# WORD PROBLEMS ON LINEAR EQUATIONS IN ONE VARIABLE

Problem 1 :

A class of 45 students is divided into two groups. If one group has 9 students less than the other, how many stidents are in the smaller group?

Solution :

Let x be the number of students in one of the two groups.

Number of students in the other group is (x - 9).

Given : Total number of students in the class is 45.

x + (x - 9) = 45

x + x - 9 = 45

2x - 9 = 45

2x = 54

Divide both sides by 2.

x = 27

x - 9 = 27 - 9 = 18

There are 18 students in the smaller group.

Problem 2 :

Subtracting 17 from four times of a number results 35. Find the number.

Solution :

Let x be the number.

Given : When 17 is subtracted 17 from four times of the number is equal to 35.

Then, we have

4x - 17 = 35

4x = 52

Divide both sides by 4.

x = 13

The number is 13.

Problem 3 :

9 less than five times of a number is equal 3 more than two times of the same number. Find the number.

Solution :

Let x be the number.

Five times of the number = 5x

9 less than five times of the number = 5x - 9 ----(1)

Two times of the number = 2x

3 more than two times of the number = 2x + 3 ----(2)

From the given information,

(1) = (2)

5x - 9 = 2x + 3

3x = 12

x = 4

The number is 4.

Problem 4 :

The length of one of the sides of a triangle is 7 cm and the lengths of the remaining two sides differ by 4 cm. If the perimeter of the triangle is 33 cm, find the lengths of the missing sides of the triangle.

Solution :

Since the lengths of the two sides of the triangle differ by 4 cm, the lengths can be assumed as x and (x + 4).

The sides of the triangle are

7 cm, x cm, (x + 4) cm

Perimeter of the triangle = 33 cm

7 + x + (x + 4) = 33

7 + x + x + 4 = 33

2x + 11 = 33

Subtract 11 from both sides.

2x = 22

Divide both sides by 2.

x = 11

x + 4 = 11 + 4 = 15

The missing sides of the triangle are 11 cm and 15 cm.

Problem 5 :

One-sixth of a number is greater than one-seventh of the same number by 1. Find the number.

Solution :

Let x be the number.

One-sixth of the number =()x = ˣ⁄₆

One-seventh of the number = ()x ˣ⁄₇

From the given information,

ˣ⁄₆ = ˣ⁄₇ + 1

Least common multiple of the denominators (6, 7) is 42.

Multiply both sides of the equation by 42 to get rid of the denominators 6 and 7.

42(ˣ⁄₆) = 42(ˣ⁄₇ + 1)

7x = 42(ˣ⁄₇) + 42

7x = 6x + 42

x = 42

The number is 42.

Problem 6 :

Three times of my age 5 years ago is equal to 5 less than two times of my age 5 years hence. How old am i now?

Solution :

Let x be the your present age.

Your age 5 years ago = x - 5

Three times of your age 5 years ago :

= 3(x - 5)

= 3x - 15----(1)

Your age 5 years hence = x + 5

Two times of my age 5 years hence = 2(x + 5)

= 2x + 10

5 less than two times of my age 5 years hence :

= 2x + 10 - 5

= 2x + 5 ----(2)

From the given information,

(1) = (2)

3x - 15 = 2x + 5

x = 20

You are 20 years old now.

Problem 7 :

One number is 4 times the another number, and they sum to -20. What is the greater of the two numbers?

Solution :

Let x be one of the numbers.

Then the other number is 4x.

Given : Sum of the two numbers is -20.

x + 4x = -20

5x = -20

Divide both sides by 5.

x = -4

4x = 4(-4) = -16

The two numbers are -4 and -16.

The greater one is -4.

Problem 8 :

Paulina owns a certain number of dishes. 24 added to 3 times the number of dishes is equal to 9 times the number of dishes. How many dishes does Palina own?

Solution :

Let x be the numbers of dishes Paulina owns.

Given : 24 added to 3 times the number of dishes is equal to 9 times the number of dishes.

3x + 24 = 9x

Subtract 3x from both sides.

24 = 6x

Divide both sides by 6.

4 = x

Paulina owns 6 dishes.

Problem 9 :

There is a monthly subscription fee of \$15 for a movie rental sevice and \$3 for each movie rented. Last month, Adam paid  \$33 for the movie rental sevice. How many movies did Adam rent last month.

Solution :

Let x be the numbers of movies Adam rented last month.

Amount paid by Adam last month = 3x + 15

It is given that Adam paid  \$33 for the movie rental sevice.

3x + 15 = 33

Subtract 15 from both sides.

3x = 18

Divide both sides by 3.

x = 6

Adam rented 6 movies last month.

Problem 10 :

In a fraction, the numerator is less than the denominator by 7. If 4 is added to the numerator and the denominator 2 is added to the denominator, the fraction becomes ¾. Find the fraction.

Solution :

Let x be the denominator.

Given : The numerator is less than the denominator by 7.

From the above information,

fraction = ⁽ˣ ⁻ ⁷⁾⁄ₓ ----(1)

Given : If 4 is added to the numerator and the denominator 2 is added to the denominator, the fraction becomes ¾.

From the above information, we have

⁽ˣ ⁻ ⁷ ⁺ ⁴⁾⁄₍ₓ ₊ ₂₎ = ¾

Simplify.

⁽ˣ ⁻ ³⁾⁄₍ₓ ₊ ₂₎ = 3/4

4(x - 3) = 3(x + 2)

4x - 12 = 3x + 6

x = 18

Substitute x =18 in (1).

fraction = ⁽¹⁸ ⁻ ⁷⁾⁄₁₈

¹¹⁄₁₈

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