Jesus Christ

For the second problem, x = 50y + 69. y must be even. If it were odd, then x would be even.

The way I see it, the only possible answers are 13 and 5, neither of which options. Are you sure it is not between W and Z? It would make more sense.

(I) Counter example. If the units digit of a is 5 and the units digit of b is 5 then the units digit of a + b will be 0. So (I) is false. (II) Since the tens digit of a is 8, this will be true. The highest the tens digit of b can be is 9 and it has to be greater than 2. Suppose it's 9. Then the tens digit of a + b will be 7 plus a possible 1 carry over from adding the ones digit. Clearly 9 is not attainable, unless the ones digit of b is 0 or 1, which is restricted. True. (III) is true. It follows that if b + a > 1000 and b > a, then b + b > 1000 and b > 500. So my answer is D

The trick here is basically to identify that 2/2^m = 1/2^(m-1). If you know that you are golden.

This is a tricky questions because it deals with time series data. For 1992 there are 6 data points which means to get the meadian we should look at the 3rd and 4th highest values, not the 3rd and 4th months. The 3rd and 4th highest values happen in Aug and October and both have a value of 6500 (approx), thus the median in 1992 is 6500. They trick you here because in 1993 there are 7 data points, so we need to look at the 4th highest value which happens to also be the 4th month in the year, or 6000. So the anwere is 500 less. When calculating medians, time is irrelevent.

This looks a little suspect. They first leave the radical out and then in the second line they put it back in, but they have a 2 under the radical. I'd say it is an error.

Interesting. The prime factorization of 30 is 2*3*5. To be divisible by 12 we need another 2* in the front of this thing, which will only happen with the even multiples. Since there are 12 total integers that satisfy the first condition, I deduce that half of them are divisible by 12, namely 60, 120, 180, 240, 300, and 360. Another way to do this would be brute force, since there are only 12 options.

7x + 3y = 12 3x + 7y = 6 Subtract the second from the first... (7x + 3y) - (3x + 7y) = 12 - 6 Notice since 3x + 7y = 6, we subtracted the same number from each side so the equality still holds. Simplifying gives you 4x - 4y = 6... dividing both sides by 4 gives you x - y = 3/2.

Just use an example. Let c = 12 and d = 18. Then m = 6. gcf of c + d = 30 and 12 is 6. gcf of 2 + d = 20 and 12 is 4. gcf of cd = 216 and 12 is 12. gcf of 2d = 36 and 12 is 12. gcf of d^2 = 324 and 12 is 12. So the only answer is A.

You need the prime factorization of each. 1000 = 2*2*2*5*5*5 68 = 2*2*17 So the greatest prime factor of 1000 is 5 and the greastes prime factor of 68 is 17.

If BC went through the center of the circle then x would = 90. It specifically says that it's not the diameter, so we know x 90. We do not know the position of BC. if the center of the circle is between A and BC then x 90. What is important in this question is to not make any assumptions based on the scale of the drawing. From the picture, it certainly appears that BC lies between A and the center which would lead us to believe that x > 90, but notice that underneath the drawing it says "Figure not drawn to scale" so we cannot make this assumption.

Good catch! They fooled me. 43/60 is correct. In my formula, just switch the (2/3) and the (1/3).

(2/3)*(75/100)+(1/3)*(70/100) = 220/300 = 11/15

Adding to the second part of my reply, the question is probably not best looked at as a combination question, but you COULD do the following if you really wanted to: Total combinations of 2 students from 2-year community colleges: 34!/(2!(32!)) = 561 Total combinations of 2 students from all 100 students: 100!/(2!(98!)) = 4950 Then the probability is 561/4950 = 17/150

((1/10)+1/(1/10))(1/(1/10))= 10.1 * 10 = 101 34/100 * 33/99 = 1122/9900 = 17/150 34/100 is the probablilty that the first selected student came from a 2-year community college. Now that one student has been selected, there are only 99 left and 33 2-year community college transfers, so the probability of the second selected student being from a 2-year community college is 33/99. These are independent events, so you can mulitply them to get the total probability.