S.chand publication New Learning Composite mathematics solution of class 8 Chapter 8 Percentage and Its Application Exercise 8B

 Exercise 8B

---

Q1 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 1

 If 45% people in a city like football, 40% like cricket and the remaining like other games, then what per cent of the people like other games? If the city has a population of 4,00,000 people, then find the number of people who like one of these games.

Sol :

Population of city=400000

Peoples like football$=400000\times \frac{45}{100}$

=180000

Peoples like cricket$=400000\times \frac{40}{100}$

=160000

Remaining peoples who like other games=400000-(180000+160000)

=400000-340000

=60000

Remaining peoples who like other games(in %) $=\frac{60000\times 45}{180000}$

=15%

---

Q2 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 2

In an examination, 96% of the students passed and 500 failed. How many students appeared at the examination?

Sol :

If total students be x

Failed student =500 in % (100-96)=4%

So, 4% of total student =500

$\frac{4}{100}\times x=500$

$x=\frac{500\times 100}{4}$

x=12500

Total students appeared 12500

---

Q3 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 3

Two numbers are respectively $12\frac{1}{2}$% and 25% more than a third number. Find what percent is first number of the second number?

Sol :

Let the 3rd number be 100

The 1st number be $=100+12\frac{1}{2}\%$

$=100+\frac{25}{100}=\frac{225}{200}=\frac{9}{8}$

The 2nd number be $=100+\frac{25}{100}\%$

$=\frac{100+25}{100}=\frac{125}{100}=\frac{5}{4}$

Percent$=\frac{\text{1st number }}{2nd number}\times 100$

$=\left(\frac{9}{8}\times 100\right) \div \frac{5}{4}$

$=\frac{9}{8}\times \frac{4}{5}\times 100$

=90%

---

Q4 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 4

A person invests Rs. 89856 in mutual funds, which is 26% of his annual income. What is the monthly income.

Sol :

Let, Annual income be x

Amount invested in mutual fund$=x\times \frac{26}{100}$

ATQ,

$\frac{26x}{100}=89856$

$x=\frac{89856\times 100}{26}$

=345600

∴Annual income=345600

∴Monthly income$=\frac{345600}{12}$

=28800

---

Q5 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 5

The difference between 54% of a number and 26% of the same number is 22526. What is 66% of that number.

Sol :

Let , the number be x

ATQ,

$\frac{54x}{100}-\frac{26x}{100}=22526$

$\frac{54x-26x}{100}=22526$

28x=22526×100

$x=\frac{22526\times 100}{28}$=80450

∴The number=80450

∴66% of the number(80450) $=\frac{80450\times 66}{100}$

=53097

---

Q6 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 6

When 125 is subtracted from a number, it reduces to its 37.5 per cent. What is 25 per cent of this number.

Sol :

Let the number be x

ATQ,

x-125$=\frac{37.5x}{100}$

$x-\frac{375x}{1000}=125$

1000x-375x=125×1000

$x=\frac{125\times 1000}{625}=200$

∴25% of 200$=200\times \frac{25}{100}$

=50

---

Q7 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 7

A person spends 75% of his income, if his income increases by 20% and expenses increase by 15% then find the per cent increase in his savings.

Sol :

Let income of the person=100

∴Spending=75%

New income=120

Expenses$=75+75\times \frac{20}{100}$

=90

New saving=(120-90)=30 

Difference=(30-25)=5

∴Increase %$=\frac{5}{25}\times 100$

=20%

---

Q8 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 8

A shopkeeper first increased the price of an article by 25% and then by 20%. What is the total percent increase.

Sol :

Price of article=100

First increased the price of an article$=100\times \frac{25}{100}$

=25

∴Increased price of article=100+25=125

∴Second increased the price of an article$=125 \times \frac{20}{100}$=25

∴Increased price of article=125+25=150

Total price increase=150-100=50

(In %)$=\frac{50}{100}\times 100$

=50

---

Q9 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 9

A sum of Rs. 2236 is divided among A, B and C in such a way that A receives 25% more than C and C receives 25% less than B. What is A’s sharer in the amount?

Sol :

Let, C's share=x

A's share$=x+\frac{25x}{100}=x+\frac{x}{4}$

B's share$=x+\frac{25x}{100}=x+\frac{x}{4}$

ATQ,

$x+x+\frac{x}{4}+x+\frac{x}{4}=2226$

$3x+\frac{2x}{4}=2226$

$\frac{12x+2x}{4}=2226$

$x=\frac{2226}{14}\times 4$

∴A's share$=610+\frac{610}{4}=770$

---

Q10 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 10

In a company, there are 75% skilled workers and the remaining are unskilled. 80% of skilled workers and 20% of unskilled workers are permanent. If the number of temporary workers is 126, then what is the number of workers?

Sol :

Let total workers be x

∴Skilled workers$=x\times \frac{75}{100}=\frac{3x}{4}$

∴Unskilled workers$=x-\frac{3x}{4}=\frac{4x-3x}{4}=\frac{x}{4}$

∴Skilled permanent workers$=\frac{3x}{4}\times \frac{80}{100}=\frac{3x}{5}$

∴Unskilled permanent workers$=\frac{x}{4}\times \frac{20}{100}=\frac{x}{20}$

∴Temporary workers$=x-\left(\frac{3x}{5}+\frac{x}{20}\right)=\frac{7x}{20}$

ATQ,

$\frac{7x}{20}=126$

$x=\frac{126\times 20}{7}$

=360

∴Total workers=360

---

Q11 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 11

Ravi scored 30% marks and failed by 15 marks. Deepak scored 40% marks and obtained 35 marks more than those required to pass. Find the minimum per cent pass marks.

Sol :

Let, Ravi get$=x\times \frac{30}{100}=\frac{3x}{10}$

Deepak get$=x\times \frac{40}{100}=\frac{4x}{10}$

ATQ,

$\frac{4x}{10}-35=\frac{3x}{10}+15$

$\frac{4x}{10}-\frac{3x}{10}=35+15=50$

$\frac{4x-3x}{10}=50$

x=50×10=500

∴Total number of marks=500

∴Ravi get$=\frac{3}{10}\times 500$=150

Deepak get$=\frac{4}{10}\times 500$=200

∴Min pass marks=150+15=165=200-35

∴Percentage of passing marks$=\frac{165\times 100}{500}$

=33%

---

Q12 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 12

Two candidates fought an election. One of them got 62% of the total votes and won by 432 votes. What is the total number of votes polled?

Sol :

Let, total number of votes polled=x

Won person get$=x\times \frac{62}{100}=\frac{62x}{100}$

Other person get$=x-\frac{62x}{100}=\frac{38x}{100}$

ATQ,

$\frac{62x}{100}-\frac{38x}{100}=432$

$\frac{24x}{100}=432$

$x=\frac{432\times 100}{24}=1800$

---

Q13 | Ex-8B | Class 8 | S.Chand | New Learning Composite maths |Percentage and its application | myhelper

OPEN IN YOUTUBE

Question 13

In an election between two persons, 68 votes were declared invalid. The winning persons for 52% and won 98 votes. Find the total number of votes polled.

Sol :

Let, total number of votes=x

winner get 52%

∴Losing team=(100-52)%

=48%

Winner won by 98 votes

ATQ,

(52%-48%)of x=98

4% of x=98

$x=\frac{98\times 100}{4}=2450$

∴68 votes were invalid

∴Total=(2450+68)=2518