SOLVING MULTI STEP WORD PROBLEMS

Problem 1 :

In a fraction, the numerator is 3 less than its denominator. If 1 is added to both numerator and denominator, the fraction becomes ½. Find the fraction. 

Solution :

Step 1 :

Let x be the denominator.

"The numerator is 3 less than the denominator"

Then, the numerator = x - 3.

Fraction = ⁽ˣ ⁻ ³⁾⁄ₓ ----(1)

Step 2 :

"If 1 is added to both numerator and denominator, the fraction becomes 3/4"

⁽ˣ ⁻ ³ ⁺ ¹⁾⁄₍ₓ ₊ ₁₎ = ½

⁽ˣ ⁻ ²⁾⁄₍ₓ ₊ ₁₎ = ½

Step 3 :

Solve for x.

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

2x - 4 = x + 1

x = 5

Step 4 :

Substitute x = 5 in (1). 

 ⁽ˣ ⁻ ³⁾⁄ₓ = ⁽⁵ ⁻ ³⁾⁄₅

 ⁽ˣ ⁻ ³⁾⁄ₓ = 

The required fraction is .

Problem 2 :

If twice of Jessica's age 5 years ago is added to thrice of her age 6 years ago, the result would be equal to her present age. Find Jessica's present age. 

Solution :

Step 1 :

Let x be Jessica's present age.

Jessica's age 5 years ago = x - 6

Twice of Jessica's age 5 years ago = 2(x - 5) ----(1)

Step 3 :

Jessica's age 6 years ago = x - 6

Thrice of Jessica's age 6 years ago = 3(x - 6) ----(2)

Step 4 :

According to the question,

(1) + (2) = Jessica's present age

2(x - 5) + 3(x - 6) = x

Step 4 :

Solve for x :

2x - 10 + 3x - 18 = x

5x - 28 = x

4x - 28 = 0

4x = 28

x = 7

Jessica's present age is 7 years.

Problem 3 :

In a three digit number, the middle digit is 2 and the digit at hundreds place is thrice the digit at ones place. The number formed by switching the digits at hundreds place and ones place is less than the original number by 594. Find the number.    

Solution :

Step 1 :

Let x be the digit at ones place.

Then the digit at tens place = 3x and the three digit number is

(3x)2(x)

Step 2 :

It is given that the number formed by switching the digits at hundreds place and ones place is less than the original number by 594.

(x)2(3x) = (3x)2(x) - 594

100(x) + 10(2) + 1(3x) = 100(3x) + 10(2) + 1(x) - 594

100x + 20 + 3x = 300x + 20 + x - 594

Subtract 20 from both sides.

100x + 3x = 300x + x - 594

103x = 301x - 594

0 = 198x - 594

594 = 198x

3 = x

Step 3 :

So, the digit at ones place is 3.

The digit at tens place is

= 3(3)

= 9

So, the required number is 923.

Problem 4 :

In a rectangle, the length is 3 more than thrice the width. If the sum of the length and twice the width is 33 cm, find the perimeter of the rectangle.

Solution :

Step 1 :

Let x and y be the length and width of the rectangle respectively.

Step 2 :

It is given that the length is 3 more than thrice the width.

x = 3y + 3 ----(1)

Step 3 :

It is given that sum of the length and twice the width is 33 cm.

x + 2y = 33

x = 33 - 2y ----(2)

Step 4 :

From (1) and (2),

3y + 3 = 33 - 2y

5y + 3 = 33

5y = 30

y = 6

width = 6 cm.

Step 3 :

Substitute y = 6 in (1).

x = 3(6) + 3

x = 18 + 3

x = 21

length = 21 cm.

Step 4 :

Perimeter of the rectangle :

= 2(length + width)

= 2(21 + 6)

= 2(27)

= 54 cm.

Problem 5 :

The ratio between the number of boys and girls in a school is 15 : 31. If the number of boys is 10 more than twice the number of girls, find the number of boys and girls in the school.

Solution :

Step 1 :

:Let x be the number of girls in the school.

Then the number of boys in the school = 2x + 10.

Step 2 :

It is given that the ratio between the number of boys and number of girls is 15 : 31.

x : (2x + 10) = 15 : 31

ˣ⁄₍₂ₓ ₊ ₁₀₎ ¹⁵⁄₃₁

Step 3 :

Solve for x.

31x = 15(2x + 10)

31x = 30x + 150

x = 150

Step 4 :

2x + 10 = 2(150) + 10

= 300 + 10

= 310

Step 4 :

Number of boys = 310

Number of girls = 150

Problem 6 :

In a triangle, if all the three angles are different and each angle is a multiple of 3, find the three angles.

Solution :

Step 1 :

Since all the three angles are different and each angle is a multiple of 3, the three angles of the triangle can be assumed as follows.

3x, 6x, 9x

Step 2 :

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

3x + 6x + 9x = 180°

18x = 180°

x = 10°

3x = 3(10°) = 30°

6x = 6(10°) = 60°

9x = 6(10°) = 90°

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

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