ALGEBRA WORD PROBLEMS

Problem 1 :

Thrice of a number decreased by 17 results 28. Find the number.

Answer :

Step 1 :

Let x be the required number.

Step 2 :

Given : When 17 is decreased from thrice of the number is equal to 28.

3x - 17 = 28

Add 17 to both sides.

3x = 45

Divide both sides by 3.

x = 15

The number is 15.

Problem 2 :

In a fraction, the denominator is 2 more than thrice the numerator. If 3 be added to both numerator and denominator, the fraction becomes ⅖. Find the fraction.

Answer :

Plan to solve the problem :

In this problem, the denominator of the fraction is linked to the value in the numerator. 

That is, the denominator is 2 more than thrice the numerator. If we find the value of the numerator, we can find the value of the denominator and find the required fraction.

Assume a variable for the numerator. Since the denominator is linked to numerator, the denominator can be written in terms of the same variable. Write the equation using the information and solve for the variable.

Step 1 : 

Let x be the numerator.

Denominator of the fraction is (3x + 2).

Then, the required fraction is 

= ˣ⁄₍₃ₓ ₊ ₂₎ ----(1)

Step 2 : 

It is given that if 3 be added to both numerator and denominator, the fraction becomes ⅖. 

⁽ˣ ⁺ ³⁾⁄₍₃ₓ ₊ ₂ ₊ ₃₎ = ⅖

Simplify and solve fore x.

⁽ˣ ⁺ ³⁾⁄₍₃ₓ ₊ ₅₎ = ⅖

5(x + 3) = 2(3x + 5)

5x + 15 = 6x + 10

15 = x + 10

5 = x

Step 3 : 

Denominator of the fraction :

3x + 2 = 3(5) + 2

= 15 + 2

= 17

Step 4 : 

Substitute x = 5 and 3x + 2 = 17 in (1).

(1)---->  ˣ⁄₍₃ₓ ₊ ₂₎ = ⁵⁄1₇

Therefore, the required fraction is ⁵⁄1₇.

Problem 3 :

Sum of the age of a father and 2 more than twice the age of his daughter is equal 55 years. Age of the father is 5 more than four times the age of the daughter. Find the age of the father and son.

Answer :

Step 1 :

Let x and y be the ages of the father and daughter respectively.

Given : Sum of the age of a father and 2 more than twice the age of his daughter is equal 55 years.

x + (2y + 2) = 55

x + 2y + 2 = 55

x + 2y = 53 ----(1)

Step 2 :

Given : Age of the father is 5 more than four times the age of the daughter. Find the age of the father and son.

x = 4y + 5 ----(2)

Substitute x = 4y + 5 in (1).

(4y + 5) + 2y = 53

4y + 5 + 2y = 53

6y + 5 = 53

6y = 48

y = 8

Age of the daughter = 8 years

Step 3 :

Sum y = 8 in (2).

x = 4(8) + 5

x = 32 + 5

x = 37

Age of the father = 37 years

Problem 4 :

In a triangle, the second angle is 100% more than the first angle and the third angle is 50% more than the second angle. Find the three angles of the triangle.

Answer :

Step 1 :

Let x be the first angle of the triangle.

Step 2 :

Second Angle : 

= (100 + 100)% of first angle

= 200% of x

= 2x

Step 3 :

Third angle = (100 + 50)% of second angle

= 150% of 2x

= 1.5 ⋅ 2x

= 3x

Step 4 :

In a triangle, three angles add up to 180°.

first angle + second angle + third angle = 180°

x + 2x + 3x = 180°

6x = 180°

x = 30°

2x = 2(30°) = 60°

3x = 3(30°) = 90°

The three angles of the triangle are 30°, 60° and 90°.

Topic Wise Word Problems in Algebra

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