A shopkeeper mixed coffee powder worth $2.50 per kg with coffee powder worth $3.50 per kg, and sold 20 kg of the mixture at $2.80 per kg. Find the weights of the 2 grades of coffee powder that he mixed together. Explain clearly your steps of solving it.
Let the amount of the $2.50 coffee be x kg
then the amount of the $3.50 coffee will be 20-x kg
2.5x + 3.5(20-x) = 2.8(20) = 56
times 10
25x + 35x(20-x) = 560
25x + 700 - 35x = 560
-10x = -140
x = 14
He should mix 14 kg of the $2.50 and 6 kr of the $3.50 coffee.
Weight of $2.50 coffee be x kg
Weight of $3.50 coffee be y kg
2.5x+3.5(20-x)= 2.8(20)
2.5x+700-3.5x=56
-x=56-70
x=14
14 kg of $2.50 coffee
20-14kg of $3.50 coffee
Thus
14 kg of $2.50 coffee
6kg of $ 3.50 coffee
Thank you. It was a great help.
THANKS
y did we multiply it by ten
Well, it seems like the shopkeeper got himself into quite a mix-up! But don't worry, I'll help him out with a dash of humor.
Let's use a little algebraic magic to solve this coffee conundrum.
Let's say the weight of the cheaper coffee powder is x kilograms, and the weight of the expensive coffee powder is (20 - x) kilograms. Here's why: if he mixed x kilograms of the cheaper coffee powder, then the total weight of the mixture would be 20 kilograms. We subtract x from 20 because the weight of the more expensive coffee powder will be whatever is left over.
Now let's work on the cost equation. The shopkeeper mixed the cheaper coffee powder worth $2.50 per kg with the expensive coffee powder worth $3.50 per kg. The resulting mixture was sold at $2.80 per kg. So, we can set up the equation:
(x * $2.50) + [(20 - x) * $3.50] = 20 * $2.80
Now let's have a little fun and simplify this equation:
($2.50x) + ($3.50 * 20) - ($3.50x) = $2.80 * 20
($2.50x) - ($3.50x) + $70 = $56
Now let's solve this equation:
-$1.00x + $70 = $56
Subtract $70 from both sides:
-$1.00x = $56 - $70
-$1.00x = -$14
Divide both sides by -1.00:
x = 14
So, the shopkeeper mixed 14 kilograms of the cheaper coffee powder and (20 - 14) = 6 kilograms of the more expensive coffee powder.
There you have it! The shopkeeper should have used 14 kilograms of the cheaper coffee powder and 6 kilograms of the more expensive coffee powder for his mixture. Happy brewing!
To solve this problem, we can use a method called the "weighted average" or "mixture problem."
Step 1: Define the variables: Let x be the weight (in kg) of the coffee powder worth $2.50 per kg, and let y be the weight (in kg) of the coffee powder worth $3.50 per kg.
Step 2: Determine the total cost of the mixture: The shopkeeper sold 20 kg of the mixture at $2.80 per kg, so the total cost of the mixture is 20 times $2.80, which is $56.
Step 3: Set up the equations: We know that the total cost of the mixture is the sum of the individual costs. Since the cost of the coffee powder worth $2.50 per kg is $2.50 times x, and the cost of the coffee powder worth $3.50 per kg is $3.50 times y, we can write the equation as:
2.50x + 3.50y = 56
Step 4: Use another constraint: We also know that the total weight of the mixture is 20 kg. Therefore, the weights of the two grades of coffee powder must add up to 20 kg:
x + y = 20
Step 5: Solve the equations simultaneously: We can solve this system of equations using any suitable method, such as substitution or elimination. Let's use substitution:
From the second equation, we can write y = 20 - x. Substitute this value for y in the first equation:
2.50x + 3.50(20 - x) = 56
Simplify the equation:
2.50x + 70 - 3.50x = 56
Combine like terms:
-1x + 70 = 56
Solve for x:
-1x = 56 - 70
-1x = -14
Divide by -1 on both sides:
x = 14
Step 6: Find the other variable: Now that we have the value of x, we can substitute it back into one of the original equations to find y:
y = 20 - x = 20 - 14 = 6
Step 7: Answer the question: The shopkeeper mixed 14 kg of coffee powder worth $2.50 per kg with 6 kg of coffee powder worth $3.50 per kg.
Explanation summary:
To solve this problem, we defined the variables x and y to represent the weights of the two grades of coffee powder. We used the information given to set up two equations: one for the total cost of the mixture and one for the total weight of the mixture. By solving these equations simultaneously, we found that the weights of the two grades of coffee powder are 14 kg and 6 kg, respectively.