8th Maths 2.2

Chapter 2 Linear Equations in One Variable

NCERT Class 8th solution of Exercise 2.1

NCERT Class 8th solution of Exercise 2.3

NCERT Class 8th solution of Exercise 2.4

NCERT Class 8th solution of Exercise 2.5

NCERT Class 8th solution of Exercise 2.6

Exercise 2.2

Q1. If you subtract $1/2$ from a number and multiply the result by $1/2$ , you get $1/8$ . What is the number?

Solution :

Let a number be

$x$

According to question

$(x-1/2)times1/2=1/8$

Multiply

$2$ on both sides.

$(x-1/2)times1/2times2=1/8times2$

$x-1/2=1/4$

Add

$1/2$ on both sides

$x-1/2+1/2=1/4+1/2$

$x=(2+4)/4times2$

$x=6/8$

$x=3/4$

Answer :

The number is

$3/4$ .

Q2. The perimeter of a rectangular swimming pool is $154m$ . Its length is $2m$ more than twice its breadth. What are the length and the breadth of the pool?

Solution :

Let breadth be

$x m$ ,

then length be

$2x+2$

According to question

The perimeter of a rectangle swimming pool

$=2(L+B)$

$154=2(2x+2+x)$

$154=2(3x+2)$

Dividing both sides by

$2$

$154/2=3x+2$

$77=3x+2$

Subtract

$2$ on both sides.

$77-2=3x$

$75=3x$

Dividing both sides by

$3$

$75/3=x$

$x=25$

$2x+2=2times25+2=52$

Answer :

The length

$52m$ and breadth

$25m$ .

Q3. The base of an isosceles triangle is $4/3cm$ .The perimeter of the triangle is $4(2)/15cm$ . What is the length of either of the remaining equal sides?

Solution :

Let the length of equal sides be

$x cm$

According to question

The perimeter of an isosceles triangle = sum of all side

$4(2)/15=x+x+4/3$

$62/15=2x+4/3$

Subtract

$4/3$ on both side

$62/15-4/3=2x$

Dividing both sides by

$2$

$31/15-2/3=x$

$(93-30)/(15times3)=x$

$63/45=x$

$7/5=x$

$1(2)/5=x$

Answer :

The length of equal sides is

$1(2)/5cm$ .

Q4. Sum of two numbers is $95$ . If one exceeds the other by $15$ , find the numbers?

Solution :

Let the first number be

$x$

the second number be

$x+15$

According to question

$x+x+15=95$

$2x+15=95$

Subtract

$15$ on both sides

$2x=95-15$

$2x=80$

Dividing both sides by

$2$

$x=80/2$

$x=40$

$x+15=40+15=55$

Answer :

The first number is

$40$ and the second number is

$55$ .

Q5. Two numbers are in the ratio $5:3$ . If they differ by $18$ , what are the numbers?

Solution :

$text{Let first number is 5x}$

$text{and second number is 3x.}$

$text{According to question}$

$text{5x - 3x = 18}$

$text{2x = 18}$

$text{x} = 18/2$

$text{x} = 9$

$text{5x} = 5times9 = 45$

$text{3x} = 3times9 = 27$

Answer :

$text{The numbers are 45 and 27.}$

Q6. Three consecutive integers add up to $51$ . What are these integers?

Solution :

$text{Let first number is x}$

$text{and second number is (x+1)}$

$text{and third number is (x+2)}$

$text{According to question}$

$text{x + (x+1) + (x+2) = 51}$

$text{3x + 3 = 51}$

$text{3x = 51 - 3}$

$text{x} = 48/3$

$text{x} = 16$

$text{x + 1 = 16 + 1 = 17}$

$text{x + 2 = 16 + 2 = 18}$

Answer :

$text{The integers are 16, 17, and 18}$ .

Q7. The sum of three consecutive multiples of $8$ is $888$ . Find the multiples?

Solution :

$text{Let first number is 8x}$

$text{and second number is 8x+8}$

$text{and third number is 8x+16}$

$text{According to question}$

$text{8x + 8x + 8 + 8x + 16 = 888}$

$text{24x + 24 = 888}$

$text{24x = 888 - 24}$

$text{24x = 864}$

$text{x} = 864/24$

$text{x} = 36$

$text{8x = 8 times36 = 288}$

$text{8x + 8 = 288 + 8 = 296}$

$text{8x + 16 = 288 + 16 = 304}$

Answer :

$text{The multiples are 288, 296, and 304}.$

Q8. Three consecutive integers are such that when they are taken in increasing order and multiplied by $2, 3$ and $4$ respectively, they add up to $74$ . Find these numbers.

Solution :

$text{Let first number is x}$

$text{and second number is (x + 1)}$

$text{and third number is (x + 2)}$

$text{According to question}$

$text{2x + 3(x + 1) + 4(x + 2) = 74}$

$text{2x + 3x + 3 + 4x + 8 = 74}$

$text{9x + 11 = 74}$

$text{9x = 74 - 11}$

$text{x} = 63/9$

$text{x} = 7$

$text{x + 1}= 7 + 1 = 8$

$text{x + 2} = 7 + 2 = 9$

Answer :

$text{The numbers are 7, 8, and 9}$ .

Q9. The age of Rahul and Haroon are in the ratio $5:7$ . Four years later the sum of their ages will be $56$ years. What are their present ages?

Solution :

$text{Let Rahul's present age 5x years}$

$text{and Haroon's present age 7x years}$

$text{After 4 years}$

$text{Rahul's age 5x + 4 years}$

$text{Haroon's age 7x + 4 years}$

$text{According to questtion}$

$text{5x + 4 + 7x + 4 = 56}$

$text{12x + 8 = 56}$

$text{12x = 56 - 8}$

$text{x} = 48/12$

$text{x} = 4$

$text{5x} = 5times4 =20$

$text{7x} = 7times4 = 28$

Answer :

$text{The present age of Rahul is 20 years}$

$text{and Haroon is 28 years}$ .

Q10. The number of boys and girls in a class are in the ratio $7:5$ . The number of boys is $8$ more than the number of girls. What is the total class strength?

Solution :

$text{Let the no. of boys are 7x}$

$text{and the no. of girls are 5x}$

$text{According to question}$

$text{7x = 5x + 8}$

$text{7x - 5x = 8 }$

$text{2x = 8}$

$text{x} = 8/2$

$text{x} = 4$

$text{Boys 7x} = 7 times 4 = 28$

$text{Girls 5x} = 5 times 4 = 20$

$text{Total strength }= 48$

Answer :

$text{The total strength of the class is 48}$ .

Q11. Baichung's father is $26$ years younger than Baichung's grandfather and $29$ years older than Baichung. The sum of the ages of all the three is $135$ years. What is the age of each one of them?

Solution :

$text{Let Baichung's age is x years}$

$text{and father's age is x + 29 years}$

$text{and grandfather's age x + 29 + 26 = x + 55 years}$

$text{According to question}$

$text{x + x + 29 + x + 55 = 135}$

$text{3x + 84 = 135}$

$text{3x = 135 - 84}$

$text{x} = 51/3$

$text{x} = 17$

$text{x + 29 = 17 + 29 = 46}$

$text{x + 55 = 17 + 55 = 72}$

Answer :

$text{Baichung's age 17 years, father's age 46 years}$

$text{and grandfather's age 72 years}$ .

Q12. Fifteen years from now Ravi's age will be four times his present age. What is Ravi's present age?

Solution :

$text{Let Ravi's present age is x years}$

$text{According to question}$

$text{x + 15 = } 4 times text{x}$

$text{4x - x = 15}$

$text{3x = 15}$

$text{x} = 15/3$

$text{x} = 5$

Answer :

$text{Ravi's present age 5 years.}$

Q13. A rational number is such that when you multiply it by $5/2$ and add $2/3$ to the product, you get $-7/(12)$ . What is the number?

Solution :

$text{Let the number is x}$

$text{According to question}$

$text{x}times5/2 + 2/3 = -7/12$

$text{x}times5/2 = -7/12 - 2/3$

$text{x}times5/2 = (- 7 - 8)/12$

$text{x} = -15/12times 2/5$

$text{x} = - 3/6 = - 1/2$

Answer :

$text{The number is }- 1/2$ .

Q14. Lakshmi is a cashier in a bank. She has currency notes of denominations $₹100, ₹50$ and $₹10$ , respectively. The ratio of the number of these notes is $2:3:5$ . The total cash with Lakshmi is $₹4,00,000$ . How many notes of each denomination does she have?

Solution :

$text{Let ₹ 100 notes 2x = value ₹ 200x}$

$text{and ₹ 50 notes 3x = value ₹ 150x}$

$text{and ₹ 10 notes 5x = value ₹ 50x}$

$text{According to question}$

$text{200x + 150x + 50x = 400,000}$

$text{400x = 400,000}$

$text{x} = (400,000)/400$

$text{x} = 1000$

$text{2x } = 2times1000 = 2000$

$text{3x } = 3times1000 = 3000$

$text{5x } = 5times1000 = 5000$

Answer :

$text{₹ 100 notes 2000, ₹ 50 notes 3000,}$

$text{and ₹ 10 notes 5000}$ .

Q15. I have a total of $₹300$ in coins of denomination $₹1, ₹2$ and $₹5$ . The number of $₹2$ coins is 3 times the number of $₹5$ coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution :

$text{Let ₹ 5 coins x = value ₹ 5x}$

$text{and ₹ 2 coins 3x = value ₹ 6x}$

$text{and ₹ 1 coins 160 - (x + 3x) = value ₹ (160 - 4x)}$

$text{According to question}$

$text{5x + 6x + 160 - 4x = 300}$

$text{7x = 300 - 160}$

$text{7x = 140}$

$text{x} = 140/7 = 20$

$text{3x} = 3times20 = 60$

$text{160 - 4x} = 160 - 4times20 = 80$

Answer :

$text{₹ 5 coins 20, ₹ 2 coins 60 and ₹ 1 coin 80}.$

Q16. The organisers of an essay competition decide that a winner in the competition gets a prize of $₹100$ and a participant who does not win gets a prize of $₹25$ . The total prize money distributed is $₹3,000$ . Find the number of winners, if the total number of participants is $63$

Solution :

$text{Let no. of winner x = value ₹ 100x}$

$text{and no. of not win (63 - x) = value ₹ 25(63 - x) }$

$text{According to question}$

$text{100x + 25(63 - x) = 3000}$

$text{100x + 1575 - 25x = 3000}$

$text{75x} = 3000 - 1575$

$text{x} = 1425/75 = 19$

Answer :

$text{The number of winners 19}$

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8th Maths 2.2

Chapter 2

Linear Equations in One Variable

NCERT Class 8th solution of Exercise 2.1

NCERT Class 8th solution of Exercise 2.3

NCERT Class 8th solution of Exercise 2.4

NCERT Class 8th solution of Exercise 2.5

NCERT Class 8th solution of Exercise 2.6

Exercise 2.2

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