Feb 17
Algebra. [please open]?
If you do not want to help me with the following algebra questions, then don’t bother to read on.
[1] At its grand opening, the Guthrie Market distributed two different types of souvenirs. One type costs 20 cents each, and the other type costs 25 cents each. One thousand souvenirs were distributed in all. If the cost of these souvenirs were $220, ho many of each type was distributed?
[2] Find three consecutive integers such that five times the thrid interger equals twice the first integer plus seven more than the sevond integer.
[3] Match each statement in the following proof with the property that justifies the step (in other words, match the roman numeral with the letter). (Note: You will not use all the properties.) Begin with the statement that follows:
3x = -9
Final Statement
x = -3 (multiplication fact)
(a) x = 1/3 (-9)
(b) 1/3 (3x) = 1/3 (-9)
(c) (1/3 * 3)x = 1/3 (-9)
(d) 1x = 1/3 (-9)
(i) identify property for multiplication
(ii) associative property for multiplication
(iii) multiplication property of equality
(iv) property of reciprocals
(v) property of opposites
[5] A health-food shop sells three grades of fiber product, grade A for $1.80 a pound, grade B for $ 1.30 a pound, and grade C for 65 cents a pound. The shopkeeper uses all three grades of fiber products to make 21 pounds of a mixture that will sell for $1 a pound. He uses twice as many pounds of grade C fiber as grade B fiber. How many pounds of each grade of fiber does he use?
Please show how you got the answer for each question.
Thanks in advance.
(>^_^<)
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One Response to “Algebra. [please open]?”
[1] Let the number of 25-cent souvenirs be x. Then the number of 20-cent ones is (1000-x). Now we can write an equation to add up the cost:
.25x + .20(1000-x) = 220
.25x + 200 - .20x = 220
.05x = 220 - 200 = 20
x = 20 / .05 = 400
1000 - x = 600
So there were 400 25-cent souvenirs (total cost $100)
and 600 20-cent souvenirs (total cost $120).
[2] Let n be the first integer. Since they are consecutive, the others are n+1 and n+2. Again, that lets us write an equation from what we’re given:
5(n+2) = 2n + [(n+1) + 7]
5n + 10 = 2n + n + 1 + 7
5n - 3n = 8 - 10
2n = -2
n = -1
n+1 = 0
n+2 = 1
[3] They’ve made this a bit nastier by not putting the steps in order. Here they are, in order, with the reasons:
(b) (iii)
(c) (ii)
(d) (iv)
(a) (i)
[5] Let b be the amount of grade B fiber used. Then we are given that the amount of grade C fiber is 2b. Since the total is 21 pounds, the rest must be grade A fiber, and the amount of that must be
21 - (b + 2b)
Since the resulting mixture will sell for a dollar a pound, the total value will be $21, and we can use what we know about the amounts and costs of the components to write the equation and solve:
1.80[21 - (b + 2b)] + 1.30b + .65*2b = 21
37.8 - 5.4b + 1.3b + 1.3b = 21
16.8 = 2.8b
b = 16.8 / 2.8 = 6
2b = 12
21 - (b+2b) = 3
So there are
3 pounds of grade A (value $5.40)
6 pounds of grade B (value $7.80)
12 pounds of grade C (value $7.80)
and the total value does add up to $21.