ALGEBRA WORD PROBLEMS WORKSHEET WITH ANSWERS

Problem 1 :

When 23 is decreased from 5 times a number. the result is 57. Find the number.

Problem 2 :

If four times of Liam's age 9 years ago be subtracted from thrice of his age 4 years hence, the result would be equal to his present age. Find the present age of Liam.

Problem 3 :

The denominator of a fraction is 1 less than twice the numerator. If 1 be added to both numerator and denominator, the fraction becomes ⅗. Find the fraction.

Problem 4 :

Sum of the digits in a two digit number is 11. Switching the digits results a number which is less than the original number by 45. Find the number.

Problem 5 :

The demand equation for a certain item is 'd = 15(p - 10)', where d is demand in lbs and p is price in dollars. The supply equation giving the supply s in lbs. for a price p in dollars per lb. is 's = 20(17 - p)'. What is the equilibrium price?

Problem 6 :

If one third of a number exceeds its fifth part by 8, find the number.

Problem 7 :

The perimeter of a rectangle is 28 cm. If the width is two-fifth of the length, find the area of the rectangle.

Problem 8 :

In triangle ABC, the average of m∠A and m∠B is 75°. Find m∠C.

Problem 9 :

In a fraction, sum of the numerator and denominator is 6. If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1/2. Find the fraction.

Problem 10 :

For a concert, it was charged $12 for an adult and $7 for a kid. ids. If a total of 600 tickets were sold for a total of $5950, find the number of adults tickets and kids tickets were sold.

Answers

1. Answer :

Let x be the number.

It is given that decreasing 23 from 5 times of a number results 57.

5x - 23 = 57

Add 23 to both sides.

5x = 80

Divide both sides by 5.

x = 16

Therefore, the number is 16.

2. Answer :

Let x be the present age of Liam.

Age of Liam 9 years ago = x - 9

Age of Liam 4 years hence = x + 4

From the given information,

3(x + 4) - 4(x - 9) = x

3x + 12 - 4x + 36 = x

-x + 48 = x

48 = 2x

24 = x

Therefore, the present age of Liam is 24 years.

3. Answer :

Let x be the numerator.

Denominator = 2x - 1

Then, the fraction is

= ˣ⁄₍₂ₓ ₋ ₁₎ ----(1)

It is given that adding 1 to both numerator and denominator makes the fraction ⅖.

⁽ˣ ⁺ ¹⁾⁄₍₂ₓ ₋ ₁ ₊ ₁₎ = ⅗

⁽ˣ ⁺ ¹⁾⁄₂ₓ = ⅗

5(x + 1) = 3(2x)

5x + 5 = 6x

Subtract 5x from both sides.

5 = x

2x - 1 = 2(5) - 1

= 10 - 1

= 9

ˣ⁄₍₂ₓ ₋ ₁₎ = ⁵⁄₉

Therefore, the fraction is ⁵⁄₉.

4. Answer :

Let x and y be the digits at tens place and ones place respectively.

The two digit number is xy.

Given : The two digits add up to 11.

x + y = 11 ----(1)

Given : Switching the digits results a number which is less than the original number by 45.

yx = xy - 45

10(y) + 1(x) = 10(x) + 1(y) - 45

10y + x = 10x + y - 45

-9x + 9y = -45

Divide both sides by -9.

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

(1) + (2) :

2x = 16

x = 8

Substitute x = 8 in (1).

8 + y = 11

y = 3

xy = 83

Therefore, the two digit number is 83.

5. Answer :

The equilibrium price is the price in which the quantity of goods demanded is equal to the quantity of goods supplied.

That is,

d = s

15(p - 10) = 20(17 - p)

15p - 150 = 340 - 20p

Add 20p to both sides.

35p - 150 = 340

Add 150 to both sides.

35p = 490

Divide both sides by 35.

p = 14

Therefore, the equilibrium price is $14.

6. Answer :

Let x be the required number.

One-third of the number = ˣ⁄₃

Fifth part of the number = ˣ⁄5 

Given : One-third of the number exceeds its sixth part by 4. 

ˣ⁄₃ = ˣ⁄5 + 8

Least common multiple of the two denominators (3, 5) is 15.

Multiply both sides of the equation above by 15 to get rid of the denominators 3 and 5.

15(ˣ⁄₃) = 15(ˣ⁄5 + 8)

15(ˣ⁄₃) + = 15(ˣ⁄5) + 15(8)

5x = 3x + 120

Subtract 3x from both sides.

2x = 120

Divide both sides by 2.

x = 60

Therefore, the number is 60.

7. Answer :

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

Given : Perimeter of the rectangle is 28 cm.

Perimeter = 28 cm

2(x + y) = 28

Divide both sides by 2.

x + y = 14 ----(1)

Given : The width of the rectangle is two-fifth of its length.

y = ⅖ ⋅ x

y = ²ˣ⁄₅ ----(2)

Substitute y = ²ˣ⁄₅ in (1).

x + ²ˣ⁄₅ = 14

Multiply both sides by 5.

5(x + ²ˣ⁄₅)5 = 5(14)

5(x) + 5(²ˣ⁄₅)5 = 70

5x + 2x = 70

7x = 70

Divide both sides by 7.

x = 10

length = 10 cm

Substitute x = 10 in (2).

y = ²⁽¹⁰⁾⁄₅

y = 4

width = 4 cm

Area of the rectangle :

= length x width

= 10 cm x 4 cm

= 40 cm²

8. Answer :

In triangle ABC, it is given that the average of m∠A and m∠B is 75°.

(m∠A + m∠B)/2 = 75°

Multiply both sides by 2.

m∠A + m∠B = 150°

In triangle ABC,

m∠A + m∠B + m∠C = 180°

Substitute m∠A + m∠B = 150°.

150° + m∠C = 180°

Subtract 150° from both sides.

m∠C = 30°

9. Answer :

let ˣ⁄y be the required fraction.

Given : Sum of the numerator and denominator is 6.

x + y = 6 ----(1)

Given : If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1/2

⁽ˣ ⁺ ²⁾⁄₍y ₊ ₁₎ = ½

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

2x + 4 = y + 1

2x - y = -3 ----(2)

(1) + (2) :

3x = 3

x = 1

Substitute x = 1 in (1).

1 + y = 6

y = 5

 ˣ⁄y = ⅕

Therefore, the required fraction is ⅕.

10. Answer :

Let x be the number of adult tickets and y be the number of kids tickets sold.

Given : A total of 600 tickets were sold.

x + y = 600 ----(1)

Given : Cost of each adult ticket was $12 and kid ticket was $7 and tickets were sold for a total of $5950.

12x + 7y = 5950 ----(2)

(2) - 7(1)

5x = 1750

Divide both sides by 5.

x = 350

Substitute x = 350 in (1).

350 + y = 600

Subtract 350 from both sides.

y = 250

Therefore, the number of adults tickets sold was 350 and kids tickets was 250.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.