2 . Linear Equations in One Variable

Chapter 2 : Linear Equations in One Variable

2. Linear Equations in One Variable

Exercise 2.1

Solve the following equations.

Solution: We have ,

Solution: We have,

Solution: We have,

Solution: We have,

Solution: We have ,

Solution : We have,

Solution: We have ,

10.

Solution: We have,

11.

Solution: We have,

12.

Solution: We have,

Exercise 2.2

1. If you subtract

from a number and multiply the result by

, you get

. What is the number?

Solution : let, be the number .

A/Q ,

Therefore, the number is .

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

Solution : let (in metres) be the breadth of the rectangular swimming pool and the length of the rectangular swimming pool will be .

Here , breadth and length

A/Q ,

Therefore, the breadth is 25 m and the length is m

3. The base of an isosceles triangle is

cm . The perimeter of the triangle is

cm . What is the length of either of the remaining equal sides?

Solution: let (in cm) be the length of equal sides of an isosceles triangle .

A/Q ,

Therefore, the length of the side is cm .

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

Solution : let be the first number and the second number will be .

A/Q,

Therefore, the numbers are 40 and 55 (= 40+15).

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

Solution : let the two numbers are and .

A/Q ,

Therefore, the numbers are and .

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

Solution : let and are the three consecutive integers .

A/Q ,

Therefore , the three consecutive integers are 16 , 17(= 16+1) and 18 (= 16+2) .

7. The sum of three consecutive multiples of 8 is 888. Find the multiples.

Solution : let the three consecutive multiples of 8 are and .

A/Q ,

Therefore , the three consecutive number are 288 , 296 (= 288+8) and 304 (= 296+8) .

8. 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 : let and are three consecutive numbers .

A/Q,

Therefore , the numbers are 7 , 8 (= 7+1) and 9 (= 7+2) .

9. The ages 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 : let the ages of Rahul and Haroon are and respectively .

Four years later, Rahul and Haroon age will be and years respectively .

A/Q ,

Therefore, the ages of Rahul and Haroon are years and years respectively .

10. 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 : let the number of boys and girls in a class are and respectively .

A/Q ,

The numbers of boy and the numbers of girls .

Therefore, the total numbers of student in the class is 28 + 20 = 48 .

11. 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 : let be the ages of Baichung .

Then , Baichung’s father ages is years and Baichung’s grandfather ages is years

A/Q ,

Therefore, the ages of Baichung is 17 years .

Baichung’s father ages is years

and Baichung’s grandfather ages is years .

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

Solution: let be Ravi’s present age .

Fifteen years from now , Ravi’s age will be years .

A/Q ,

Therefore, Ravi’s present ageis 5 years .

13. A rational number is such that when you multiply it by

and add

to the product,you get

. What is the number?

Solution : let be the number .

A/Q ,

Therefore, the number is

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

Solution: let and are numbers of notes .

A/Q ,

The number of Rs 100 notes

The number of Rs 50 notes

and the number of Rs 10 notes .

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

Solution : let be the number of Rs 5 coins . The number of Rs 2 coins is and the number of Rs 1 coins is .

A/Q ,

Therefore , the number of Rs 5 coins is 20 , the number of Rs 2 coins is and the number of Rs 1 coins is .

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

Solution: let be the number of winners and the number of participant who does not win the competition will be .

A/Q ,

Therefore, the number of winner is 19 .

Exercises 2.3

Solve the following equations and check your results .

Solution: We have,

LHS:

RHS:

LHS = RHS

Solution: We have,

(Solution)

LHS:

RHS:

LHS = RHS

Solution: We have,

LHS:

RHS:

LHS = RHS

Solution: We have,

LHS:

RHS:

Solution: We have,

LHS:

RHS:

LHS = RHS

Solution: We have,

LHS:

RHS:

LHS = RHS

Solution: We have,

LHS:

RHS:

LHS = RHS

Solution: We have,

LHS:

RHS:

Solution: We have,

LHS:

RHS :

10.

Solution: We have,

LHS:

RHS:

Exercise 2.4

1. Amina thinks of a number and subtracts

from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Solution: let be the number .

A/Q,

Therefore, the number is 4 .

2. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution: let be the first number and be the second number .

If 21 is added to both the numbers, then the numbers are and respectively .

A/Q,

Therefore, the numbers are 7 and 35 (= 5×7) .

3. Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Solution: let be the unit place and be the ten place of the two digit number respectively.

Then, the number

When we interchange the digits , then the number

A/Q,

Therefore, the number

4. One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution: let be the unit place and be the ten’s of the two digit number respectively.

The original number

If we interchange the digits , then the number

A/Q,

Therefore, The original number

The original number

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Solution: let be the present age of Shobo and his mother’s age will be years .

Five years from now , Shobo’s age will be years .

A/Q,

Therefore, the present age of Shobo is 5 years and his mother’s age is 30 (= 6 × 5) years .

6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per metre it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution: let and (in metres) be the length and breadth of the rectangular plot .

The perimeter of rectangular plot

The cost of the rectangular plot

A/Q,

The length of the plot

The breadth of the plot .

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

Solution: let and be the shirt and trouser materials buys by Hasan .

So, the cost of shirt material

And the cost of trouser material

The price of shirt material

The price of trouser material

A/Q,

Therefore, the trouser material is metres

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution: let be the number of deer in the herd.

The number of deer are grazing in the field

And the number of deer playing in the field

A/Q,

Therefore, the number of deer is 72 .

9. A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution: let be the present age of granddaughter and the grandfather age will be years .

A/Q,

Therefore, the present age of granddaughter is 6 years and the grandfather age is 60 (= 10×6) years .

10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Solution: let and be the age of Aman’s son and Aman respectively.

Ten years ago , Aman’s son and Aman age will be and years respectively .

A/Q,

Therefore, Aman’s son present age is 20 years and Aman age is 60 (= 3 × 20) years .

Exercise 2.5

Solve the following linear equations.

Solution: We have,

Therefore, the solution is

Solution: We have,

Therefore, the solution is

Solution: We have,

Therefore, the solution is .

Solution: We have,

Therefore, the solution is .

Solution: We have,

Therefore, the solution is .

Solution: We have,

Therefore, the solution is

Simplify and solve the following linear equations.

Solution: We have,

LHS :

RHS:

The equation is ,

Therefore, the solution is .

Solution: We have,

LHS :

The equation is

Therefore, the solution is .

Solution: We have,

LHS:

RHS:

The equation is

Therefore, the solution is .

10.

Solution: We have,

LHS:

RHS:

The equation is

Therefore, the solution is .

Exercise 2.6

Solve the following equations :

Solution: We have,

Therefore, the solution is

Solution: We have,

Therefore, the solution is

Solution: We have,

Therefore, the solution is .

Solution: We have,

Therefore, the solution is .

Solution: We have,

Therefore, the solution is

6. The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Solution: let and are the present age of Hari and Harry respectively .

Four years from now ,the age Hari and Harry will be years and years respectively .

A/Q,

Therefore , the present age of Hari years

And the present age of Harry years .

7. The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is

. Find the rational number.

Solution: let be the numerator of the rational number ,then the denominator will be .

Therefore, the rational number(fraction) .

If the numerator is increased by 17 and the denominator is decreased by 1, then the rational number is

A/Q ,

Therefore, the rational number .