WBBSE Solutions For Class 8 Maths Chapter 11 Percentage

Percentage

Today we will go to the fair at Khadinan village. My sister and I will go there with our elder brother. There is Rs. 75 with my elder brother, Rs. 50 with me and Rs. 35 with my sister.

Question 1.  Let’s calculate in percentage the amount which my elder brother has more than me.

Solution:

My brother has more than me Rs. 75 – Rs. 50 = Rs. 25

The excess amount he has in percentage is =\(\frac{25}{50}\) x100 = Rs. 50

My brother has 50% more amount than me. In other language,

On Rs. 50, Rs. 25 is more

On Rs.1, Rs. \(\frac{25}{50}\) is more 50

On Rs. 100, Rs.\(\frac{25}{50}\) x 100 = Rs. 50 is more.

My brother has 50% more amount than me.

Question 2. Let’s calculate in percentage the amount which I have less than my elder brother.

Solution:

My elder brother’s money in comparison with me, I have less Rs. 25

For Rs. 1, I have less amount of Rs. \(\frac{25}{75}\)

For Rs. 100, I have less amount of Rs. \(\frac{25}{75} \times 100=\text { Rs. } 33 \frac{1}{3}\)

I have \(33 \frac{1}{3} \%\) less amount than my elder brother.

The sum total of amount we have = Rs. (75+50+35) = Rs. 160

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My elder brother calculated that we spent 10% of the amount at the time of going to the fair.

Let’s work out the sum we spent at the time of going to the fair.

= \(10 \%=\frac{.10}{100} \text { part }=\frac{1}{10} \text { part }\)

∴ Of the total amount the part was spent at the time of going to the fair.

So, of Rs. 160, 10% = Rs. \(\left(160 \times \frac{10}{100}\right)\) = Rs. 16

∴ Rs. 16 was spent at the time of going to the fair.

I came across Sunit in the fair.

Four of us rode on merry-go-round. We spent Rs. 40 for this ride.

Question 3. Let’s work out the amount we spent for riding on the merry-go-round.

Solution:

Out of Rs. 160, we spent Rs. 40

Out of Rs. 1 the amount spent Rs. \(\frac{40}{160}\)

Out of 100 the amount spent Rs. \(\frac{40}{160} \times 100=\text { Rs. } 25\)

∴ Of the total amount 25% was spent on the ride of merry-go-round.

We have decided to spend 35% of the total amount for fooding.

Question 4. Let’s calculate the sum for this purpose.

Solution:

Out of Rs.160, I spent 35% = Rs.\(\left(160 \times \frac{35}{100}\right)\) = Rs. 56

∴ We shall spend Rs. 56 in the fair for our food.

I spent Rs. 24 to buy glass bangles in the fair.

Question 5. Let’s calculate in percentage the amount I spent for purchasing glass bangles in the fair.

Solution:

Out of Rs. 160, I spent to purchase glass bangles Rs, 24

Out of Rs. 1,1 spent to purchase glass bangles Rs. \(\frac{24}{160}\)

Out of Rs. 100, I spent to purchase glass bangles Rs. =\(\frac{24}{160} \times 100\)= Rs. 15

∴ Out of the total amount I spent for glass bangles =15%

Question 6.  I painted in red 15% of the bamboo and \(\frac{1}{5}\) part in green.

Solution:

∴ Of the total length I painted \(\frac{15}{100}\) part = \(\frac{3}{20}\) part in red, but how much percent of total length is painted in yellow, let’s calculate.

Question 7.  If the bamboo is 2 m long and of which 38 cm is coloured with yellow then what is the percentage of yellow in relation to the whole length of the bamboo?

Solution:

Given

If the bamboo is 2 m long and of which 38 cm is coloured with yellow

Length of bamboo = 2 m = 200 cm.

∵ Of 200 cm long bamboo yellow coloured is 38.

∴ Of 1 cm long bamboo yellow coloured is \(\frac{38}{200}\)

∴ Of 100 cm long bamboo yellow coloured is \(\frac{38}{200} \times 100 \mathrm{~cm}=19\)

∴ Of the total length of the bamboo,19% is yellow coloured.

Question 8.  A train will go to Budwan from Howrah station. The distance of cord line between Burdwan and Howrah is 85 km. But .the distance in the main line is 5% more than in the cord line. Let’s calculate the distance between Howrah and Burdwan in the main line by the method of proportion.

Solution:

Given

A train will go to Budwan from Howrah station. The distance of cord line between Burdwan and Howrah is 85 km. But .the distance in the main line is 5% more than in the cord line.

That distance is 5% more through main line.

That means, if the distance in cord line is 100 km then in the main line it is 5 km more.

So the distance through main line is (100 + 5) km = 105 km.

In mathematical language the problem is –

If the distance between Howrah and Burdwan is more / less in the cord line, in the main line it is inversely or less. Both those distances are Directly (directly/inversely) related.

So, 100:85::105:?

The distance in mainline is = \(\frac{85 \times 105}{100}\)

Let’s work out by unitary method.

If the distance in the cord line is 100 km then in the main line is 105 km.

If the distance in the cord line is 1 km then in the main line it is

= \(\frac{105}{100} \mathrm{~km}.\)

If the distance in the cord line is 85 km then in the main line it is will be

= \(85 \times \frac{105}{100}\) km. = 89.25km

Question 9. Niyamat uncle of Faridpur has used high – yielding paddy seed in his field. For this, the cost of production of paddy has increased by 35% but the cost of cultivation has increased by 35%. Let’s workout the present profit of Niyamat uncle in comparison with the previous return of Rs. 1220 by investing Rs. 450 in the same field.

Solution:

Given

Niyamat uncle of Faridpur has used high – yielding paddy seed in his field. For this, the cost of production of paddy has increased by 35% but the cost of cultivation has increased by 35%.

Let’s work out the increase in the cost of cultivation after using light- yielding paddy seeds by the rule of three.

In mathematical language the problem is –

For using high – yielding paddy seeds the expense earlier and the expense at present are there in the Directly (directly / inverse) relation.

∴ For using high-yielding paddy seed the present is

=  Rs. 135 × \(\frac{450}{100}\) = Rs.607.50

Let’s work out by unitary method.

Before the use of high-yielding seed

If Rs.100 was the expense, now it is Rs.135

If Re. 1 was the expense,now it is Rs. \(\frac{135}{100}\)

If Rs. 450 was the expense, now it is Rs.\(\frac{135×450}{100}\) = Rs. 607.50

Question 10. Let’s workout by the rule of three the production of paddy after using high-yielding seeds.

Solution:

The production of paddy increased by 30%, i.e., if the production of paddy was Rs. 100, it is now Rs. (100 + 30) = Rs 130.

In mathematical language the problem is –

The production before the use of high-yielding paddy is with the production after that use Direct (direct / inverse) related.

As the production of paddy in earlier has increased (i.e., from Rs. 100 to Rs. 1220)

So, the production of paddy will now Increase (increase/decrease).

Due to the use of high-yielding seed the production will be

= Rs. \(130 \times \frac{1220}{100}=\text { Rs. } 1586\)

Let’s work out by unitary method.

Before the use of high-yielding seed :

If the production was Rs.100, it is now Rs. 130.

If the production was Rs. 1, it is Rs. \(\frac{130}{100}\)

If the production was Rs. 1220, it is now \(\frac{130 \times 1220}{100}\)

Before the use of high-yielding seeds Niyamat uncle got an income of Rs. 1220- Rs. 450 = Rs. 770

Now the income is = Rs. (1586 – 607.50 ) = Rs. 978.50

More income is = Rs. ( 978.50 – 770 ) = Rs. 208.50

To reach a certain distance earlier the speed needs to increase (increase/decrease)

The speed is in increased (direct / inverse) relation with the time to reach a certain distance.

Let’s work out by the rule of three to reach in 100-20 = 80 seconds the speed of the vehicle will be 125 (unit / second).

Let’s work out by unitary method.

To reach in 100 seconds the speed of the vehicle will be 100 unit /second.

To reach in 1 second the speed of the vehicle will be 100 x 100 unit/second.

To reach in 80 seconds the speed of the vehicle will be \( \frac{100 \times 100}{80}\)unit/second.

= 125 unit/second

∴ To reduce the time of going to station from house by 20% the speed of the vehicle will be increased by –

(125 unit / second – 100 unit / second ) = 25 unit/ second So the speed needs to be increased by 25%.

Let’s work out how much the speed of my vehicle will have to be increased if I want to reduce time by 10% for the same distance.

Let by going from home to station at 100 unit/second speed 100 seconds time is taken.

In mathematical language, the problem is –

Speed is increased to cover same distance in less time, i.e., time and speed are inversely related.

∴ To reach in 90 seconds the speed of car will be = \(\frac{100 \times 100}{90}\) unit/second = 111.11 unit/second

∴  Increase in the speed of car = (111.11 – 100) unit/second

=11.11 unit/second

2. 12% of Rs. 215

Solution:

Given Rs. 215 x 12%

= Rs. \(\frac{215 \times 12}{100}\)

= Rs. \(\frac{129}{5}\)

= Rs. 25.80

Rs. 215 x 12% = Rs. 25.80

3. 110% of 37.8 meters

Solution:

Given 110% of 37.8 meters

= \(\frac{37.8 \times 110}{100} \mathrm{~m}\)

= 41.58m

110% of 37.8 meters = 41.58m

4. 200% of 480 gm

Solution:

Given 200% of 480 gm

= \(\frac{480 \times 200}{100} \mathrm{gm}\)

= 960 gm.

200% of 480 gm = 960 gm.

Question 4. (1)How much is Rs. 2.25 of Rs. 5?

Solution:

Given Rs.2.24 of Rs.5

= \(\frac{2.25}{5} \times 100 \%\)

= 45%

Rs.2.24 of Rs.5 = 45%

2. How much is 85 gm of 17 gm?

Solution:

Given 85 gm of 17 gm

= \(\frac{85}{17 \times 1000} \times 100 \%\)

= 0.1%

85 gm of 17 gm = 0.1%

3. How much is 2 kg 250 gm of 0.72 quintal?

Solution:

Given 250 gm of 0.72 quintal

2 kg 250 gm = 2250 gm

0.72 quintal = .72 x 100 x 1000 gm.

= 72000 gm.

∴ \(\frac{2250}{72000} \times 100 \%\)

= \(3 \frac{1}{8} \%=3.125 \%\)

250 gm of 0.72 quintal = \(3 \frac{1}{8} \%=3.125 \%\)

Question 5. Let’s fill up the table below:

Solution:

Question 6. The ratio of hydrogen and oxygen in water is 2:1. Let’s workout the percentage of hydrogen and oxygen in water.

Solution:

Given

Ratio of hydrogen and oxygen in water = 2:1

Sum of ratio = 2 + 1 =3

Hydrogen in water = \(\frac{2}{3}\) part

= \(\frac{2}{3} \times 100 \%\)

= \(66 \frac{2}{3} \%\)

Oxygen in water = \(\frac{1}{3}\) part

= \(\frac{1}{3} \times 100 \%\)

= \(33 \frac{1}{3} \%\)

Of the quantity of water, hydrogen is \(66 \frac{2}{3} \%\) and oxygen is \(33 \frac{1}{3} \%\).

Question 7. 1500 bottles were produced in a factory at Hridaypur. New 1695 bottles are produced there. Let’s workout the percentage increase in production in that factory.

Solution:

Given

1500 bottles were produced in a factory at Hridaypur. New 1695 bottles are produced there.

Bottles were produced in a factory = 1500

New bottles are produced = 1695

Increase in production = (1695 – 4500) = 195

% increase in production = \(\frac{195}{1500} \times 100 \%\) = 13%

The production has increased by 13%.

Question 8. The quantity of Nitrogen, Oxygen and Carbon dioxide in air is 75.6%, 23.04%, and 1.36%. Let’s work out the quantity of each in 25 liters of air.

Solution:

Given

The quantity of Nitrogen, Oxygen and Carbon dioxide in air is 75.6%, 23.04%, and 1.36%.

In 25 I air nitrogen is = 25 I x 75.6 %

= \(\frac{25 \times 756}{100 \times 10}\)

= 18.9 liters

In 25 I air oxygen is = 25 I x 23.04%

= \(\frac{25 \times 23.04}{100}\)

= 5.76 liters

In 25 I air carbondioxide is = 251x 1.36%

= \(\frac{25 \times 1.36}{100}\)

= 0.34 I

∴ In 25 I air nitrogen is 18.9 I, oxygen is 5.6 I, and carbondioxide is 0.34.

Question 9. Trisha has bought a book from the book stall of Milan dada. He gave her discounts of 10% and 5% respectively. Let’s work out how much Trisha paid to Milan dada if the M.R.P of the book was Rs. 200.

Solution:

Given

Trisha has bought a book from the book stall of Milan dada. He gave her discounts of 10% and 5% respectively.

Price of book = Rs. 200

First discount = Rs. 200 x 10%

= Rs. \(\frac{200 \times 10}{100}\)

= Rs. 20

Price of book after giving first discount = Rs.(200-20)

= 180

Second discount = Rs. 180×5

= Rs. \(\frac{180 \times 5}{100}\)

= Rs. 9

Price of book after giving second discount = Rs. (180-9)

= 171

Trisha gave Milan Dada Rs. 171.

Question 10. The length of each arm of a rectangle was increased by 10%. Let’s work out the percentage of increase in area by the rule of three.

Solution:

Given

The length of each arm of a rectangle was increased by 10%.

Let the length of each side of rectangle is 100 units.

∴ Area of the rectangle = (100)2sq.units.

= 10,000sq.units.

Increase in each side of the rectangle = 10%of100

= \(\frac{100 \times 10}{100} \text { units. }\)

= 10 units

∴ After increase, length of each side of the rectangle = (100+10) units

= 110units.

∴ Area of rectangle = (110)2 units

= 12100 units

Increase in area = (12100-10000)

= 2100 units.

% Increase in area of rectangle = \(\frac{2100 \times 100}{10000}\)

= 21%

The area of that rectangle increased by 21%.

There is a direct relation.

Percentage increase in area of rectangle = \(2100 \times \frac{100}{10000}=21 \%\)

Question 11. We get 15% discount if electricity bill is paid in time. My aunt got a rebate of Rs. 54 by paying the bill in time. Let’s v/ork out the total amount of the bill.

Solution:

Given

We get 15% discount if electricity bill is paid in time. My aunt got a rebate of Rs. 54 by paying the bill in time.

Let the bill was Rs. x.

Discount = 15% X. Rs. x

= Rs. \(\frac{X \times 15}{100}\) = Rs. \(\frac{3 x}{20}\)

B.T.P.,

= \(\frac{3 x}{20}=54\)

or x = \(\frac{54 \times 20}{3}\)

or, x = 360

∴ The electricity bill was of Rs. 360.

Question 12. The price of sugar has increased by 20%. Let’s work out the percentage of sugar if the monthly expenses of sugar remained same.

Solution:

Given

The price of sugar has increased by 20%.

Let in Rs. 100r 100 kg sugar is bought and its price is Rs. 100. Increase in the price of sugar = 20% of Rs. 100.

= Rs. \(\frac{100 \times 20}{100}\)

= Rs. 20

∴ After increase new price of 100 kg sugar

= Rs. (100 + 20) = Rs. 120

∴ After increase, in Rs. 120, 100 kg sugar will be bought.

∴ After increase, in Rs. 1, \(\frac{100}{120}\) kg sugar will be bought.

∴ After increase, in Rs. 100, \(\frac{100}{120} \times 100\) kg sugar will be bought.

∴ After increase, in Rs. 100, \(\frac{250}{3}\) kg sugar will be bought.

∴ Decrease in sugar consumption \(=\left(100-\frac{250}{3}\right) \mathrm{kg} .\)

= \(\frac{50}{3} \mathrm{~kg}\)

∴ % decrease in sugar consumption = \(\frac{50 \times 100}{3 \times 100}\)

= \(16 \frac{2}{3} \%\)

∴ If the monthly expenses of sugar remains same, its consumption is to be decreased by 16

= \(16 \frac{2}{3} \%\)

Question 13. When water freezes into ice, it increases in volume by 10%. Let’s workout in percentage how much it will decrease in volume if the ice melts into water.

Solution:

Given

When water freezes an increase of 10% occurs in volume. If 100 cube units be the volume of ice = (100 + 10) cube units = 110 cubic units

In mathematical language the problem is –

Relation will be – Directly proportional.

∴ Volume of water ( ?) = \(100 \times \frac{100}{110}=\frac{1000}{11}=90 \frac{10}{11}\)

∴ Decrease in volume due to melting = \(\left(100-90 \frac{10}{11}\right)=9 \frac{1}{11} \%\)

∴ There will be \(9 \frac{1}{11} \%\) decrease in volume.

Question14. Due to use of high-yielding seed Utpal babu has got 55% production hike in paddy cultivation. But for this the cost of cultivation has increased by 40%. Previously a yield of Rs. 3000 was produced by investing Rs. 1200. Let’s work out whether his income will be increased or decreased after using high-yielding seeds.

Solution:

Given

Due to use of high-yielding seed Utpal babu has got 55% production hike in paddy cultivation. But for this the cost of cultivation has increased by 40%. Previously a yield of Rs. 3000 was produced by investing Rs. 1200.

Earlier cost in paddy farming = Rs. 1200

Earlier price of yield = Rs. 3000

Increase in cost = 40% of Rs. 1200

= Rs. \(\frac{1200 \times 40}{100}\)

= Rs. 480

New cost after using high-yielding seed = Rs. (1200 + 480)

= Rs. 1680

Increase in yield = 55% of Rs. 3000

= Rs. \(\frac{3000 \times 55}{100}\)

= Rs. 1650

New, the price of yield is = Rs. (3000 + 1650)

= Rs. 4650

Earlier income = Rs. (3000 – 1200)

= Rs. 1800

New income = Rs. (4650 – 1680) = Rs. 2970

Increase in income = Rs. (2970- 1800) = Rs. 1170

∴ Income increased by Rs. 1170

Question 15. In a legislative election 80% voters cast their votes and the winning candidate got 65% of the cast votesi Let’s workout the percentage of total voters who supported him.

Solution:

Given

In a legislative election 80% voters cast their votes and the winning candidate got 65% of the cast votes

Let the total number of voters are 100.

∴ Votes cast = 80% of 100

= \(\frac{100 \times 80}{100}=80\)

Votes got by winning candidate = 65% of 80

= \(\frac{80 \times 65}{110}=52\)

∴ The winning candidate got 52% of the total votes cast.

Question 16. This year the students of Waianda H.S. school have passed 85% in Hindi, 70% in maths and 65% in both subjects A+ scoring. If the number of students are 120 then let’s work out how many students :

Solution:

1. Who got A+ in both subjects
2. Who got A+ only in maths
3. Who got A+ only in Hindi
4. Who got A+ in no subject

Number of students = 120

1. In both subjects got A+

Solution:

= 65% of 120

= \(\frac{120 \times 65}{100}\) = 78 Students

2. Students who got A+ in Maths Only

Solution:

= 70% of 120

= \(\frac{120 \times 70}{100}\) = 84 Students

Who got A+ only in Maths are = (84 – 78) = 6

3. Got A+ in Hindi only

Solution:

Students who got A+ in Hindi = 85% of 120

= \(\frac{120 \times 85}{100}\)

Who got A+ only in Hindi are = (102 – 78) = 24

4. Did not get A+ in both subjects

Solution:

= 120 – (78 + 6 + 24)

= 120-108=12

Question 17. The income of Amina bibi was increased by 20% and later decreased by 20%. Let’s workout the total percentage change of her income.

Solution:

Given

The income of Amina bibi was increased by 20% and later decreased by 20%.

Let the original income was Rs. 100.

1st increased = 20% of Rs. 100

= Rs. \(\frac{100 \times 20}{100}\)

= Rs. 20

Income after 1st increase = Rs. (100 + 20)

= Rs. 120

Then decrease = 20% of Rs. 120

= Rs. \(\frac{120 \times 20}{100}\)

= Rs. 24

Income after decrease = Rs. (120 – 24)

= Rs. 96

Decrease in her income = Rs. (100 – 96) = Rs. 4

% decrease in her income = \(\frac{4 \times 100}{100}\)= 4%.

The original income of Amina bibi has decreased by 4%.

Question 18. The length of a rectangle is increased by 15% and the breadth is reduced by 15%. Let’s work out the percentage increase or decrease of area.

Solution:

Given

The length of a rectangle is increased by 15% and the breadth is reduced by 15%.

Let the length of rectangle is x units & breadth is y units.

∴ Area of the rectangle = L x B

= x×y sq. units

= xy sq. units

Length of rectangle after 15% increase

= x+xX15%

= \(x+\frac{x \times 15}{100}\)

= \(\frac{23 x}{20}\) units

Breadth of rectangle after 15% decrease = y-y×15%

= \(y-\frac{y \times 15}{100}\)

= \(\frac{17y}{20}\) units

∴ New area of the rectangle = \(\frac{23 x}{20} \times \frac{17 y}{20}\) sq.units

= \(\frac{391 x y}{400}\) sq.units

∴ Decreases in the area of the rectangle = xy – \(\frac{391 x y}{400}\) sq.units

= \(\frac{400 x y-391 x y}{400}\) sq.units

= \(\frac{9 x y}{400}\) sq.units

∴ % decrease in the area of the rectangle = \(\frac{9 x y \times 100}{400 ×x y}=2 \frac{1}{4} \%\)

∴ The area of the rectangle decreased by \(2 \frac{1}{4} \%\).

Question 19. The length, breadth and height of a room are 15 m, 10m and 5 m. If the length, breadth and height are increased by 10 % each then work out the increase of the area of the 4 walls of the room.

Solution:

Given

The length, breadth and height of a room are 15 m, 10m and 5 m. If the length, breadth and height are increased by 10 %

Area of the 4 walls of the room = 2 (15 + 10) x 5 sq.m = 250 sq.m

After 10% increase, length of the room = \(\frac{110}{100} \times 15 \mathrm{~m}=\frac{33}{2} \mathrm{~m}\)

After 10% increase, breadth of the room =\(\frac{110}{100} \times 10\) =11 m

After 10% increase, hight of the room = \(\frac{110}{100} \times 5 \mathrm{~m}\) = \(\frac{11}{2}\) m

After 10% increase, the area of the 4 walls of the room = \(2\left(\frac{33}{2}+11\right) \times \frac{11}{2}\) sq.m

= \(2\left(\frac{33+22}{2}\right) \times \frac{11}{2}\) sq.m

= \(\frac{605}{2}\) sq.m

Increase in the area of the 4 walls of the room

= \(\left(\frac{605}{2}-250\right)\)

= \(\left(\frac{605-500}{2}\right) \mathrm{sq} \cdot \mathrm{m}=\frac{105}{2} \mathrm{sq} \cdot \mathrm{m}\)

Percentage increase in the area of the 4 walls of the room = = \(\frac{105 \times 100}{2 \times 250}\) = 21%

There will be 21 % increase in the area of the 4 walls of the room.

Question 20. In annual sports 20% of the students took part in 100 m race 15% of the students in 200 m sprint and 10% of the students took part in the long jump event. 5% of the students took part in these three events. Let’s work out the number of students who did not take part in any of the events, if the total number of the students were 780.

Solution:

Given

In annual sports 20% of the students took part in 100 m race 15% of the students in 200 m sprint and 10% of the students took part in the long jump event. 5% of the students took part in these three events.

Total number of students = 780

Number of students taking part in 100 m race = 20% of 780

= \(\frac{780 \times 20}{100}\)

= 156

Number of students taking part in 200 m race = 15% of 780

= \(\frac{780 \times 15}{100}\)

= 117

Number of students taking part in long jump race = 10% of 780

= \(\frac{780 \times 150}{100}\)

= 78

Number of students taking part in all three games = 5% of 780

= \(\frac{780 \times 5}{100}\)

= 39

Number of students taking part only in 100 m race = (156- 39) = 117

Number of students taking part only in 200 m race = (117- 39) = 78

Number of students taking part only in long jump = (78 – 39) = 39

Number of students who did not take part in any of the three games

= 780-(39+ 117 + 78 + 39)

= 780-273 = 507

Number of students who did not take part in any of the three games = 507