# Algebra Word Problems - Page 2

Algebra Word Problems
• Page 2
11.
The perimeter of the base of a triangular shelf is more than 14 cm. The first side is 6 cm longer than the shortest side. The length of the third side is 4 cm less than thrice the length of the shortest side. Write an equation or an inequality to find the least possible length of the shortest side. Let $x$ represent the length of the shortest side in cm.
 a. 5$x$ + 2 > 14 b. 5$x$ - 2 < 14 c. 6$x$ + 2 < 14 d. 6$x$ + 4$x$ > 14

#### Solution:

The length of the shortest side = x cm.
Then (x + 6) represent the length of the first side and (3x - 4) represent the length of the third side.
[As per the question.]

Perimeter of the shelf > 14
[Write a word inequality.]

x + x + 6 + 3x - 4 > 14
[Translate into an algebraic inequality.]

5x + 2 > 14

12.
The unequal side of an isosceles triangle is 9 cm more than the length of its equal sides. Find the length of the three sides if the perimeter of the triangle is 27 cm.
 a. 6 cm, 15 cm, 15cm. b. 6 cm, 6 cm, 15 cm. c. 15 cm, 15 cm, 6 cm. d. 6 cm, 6 cm, 6 cm.

#### Solution:

Let x represent the length of each equal side of the isosceles triangle in cm. Then x + 9 represents the length of the unequal side in cm.

Perimeter of the triangle = 27 cm.

x + x + x + 9 = 27

3x + 9 = 27

3(x + 3) = 27

x + 3 = 9

x = 6
[Solve for x.]

The three sides of the triangle have lengths 6 cm, 6 cm, 15 cm.

13.
The age of a mother is 29 years more than that of her daughter. 4 years later, mother's age will be twice that of the daughter's age. Find their present ages.
 a. Daughter: 8 years, Mother: 46 years b. Daughter: 25 years, Mother: 54 years c. Daughter: 4 years, Mother: 33 years d. Daughter: 29 years, Mother: 58 years

#### Solution:

Let x represent the daughterÃ¢â‚¬â„¢s present age. Then (x + 29) represents the motherÃ¢â‚¬â„¢s present age.
[As per the question.]

4 years later, the daughter's age becomes (x + 4) and mother's age becomes (x + 33) years.

4 years later motherÃ¢â‚¬â„¢s age = 2(daughterÃ¢â‚¬â„¢s age 4 years later).
[As per question.]

x + 33 = 2(x + 4)

x + 33 = 2x + 8

x = 25, so x + 29 = 25 + 29 = 54

So, DaughterÃ¢â‚¬â„¢s present age is 25 years and motherÃ¢â‚¬â„¢s age 54 years.

14.
A girl bought 14 stamps of denomination $0.05, and$0.20 for one dollar. How many stamps of each denomination did she buy?
 a. 14 stamps of $0.05, 0 stamps of$0.20 b. 12 stamps of $0.05, 2 stamps of$.20 c. 2 stamps of $0.05, 12 stamps of$0.20 d. 1 stamp of $0.05, 13 stamps of$0.20

#### Solution:

Let x represent the number of $0.05 stamps she bought. Then 14 - x represents the number of$0.20 stamps bought by the girl.
[As per the question.]

Value of $0.05 stamps + value of$0.20 stamps = $1 0.05x + 0.20(14 - x) = 1 0.05x + 2.80 - 0.20x = 1 - 0.15x = - 1.8 x = 12, so 14 - x = 14 - 12 = 2 So, the girl bought 12 stamps of$0.05 denomination and 2 stamps of \$0.20 denomination.

15.
Find three numbers if their sum is 308, and the second number is four times the first and the third is half the second.
 a. 44, 22, 88 b. 44, 176, 88 c. 44, 180, 88 d. 176, 44, 88

#### Solution:

Let n represent the first number.
Then second number represent 4n and third number represent 4n2.
[As per the question.]

1st number + 2nd number + 3rd number = 308

n + 4n + 4n2 = 308

n + 4n + 2n = 308
[Simplify.]

7n = 308

n = 44

1st number = 44
[Substitute.]

2nd number = 44 × 4 = 176
[Substitute.]

3rd number = 176 / 2 = 88
[Substitute.]

So, the three numbers are 44, 176, and 88.

16.
On a certain day, more than 6801 people visited an exhibition. The number of men exceeded the number of women by 677. The number of children was 1022 less than the number of women. Find the least possible number of women visitors on that day.
 a. 1699 b. 683 c. 345 d. 2382

#### Solution:

Let x represent the number of women.
Then x + 677 represents the number of men
and x - 1022 represents the number of children.
[As per the question.]

Number of men + number of women + number of children > 6801
[Word inequality.]

x + 677 + x + x - 1022 > 6801
[Algebraic inequality.]

3x - 345 > 6801

3x > 7146

x > 2382

The least possible number of women visitors on that day was 2382.

17.
The weight of 2 packets of bread is 5 lb. Find the weight of the other packet, if the weight of one packet is 1 lb.
 a. 7 lb b. 5 lb c. 6 lb d. 4 lb

#### Solution:

Let a be the weight of the other packet.

The weight of two packets = weight of one packet + weight of the other packet.

5 lb = 1 lb + a
[Substitute the weights.]

5 - 1 = a
[Subtract 1 from each side.]

4 lb = a

So, the weight of the other packet of bread is 4 lb.

18.
Kate has 10 stamps and her friend Dina has 4 stamps less than the stamps Kate has. Find the number of stamps that Dina has.
 a. 7 b. 6 c. 9 d. 8

#### Solution:

Let b be the number of stamps Dina has.

Number of stamps Dina has = number of stamps Kate has - the number of stamps less than that Kate has.

b = 10 - 4
[Substitute the number of stamps.]

b = 6
[Subtract.]

So, Dina has 6 stamps.

19.
Dolly bought 3 mushroom pizzas and 2 corn pizzas. How many pizzas did she buy?
 a. 5 b. 6 c. 4 d. 3

#### Solution:

Let p be the total number of pizzas Dolly bought.

Total Number of pizzas Dolly bought = number of mushroom pizzas + the number of corn pizzas.

p = 3 + 2
[Substitute the number of pizzas.]

p = 5

So, Dolly bought 5 pizzas in all.

20.
Quincy spends 8 hours on computer every day. Find how long Quincy needs to spend on the computer further, if she has already spent 4 hours on a particular day.
 a. 5 b. 6 c. 7 d. 4

#### Solution:

Let t be the number of hours Quincy has to spend on the computer.

Number of hours daily Quincy spends on the computer = number of hours she already spent + number of hours she needs to spend.

8 = 4 + t
[Substitute the number of hours.]

8 - 4 = t
[Subtract 4 from each side.]

4 = t

So, Quincy has to spend 4 hours on the computer.