## WBBSE Solutions For Class 8 Maths Arithmetic Chapter 5 Percentage

**Introduction**

You are familiar with the word ‘per cent’ from your childhood. Perhaps you have heard your parents say that ‘my son has scored 95 per cent marks’, ‘at present bank interest is only 6 per cent per annum, ‘a bookseller allows 15 per cent commission’ etc. In fact, the per cent is an abbreviation of the Latin word per centum meaning per hundred. A fraction with a denominator of 100 is called

per cent. For example, 7/100 = 7 per cent.

Per cent is denoted by the symbol “%”.

**Necessity of percentage**

Sometimes it becomes necessary to compare two fractions. For this purpose, we convert the fractions so that both of them have a common denominator. Then we can say that the fraction having a greater numerator, is greater than the two fractions. Let us consider an example. Suppose, there are two schools in your locality. In school A, 225 students have passed out of 300 students and in school B, 288 students have passed out of 360 students. In order to compare the performance of the two schools, we have to compare the ratios 225/300

and 288/360

Now, 225/300= 3/4

= 3 x 25 / 4 x 25

= 75/100

and, 288/360 = 4/5

4 x 20 / 5 x 20

= 80/100

Now, it is obvious that 80/100>75/100

Therefore, the performance of school B is better than that of A. Thus it becomes easy for us to compare two results when two fractions are converted to fractions having a common denominator of 100.

**Read And Learn More WBBSE Solutions For Class 8 Maths**

**Important Formulae**

**1. To convert a percentage into a corresponding fraction:** The number is to be written as the numerator and 100 as the denominator. Subsequently, the fraction is to be reduced to its simplest form.

**Example:** 10% = 10/100

= 1/10

**2. To convert a fraction into a corresponding percentage:** The fraction is to be multiplied by 100 with an addition of the % symbol at the end.

**Example:** 2/5 = (2/5 x100)%

= 40%

**3. To convert a percentage into a corresponding decimal:** The percentage sign is to be removed and the decimal point is to be shifted in the number by two places to the left.

**Example:** 5% = 0.05

**4. To convert a decimal into a corresponding percentage:** The decimal point in the number is to be shifted by two places to the right and the % symbol is to be added at the end.

**Example:** 0.453 = 45.3%

**5. To find a given percentage of a quantity:** The percentage is to be written as a fraction followed by multiplication by the quantity given.

**Example:** 1. x% of y = x/100 x y

2. 10% of 500 = 10/100 x 500

= 50

**6. To express a given quantity as a percentage of another quantity of the same kind:** The given quantity is to be divided by the other quantity followed by multiplication of the result by 100 and subsequent addition of the % symbol.

**Example:**

1. x as percentage of y = (x/y x 100)%

2. 20 as a percentage of ₹500

= (20/500 x 100)%

**7. To find a quantity from a given percentage of the quantity:** The given quantity is to be divided by the percentage after expressing it as a fraction.

**Example** :

If 20% of full marks is 40 then, full marks = 40++20

=40×100/20

= 200.

**8. To find the percentage change in a quantity:** The change (either increase or decrease) in the quantity is to be written as the numerator and the original quantity as the denominator followed by multiplication of the fraction by 100

1. If a quantity increases by x, percentage increase = (x/original quantity x 100 )%.

**Example:** An increase of 5 over the original price of a commodity of 20 means, the percentage increase

= (5/20 x 100) %

= 25%

2. If a quantity increases by x %, the new quantity

= (1+ x / 100) x original quantity

or, original quantity = new quantity / 1+ x/100

**Example:** An increase of 5% over the original price of a commodity of ₹200 means,

new price=₹ [(1+5/100)×200]= ₹210.

3. If a quantity decreases by x, the percentage decrease

= (x/original quantity x 100) %

**Example:** A decrease of 5 over the original price of a commodity of * 50 means, a percentage decrease

=(5×100)% = 10%.

4. If a quantity decreases by x%, the new quantity

= (1-x/100)x original quantity

or, original quantity = new quantity / 1- x/100

**Example:** A decrease of 10% over the original price of a commodity of 70 means,

new price =₹[(1-10/100)×70]= ₹63.

**9. To compare between two quantities X and Y when X>Y (given) :**

1. The percentage by which greater quantity (X) is greater than smaller quantity (Y),

% increase = (X-Y/Y x 100)%.

**Example:** The percentage by which the unit cost of 50 per kg of a certain variety of sugar is greater than that of 40 per kg of another variety is,

% increase = (50 – 40×100)%

= 25%

2. The percentage by which the smaller quantity (Y) is less than the bigger quantity (X),

% decrease = (X-Y/ Y x 100) %.

**Example:** For the above-quoted example of the unit cost of two varieties of sugars,

% decreases = (50-40 / 50 x 100)% = 20%

3. When X exceeds Yby p%, then Y is Less than X by (p / 100+p x 100)%.

4. when Y is less than X by p%, then X is more than Y by (p / 100-p x 100)%

## Chapter 5 Percentage Some examples of percentage

**Example 1**

**Convert each of the following fractions into a per cent:**

**1. 1/4,**

**2. 1/2,**

**3. 3/4,**

**4. 3/8.**

**Solution:**

1. 1/4 = 1/4 x 100%

= 25%

2. 1/2 = 1/2 x 100%

= 50%

3. 3/4 = 3/4 x 100%

= 75%

4. 3/8 = 3/8 x 100%

= 75/2%

= 37 1/2%

**Example 2**

**Convert each of the following per cent into a vulgar fraction :**

**1. 25%,**

**2.6 2/3%,**

**3. 12 1/2 %,**

**4. 16 2/3 %.**

**Solution:**

1. 25% = 25/100

= 1/4

2. 6 2/3% = 20/3 x 100

= 1/15

3. 12 1/2 % = 25/ 2 x 100

= 1/8

4. 16 2/3% = 50 / 3 x 100

= 1/6

**Example 3**

**Convert each of the following percents into a decimal:**

**1. 25%,**

**2. 22.5%,**

**3. 41 2/3%,**

**4. 75%**

**Solution:**

1. 25% = 25/100

= 0.25

2. 22.5% = 22.5/100

0.225

25% = 0.225

3. 41 2/3 % = 125 / 3 x 100

= 5/3 x 4

= 5/12

= 0.416

41 2/3 % = 0.416

4. 75% = 75/100

= 0.75

75% = 0.75

**Example 4**

**Convert each of the following decimals into a per cent:**

**1. 0.28,**

**2. 0.125,**

**3. 1.25,**

**4. 2.75.**

**Solution:**

1. 0.28 = 28/100

= 28%

0.28 = 28%

2. 0.125 = 125/100

= 125 / 10 x 100

= 2.5%

0.125 = 2.5%

3. 1.25 = 125/100

= 125%

1.25 = 125%

4. 2.75 = 275/100

= 275%

2.75 = 275%

**Example 5**

**Find 25% of ₹300.**

**Solution:**

**Given:**

** 25% And ₹300**

25% of ₹300 = ₹300 x 25/100

= ₹75

25% of ₹300 = ₹75

**Example 6**

**The ratio of hydrogen and oxygen in the water is 2:1. Find the percentage of hydrogen and oxygen in the water.**

**Solution:**

Given:

The ratio of hydrogen and oxygen in the water is 2:1.

Since the ratio of hydrogen and oxygen in water is 2:1 therefore,

amount of hydrogen

= 2/3 part

= 2/3 x 100%

= 200/3 %

= 66 2/3 %

amount of oxygen

= 1/3 part

= 1/3 x 100%

= 100/3 %

= 33 1/3%

Hydrogen 66 2/3%, oxygen 33 1/3%.

**Example 7**

**What per cent is 2 kg 250 gms of 0.72 quintals? **

**Solution:**

Given:

2 kg 250 gms And 0.72 Quintals.

2 kg 250 gms = 2.25 kg

0.72 quintals = 72 kg

∴ The required percent = 2.25/72 x 100 %

= 3.125%

3.125% per cent is 2 kg 250 gms of 0.72 quintals

**Example 8**

**A factory used to produce 1500 bottles per month. Now 1695 bottles are produced there per month. Find the percentage of increase of production in that factory. **

**Solution :**

Given:

A factory used to produce 1500 bottles per month. Now 1695 bottles are produced there per month.

Previously, 1500 bottles were produced. Now 1695 bottles are produced. Production of bottles has increased by (1695 – 1500) 195 per month.

In 1500 the increase is 195

In 1 the increase is 195/1500

In 100 the increase is 195/1500 x 100

= 13

The percentage of increase in production is 13%

**Example 9**

**What per cent is ₹ 20 of ₹ 50?**

**Solution:**

Given:

₹ 20 And ₹ 50.

The required percent = ₹20/₹50 x 100%

= 40%

The required percent = 40%

**Example 10**

**The quantity of Nitrogen, Oxygen and Carbon dioxide in the air is 75.6%, 23.04% and 1.36%. Find the quantity of each in 25 litres of air.**

**Solution:**

Given:

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

In 25 litres of air,

quantity of Nitrogen = 25 litres × 75.6%

= 18.9 litres

quantity of Oxygen = 25 litres x 23.04%

= 5.76 litres

quantity of Carbon dioxide = 25 litres × 1.36%

= 0.34 litre

Nitrogen 18.9 litres,

Oxygen 5.76 litres,

Carbon dioxide 0.34 litre.

**Example 11**

**40% of the number is 48. What is the number?**

**Solution:**

Given:

40% of the number is 48.

40% of the number = 48

or, 1% of the number = 48/40

or, 100% of the number = 48/40 x 100

= 120

The number is 120.

**Example 12**

**A man purchased a book from a book stall. He got discounts of 10% and 5% respectively. How much did the man pay if the price of the book was printed as 200? **

**Solution:**

Given:

A man purchased a book from a book stall. He got discounts of 10% and 5% respectively.

The printed price of the book is 200.

Discount on the first time

=200 x 10/100

=20

Price of the book after the first discount

=(200-20)

=180

Discount at the second time

= 180 × 5/100

=9

Price of the book after the second discount

= (180 – 9)

= 171

The price of the book is 171.

**Example 13**

**Out of 40 students in a school, 16 failed. Find the percentage of successful candidates.**

**Solution :**

Given:

Out of 40 students in a school, 16 failed.

The number of students who failed = 16

∴ The number of students passed = (40 – 16) = 24

.’. Out of 40 students the number of successful candidates = 24

Out of 1 student number of successful candidates = 24/40

Out of 100 students a successful 24

candidates = 24/40 x 100

= 60

The percentage of successful candidates is 60%.

**Example 14**

**The length of each side of a square is increased by 10%. Find the percentage of increase in its area.**

**Solution:**

Given:

The length of each side of a square is increased by 10%.

Let, the length of the side of the square = * a* units

Area of the square = a^{2} sq units.

After increasing 10% the length of the beach

side = (a+a x 10/100) units

= (a + a/10)units

= 11a / 10 units

∴ Area = (11a / 10)² sq units

= 121a²/100 sq units

∴ Increase in area = (121a² / 100 – a²)sq units = 21a² / 100 sq units

In a^{2} sq units the area increases by

21 a^{2 }/ 100 sq units

In 1 sq unit, the area increases by

21 a² / a^{2} x 100 sq units

In 100 sq units, the area increases by

21/100 x 100 sq units = 21 aq units

It will increase by 21%

**Example 15**

**Out of 350 mangoes in a basket, 210 mangoes were distributed among some students. Find the percentage of mangoes left in the basket.**

**Solution :**

Given:

Out of 350 mangoes in a basket, 210 mangoes were distributed among some students.

The original number of mangoes = 350

Number of mangoes distributed = 210

Number of mangoes left in the basket = (350-210)

= 140

Therefore, percentage of the mangoes left

= 140/350 x 100 = 40 350

40% of the mangoes were left in the basket.

**Example 16**

**15% discount L obtained if the bill of electricity is paid on time. A man got a discount of ₹ 54 by paying the bill on time. What was the amount of the bill**

**Solution :**

Given:

15% discount L obtained if the bill of electricity is paid on time. A man got a discount of ₹ 54 by paying the bill on time.

If a discount of ₹ 15 is obtained then the amount of the bill is ₹ 10

If a discount of ₹ 1 is obtained then the amount of the bill is ₹ 100 / 15

If a discount of ₹ 54 is obtained then the

amount of the bill is ₹ 100 x 54 / 15

= ₹ 360

The amount of the bill was ₹ 360.

**Example 17**

**Ram scored 642 marks out of 800 and Shyam scored 515 marks out of 700 in an examination. Whose performance was better ?**

**Solution :**

Given:

Ram scored 642 marks out of 800 and Shyam scored 515 marks out of 700 in an examination.

Out of 800 marks, Ram scored 642 marks

.’. Out of 100 marks, Ram scored

642 x100 / 800 = 80.25 marks

Out of 700 marks, Shyam scored 5115 marks

∴ Out of 100 marks, Shyam scored

515 x 100 / 700

= 73.57 marks

Ram’s performance was better.

**Example 18**

**The price of sugar has increased by 20%. Find the percentage decrease in the monthly use of sugar to keep the expenses of sugar unaltered.**

**Solution :**

Given

The price of sugar has increased by 20%.

What price of sugar of ₹ 100 has become ₹ 120.

Therefore, on ₹ 120 the expenses should be decreased by ₹ (120 – 100)

= ₹ 20.

On ₹ 120 expenses should be decreased by ₹ 20

On ₹ 1 expenses should be decreased by

₹20 / 120

On ₹ 100 expenses should be decreased by

₹ 20/120 x 100

= ₹ 100/6

= ₹ 50/3

= ₹ 16 2/3

Monthly use of sugar decreased by 16 2/3%

**Example 19**

**Ram’s income is greater than Shyam’s income by 20%. Find the per cent by which Shyam’s income is less than Ram’s income.**

**Solution :**

Given:

Ram’s income is greater than Shyam’s income by 20%.

If Shyam’s income is ₹ 100, then Ram’s income is ₹ 120.

.’. Shyam’s income is less than Ram’s income by ₹ (120 – 100)

= ₹ 20.

.’. In ₹ 120, Shyam’s income is less than Ram’s income by ₹ 20

In ₹ 1, Shyam’s income is less than Ram’s income by ₹ 20/120

In ₹ 100, Shyam’s income is less than Ram’s income ₹ 20/100 x 100

= ₹ 50/3

= ₹ 16 2/3

Shyam’s income is 16 2/3% less than Ram’s income.

**Example 20**

**When water freezes into ice, its volume is increased by 10%. By what per cent the volume will be decreased if ice melts into water.**

**Solution :**

Given:

When water freezes into ice, its volume is increased by 10%.

100 c.c. water freezes into 110 c.c. ice.

When ice melts into the water the volume decreases in 110 cc volume = (110 – 100) c.c.

= 10 c.c.

In 110 c.c., volume decreases by 10 c.c.

In 1 c.c., volume decreases by 10/110 c.c.

In 100 c.c., volume decreases by 10/110 x 100 c.c.

= 100/11 c.c.

= 9 1/11 c.c.

The volume will be decreased by 9 1/11%

**Example 21**

**Due to the increase in the price of sugar by 25%, a family decreases the consumption of sugar by 15%. As a result of this, what will be the percentage increase or decrease in the expenditure of sugar of that family?**

**Solution :**

Given

Due to the increase in the price of sugar by 25%, a family decreases the consumption of sugar by 15%.

Let, the price of sugar be ₹ * x* per kg and the consumption of sugar be

*kg.*

*y*At present

price of 1 kg of sugar is ₹ *x *

∴ The price of * y* kg of sugar is ₹

*xy*Due to a 25% increase, the price of sugar per kg

= ₹ (x + x x 25/100)

= ₹ (x + x/4)

= ₹ 5x / 4

Due to a 15% decrease, in the consumption of sugar

= (y – y x 15/100)kg

= (y – 3y/20) kg

= 17y/20 kg

After rising the price

Price of 1 kg of sugar = ₹ 5x/4

Price of 17y/20 kg of sugar = ₹ 5x/4 x 17y/20

= ₹ 17xy/16

.’. Expenditure increases by

₹ (17xy / 16 – xy)

= ₹ xy/16

On ₹ xy, expenditure increases by ₹ xy/16

On ₹ 1, expenditure increases by ₹ xy / 16 x 1/xy

On ₹ 100, expenditure increases by ₹ 100/16

= ₹25/4

= ₹ 6 1/4

Expenditure will increase by 6 1/4%

**Example 22**

**Due to the use of the high-yielding seed, a man has got a 55% production hike in paddy cultivation. But for this production, the cost of cultivation has increased by 40%. Previously a yield of ₹ 3000 was produced by investing ₹1200. Find whether his income will be increased or decreased and by how much after using a high-yielding seed. **

**Solution :**

Given:

Due to the use of the high-yielding seed, a man has got a 55% production hike in paddy cultivation.

But for this production, the cost of cultivation has increased by 40%. Previously a yield of ₹ 3000 was produced by investing ₹1200.

Previously, the yield of ₹ 3000 was produced by investing ₹ 1200 (cost).

.’. Income was =* ₹* (3000 – 1200)

= ₹ 1800

After increasing 40% the present cost of cultivation = ₹ 1200 x 140/100 = ₹ 1680

After increasing by 55% the present yield

= ₹ 3000 x 155/100

= ₹ 4650

Therefore, at present income

= ₹ (4650 – 1680)

= ₹ 2970

Therefore, income has increased by = ₹ (2970 – 1800)

= ₹ 1170

Income has increased by ₹ 1170.

**Example 23**

**The length, breadth and height of a room are 15 m, 10 m and 5 m respectively, If the length, breadth and height be increased by 10%, find the percentage increase in the area of four walls.**

**Solution :**

Given:

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

Area of the four walls of the room

= 2 x (15 + 10) x 5 sq m

= 2x25x5sqm

= 250 sq m

After a 10% increment

length becomes (15 + 15 x 10 / 100)m

= 16.5m

breadth becomes (10 + 10 x 10 / 100) m

= 11 m

height becomes (5 + 5 x 10 / 100) m

= 5.5 m

∴ Area of the four walls becomes 2 x (16.5 + 11) x 5.5 sq m

= 2 x 27.5 x 5.5 sq m

= 302.5 sq m

.’. The area of four walls increases by (302.5 – 250) sq m

= 52.5 sq m.

In 250 sq m area increases by 52.5 sq m

In 1 sq m area increases by 52.5/250 sq m

In 100 sq m area increases by 52.5 x 100 / 250 sq m

= 21 sq m.

The area of the four walls increases by 21%.

**Example 24**

**In a legislative election, 80% of voters cast their votes and the winning candidate got 65% of the cast votes. Find the percentage of total votes he got.**

**Solution:**

Given:

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

Let, the total number of voters = *x*

∴ Cast votes = x x 80/100

= 4x/5

The Winning candidate has got votes

= 4x/5 x 65/100

= 13x/25

∴ Out of x votes, he has got 13x/25 votes

Out of 1 vote, he has got 13x / 25 x x votes

Out of 100 votes, he has got 13 x 100 / 25 votes

= 52 votes

the winning candidate got 52% of the total votes.

**Example 25**

**In an examination 52% of the examinees failed in English, 42% failed in Mathematics and 17% failed in both the subjects. If 115 examinees passed in both subjects, find the total number of examinees.**

**Solution:**

Given:

In an examination 52% of the examinees failed in English, 42% failed in Mathematics and 17% failed in both the subjects.

If 115 examinees passed in both subjects.

(52 – 17)% or, 35% of examinees failed only in English.

(42 – 17)% or 25% of examinees failed only in Mathematics.

17% of examinees failed in both subjects.

Therefore, altogether (35 + 25 + 17)% or, 77% of examinees failed.

Hence, (100 – 77)% or, 23% of examinees passed in all.

If 23 examinees pass then the total number of examinees is 100

If 1 examinee pass then the total number of examinees is 100/23

If 115 examinees pass then the total number of examinees is 100 x 115 / 23

= 500

The total number of examinees is 500.

**Example 26**

**The students of the school have passed 85% in Bengali, 70% in Mathematics and 65% in both subjects scoring A+. If the number of students is 120 then how many students**

**1. got A+ in both subjects,**

**2. got A+ only in Mathematics,**

**3. got A+ only in Bengali,**

**4. did not get an A+ in any subject.**

**Solution :**

Given:

The students of the school have passed 85% in Bengali, 70% in Mathematics and 65% in both subjects scoring A+.

If the number of students is 120.

1. Number of students who got A+ in both the 65

subjects = 120 x 65/100

= 78

2. Got A+ only in Mathematics (70 – 65)% = 5%

.’. Number of students who got A+ only in Mathematics = 12 x 5/100

= 6

3. Got A+ only in Bengali = (85 – 65)% = 20%

∴ number of students who got A+ in, 20

Bengali only = 120 x 20/100

= 24.

4. Got A+ in neither subject = {120 – (78 + 6 + 24)}

= 12 students.

Got A+ in neither subject = 12 students.

**Example 27**

**The income of a man was increased by 20% and later decreased by 20%. Find the percentage of change in his income.**

**Solution:**

Given:

The income of a man was increased by 20% and later decreased by 20%.

If the initial income is ₹ 100 then it increases to ₹ 120.

Later, on ₹ 100 income decreases by ₹ 20

Later, on ₹ 1 income decreases by ₹ 20/100

Later, on ₹ 120 income decreases by ₹ 20/100 x 120

= ₹ 24

Therefore, income increases by ₹ 100 by ₹ 20 and then decreases by ₹ 24.

∴ Income decreases by 4%.

**Example 28**

**The length of a rectangle is increased by 15% and the breadth is decreased by 15%. Find the percentage increase or decrease in area.**

**Solution :**

Given:

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

Let, the length be * x* units and breadth be

*units, then area =*

*y**sq units.*

*xy*After increasing by 15%, the length becomes

= (x + x x 115/100) units = (x + 3x/20) units

= 23x/20 units.

After decreasing by 15%, the breadth becomes

= (y – y x 15/100)units

= (y – 3y/20)units

= 17y/20 units

∴ New area = 23x/20 x 17y/20 sq units

= 391 xy/400 sq units

∴ Area decreases by (xy – 391 xy/400) sq units = 9xy/400 sq units

on xy sq units area decreases by = 9xy/400 sq units

on 1 sq units area decreases by = 9xy/400xy sq units

on 100 sq units, the area decreases by = 9/400 x 100 sq units

= 9/4 sq units

The area will decrease by 2 1/4 %.

**Example 29**

**In annual sports, 20% of the students took part in the 100 m race. 15% of the students in the 200 m sprint and 10% of the students in the long jump event. 5% of the students took part in all these three events. Find the number of students who did not take part in any of the events if the total number of students in the school was 780. (No students took part in two events).**

**Solution :**

Given:

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

5% of the students took part in all these three events.

Took part only in 100 m race (20 – 5)% or, 15% students

Took part only in 200 m race (15 – 5)% or, 10% students

Took part only in long jump (10 – 5)% or, 5% students

∴ Took part in one or three events = (15 + 10 + 5 + 5)% or, 35% of students

∴ Took part in neither event (100 – 35)%

or, 65% of students.

∴ Took part in neither event 780 x 65/100

= 507 students.

507 students did not take part in any event.

**Example 30.**

**A panchayat family spends 40% of the government grant on health, 35% on education and 30% of the grant for education for literacy projects. If the expenditure for the literacy project is ₹ 210000 then find**

**1. Total amount of grant,**

**2. Expenditure on education and**

**3. Expenditure for health.**

**Solution :**

Given:

A panchayat family spends 40% of the government grant on health, 35% on education and 30% of the grant for education for literacy projects.

If the expenditure for the literacy project is ₹ 210000.

Let, ‘s total government grant = ₹ *x*

Expenditure for health = ₹ x x 40/100

= ₹ 2x / 5

Expenditure for education = ₹ x x 35/100

= ₹ 7x/20

Expenditure for a literacy project = ₹ 7x/20 x 30/100

= ₹21 x / 200

According to the question, 21x/200 = 210000

or, x = 21000 x 200/21

= 2000000

Expenditure for education = ₹ 7 x 2000000 / 20

= 700000

Expenditure for education = ₹ 2 x 2000000 / 5

= ₹ 800000

1. Total grant = ₹ 2000000,

2. expenditure for education ₹ 700000 and

3. expenditure for health = ₹ 800000.

**Example 31**

**Due to the use of the high-yielding seed, a man has got a 30% production hike in paddy cultivation. But for this production, the cost of cultivation has increased by 35%. Previously a yield of ₹ 1220 was produced by investing ₹ 450. Find by how much his income will be increased after using high-yielding seeds.**

**Solution :**

Given:

Due to the use of the high-yielding seed, a man has got a 30% production hike in paddy cultivation.

But for this production, the cost of cultivation has increased by 35%. Previously a yield of ₹ 1220 was produced by investing ₹ 450.

Previously yield of ₹ 1220 was produced by investing ₹ 450.

Income was = ₹ (1220 – 450) = ₹ 770

After increasing by 35% the present cost of cultivation = ₹ 450 x 135/100

= ₹ 607.50

After increasing by 30% the present yield = ₹ 1220 x 130/100

= ₹ 1586

∴ After present income = ₹ (1586 – 607.50)

= ₹ 978.50

∴ Income has increased by = ₹ (378.50 – 770)

= ₹ 208.50

Income has increased by ₹ 208.50.

**Example 32**

**20000 examinees were to appear in an examination. But 5% of them were absent. 60% of the examinees who had appeared in the examination passed. If the ratio of the examinees passed in the first, second and third divisions is 1:2:3, find the number of examinees passed in each division.**

**Solution :**

Given:

20000 examinees were to appear in an examination.

But 5% of them were absent. 60% of the examinees who had appeared in the examination passed.

If the ratio of the examinees passed in the first, second and third divisions is 1:2:3.

The total number of examinees = 20000

∴ Number of absentees

= 20000 x 5%

= 1000

∴ Number of examinees present

= 20000 – 1000

= 19000

∴ The number of examinees passed

= 19000 x 60/100

= 11400

Now, the ratio of examinees passed in the first, second and third divisions is 1:2:3.

∴ Passed in the first division = 11400/6 x 1

= 1900

Passed in second division = 11400/6 x 2

= 3800

Passed in third division = 11400/6 x 3

= 5700

First division = 1900,

second division = 3800

and third division = 5700.