Term
10 books total
4 are biographies
6 are novels
how many combinations of 4 books with at least 2 biographies? |
|
Definition
Groups:
-4 biographies : combinations of 4 of 4 = 1
-3 biographies: combos of 3outof4 * combos of 1 out of 6
-2 biographies:combos of 2 outof 4 * combos of 2 outof 6
add these together: 115 combinations |
|
|
Term
of 236 students surveyed, 142 took algebra and 121 took chemistry
what is the greatest possible number of students that could have taken both algebra and chemistry? |
|
Definition
first, note that the numbers don't have to add up to 236, since there is a group that took neither subject.
the number of students who took both algebra and chemistry can be X, so the number who took algebra only is 142-X, and the number who took chemistry only is 121-X. |
|
|
Term
Total paid for a book was the price of the book plus 4% tax. The book was paid for with a $10 bill, receiving less than $3.00 change. Which statements MUST be true?
A: Price was less than $9.50
B: Price was greater than $6.90
C: Sales tax was less than $0.45 |
|
Definition
Define price of the book P
Define total inc. tax as 1.04P
Since change was less than $3, we know that 10-1.04P<3
from this, we get that P>6.73
ALSO, since it was paid for with a $10, we know that 1.04Pfrom this, we also find that tax<=.39
A: Price can be greater than $9.50, but MUST be less than 9.61
B: Price can be less than $6.90, but MUST be greater than $6.73
C. sales tax MUST be less than $.38, and therefor MUST be less than $.45
C is correct |
|
|
Term
| if 1/(2^11)(5^17) is expressed as a terminating decimal, how many nonzero digits will the decimal have? |
|
Definition
any decimal can be expressed in powers of 10, we we should seek those out.
1/(2^11)(5^11)(5^6) = 1/(10^11)(5^6)
(10^-11)((1/5)^6) = (10^-11)(.2^6)
(10^-11)(((10^-1)2)^6)
(10^-11)(10^-6)(2^6) = 64*(10^-17)
-------6 & 4 are our non-zero digits
the answer is 2 |
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
| In a sequence a1, a2, a3, a4, a5...aN, every term after the first in equal to the preceding term plus the constant C. If a1+a3+a5 = 27, what is the value of a2+a4? |
|
Definition
Find everything in terms of 'a1'
a1
a2=a1+c
a3=a1+2c
a4=a1+3c
a5=a1+4c
now solve for the 27 equationq with our 'a1's' to get a1+2c=9
knowing that, we can solve the a2+a4 equation with our 'a1's' to get 2(a1+2c). This is double the previously attained equation, so the answer is 18. |
|
|
Term
|
Definition
using algebra, we get to the equation x^2=1.
this gives us 2 answers: (1,-1), since both squared equal the positive 1. Anytime this happens, we should immediately look back to the original equation to see if any of these answers do not fit. in this case, our original denominators can not be 1 or 0, since for both we are left with a 0 denominator, which is undefined. Now we can narrow down our solution for X: -1 |
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
|
Definition
|
|
Term
the integer V is greater than 1. If V is the square of an integer, which of the following numbers must also be the square of an integer?
a. 81V
b.25V + 10√ V + 1
c. 4V^2 + 4√ V + 1 |
|
Definition
Note, √ V is an integer, and V is an integer. The answer must be something squared, and the square root of the answer can include √ V
a. 9√ V : those are both integers, and the square root of the answer
b. this is a quadratic equation, which can be expressed as (5√ V + 1)^2. Whatever V is, the sum will be squared, and therefore its square root is an integer.
c. We should try to factor this, but it does not, and so we look for a counter example, and it can be found in V=4 |
|
|
Term
S & T are positive integers, and 32^S = 2^T
Quantity A: S/T
Quantity B: 1/5 |
|
Definition
when comparing numbers with different variables as exponents, if we can make the bases identical, then the exponents must be equal.
In this case, we either want to make both bases 2 or 32.
We try to make 32 into 2. since the prime factorization of 32 is 2^5 = 32, that side of the equation can be re-written as (2^5)^S, or 2^(5S).
With the bases identical, we can conclude that 5S = T.
T is five times S, so S/T = 1/5: they are equal. |
|
|
Term
The operation ? is defined for all integers X and Y as X?Y = XY-Y. If X and Y are positive integers, which of the following cannot be zero?
a. X?Y
b. Y?X
c. (X-1)?Y
d. (X+1)?Y
e. X?(Y-1) |
|
Definition
Since we are trying to find out what CANNOT equal zero, we need to set up equations equal to zero and eliminate answers. Remember the variables themselves are positive, therefore non-zero.
a. Just try the equation as is, XY-Y = 0
X=1 can equal 0
b. switch positions of X and Y, so YX-X=0
Y=1 can equal zero
c. Substitute X with (X-1), to set up the equation (X-1)Y-Y=0. X=2 can equal zero
d. Substitute X with (X+1) to set up the equation (X+1)Y-Y=0. This simplifies to XY=0. Since there is no non-zero substitution of XorY in this, the equation is false. Cannot equal zero
e. Substitute Y with (Y-1) to set up the equation X(Y-1)-(Y-1=0. This simplifies to (Y-1)(X-1)=0
X=1 and/or Y=1 gets us to zero. |
|
|
Term
| If Xa. Y+1b. Y-1c. XY^2d. XYe. XY < X^2 |
|
Definition
Fist, note the X and Y are negative numbers
in this type of problem, see if any of the answers are multiples of the given equation. If one is, then that's the answer. otherwise, eliminate by finding counter examples.
In this problem, answers d. and e. have elements that are all multiples of the elements of the given equation. By paying attention to this, we find that d. is impossible, and e. must be true. Keep in mind that since these are negative numbers, multiplying both sides by either reverses the direction of the inequality. |
|
|
Term
| pythagorean theorem can be used when |
|
Definition
| when you know the length of 2 sides of a right triangle, and want to know the length of the 3rd. |
|
|
Term
Set S is all odd integers. If A and B are in S, which of the following must also be in S?
a. A+B
b.A-B
c.AB
d. A/B
e. A^B |
|
Definition
First, always pay attention to when INTEGER is used. In this case, the answer MUST be an integer, as well as odd.
a. no
b. no
c. always odd
d. always odd, but not always an integer. (5/3)
e. always an integer by the same logic as c. |
|
|
Term
|
Definition
| can be the sides of a right triangle. if you know 2, you know the last one. |
|
|
