smile06

CB blue book math test one

p409 question #7 (section 6)
dwayne has a newspaper route for which he collects k dollars each day. from this amount he pays out k/3 dollars per day for the cost of papers, and he saves the rest of the money. in terms of k, how many days will it take dwayne to save $1,000?
a) k/1500
b) k/1000
c) 1000/k
d) 1500/k
e) 1500k

p425 question #9 section 9
in the figure above, AD=1 and DC= radical 3. what is the value of z?

p426 question 13 section 9
Carlos delivered n packages on Monday, 4 times as many packages on tuesday as on monday, and 3 more packages on wednesday than on monday. what is the average number of packages he delivered per day over the three days?
a) 2n-3
b) 2n-1
c) 2n+1
d) 2n+3
e) 6n+1

thanx

CDN_dancer

p426 question 13 section 9

n + 4n + (n + 3)
(6n + 3)/3 -> Add up the packages for all 3 days, then divide by 3 to get the average.

2n + 1 (C)

setzwxman

p409 question #7 (section 6)
I try to use the units to help me w/ these problems as much as I can...

k[$/day] * x[days] - k/3[$/day] * x[days] = 1000[$]

A more cleaned up version: kx - (kx)/3 = 1000. Now just solve for x.
(3kx - kx)/3 = 1000
3kx - kx = 3000
2kx = 3000
x = 3000/(2k) = 1500/k

p425 question #9 section 9
Just use recognition of the special right triangle ratio associated w/ 30-60-90 triangles. DB = 1 b/c ADB is a 45-45-90 triangle. The legs are congruent to each other. So, we know that the two legs of BDC are 1 and 3^(1/2). Angle z corresponds to the shorter leg (1), so it has a measure of 30°.

p426 question 13 section 9
Monday's number of books: n
Tuesday's: 4n
Wednesday's: 3 + n

Average = (n + 4n + 3 + n)[books]/3[days]
= (6n + 3)/3 = 2n + 1 [books/day]

Hope this helps.

smile06

thanx, it really helped.

gcf101

1.
SAT-specific shortcut.
Let's assume k equal some reasonable, say, $10.
Answers a, b, and e don't make sense,
c should be out because it features 1,000 - a number from the question.
That leaves d.

2.
Slightly more "scientific" approach.
Every day: $k in, $k/3 out, so Dwayne pockets $(k-k/3) per day.
In x days (k-k/3)*x = 1000, so x = 1000/something, where something is not k.
Only 1500/k fits this criteria, c is the answer.

3.
Math-ly correct:
(k-k/3)*x = 1000
k(1 - 1/3)*x = 1000
x * k * 2/3 = 1000
x = (1000 * 3 / 2) / 2k = 1500 / k.

gcf101

Correcting a typo:
3.
(k-k/3) (x) = 1000
(2k/3) (x) = 1000
x = (1000) (3) / (2k)
x = 1500 / k.