Mathematics Part ii Solutions Solutions for Class 8 Math Chapter 2 Equations are provided here with simple step-by-step explanations. These solutions for Equations are extremely popular among class 8 students for Math Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 8 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

Page No 101:

Question 1:

“I have 52 stamps in my collection”, Ruby said. Making some mental calculations, Razia said, ”I need 15 more to catch up with you”. How many stamps does Razia have?

 

Answer:

Number of stamps Ruby has = 52

Number of more stamps Razia needs to catch up with Ruby = 15 

∴ Number of stamps Razia has = 52 − 15 = 37

 

Page No 101:

Question 2:

Gopalan bought a bunch of bananas for his shop. 7 of them had slightly turned bad. After removing them, he had 46 left. How many were there in the bunch at first?

 

Answer:

Number of bad bananas in the bunch = 7

Number of good bananas left = 46

∴ Total number of bananas in the bunch = 46 + 7 = 53

 

Page No 101:

Question 3:

A number subtracted from 500 gave 234. What is the number subtracted?

 

Answer:

Original number = 500

Number obtained after subtraction = 234

∴ Subtracted number = 500 − 234 = 266

 



Page No 102:

Question 1:

The manager of a company gets four times the salary of a peon. A peon gets 7000 rupees a month. How much does the manager get every month?

 

Answer:

Salary of peon = Rs.7000

Salary of manager = 4 × Salary of peon

= 4 × Rs.7000

= Rs.28000

 

Page No 102:

Question 2:

The perimeter of a pentagon with equal sides is 65 centimetres. What is the length of each side?

 

Answer:

Perimeter of the pentagon = 65 cm

We know that the number of equal sides of a pentagon is 5.

∴ Length of each side =

 

Page No 102:

Question 3:

A number multiplied by 17 gives 204. What is the number?

 

Answer:

Required number × 17 = 204

⇒ Required number =

 

Page No 102:

Question 4:

A number divided by 12 gives 25. What is the number?

 

Answer:

⇒ Required number = 25 × 12 = 300

 



Page No 103:

Question 1:

Anita and her friends bought some pens. For a packet of five pens, they got two rupees reduction in price. They had to pay only 18 rupees. Had they bought the pens separately, how much would have been the price for each pen?

 

Answer:

Cost of five pens after reduction = Rs.18

Reduction amount for five pens = Rs.2

∴ Original cost of five pens = Rs.18 + Rs.2 = Rs.20

Cost of each pen separately = Original cost of five pens divided by five

 

Page No 103:

Question 2:

Three added to half a number gives 23. What is the number?

 

Answer:

The number before 3 was added = 23 − 3 = 20

20 is half the required number.

∴ Required number = 2 × 20 = 40

Thus, 3 added to half of 40 gives 23. 

 

Page No 103:

Question 3:

Two subtracted from one-third of a number gives 40. What is the number?

 

Answer:

The number before 2 was subtracted = 40 + 2 = 42

42 is one-third of the required number.

∴ Required number = 3 × 42 = 126

Thus, 2 subtracted from one-third of 126 gives 40.

 



Page No 104:

Question 1:

The sum of a number and 2, multiplied by 3 gives 18. What is the number.

 

Answer:

The number before 3 was multiplied = 18 ÷ 3 = 6

6 is the sum of the number and 2.

∴ Required number = 6 − 2 = 4

Thus, the sum of 4 and 2, when multiplied by 3, gives 18.

 

Page No 104:

Question 2:

3 subtracted from a number and the result multiplied by 2, gives 10. What is the number?

 

Answer:

The number before 2 was multiplied = 10 ÷ 2 = 5

5 is obtained by subtracting 3 from the required number.

∴ Required number = 5 + 3 = 8

Thus, 3 subtracted from 8 and the result, when multiplied by 2, gives 10.

 



Page No 106:

Question 1:

The price of a table and chair together is 1500 rupees. The price of the table is four times that of the chair. What is the price of each?

 

Answer:

Let the price of the chair be Rs.x.

The price of the table = 4 × Price of the chair = Rs.4x

Algebraic form of the problem is:

x + 4x = 1500

⇒ 5x = 1500

∴ Price of the chair = Rs.300

Price of the table = Rs.4 × 300 = Rs.1200

 

Page No 106:

Question 2:

Let the price of the chair be Rs.x.

The price of the table = 4 × Price of the chair = Rs.4x

Algebraic form of the problem is:

x + 4x = 1500

⇒ 5x = 1500

∴ Price of the chair = Rs.300

Price of the table = Rs.4 × 300 = Rs.1200

 

Answer:

Let the measure of the smaller angle be x°.

The measure of the larger angle = x° + 50°

As per the question, from a point on a line, another line is to be drawn such that two angles are formed. So, the two angles form a linear pair.

We know that sum of the angles forming a linear pair is 180°.

Algebraic form of the problem is:

x° + x° + 50° = 180°

⇒ 2x° + 50° = 180°

⇒ 2x° = 180° − 50°

⇒ 2x° = 130°

Therefore, the measure of the smaller angle is 65°.

 



Page No 107:

Question 1:

A hundred rupee note was changed into ten rupee notes and twenty rupee notes. There were seven notes in all. How many of each denomination were there?

 

Answer:

Total number of notes = 7

Let the number of 10 rupee notes be x.

The number of 20 rupee notes = 7 − x 

Algebraic form of the problem is:

10x + 20(7 − x) = 100

⇒ 10x + 140 − 20x = 100

⇒ −10x + 140 = 100

⇒ 10x = 140 − 100 

⇒ 10x = 40

Therefore, there are 4 notes of 10 rupees and 3 (= 7 − 4) notes of 20 rupees.

 

Page No 107:

Question 2:

Jaffer took a test in which there were 25 questions. Two marks would be given for each correct answer and one mark would be subtracted for each wrong one. Jaffer answered all questions and got 35 marks. How many of his answers were correct?

 

Answer:

Total number of questions = 25

Let the number of questions correctly answered be x.

The number of questions wrongly answered = 25 − x 

Marks given for each correct answer = 2

Mark subtracted for each wrong answer = 1

Total marks obtained = 35

Algebraic form of the problem is:

2x − 1(25 − x) = 35

⇒ 2x − 25 + x = 35

⇒ 3x − 25 = 35

⇒ 3x = 35 + 25

⇒ 3x = 60

Therefore, Jaffer’s 20 answers were correct.

 

Page No 107:

Question 3:

The price of a book is 4 rupees more than the price of a pen; and the price of a pencil is 2 rupees less than the price of this pen. A man bought 5 books, 2 pens and 3 pencils and paid 54 rupees for the lot. What is the price of each?

 

Answer:

Let the price of the pen be Rs.x

The price of the book = Rs (x + 4)

The price of the pencil = Rs (x − 2) 

Cost of 5 books = Rs 5(x + 4) = Rs (5x + 20)

Cost of 2 pens = Rs.2x 

Cost of 3 pencils = Rs 3(x − 2) = Rs (3x − 6) 

Total amount paid = Rs.54

Algebraic form of the problem is:

5x + 20 + 2x + 3x − 6 = 54

⇒ 10x + 14 = 54

⇒ 10x = 54 − 14 

⇒ 10x = 40

Cost of one pen = Rs.4 

Cost of one book = Rs (4 + 4) = Rs.8

Cost of one pencil = Rs (4 − 2) = Rs.2

 

Page No 107:

Question 4:

The sum of the digits of a two-digit number is 6 and the difference of the digits is 2. How many such numbers are there? What are they?

 

Answer:

Sum of the digits of a two digit number = 6

Let the ones digit of the two digit number be x.

∴ Tens digit of the two digit number = 6 − x

Let us suppose that the ones digit is greater than the tens digit.

Algebraic form of the problem is:

x − (6 − x) = 2

⇒ x − 6 + x = 2

⇒ 2x − 6 = 2

⇒ 2x = 2 + 6

⇒ 2x = 8

∴ Ones digit = 4

Ten’s digit = 6 − 4 = 2

Two-digit number = 2 × 10 + 4 = 24

When the tens digit is greater than the one digit, the difference is not 2.

Thus, only one such number is possible, which is 24.

 



Page No 110:

Question 1:

For a science exhibition, tickets for children cost 2 rupees and for adults, 5 rupees. 100 tickets were sold for one show, and 290 rupees was got. How many children were there for this show?

 

Answer:

Cost of ticket for children = Rs.2

Cost of ticket for adults = Rs.5

Total number of tickets sold = 100

Total money collected = Rs.290

Let the number of tickets sold for children be x.

The number of tickets sold for adults = 100 − x

Total cost collected from the tickets sold for children = Rs.2x

Total cost collected from the tickets sold for adults = Rs 5(100 − x) = Rs (500 − 5x)

Algebraic form of the problem is:

2x + 500 − 5x = 290

⇒ 500 − 3x = 290

Adding 3x to both the sides:

500 − 3x + 3x = 290 + 3x

⇒ 500 = 290 + 3x

Subtracting 290 from both the sides:

⇒ 500 − 290 = 290 + 3x − 290

⇒ 210 = 3x

Dividing both the sides by 3:

Therefore, there were 70 children in the show.

 

Page No 110:

Question 2:

The number of boys and girls are equal in a class. On a day when eight boys did not attend, the number of girls was double the number of boys in this class. What is the actual number of boys and girls?

 

Answer:

Let the number of boys in the class be x

The number of boys and girls are equal. So, the number of girls in the class is also x.

The number of boys who did not attend the class = 8

∴ Number of boys who attended the class = x − 8

Number of girls in the class = 2 × (x − 8) = 2x − 16 

Algebraic form of the problem is:

2x − 16 = x

Subtracting x from both the sides: 

⇒ 2x − 16 − x = x − x

⇒ x − 16 = 0

Adding 16 to both the sides:

⇒ x − 16 + 16 = 0 + 16

⇒ x = 16

Therefore, the actual number of boys and girls in the class is 16.

 

Page No 110:

Question 3:

Appu’s mother is nine times as old as Appu. After nine years, her age would be three times that of Appu. What are their ages now?

 

Answer:

Let Appu’s present age be x years

Appu’s mother’s present age = 9x years

After 9 years:

Appu’s age = (x + 9) years

Appu’s mother’s age = (9x + 9)

Algebraic form of the problem is:

9x + 9 = 3(x + 9)

⇒ 9x + 9 = 3x + 27

Subtracting 3x from both the sides: 

⇒ 9x + 9 − 3x = 3x + 27 − 3x

⇒ 6x + 9 = 27 

Subtracting 9 from both the sides: 

⇒ 6x + 9 − 9 = 27 − 9

⇒ 6x = 18

Dividing both the sides by 6:

Therefore, Appu’s present age is 3 years and her mother’s present age is 27 (= 9 × 3) years.

 

Page No 110:

Question 4:

In a class of 40 children, 40% are girls. How many more girls coming to this class would make this 50%?

 

Answer:

Total number of children in the class = 40

Number of girls in the class = 40% of 40 

Let x girls be added to the class.

∴ Total number of children in the class = 40 + x

Total number of girls in the class = 16 + x

Algebraic form of the problem is:

50% of (40 + x) = 16 + x

Subtracting 2x from both the sides: 

Subtracting 40 from both the sides:

Therefore, adding 8 more girls in the class would make the number of girls 50% of the total children in the class.

 

Page No 110:

Question 5:

The breadth and length of a rectangle are in the ratio 2:3. In a rectangle with breadth 1 centimetre less than this and length 3 centimetres less, the breadth to length ratio is 3:4. Find the lengths and breadths of both rectangles.

 

Answer:

Let the breadth of the first rectangle be 2x cm.

Length of the first rectangle = 3x cm

Breadth of the second rectangle = (2x − 1) cm 

New length of the second rectangle = (3x − 3) cm

Algebraic form of the problem is:

Subtracting 9x from both the sides: 

Adding 4 to both the sides:

∴ Breadth of the first rectangle = 2 × 5 cm = 10 cm

Length of the first rectangle = 3 × 5 = 15 cm

Breadth of the second rectangle = (2 × 5 − 1) cm = 9 cm

Length of the second rectangle = (3 × 5 − 3) cm = 12 cm

 



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