left(2×\frac{400}{25}\right)=32$
Hence, the required number is 32.

#### Question 11:

What amount is 10% more than Rs 90?

10% of Rs 90 = Rs $\left(\frac{10}{100}×90\right)$ = Rs 9
∴ Amount that is 10% more than Rs 90 = Rs (90 + 9) = Rs 99

Hence, the required amount is Rs 99.

#### Question 12:

What amount is 20% less than Rs 60?

20% of Rs 60 = Rs $\left(60×\frac{20}{100}\right)$= Rs 12
∴ Amount that is 20% less than Rs 60 = Rs (60 − 12) = Rs 48

Hence, the required amount is Rs 48.

#### Question 13:

If 3% of x is 9, find the value of x.

3% of x = 9
$\left(\frac{3}{100}×x\right)=9$
x = $\left(9×\frac{100}{3}\right)=300$
Hence, the value of x is 300.

#### Question 14:

If 12.5% of x is 6, find the value of x.

12.5% of x = 6
$\left(\frac{12.5}{100}×x\right)=6$
x = $\left(6×\frac{100}{12.5}\right)=\left(6×8\right)=48$
Hence, the value of x is 48.

#### Question 15:

What per cent of 84 is 14?

Let x% of 84 be 14.
Then, $\left(\frac{x}{100}×84\right)=14$
$\frac{21x}{25}=14$
x = $\left(14×\frac{25}{21}\right)=\left(\frac{2×25}{3}\right)=\frac{50}{3}=16\frac{2}{3}%$
Hence, $16\frac{2}{3}%$ of 84 is 14.

#### Question 16:

What percentage is

(i) Rs 15 of Rs 120?
(ii) 36 minutes of 2 hours?
(iii) 8 hours of 2 days
(iv) 160 metres of 4 km?
(v) 175 mL of 1 litre?
(vi) 25 paise of Rs 4?

(i) Let x% of Rs 120 be Rs 15.
Then, Rs $\left(\frac{x}{100}×120\right)$ = Rs 15
⇒ $\left(\frac{6x}{5}\right)$ = 15
∴ x = $\left(\frac{15×5}{6}\right)%$ = $\left(\frac{25}{2}\right)%$ = 12.5%
Hence, 12.5% of Rs 120 is Rs 15.

(ii) Let x% of 2 h be 36 min.
Then, $\left(\frac{x}{100}×2×60\right)$ min = 36 min
⇒ $\left(\frac{120x}{100}\right)$ = 36
∴ x = $\left(\frac{36×100}{120}\right)%$ = 30%
Hence, 30% of 2 h is 36 min.

(iii) Let x% of 2 days be 8 h.
Then, $\left(\frac{x}{100}×2×24\right)$ h = 8 h
⇒ $\left(\frac{48x}{100}\right)$ = 8
∴ x = $\left(\frac{8×100}{48}\right)%$ = $16\frac{2}{3}%$
Hence, $16\frac{2}{3}%$ of 2 days is 8 h.

(iv) Let x% of 4 km be 160 m.
Then, $\left(\frac{x}{100}×4×1000\right)$ m = 160 m
⇒ 40x = 160
∴ x = $\left(\frac{160}{40}\right)%$ = 4%
Hence, 4% of 4 km is 160 m.

(v) Let x% of 1 L be 175 mL.
Then, $\left(\frac{x}{100}×1×1000\right)$ mL = 175 mL
⇒ 10x  = 175
∴ x = $\left(\frac{175}{10}\right)%$ = 17.5%
Hence, 17.5% of 1 L is 175 mL.

(vi) Let x% of Rs 4 be 25 paise.
Then, $\left(\frac{x}{100}×4×100\right)$ paise = 25 paise
⇒ 4x  = 25
∴ x = $\left(\frac{25}{4}\right)%$ = $6\frac{1}{4}%$
Hence, $6\frac{1}{4}%$ of Rs 4 is 25 paise.

#### Question 1:

Rupesh secures 495 marks out of 750 in his annual examination. Find the percentage of marks obtained by him.

Maximum marks of the examination = 750
Marks secured by Rupesh = 495
Percentage of marks secured = $\left(\frac{495}{750}×100\right)%$ = 66%

Hence, Rupesh scored 66% in the examination.

#### Question 2:

The monthly salary of a typist is Rs 15625. If he gets an increase of 12%, find his new salary.

Total monthly salary = Rs 15625
Increase percentage = 12%
∴ Amount increase = 12% of Rs 15625
= Rs $\left(15625×\frac{12}{100}\right)$ = Rs 1875
∴ New salary = Rs 15625 + Rs 1875
= Rs 17500
Hence, the new salary of the typist is Rs 17,500.

#### Question 3:

The excise duty on a certain item has been reduced to Rs 760 from Rs 950. Find the reduction per cent in the excise duy on that item.

Original excise duty on the item = Rs 950
Amount reduced on excise duty = Rs (950 − 760) = Rs 190
∴ Reduction percent =
= $\left(\frac{190}{950}×100\right)$ = 20
Hence, the excise duty on that item is reduced by 20%.

#### Question 4:

96% of the cost of a TV is Rs 10464. What is its total cost?

Let Rs x be the total cost of the TV set.

Now, 96% of the total cost of TV = Rs 10464
⇒ 96% of Rs x = Rs 10464
$\left(\frac{96}{100}×x\right)$ = 10464
x = $\left(\frac{10464×100}{96}\right)$ =  10900
Hence, the total cost of the TV set is Rs 10900.

#### Question 5:

70% of the students in a school are boys and the number of girls is 504. Find the number of boys in the school.

Let the total number of students be 100.
Then, number of boys = 70
∴ Number of girls = (100 − 70) = 30

Now, total number of students when the number of girls is 30 = 100
Then, total number of students when the number of girls is 504 = $\left(\frac{100}{30}×504\right)$ = 1680
∴ Number of boys = (1680 − 504) = 1176

Hence, there are 1176 boys in the school.

#### Question 6:

An ore contains 12% copper. How many kilograms of the ore are required to get 69 kg of copper?

Let x kg be the amount of the required ore.

Then, 12% of x kg = 69 kg
$\left(\frac{12}{100}×x\right)$ kg = 69 kg
x = $\left(\frac{69×100}{12}\right)$ kg = 575 kg

Hence, 575 kg of ore is required to get 69 kg of copper.

#### Question 7:

36% of the maximum marks are required to pass a test. A student gets 123 marks and is declared failed by 39 marks. Find the maximum marks.

Let x be the maximum marks.
Pass marks = (123 + 39) = 162
Then, 36% of x = 162
$\left(\frac{36}{100}×x\right)=162$
x$\left(\frac{162×100}{36}\right)$ = 450
∴ Maximum marks = 450

#### Question 8:

A fruit-seller had some apples. He sells 40% of them and still has 420 apples. Find the number of apples he had originally.

Suppose that the fruit seller initially had 100 apples.
Apples sold = 40
∴ Remaining apples = (100 − 40) = 60

Initial amount of apples if 60 of them are remaining = 100
Initial amount of apples if 1 of them is remaining = $\left(\frac{100}{60}\right)$
Initial amount of apples if 420 of them are remaining = $\left(\frac{100}{60}×420\right)$ = 700
Hence, the fruit seller originally had 700 apples.

#### Question 9:

In an examination, 72% of the total examinees passed. If the number of failures is 392, find the total number of examinees.

Suppose that 100 candidates took the examination.
Number of passed candidates = 72
Number of failed candidates = (100 − 72) = 28

Total number of candidates if 28 of them failed = 100
Total number of candidates if 392 of them failed = $\left(\frac{100}{28}×392\right)$ = 1400
Hence, the total number of examinees is 1400.

#### Question 10:

After decuting a commission of 5%, a moped costs Rs 15200. What is its gross value?

Suppose that the gross value of the moped is Rs x.
Commission on the moped = 5%
Price of moped after deducting the commission = Rs ( x − 5% of x)
= Rs $\left(x-\frac{5x}{100}\right)$ = Rs $\left(\frac{100x-5x}{100}\right)$ = Rs $\left(\frac{95x}{100}\right)$
Now, price of the moped after deducting the commission = Rs 15200
Then, Rs $\left(\frac{95x}{100}\right)$= Rs 15200
x  = Rs $\left(\frac{15200×100}{95}\right)$ = Rs (160 $×$ 100) = Rs 16000
Hence, the gross value of the moped is Rs 16000.

#### Question 11:

Gunpowder contains 75% of nitre and 10% of sulphur, and the rest of it is charcoal. Find the amount of charcoal in 8 kg of gunpowder.

Total quantity of gunpowder = 8 kg = 8000 g                         (1 kg = 1000 g)
Quantity of nitre in it = 75% of 8000 g
= $\left(\frac{75}{100}×8000\right)$ g = 6000 g = 6 kg

Quantity of sulphur in it = 10% of 8000 g
= $\left(\frac{10}{100}×8000\right)$ g = 800 g = 0.8 kg
∴ Quantity of charcoal in it = {8000 − (6000 + 800)} g
= (8000 − 6800) g
= 1200 g = 1.2 kg

Hence, the amount of charcoal in 8 kg of gunpowder is 1.2 kg.

#### Question 12:

Chalk contains 3% of carbon, 10% of calcium and 12% of oxygen. Find the amount in grams of each of these substances in 1 kg of chalk.

Total quantity of chalk = 1 kg = 1000 g

Now, we have the following:

Quantity of carbon in it = 3% of 1000 g
=$\left(\frac{3}{100}×1000\right)$ = 30 g
Quantity of calcium in it = 10% of 1000 g
= $\left(\frac{10}{100}×1000\right)$ g = 100 g
Quantity of oxygen in it = 12% of 1000 g
= $\left(\frac{12}{100}×1000\right)$ g = 120 g

#### Question 13:

Sonal went to school for 219 days in a full year. If her attendance is 75%, find the number of days on which the school was open.

Let x be the total number of days on which the school was open.
Number of days when Sonal went to school = 219
Percentage of attendance = 75

Thus, 75% of x = 219
$\left(\frac{75}{100}×x\right)=219$
x = $\left(\frac{219×100}{75}\right)=292$ days
Hence, the school was open for a total of 292 days.

#### Question 14:

3% commission on the sale of a property amounts to Rs 42660. What is the total value of the property?

Let the total value of the property be Rs x.
Percentage of commission = 3
Amount of commission  = Rs 42660
Thus, 3% of Rs x = Rs 42660
$\left(\frac{3}{100}×x\right)$ = 42660
x = $\left(\frac{42660×100}{3}\right)=1422000$
Hence, the total value of the property is Rs 14,22,000.

#### Question 15:

In an election, there were two candidates A and B. The total number of voters in this constituency was 60000 and 80% of the total votes were polied. If 60% of the polied votes were cast in favour of A, how many votes were received by B?

Total number of eligible voters = 60000
Number of voters who gave their votes = 80% of 60000
= $\left(\frac{80}{100}×60000\right)$ = 48000
Number of votes in favour of candidate A = 60% of 48000
= $\left(\frac{60}{100}×48000\right)$ = 28800
∴ Number of votes received by candidate B = (48000 − 28800) = 19200

Hence, candidate B recieved 19,200 votes.

#### Question 16:

The price of a shirt is reduced by 12% in a discount sale. If its present price is Rs 1188, find its original price.

Let us assume that the original price of the shirt is Rs x.
Discount on the shirt = 12%
So, value of discount on the shirt = 12% of Rs x
= Rs $\left(\frac{12}{100}×x\right)$ = Rs $\left(\frac{12x}{100}\right)$
Value of the shirt after discount = Rs $\left(x-\frac{12x}{100}\right)$
= Rs $\left(\frac{100x-12x}{100}\right)$ = Rs $\left(\frac{88x}{100}\right)$
Present price of the shirt = Rs 1188
Then, Rs $\left(\frac{88x}{100}\right)$ = Rs 1188
⇒ 88x = (1188 $×$ 100)
⇒ 88x = 118800
∴  x = $\left(\frac{118800}{88}\right)$ = 1350

Hence, the original price of the shirt is Rs 1350.

#### Question 17:

The price of a sweater is increased by 8%. If its increased price is Rs 1566, find the original price.

Let us assume that the original price of the sweater is Rs. x
Increased percentage = 8%
So, value of increase on the sweater = 8% of Rs x
= Rs $\left(\frac{8}{100}×x\right)$ = Rs $\left(\frac{2x}{25}\right)$
Increased price of the sweater = Rs $\left(x+\frac{2x}{25}\right)$
= Rs $\left(\frac{25x+2x}{25}\right)$ = Rs $\left(\frac{27x}{25}\right)$
However, increased price of the sweater = Rs 1566
Then, Rs $\left(\frac{27x}{25}\right)$ = Rs 1566
∴  x$\left(\frac{1566×25}{27}\right)$ = 1450
Hence, the original price of the sweater is Rs 1450

#### Question 18:

After spending 80% of his income and giving 10% of the remainder in a charity, a man has Rs 46260 left with him. Find his income.

Let the income of the man be Rs x.
Then, income spent = 80% of Rs. x
=
Rs $\left(\frac{80}{100}×x\right)$ = Rs $\left(\frac{80x}{100}\right)$ = Rs $\left(\frac{4x}{5}\right)$
Amount left after all the expenditure = Rs $\left(x-\frac{4x}{5}\right)$ = Rs $\left(\frac{5x-4x}{5}\right)$ = Rs $\left(\frac{x}{5}\right)$
Amount given to the charity = 10% of Rs $\left(\frac{x}{5}\right)$
= Rs $\left(\frac{10}{100}×\frac{x}{5}\right)$ = Rs $\left(\frac{10x}{500}\right)$= Rs $\left(\frac{x}{50}\right)$
Amount left after the charity = Rs $\left(\frac{x}{5}-\frac{x}{50}\right)$
= Rs $\left(\frac{10x-x}{50}\right)$ = Rs $\left(\frac{9x}{50}\right)$
Now, we have:
Rs $\left(\frac{9x}{50}\right)$ = Rs 46260
x = Rs $\left(\frac{46260×50}{9}\right)$ = Rs 257000
Hence, the income of the man is Rs 2,57,000.

#### Question 19:

A number is increased by 20% and the increased number is decreased by 20%. Find the net increase of decrease per cent.

Let the number be 100.
Increase in the number = 20%
Increased number = (100 + 20) =120
Now, decrease in the number = (20% of 120)
= $\left(\frac{20}{100}×120\right)=24$
New number = (120 − 24) = 96
Net decrease = (100 − 96) = 4
Net decrease percentage = $\left(\frac{4}{100}×100\right)$ = 4
Hence, the net decrease is 4%.

#### Question 20:

The salary of an officer is increased by 20%. By what percentage should the new salary be reduced to restore the original salary?

Let the original salary be Rs 100.
Increase in it = 20%
Salary after increment = Rs (100 + 20) = Rs 120
To restore the original salary, reduction required = Rs (120 − 100) = Rs 20
Reduction on Rs 120 = Rs 20
∴ Reduction percentage = $\left(\frac{20}{120}×100\right)$ = $\left(\frac{100}{6}\right)$ = $16\frac{2}{3}$
Hence, the required reduction on the new salary is $16\frac{2}{3}%$.

#### Question 21:

A property dealer charges commission at the rate of 2% on the first Rs 200000, 1% on the next Rs 200000 and 0.5% on the remaining price. Find his commission on the property that has been sold for Rs 540000.

Total cost of the property = Rs 540000
Commission on the first Rs 200000 = 2% of Rs 200000
= $\left(\frac{2}{100}×200000\right)$ = Rs 4000

Commission on the next Rs 200000 = 1% of Rs 200000
= $\left(\frac{1}{100}×200000\right)$ = Rs 2000
Remaining amount = Rs (540000 − 400000) = Rs 140000
∴ Commission on Rs 140000 = 0.5% of Rs 140000
= Rs $\left(\frac{0.5}{100}×140000\right)$
= Rs $\left(\frac{5}{1000}×140000\right)$ = Rs 700
Thus, total commission on the property worth Rs 540000 = Rs (4000 + 2000 + 700)
= Rs 6700
Hence, the commission of the property dealer on the property that has been sold for Rs 540000 is Rs 6700.

#### Question 22:

Nikhil's income is 20% less than that of Akhil. How much per cent is Akhil's income more than that of Nikhil's?

Let Akhil's income be Rs 100.
∴ Nikhil's income = Rs 80
Akhil's income when Nikhil's income is Rs 80 = Rs 100
Akhil's income when Nikhil's income is Rs 100 = Rs $\left(\frac{100}{80}×100\right)$ = Rs 125
i.e., if Nikhil's income is Rs.100, then Akhil's income is Rs 125.
Hence, Akhil's income is more than that of Nikhil's by 25%.

#### Question 23:

Jhon's income is 20% more than that of Mr Thomas. How much per cent is the income of Mr Thomas less than that of John?

Let Rs 100 be the income of Mr. Thomas.
∴ John's income = Rs 120
Mr. Thomas' income when John's income is Rs 120 = Rs 100
Mr. Thomas' income when John's income is Rs 100 = Rs $\left(\frac{100}{120}×100\right)$ = Rs $83\frac{1}{3}$
Hence, Mr Thomas' income is less than that of John's by $16\frac{2}{3}%$.

#### Question 24:

The value of a machine depreciated 10% every year. If its present value is Rs 387000, what was its value 1 year ago?

Let Rs x be the value of the machine one year ago.

Then, its present value = 90% of Rs x

= Rs $\left(\frac{90}{100}×x\right)$ = Rs $\left(\frac{9x}{10}\right)$
It is given that present value of the machine = Rs 387000
x = Rs$\left(\frac{387000×10}{9}\right)$= Rs $\left(43000×10\right)$= Rs 430000

Hence, the value of the machine a year ago was Rs 430000.

#### Question 25:

The value of a car decreases annually by 20%. If the present value of the car be Rs 450000, what will be its value after 2 years?

The present value of  the car = Rs 450000
The decrease in its value after the first year = 20% of Rs 450000
= Rs $\left(\frac{20}{100}×450000\right)$= Rs 90000
The depreciated value of the car after the first year = Rs (450000 − 90000) = Rs 360000
The decrease in its value after the second year = 20% of Rs 360000
= Rs $\left(\frac{20}{100}×360000\right)$ = Rs 72000
The depreciated value of the car after the second year = Rs (360000 − 72000) = Rs 288000

Hence, the value of the car after two years will be Rs 288000.

#### Question 26:

The population of a town increases 10% annually. If its present population is 60000, what will be its population after 2 years?

Present population of the town = 60000
Increase in population of the town after the 1 year = 10% of 60000
= $\left(\frac{10}{100}×60000\right)$ = 6000
Thus, population of the town after 1 year = 60000 + 6000 = 66000
Increase in population after 2 years = 10% of 66000
= $\left(\frac{10}{100}×66000\right)$ = 6600
Thus, population after the second year = 66000 + 6600 = 72600
Hence, the population of the town after 2 years will be 72600.

#### Question 27:

Due to an increase in the price of sugar by 25%, by how much per cent must a householder decrease the consumption of sugar so that there is no increase in the expenditure on sugar?

Let the consumption of sugar originally be 1 unit and let its cost be Rs 100
New cost of 1 unit of sugar = Rs 125
Now, Rs 125 yield 1 unit of sugar.
∴ Rs 100 will yield $\left(\frac{1}{125}×100\right)$ unit = $\left(\frac{4}{5}\right)$ unit of sugar.
Reduction in consumption = $\left(1-\frac{4}{5}\right)$ = $\left(\frac{1}{5}\right)$ unit
∴ Reduction percent in consumption = $\left(\frac{1}{5}×\frac{1}{1}×100\right)$ %= $\left(\frac{100}{5}\right)$ %= 20%

#### Question 1:

Mark (âœ“) against the correct answer
$\frac{3}{4}$ as rate per cent is

(a) 7.5%
(b) 75%
(c) 0.75%
(d) none of these

(b) 75%

$\frac{3}{4}$ = $\left(\frac{3}{4}×100\right)%$ = 75%

#### Question 2:

Mark (âœ“) against the correct answer
The ratio 2 : 5 as rate per cent is

(a) 4%
(b) 0.4%
(c) 40%
(d) 14%

(c) 40%

2 : 5 = $\frac{2}{5}$ = $\left(\frac{2}{5}×100\right)%$ = 40%

#### Question 3:

Mark (âœ“) against the correct answer
$8\frac{1}{3}%$ expressed as a fraction, is

(a) $\frac{25}{3}$
(b) $\frac{3}{25}$
(c) $\frac{1}{12}$
(d) $\frac{1}{4}$

(c) $\frac{1}{12}$

$8\frac{1}{3}%$ = $\frac{25}{3}%$ = $\left(\frac{25}{3}×\frac{1}{100}\right)=\left(\frac{1}{3×4}\right)=\frac{1}{12}$

#### Question 4:

Mark (âœ“) against the correct answer
If x% of 75 = 9, then the value of x is

(a) 16
(b) 14
(c) 12
(d) 8

(c) 12
We have x% of 75 = 9
$\left(\frac{x}{100}×75\right)=9$
x = $\left(\frac{9×100}{75}\right)=12$
Hence, the value of x is 12

#### Question 5:

Mark (âœ“) against the correct answer
What per cent of $\frac{2}{7}$ is $\frac{1}{35}$?

(a) 25%
(b) 20%
(c) 15%
(d) 10%

(d) 10%

Let x be the required percent.
Then, x % of $\frac{2}{7}$ = $\frac{1}{35}$
$\left(\frac{x}{100}×\frac{2}{7}\right)=\frac{1}{35}$
x = $\left(\frac{100×7}{35×2}\right)$= 10
Hence, 10% of $\frac{2}{7}$ is $\frac{1}{35}$

#### Question 6:

Mark (âœ“) against the correct answer
What per cent of 1 day is 36 minutes?

(a) 25%
(b) 2.5%
(c) 3.6%
(d) 0.25%

(b) 2.5%

Let x % of 1 day be 36 min.
Then, $\left(\frac{x}{100}×1×24×60\right)$ min = 36 min
x = $\left(\frac{36×100}{24×60}\right)$ = $\left(\frac{3×5}{2×3}\right)%=\left(\frac{5}{2}\right)%=2.5%$

Hence, 2.5% of 1 day is 36 min.

#### Question 7:

Mark (âœ“) against the correct answer
A number increased by 20% gives 42. The number is

(a) 35
(b) 28
(c) 36
(d) 30

(a) 35
Let the required number be x.
Then, x + 20% of x = 42
$\left(x+\frac{20x}{100}\right)=42$
$\left(x+\frac{x}{5}\right)=42$
$\left(\frac{5x+x}{5}\right)=42$        [âˆµ LCM of 1 and 5 = 5]
$\left(\frac{6x}{5}\right)=42$
x = $\left(\frac{42×5}{6}\right)=35$
Hence, the required number is 35.

#### Question 8:

Mark (âœ“) against the correct answer
A number decreased by 8% gives 69. The number is

(a) 80
(b) 75
(c) 85
(d) none of these

(b) 75
Let the required number be x.
Then, x − 8% of x = 69
$\left(x-\frac{8x}{100}\right)$ = 69
$\left(x-\frac{2x}{25}\right)$ = 69
⇒ $\left(\frac{25x-2x}{25}\right)$ = 69            [Since L.C.M. of 1 and 25 = 25]
$\left(\frac{23x}{25}\right)=69$
x = $\left(\frac{69×25}{23}\right)$ = 75
Hence, the required number is 75

#### Question 9:

Mark (âœ“) against the correct answer
An ore contains 5% copper. How much ore is required to obtain 400 g of copper?

(a) 2 kg
(b) 4 kg
(c) 6 kg
(d) 8 kg

(d) 8 kg
Let x kg be the required amount of ore.
Then, 5% of x kg = 400 g = 0.4 kg        [âˆµ 1 kg = 1000 g]
$\left(\frac{5}{100}×x\right)=0.4$
x = $\left(\frac{0.4×100}{5}\right)$ = 8
Hence, 8 kg of ore is required to obtain 400 g of copper.

#### Question 10:

Mark (âœ“) against the correct answer
After deducting a commission of 10% a TV costs Rs 18000. What is its gross value?

(a) Rs 18800
(b) Rs 20000
(c) Rs 19800
(d) none of these

(b) Rs. 20000
Suppose that the gross value of the TV is Rs x.
Commission on the TV = 10%
Price of the TV after deducting the commission = Rs (x − 10% of x)
= Rs $\left(x-\frac{10}{100}x\right)$ = Rs $\left(\frac{100x-10x}{100}\right)$ = Rs $\left(\frac{9x}{10}\right)$
However, price of the TV after deducting the commission = Rs 18000
Then, Rs $\left(\frac{9x}{10}\right)$ = Rs 18000
x = $\left(\frac{18000×10}{9}\right)$ = Rs (2000 $×$ 10) = Rs 20000
Hence, the gross value of the TV is Rs 20,000

#### Question 11:

Mark (âœ“) against the correct answer
On increasing the salary of a man by 25%, it becomes Rs 20000. What was his original salary?

(a) Rs 15000
(b) Rs 16000
(c) Rs 18000
(d) Rs 25000

(b) Rs. 16000
Let us assume that the original salary of the man is Rs x.
Increase in it = 25%
Value increased in the salary = 25% of Rs. x
= Rs $\left(\frac{25}{100}×x\right)$ = Rs $\left(\frac{x}{4}\right)$
Salary after increment= Rs $\left(x+\frac{x}{4}\right)$ = Rs $\left(\frac{5x}{4}\right)$
However, increased salary = Rs 20000
Then, Rs $\left(\frac{5x}{4}\right)$ = Rs 20000
∴  x = Rs $\left(\frac{20000×4}{5}\right)$ = Rs 16000
Hence, the original salary of the man is Rs 16,000

#### Question 12:

Mark (âœ“) against the correct answer
In an examination, 95% of the total examinees passed. If the number of failures is 28, how many examinees were there?

(a) 600
(b) 480
(c) 560
(d) 840

(c) 560
Suppose that the number of examinees is 100.
Number of passed examinees = 95
Number of failed examinees = (100 − 95) = 5

Total number of examinees if 5 of them failed = 100
Total number of examinees if 28 of them failed = $\left(\frac{100}{5}×28\right)=\left(20×28\right)=560$
Hence, there were 560 examinees.

#### Question 13:

Mark (âœ“) against the correct answer
A fruit-seller had some apples. He sells 40% of them and still has 420 apples. How many apples had he in all?

(a) 588
(b) 600
(c) 700
(d) 725

(c) 700
Suppose that the fruit seller initially had 100 apples.
Number of apples sold = 40
∴ Number of remaining apples = (100 − 40) = 60

Initial number of apples if 60 of them are remaining = 100
Initial number of apples if 420 of them are remaining = $\left(\frac{100}{60}×420\right)$ = 700
Hence, the fruit seller originally had 700 apples with him.

#### Question 14:

Mark (âœ“) against the correct answer
The value of a machine depreciated 10% annually. If its present value is Rs 25000, what will be its value after 1 year?

(a) Rs 27500
(b) Rs 22500
(c) Rs 25250
(d) none of these

(c) Rs. 25250

Present value of the machine = Rs 25000
Decrease in its value after 1 year = 10% of Rs 25000
= Rs $\left(\frac{10}{100}×25000\right)$ = Rs 2500
Depreciated value after 1 year = Rs (25000 − 2500) = Rs 22500

Hence, the value of the machine after 1 year will be Rs 22500

#### Question 15:

Mark (âœ“) against the correct answer
8% of a number is 6. What is the number?

(a) 48
(b) 96
(c) 75
(d) 60

(c) 75

Let the required number be x. Then, we have:
8% of x = 6
$\left(\frac{8}{100}×x\right)=6$
x = $\left(\frac{6×100}{8}\right)=75$
Hence, the required number is 75

#### Question 16:

Mark (âœ“) against the correct answer
60% of 450 = ?

(a) 180
(b) 210
(c) 270
(d) none of these

(c) 270
60% of 450 = $\left(\frac{60}{100}×450\right)$
= $\left(\frac{3}{5}×450\right)$ = (3 $×$ 90) = 270

#### Question 17:

Mark (âœ“) against the correct answer
On reducing the value of a chair by 6% it becomes Rs 658. The original value of the chair is

(a) Rs 750
(b) Rs 720
(c) Rs 500
(d) Rs 700

(d) Rs. 700
Let us assume that the original price of the chair is Rs x.
Reduce percentage on the chair = 6%
So, value of reduction on the chair = 6% of Rs. x
= Rs $\left(\frac{6}{100}×x\right)$ = Rs $\left(\frac{3x}{50}\right)$
Reduced price of the chair = Rs $\left(x-\frac{3x}{50}\right)$
= Rs $\left(\frac{50x-3x}{50}\right)$ = Rs $\left(\frac{47x}{50}\right)$
However, present price of the chair = Rs 658
Then, Rs $\left(\frac{47x}{50}\right)$ = Rs 658
⇒ Rs $\left(\frac{47x}{50}\right)$ = Rs 658
x = Rs $\left(\frac{658×50}{47}\right)$ = Rs $\left(14×50\right)=700$
Hence, the original price of the chair is Rs 700

#### Question 18:

Mark (âœ“) against the correct answer
70% of students in a school are boys. If the number of girls is 240, how many boys are there in the school?

(a) 420
(b) 560
(c) 630
(d) 480

(b) 560

Let the total number of students be 100.
Then, number of boys = 70
∴ Number of girls = (100 − 70) = 30

Now, total number of students if there are 30 girls = 100
Total number of students if there are 240 girls = $\left(\frac{100}{30}×240\right)=800$
∴ Number of boys = (800 − 240) = 560

Hence, there are 560 boys in the school.

#### Question 19:

Mark (âœ“) against the correct answer
If 11% of a number exceeds 7% of the number by 18, the number is

(a) 72
(b) 360
(c) 450
(d) 720

(c) 450

Let x be the number.
(11% of x) − (7% of x) = 18
$\left(\frac{11x}{100}-\frac{7x}{100}\right)=18$
$\frac{4x}{100}=18$
x = $\left(\frac{18×100}{4}\right)=\left(18×25\right)=450$
Hence, the required number is 450

#### Question 20:

Mark (âœ“) against the correct answer
If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 105

(a) 60
Let x be the number.
According to question, we have:
(35% of x ) + 39 = x
$\left(\frac{35}{100}×x\right)+39=x$
$\left(\frac{7x}{20}\right)+39=x$
$\left(x-\frac{7x}{20}\right)=39$
$\left(\frac{20x-7x}{20}\right)=39$
$\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$
x = $\left(\frac{39×20}{13}\right)$ = 60
Hence, the required number is 60

#### Question 21:

Mark (âœ“) against the correct answer
In an examination it is required to get 36% to pass. A student gets 145 marks and fails by 35 marks. The maximum marks are

(a) 400
(b) 450
(c) 500
(d) 600

(c) 500
Let x be the maximum marks.
Pass marks = (145 + 35) = 180
∴ 36% of x = 180
$\left(\frac{36}{100}×x\right)=180$
x = $\left(\frac{180×100}{36}\right)=\left(5×100\right)=500$
Hence, maximum marks = 500

#### Question 22:

Mark (âœ“) against the correct answer
A number decreased by 40% gives 135. The number is

(a) 175
(b) 200
(c) 250
(d) 225

(d) 225
Let x be the number.
According to question, we have:
x − 40% of x = 135
$\left(x-\frac{40x}{100}\right)=135$
$\left(\frac{100x-40x}{100}\right)=135$
$\left(\frac{60x}{100}\right)=135$
x = $\left(\frac{135×100}{60}\right)$ = 225
Hence, the required number is 225

#### Question 1:

Convert:

(i) $\frac{4}{5}$ into a percentage
(ii) $\frac{7}{4}$ into a percentage
(iii) 45% into a percentage
(iv) 105% into a percentage
(v) 15% into a percentage
(vi) 12 : 25 into a percentage

We have:

(i) $\frac{4}{5}$= $\left(\frac{4}{5}×100\right)%=\left(4×20\right)%=80%$

(ii) $\frac{7}{4}$=$\left(\frac{7}{4}×100\right)%=\left(7×25\right)%=175%$

(iii) 45% = $\left(\frac{45}{100}\right)=\left(\frac{9}{20}\right)$

(iv) 105% =$\left(\frac{105}{100}\right)=\left(\frac{21}{20}\right)$

(v) 15% =$\frac{15}{100}=\frac{3}{20}$=  3 : 20

(vi) 12 : 25 = $\frac{12}{25}=\left(\frac{12}{25}×100\right)%=\left(12×4\right)%=48%$

#### Question 2:

(i) What per cent of 1 kg is 125 g?
(ii) What per cent of 80 m is 24 m?

(i) Let x% of 1 kg be 125g.
Then,
⇒ 10x = 125
⇒ x =$\left(\frac{125}{10}\right)%=12\frac{1}{2}%$
Hence, $12\frac{1}{2}%$ of 1 kg is 125 g.

(ii) Let x% of 80 m be 24 m.
Then,
⇒ $\left(\frac{4x}{5}\right)$ = 24
⇒ x =$\left(24×\frac{5}{4}\right)%=30%$
Hence, $30%$ of 80 m is 24 m.

#### Question 3:

(i) Find $16\frac{2}{3}%$ of 30.
(ii) Find 15% of Rs 140

(i) $16\frac{2}{3}%$ of 30 = $\frac{50}{3}%$ of 30
= $\left(\frac{50}{3×100}×30\right)$
= 5

(ii) 15% of Rs 140 = Rs $\left(\frac{15}{100}×140\right)$
= Rs (3 $×$ 7)
= Rs 21

#### Question 4:

(i) Find the number whose $6\frac{1}{4}%$ is 5.
(ii) Find 0.8% of 45.

(i) Let x be the required number.
Then, $6\frac{1}{4}%$ of x = 5
⇒ $\frac{25}{4}%$ of x = 5
⇒ $\left(\frac{25}{4×100}×x\right)=25$
⇒$\left(\frac{x}{16}\right)=5$
x = (5 $×$ 16) = 80
Hence, the required number is 80.

(i) 0.8% of 45 =$\left(\frac{0.8}{100}×45\right)$
=$\left(\frac{8}{10×100}×45\right)$
= $\left(\frac{72}{200}\right)=\left(\frac{36}{100}\right)=0.36$
Hence, 0.8% of 45 is 0.36.

#### Question 5:

A number is increased by 10% and the increased number is decreased by 10%. Show that the net decrease is 1%.

Let x be the number.
The number is increased by 10%.
∴ Increased number = 110% of x = $\left(x×\frac{110}{100}\right)=\left(\frac{11x}{10}\right)$
The number is, then, decreased by 10%.

∴ Decreased number = 90% of $\left(\frac{11x}{10}\right)$ = $\left(\frac{11x}{10}×\frac{90}{100}\right)=\left(\frac{99x}{100}\right)$

Net decrease = $\left(x-\frac{99x}{100}\right)=\frac{\left(100x-99x\right)}{100}=\frac{x}{100}$
Net decrease percentage = $\left(\frac{x}{100}×\frac{1}{x}×100\right)=1$

#### Question 6:

The value of a machine depriciates at the rate of 10% per annum. If its present value is Rs 10000, what will be its value after 2 years?

The present value of the machine = Rs 10000
The decrease in its value after the 1st year = 10% of Rs 10000
= Rs $\left(\frac{10}{100}×10000\right)$ = Rs 1000
The depreciated value of the machine after the 1st year = Rs (10000 − 1000) =Rs 9000
The decrease in its value after the 2nd year = 10% of Rs 9000
= Rs $\left(\frac{10}{100}×9000\right)$ = Rs 900
The depreciated value of the machine after the 2nd year = Rs (9000 − 900) = Rs 8100

Hence, the value of the machine after two years will be Rs 8100.

#### Question 7:

The population of a town increases at 5% per annum. Its present population is 16000. What will be its population after 2 years?

The present population of the town = 16000
Increase in population after 1 year = 5% of 16000
= $\left(\frac{5}{100}×16000\right)$ = 800
Thus, population after one year = 16000 + 800 = 16800
Increase in population after 2 years = 5% of 16800
= $\left(\frac{5}{100}×16800\right)$ = 840
Increased population after two years = 16800 + 840 = 17640

Hence, the population of the town after two years will be 17,640.

#### Question 8:

The price of a teaset is increased by 5%. If the increased price is Rs 441, what is its original price?

Let us assume that the original price of the tea set is Rs. x
Increase in it = 5%
So, value increased on the tea set = 5% of Rs. x
= Rs. $\left(\frac{5}{100}×x\right)$ = Rs. $\left(\frac{x}{20}\right)$
Then, increased price of the tea set = Rs. $\left(x+\frac{x}{20}\right)$
= Rs. $\left(\frac{20x+x}{20}\right)$ = Rs. $\left(\frac{21x}{20}\right)$
However, increased price = Rs. 441
Then, Rs. $\left(\frac{21x}{20}\right)$ = Rs. 441
∴  x = $\left(\frac{441×20}{21}\right)$ = 420
Hence, the original price of the tea set is Rs 420

#### Question 9:

Mark (âœ“) against the correct answer
$6\frac{1}{4}%$ expressed as a fraction is

(a) $\frac{1}{8}$
(b) $\frac{1}{16}$
(c) $\frac{4}{25}$
(d) $\frac{1}{25}$

(b) $\frac{1}{16}$

$6\frac{1}{4}%$ = $\frac{25}{4}%$ =$\left(\frac{25}{4×100}\right)=\frac{1}{16}$

#### Question 10:

Mark (âœ“) against the correct answer
If x% of 75 = 12, then the value of x is

(a) 8
(b) 10
(c) 12
(d) 16

(c) 12

Given that x% of 75 = 12
Then, $\left(\frac{x}{100}×75\right)=12$
x = $\left(\frac{12×100}{75}\right)$ =16
Hence, the value of x is 16

#### Question 11:

Mark (âœ“) against the correct answer
A number increased by 20% gives 30. The number is

(a) 150
(b) 6
(c) 25
(d) 60

(c) 25

Let the number be x. Then, we have:
120% of x  = increased number
⇒ 30 = $\left(x×\frac{120}{100}\right)$
⇒ 30 = $\left(\frac{6x}{5}\right)$
x = $\left(30×\frac{5}{6}\right)=25$
Hence, the required number is 25

#### Question 12:

Mark (âœ“) against the correct answer
5% of a number is 9. The number is

(a) 120
(b) 140
(c) 160
(d) 180

(d) 180

Let the required number be x. Then, we have:
5% of x = 9
$\left(\frac{5}{100}×x\right)=9$
x = $\left(9×\frac{100}{5}\right)=\left(9×20\right)=180$

#### Question 13:

Mark (âœ“) against the correct answer
If 35% of a number added to 39 is the number itself, the number is

(a) 60
(b) 65
(c) 75
(d) 70

(a) 60
Let the number be x.
According to question, we have:
(35% of x ) + 39 = x
$\left(\frac{35}{100}×x\right)+39=x$
$\left(\frac{7x}{20}\right)+39=x$
$\left(x-\frac{7x}{20}\right)=39$
$\left(\frac{20x-7x}{20}\right)=39$
$\left(\frac{13x}{20}\right)=39\phantom{\rule{0ex}{0ex}}$
x = $\left(\frac{39×20}{13}\right)$ = 60
Hence, the required number is 60.

#### Question 14:

Mark (âœ“) against the correct answer
In an examination it is required to get 36% to pass. A student gets 160 marks and fails by 20 marks. The maximum marks are

(a) 400
(b) 450
(c) 500
(d) 600

(c) 500
Let x be the maximum marks.
Pass marks = (160 + 20) = 180
∴ 36% of x = 180
$\left(\frac{36}{100}×x\right)=180$
x = $\left(\frac{180×100}{36}\right)=\left(5×100\right)=500$
Hence, maximum marks = 500

#### Question 15:

Fill in the blanks.

(i) 3 : 4 = (......)%
(ii) 0.75 = (......)%
(iii) 6% = ...... (express in decimals)
(iv) If x decreased by 40% gives 135, then x = ...... .
(v) (11% of x) − (7% of x) = 18 â‡¨ x = ......

We have the following:

(i) 3 : 4 = (75)%
Explanation: 3 : 4 = $\frac{3}{4}$ = $\left(\frac{3}{4}×100\right)%=\left(3×25\right)%=75%$

(ii) 0.75 = (75)%
Explanation: ( 0.75 $×$ 100)% = 75%

(iii) 6% = 0.06 (expressed in decimals)
Explanation: 6% = $\frac{6}{100}=0.06$

(iv) If x decreased by 40% gives 135, then x = 225
Explanation:
Let the number be x.
According to question, we have:
x − 40% of x = 135
⇒ $\left(x-\frac{40x}{100}\right)=135$
⇒ $\left(\frac{100x-40x}{100}\right)=135$
⇒ $\left(\frac{60x}{100}\right)=135$
⇒ x = $\left(\frac{135×100}{60}\right)$ = 225

(v) (11% of x) − (7% of x) = 18
⇒ x = 450
Explanation:

(11% of x) − (7% of x) = 18
⇒ $\left(\frac{11x}{100}-\frac{7x}{100}\right)=18$
⇒ $\frac{4x}{100}=18$
∴ x = $\left(\frac{18×100}{4}\right)=\left(18×25\right)=450$

#### Question 16:

Write 'T' for true and 'F' for false

(i) $\frac{3}{4}$ as rate per cent is 75%
(ii) $12\frac{1}{2}%$ expressed as a fraction is $\frac{1}{8}.$
(iii) 2 : 5 = 25%
(iv) 80 % of 450 = 360.
(v) 20% of 1 litre = 200 mL.

(i) True (T)
Justification: $\left(\frac{3}{4}×100\right)%$ = 75%

(ii) True (T)
Justification: $12\frac{1}{2}%=\frac{25}{2}%$ = $\left(\frac{25}{2×100}\right)=\frac{1}{8}$

(iii) False (F)
Justification: $\frac{2}{5}$ = $\left(\frac{2}{5}×100\right)$% = $\left(2×20\right)%$ = 40%

(iv) True (T)
Justification: 80% of 450 = $\left(\frac{80}{100}×450\right)=\left(\frac{80×9}{2}\right)=\left(40×9\right)=360$

(v) True (T)
Justification: 20% of 1 L = 20% of 1000 mL
= $\left(\frac{20}{100}×1000\right)$ mL = 200 mL

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