Algebra Word Problems

Algebra Word Problems
  • Page 1
 1.  
Two years later Matt will be twice as old as the present age of his friend Ed. Find the present age of Ed if Matt is 18 years old now.
a.
36 years
b.
20 years
c.
10 years
d.
18 years


Solution:

Let y represent Ed’s present age. Matt's present age is 18 years.
[As per the question.]

Matt’s age 2 years later = twice Ed’s present age.
[Word equation.]

20 = 2y
[Algebraic form.]

10 = y

So, Ed is 10 years old.


Correct answer : (3)
 2.  
A sum of $520 is distributed between A, B and C such that B gets $40 more than A, and C gets $50 more than B. Find the share of B.
a.
$510
b.
$90
c.
$170
d.
$480


Solution:

Let b represent B’s share in dollars. Then (b - 40) represent share of A and (b + 50) represent share of C in dollars.
[As per the question.]

A’s share + B’s share + C’s share = 520
[Word equation.]

b - 40 + b + 50 + b = 520
[Algebraic equation.]

3b + 10 = 520

3b = 510

b = 170

So, B’s share is $170.


Correct answer : (3)
 3.  
Nancy has some dimes, quarters and nickels in her piggy bank. The number of dimes is three more than twice the number of quarters, and the number of nickels is two more than the number of quarters. Find the least possible number of quarters in the piggy bank if she has more than $2.90 in it.
a.
5 quarters
b.
6 quarters
c.
7 quarters
d.
11 quarters


Solution:

Let c represent the number of quarters in the coin bank.
Then 2c + 3 represents the number of dimes,
and c + 2 represents the number of nickels.
[As per the question.]

Value of nickels + Value of dimes + Value of quarters > 2.90
[Word inequality.]

0.05(c + 2) + 0.10(2c + 3) + 0.25c > 2.90
[Convert all into $ and form algebraic inequality.]

0.05c + 0.10 + 0.20c + 0.30 + 0.25c > 2.90
[Distributive property.]

0.50c + 0.40 > 2.90
[Combine like terms.]

0.5c > 2.5
[Subtract 0.40 from both sides.]

c > 5

The least possible number of quarters in the coin bank is 6.


Correct answer : (2)
 4.  
At a fair, a boy tries his skill in shooting. He was to receive $2 for hitting the bull's eye and had to pay $0.50 for missing the bull's eye. He tried 30 shots, but received only $32.50. How many times did he hit the bull's eye?
a.
21
b.
19
c.
62.50
d.
15


Solution:

Let x represent the number of times he hit the bull’s eye. Then 30 - x represents the number of times he missed the bull’s eye.
[As per the question.]

Amount he received for hitting the bull’s eye - Amount he paid for missing the bull's eye = $32.50

2x - 0.50(30 - x) = 32.50
[Form an algebraic equation.]

2x - 15 + 0.50x = 32.50

2.50x = 47.50

x = 19

So, the boy hit the bull’s eye 19 times.


Correct answer : (2)
 5.  
A boy scored 22 points on the first test, 11 in the second, 16 in the third and 19 in the fourth. What is the least number of points he must score in the fifth test to keep his average above 16?
a.
17
b.
12
c.
7
d.
28


Solution:

Let x represent the number of points scored by the boy in the fifth test.

22 + 11 + 16 + 19 + x5 > 16
[Average = score1 +score2 +.....+scorenn.]

(68 + x)5 > 16

68 + x > 80

x > 12

The boy should score at least 12 points in the fifth test to keep his average above 16.


Correct answer : (2)
 6.  
The difference between 22 times the length of a line segment and 18 times the length of the same line segment is 8 cm. Write an equation or inequality for the problem. Let c be the length of the line segment.
a.
18c - 22c = 8
b.
8 + 18c = 22
c.
22c - 18c = 8
d.
8 - 22c = 18


Solution:

22 × length of the line segment - 18 × length of the line segment = 8
[As per the question.]

22c - 18c = 8


Correct answer : (3)
 7.  
A fraction becomes 1 2, when its denominator is increased by 7, and 1 3 when its numerator is decreased by 9. Write an equation or inequation that will help you find the denominator.
a.
3(y + 27) = 2(y + 7)
b.
y2 + 7 = y3 - 9
c.
2(y + 27) = 2(y + 7)
d.
y + 7 = y - 9


Solution:

NumeratorDenominator + 7 = 1 / 2 and numerator - 9denominator = 1 / 3
[As per the question.]

Let the fraction be xy.

Then, xy+7 = 1 / 2 - - - - - - - - - - - (1)

and x-9y = 1 / 3- - - - - - - - - - - - - (2)

x = y3 + 9
[Simplify.]

y3 + 9y + 7 = 1 / 2
[Substitute the value of x in equation (1).]

y + 273(y + 7) = 1 / 2
[Simplify.]

2(y + 27) = 3(y + 7)
[Cross multiply.]


Correct answer : (3)
 8.  
Quick Lime contains 44.3% of Oxygen by weight. Determine the weight of Oxygen in 800 g of Quick Lime.
a.
18.1 g
b.
354.4 g
c.
100 g
d.
800 g


Solution:

Let y represent the weight, in grams, of Oxygen in 800 g of Quick Lime.
[As per the question.]

44.3 / 100 = y800
[Equate the ratio of weight of oxygen to weight of quick lime.]

y = (44.3 / 100) × 800 = 354.4
[Solve and simplify.]

800 g of Quick Lime contains 354.4 g of Oxygen in it.


Correct answer : (2)
 9.  
The perimeter of a rectangle is 84 m. Its length is 8 m more than 4 times its width. Write an equation or an inequality to find the dimensions of the rectangle. Let x represent the width of the rectangle.
a.
8x + 4 = 2 × 84
b.
2(8 + 4x) = 84
c.
2(5x + 8) = 84
d.
8 + 4x = 84


Solution:

If x represent the width of the rectangle then 8 + 4x represent its length.
[As per the question.]

Perimeter of a rectangle = 2(length + width).

2(8 + 4x + x) = 84

2(5x + 8) = 84


Correct answer : (3)
 10.  
Edward invested part of his savings at 2% simple interest per annum and half this amount at 4% simple interest per annum. He wants to earn a total annual interest of at most $200. Write an equation or an inequality to find the greatest possible amount Edward invested at 4% interest. Let x represent the amount Edward invested at 2% simple interest.
a.
50x + 50x ≥ 200
b.
x100 + x25 ≤ 200
c.
x50 + x50 ≥ 200
d.
x50 + x50 ≤ 200


Solution:

The amount Edward invested at 2% simple interest per annum = x.
Then the amount invested at 4% simple interest per annum = x2

Simple interest at 2% + simple interest at 4% ≤ 200
[Maximum interest shall be $200.]

2x100 + 4x200 ≤ 200
[Simple interest per annum = amount invested × rate of interest.]

x50 + x50 ≤ 200


Correct answer : (4)

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