1) A party mi is made from pretzels that cost $2 lb. and cereal that costs $2.40 lb. If a total of 4 pounds is made and the cost for the mix is $2.15 a pound, how many pounds of each ingredient are used?
2) When the tens digit of a two digit number is added to three times the ones digit, the sum is 31. If the digits are added, the sum is 15. what is the original number?
Did you find these two problems on a cumulative review?
Please Check it - I think I may have figured it out.
1) 1/2 lb cereal and .48 lb. of pretzels
2) 78
Am I right? :-)
#1 is obviously wrong, since your weights don't add up to the desired 4 lbs. If there are x lbs of pretzels,
2x + 2.4(4-x) = 2.15*4
x = 2.5
so, 2.5 lbs pretzels and 1.5 lbs cereal
#2 correct
1) To solve this problem, we can set up a system of two equations using the information given. Let's define the following variables:
Let p = pounds of pretzels used
Let c = pounds of cereal used
According to the problem, a total of 4 pounds of the mix is made, so we have the equation:
p + c = 4 -- equation (1)
Also, it is given that the cost of the mix is $2.15 per pound. The cost of the pretzels is $2 per pound and the cost of the cereal is $2.40 per pound. We can use this information to set up another equation for the cost:
2p + 2.40c = 2.15(4) -- equation (2)
Now, we can solve the system of equations (1) and (2) to find the values of p and c.
To do this, let's multiply equation (1) by 2 to eliminate the variable p:
2p + 2c = 8 -- equation (3)
We can rewrite equation (2) by distributing 2.15 to simplify it:
2p + 2.40c = 8.60 -- equation (4)
Now, we can subtract equation (3) from equation (4) to eliminate the variable p:
(2p + 2.40c) - (2p + 2c) = 8.60 - 8
Simplifying further, we get:
0.40c = 0.60
Dividing both sides of the equation by 0.40, we find:
c = 1.5
Now, substitute the value of c into equation (1) to find the value of p:
p + 1.5 = 4
p = 4 - 1.5
p = 2.5
Therefore, 2.5 pounds of pretzels and 1.5 pounds of cereal are used in the party mix.
2) Let's represent the two-digit number as 10a + b, where a is the tens digit and b is the ones digit.
According to the problem, the sum of the tens digit and three times the ones digit is 31. We can write this as an equation:
a + 3b = 31 -- equation (1)
It is also given that the sum of the digits is 15:
a + b = 15 -- equation (2)
To solve this system of equations, we can use the substitution method. Rearrange equation (2) to solve for a:
a = 15 - b
Substitute this expression for a into equation (1):
15 - b + 3b = 31
Simplifying further, we get:
15 + 2b = 31
Subtracting 15 from both sides of the equation, we find:
2b = 16
Dividing both sides of the equation by 2, we get:
b = 8
Now, substitute the value of b into equation (2) to find the value of a:
a + 8 = 15
a = 15 - 8
a = 7
Therefore, the original number is 78.