Question
A store discounted the price of Doritos $0.35 and then a man bought 5 bags. If he paid a total of $12.70 for the bags of chips, how much was each bag originally?
Hint:
○ Form the equation using the given information.
○ Take variable quantity as x.
○ After discount the purchasing price will be
(original price - discount)
The correct answer is: $2.89
○ Given:
Discount on 1 pack of Doritos = $.0.35.
Total pack brought = 5.
Total money paid = $12.70.
○ Step 1:
○ Total discount;
Discount on 1 pack of Doritos= $.0.35
So,
Discount on 5 pack of Doritos= $.0.35 5 = $ 1.75
○ Step 2:
○ Original price of 5 bags Doritos:
It is given that price of 5 bags after discount is $12.70
So, original price of 5 bags = 12.70 + 1.75 = 14.45
○ Step 3:
○ Original price of 1 bag:
original price of 5 bags = $ 14.45
original price of 1 bags = $ = 2.89
- Final Answer:
Hence, the original price of one bag of Doritos is $2.89.
Related Questions to study
Use the Law of Detachment to make a valid conclusion in the true situation.
If the measure of an angle is less than 90°, then it is an acute angle.
Use the Law of Detachment to make a valid conclusion in the true situation.
If the measure of an angle is less than 90°, then it is an acute angle.
Find JK
Find JK
A girl bought two packs of gum from the store. Later she thought she would need some more gum and went back to the store to buy three more packs of gum. If she paid a total of $5.70 for gum, how much was each pack of gum?
A girl bought two packs of gum from the store. Later she thought she would need some more gum and went back to the store to buy three more packs of gum. If she paid a total of $5.70 for gum, how much was each pack of gum?
Use the Law of Detachment to make a valid conclusion in the true situation.
If angles form a linear pair, then they are supplementary.
∠𝐴 𝑎𝑛𝑑 ∠𝐵 form a linear pair.
Use the Law of Detachment to make a valid conclusion in the true situation.
If angles form a linear pair, then they are supplementary.
∠𝐴 𝑎𝑛𝑑 ∠𝐵 form a linear pair.
A man buys four books from the store and a Preferred Reader discount card for $20. Later that day, he goes back and buys five more books. He also got a $5 discount using his new card. If he spent a total of $87 at the bookstore, how much did each book cost assuming every book cost the same amount?
A man buys four books from the store and a Preferred Reader discount card for $20. Later that day, he goes back and buys five more books. He also got a $5 discount using his new card. If he spent a total of $87 at the bookstore, how much did each book cost assuming every book cost the same amount?
Show the conjecture is false by finding a counterexample. If the product of two numbers is even, then the two numbers must be even.
Show the conjecture is false by finding a counterexample. If the product of two numbers is even, then the two numbers must be even.
Dimple Bought a Calculator and binder that were both 15% off the original price. The
original price of binder was Rs 6.20. Justin spent a total of Rs 107. 27 . What was the
original price of the calculator?
The x + 2 = 6 x+2=6x, plus, 2, equals 6 contains a variable. We call this type of equation with a variable an algebraic equation. Finding the variable value that will result in a true equation is typically our aim when solving an algebraic equation.
¶Variables or constants are the two types of measurable quantities. A variable is a quantity with a varying value, and the constant value is nothing but a constant.
¶Steps to writing Variable Equation
1) Identify the variables that represent the unknowns.
2) Convert the issue into variable expressions in algebra.
3) Determine the variables' values to solve the equations for their true values.
Dimple Bought a Calculator and binder that were both 15% off the original price. The
original price of binder was Rs 6.20. Justin spent a total of Rs 107. 27 . What was the
original price of the calculator?
The x + 2 = 6 x+2=6x, plus, 2, equals 6 contains a variable. We call this type of equation with a variable an algebraic equation. Finding the variable value that will result in a true equation is typically our aim when solving an algebraic equation.
¶Variables or constants are the two types of measurable quantities. A variable is a quantity with a varying value, and the constant value is nothing but a constant.
¶Steps to writing Variable Equation
1) Identify the variables that represent the unknowns.
2) Convert the issue into variable expressions in algebra.
3) Determine the variables' values to solve the equations for their true values.