Problem 1 :
A large pizza at Palanzio’s Pizzeria costs $6.80 plus $0.90 for each topping. The cost of a large cheese pizza at Guido’s Pizza is $7.30 plus $0.65 for each topping. How many toppings need to be added to a large cheese pizza from Palanzio’s Pizzeria and Guido’s Pizza in order for the pizzas to cost the same, not including tax?
Solution :
Let x be the number of toppings need to be added to a pizza from Palanzio’s Pizzeria and Guido’s Pizza.
After adding x number of topiings, let y be the cost of each pizza from Palanzio’s Pizzeria and Guido’s Pizza.
Cost of a pizza from Palanzio’s Pizzeria :
y = 6.80 + 0.90x ----(1)
Cost of a pizza from Guido’s Pizza :
y = 7.30 + 0.65x ----(2)
Solve (1) and (2) :
(1) = (2)
y = y
6.80 + 0.90x = 7.30 + 0.65x
Subtract 0.65x from both sides.
6.80 + 0.25x = 7.30
Subtyract 6.80 from both sides.
0.25x = 0.50
Divide both sides by 0.25.
x = 2
Required number of toppings is 2.
Problem 2 :
Ms. Kitts works at a music store. Last week she sold 6 more than 3 times the number of CDs that she sold this week. Ms. Kitts sold a total of 110 CDs over the 2 weeks. Find the number of CDs sold last week and this week.
Solution :
x and y can be assumed as follows.
x ----> number of CDs sold last week
y ----> number of CDs sold this week
Given : Number of CDs sold last week is equal to 6 more than 3 times the number of CDs sold this week.
x = 3y + 6 ----(1)
Given : Total number of CDs sold over the 2 weeks is 110.
x + y = 110 ----(2)
Substituet x = 3y + 6 into (2).
(3y + 6) + y = 110
3y + 6 + y = 110
4y + 6 = 110
Subtract 6 from both sides.
4y = 104
Divide both sides by 4.
y = 26
Substitute y = 26 into (1).
x = 3(26) + 6
x = 78 + 6
x = 84
Ms. Kitts sold 84 CDs last week and 26 CDs this week.
Problem 3 :
The length of a rectangle is equal to triple the width. Find the dimensions of the rectangle if the perimeter is 86 centimeters.
Solution :
Let x be the length and y be the widh of the rectangle.
Given : The length of the rectangle is equal to triple the width
x = 3y ----(1)
Given : Perimeter of the rectabngle is 86 centimeters.
Perimeter = 86
2x + 2y = 86
Divide both sides by 2.
x + y = 43 ----(2)
Substite x = 3y into (2).
3y + y = 43
4y = 43
Divide both sides by 4.
y = 10¾
Substitute y = 10¾ into (1).
x = 3(10¾)
x = 3(⁴³⁄₄)
x = ¹²⁹⁄₄
x = 32¼
The dimensions of the rectangle are 32¼ cm and 10¾ cm.
Problem 4 :
At a restaurant the cost for a breakfast taco and a small glass of milk is $2.10. The cost for 2 tacos and 3 small glasses of milk is $5.15. Find the cost of a taco, and the cost of a small glass of milk.
Solution :
x and y can be assumed as follows.
x ----> cost of a taco
y ----> cost of a small glass of milk
Given : The cost for a breakfast taco and a small glass of milk is $2.10.
x + y = 2.10 ----(1)
Given : The cost for 2 tacos and 3 small glasses of milk is $5.15.
2x + 3y = 5.15 ----(2)
Multiply (1) by -3.
-3(x + y) = -3(2.10)
-3x - 3y = -6.30 ----(3)
Add (2) and (3) to eliminate y.
(2x + 3y) + (-3x - 3y) = 5.15 + (-6.30)
2x + 3y - 3x - 3y = 5.15 - 6.30
-x = -1.15
Multiply both sides by -1.
x = 1.15
Substitute 1.15 into (1).
1.15 + y = 2.10
Subtract 1.15 from both sides.
y = 0.95
The cost of a taco is $1.15 and that of a small glass of milk is $0.95.
Problem 5 :
The Frosty Ice-Cream Shop sells sundaes for $2 and banana splits for $3. On a hot summer day, if the shop had sold 8 more sundaes than banana splits and made $156, find the number of sundaes and banana splits sold.
Solution :
x and y can be assumed as follows.
x ----> number of sundaes
y ----> number of banana splits
Given : The shop sold 8 more sundaes than banana splits
x = y + 8 ----(1)
Given : The shop made $156 by selling sundaes and banana splits.
2x + 3y = 156 ----(2)
Substitute x = y + 8 into (2).
2(y + 8) + 3y = 156
2y + 16 + 3y = 156
5y + 16 = 156
Subtract 16 from both sides.
5y = 140
Divide both sides by 5.
y = 28
Substitute y = 28 into (1).
x = 28 + 8
x = 36
The number of sundaes sold is 36 and that of banana splits is 28.
Problem 6 :
The sum of two numbers is 13. If one of the numbers is 7 less than three times the other number, find the two numbers.
Solution :
Let x and y be the two numbers.
Given : The sum of two numbers is 13.
x + y = 13 ----(1)
Given : One of the numbers is 7 less than three times the other number.
x = 3y - 7 ----(2)
Substitute x = 3y - 7 into (1).
(3y - 7) + y = 13
3y - 7 + y = 13
4y - 7 = 13
Add 7 to both sides.
4y = 20
Divide both sides by 4.
y = 5
Substitute y = 5 into (2).
x = 3(5) - 7
x = 15 - 7
x = 8
The two numbers are 8 and 5.
Problem 7 :
The sum of numerator and denominator in a fraction is 8. If 3 be added to both numerator and denominator, the fraction becomes ¾. Find the fraction.
Solution :
x and y can be assumed as follows.
x ----> numerator
y ----> denominator
The fraction is ˣ⁄y.
Given : The sum numerator and denominator in a fraction is 8.
x + y = 8 ----(1)
Given : If 3 be added to both numerator and denominator, the fraction becomes 3/4.
⁽ˣ ⁺ ³⁾⁄₍y ₊ ₃₎ = ¾
By corss multiplying,
4(x + 3) = 3(y + 3)
4x + 12 = 3y + 9
Subtract 3y from both sides.
4x - 3y + 12 = 9
Substitute 12 from both sides.
4x - 3y = -3 ----(2)
Multiply (1) by 3.
3(x + y) = 3(8)
3x + 3y = 24 ----(3)
Add (2) and (3) to eliminate y.
(4x - 3y) + (3x + 3y) = -3 + 24
4x - 3y + 3x + 3y = 21
7x = 21
Divide both sides by 7.
x = 3
Substitute x = 3 into (1).
3 + y = 8
Substract 3 from both sides.
y = 5
The fraction is
ˣ⁄y = ⅗
Problem 8 :
The sum of the present ages of a father and his son is 50 years. Five years hence, the age of the father will be three times the age of the son. Find the present ages of father and son.
Solution :
f and s can be assumed as follows.
f ----> present age of father
s ----> present age of son.
Given : The sum of the present ages of a father and his son is 50 years.
f + s = 50 ----(1)
Given : Five years hence, the age of the father will be three times the age of the son.
f + 5 = 3(s + 5)
f + 5 = 3s + 15
Subtract 5 from both sides.
f = 3s + 10 ----(2)
Substitute f = 3s + 10 into (1).
(3s + 10) + s = 50
3s + 10 + s = 50
4s + 10 = 50
Subtract 10 from both sides.
4s = 40
Divide both sides by 4.
s = 10
Substitute s = 10 into (2).
f = 3(10) + 10
f = 30 + 10
f = 40
The present age of father is 40 years and that of son is 10 years.
Problem 9 :
Marcos has 15 coins in nickels and quarters. If the total value of the coins is $1.75, find the number of nickels and quarters.
Solution :
1 nickel = 5 cents
1 quarter = 25 cents
Convert $1.75 to cents.
$1.75 x 100 = 175 cents
Let n be the number of nickels and q be the numer of quarters.
Given : There are 15 coins in nickels and quarters.
n + q = 15 ----(1)
Given : The total value of the coins is $1.75.
5n + 25q = 175
Divide both sides by 5.
n + 5q = 35 ----(2)
(2) - (1) :
(n + 5q) - (n + q) = 35 - 15
n + 5q - n - q = 20
4q = 20
Divide both sides by.
q = 5
Substitute q = 5 into (1).
n + 5 = 15
Subtract 5 from both sides.
n = 10
There are 10 nickels and 5 quarters.
Problem 10 :
Some students want to order shirts with their school logo. One company charges $9.65 per shirt plus a setup fee of $43. Another company charges $8.40 per shirt plus a $58 fee. For what number of shirts would the cost be the same?
Solution :
Let x be the number of shirts for which the cost be the same in both companies and y be the total cost for x number shirts in each company.
Total cost of x number of shirts in the first company :
y = 9.65x + 43 ----(1)
Total cost of x number of shirts in the first company :
y = 8.40x + 58 ----(2)
Since, the cost is the same in both companies, (1) and (2) must be equal.
(1) = (2)
y = y
9.65x + 43 = 8.40x + 58
Subtract 8.40x from both sides.
1.25x + 43 = 58
Subtract 43 from both sides.
1.25x = 15
Divide both sides by 1.25.
x = 12
When the number of shirts is 12, the cost would be the same in both companies.
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