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27 percent of the students at a certain school are boys who are enrolled in an advanced math class. If 40 percent of the boys at the school are enrolled in the advanced math class, what percentage of all students at the school are boys?

If 0<2x+3y<50 and -50<3x+2y<0, then which of the following must be true?

I. x>0 II. y>0 III. x<y

A. I only B . II only C. III only D. I and III E. I, II, and III

=>

When we add the two inequalities 0<2x+3y<50 and -50<3x+2y<0, we obtain -50<5x+5y<50, or -20<-2x-2y< 20.

Statement I. Adding the two inequalities -50<3x+2y<0 and -20<-2x-2y< 20 yields -70<x<20. So x may not be greater than zero. Statement I may not be true.

Statement II. Adding the two inequalities 0<2x+3y<50 and -20<-2x-2y< 20 yields -20<y<70. So y may not be greater than zero. Statement II may not be true, either.

Statement III. Since 0<2x+3y<50 is equivalent to -50<-2x-3y<0 and -50<3x+2y<0, adding the two inequalities yields -100<x-y<0. This implies that x < y. Statement III must be true.

i, j, and k are non-negative integers such that i+j+k=3. If p, q, and r are three fixed, but different, prime numbers, how many different values of p^iq^jr^k are possible?

A. 8 B. 9 C. 10 D. 11 E. 12

=>

The number of possible values of p^iq^jr^k is equal to the number of solutions of the equation i + j + k = 3. The solution set of the equation i + j + k = 3 includes all permutations of (3,0,0), (2,1,0), and (1,1,1). The number of permutations of (3,0,0) is 3!2! = 3. The number of permutations of (2,1,0) is 3! = 6. The number of permutations of (1,1,1) is 1.

Therefore, the number of solutions of the equation i+j+k=3 is 3 + 6 + 1 = 10.

A circle with center (6,8) and radius 2 lies in the standard xy-plane. What is the shortest distance between the origin and the points of the circle?

A. 4 B. 6 C. 8 D. 10 E. 12

Attachment:

A.png [ 7.44 KiB | Viewed 116 times ]

The distance between the origin and the center of the circle is 10=√(6^2+4^2). Since the radius of the circle is 2, the point of the circle closest to the origin lies along the line joining the center to the origin, and so has distance 10 – 2 = 8 from the origin.

In the number line above, the ticks are evenly spaced. Which point represents 2^{10}?

A. A B. B C. C D. D E. E

=>

The distance between each pair of consecutive ticks is 2^9 – 2^8 = 2^8(2-1) = 2^8. Find the values of each point: A = 2^9 + 2^8. B = A + 2^8 = 2^9 + 2^8 + 2^8 = 2^9 + 2*2^8 = 2^9 + 2^9 = 2*2^9 = 2^{10}.

For positive integers x and y, x@y is defined by x@y=(x+y)/xy. If a, b, and c are positive integers, what is the value of 1/a@1/(1/b@1/c)?

A. a+b+c B. 1/abc C. 1/(a+b+c) D. 1/(ab+bc+ca) E. 3/abc

=>

We can simplify the definition of the operation in the following way: x@y=(x+y)/xy = x/(xy) + y/(xy) = 1/x+1/y. So, 1/a@1/(1/b@1/c) = a + (1/b @1/ c) = a + ( b + c ).

Therefore, the answer is A. Answer: A
_________________

Nov. 18th fell on a Thursday in 1999. On which day did Nov. 18th fall in 2005?

A. Tuesday B. Wednesday C. Thursday D. Friday E. Saturday

=>

If a year is divisible by 400, it is a leap year. If a year is divisible by 100, but not divisible by 400, it is not a leap year. If a year is divisible by 4, but not divisible by 100, it is a leap year.

The day on which a particular date falls will be shifted by two days from one year to the next if the next year is a leap year, and by one day from one year to the next if the next year is not a leap year. For example, Nov. 18th fell on a Thursday in 1999, and a Saturday in 2000 since 2000 was a leap year. It fell on a Saturday in 2000, and a Sunday in 2001 since 2001 was not a leap year.

As there were two leap years (2000 and 2004) between 1999 and 2005, Nov. 18th shifted by 8 days over the 6-year period. If a date is shifted by 7 days, it will fall on the same day of the week. So, the net effect was to shift Nov. 18th by one day. Therefore, Nov 18th fell on a Friday in 2005, and the answer is D.

If m>n>0, x=m^2+n^2, and y=2mn, what is the value of √x^2-√y^2 in terms of m and n?

A. m^2+n^2 B. 2m^2+n^2 C.m^2+2n^2 D. n^2-m^2 E. m^2-n^2

=> Plugging in the expressions for x and y, and expanding gives: x^2 – y^2 = (m^2+n^2)^2 – (2mn)^2 = m^4 -2m^2n^2 + n^4 = (m^2-n^2)^2. So, since m^2 > n^2, √x^2-√y^2=√(m^2-n^2)^2=|m^2-n^2|=m^2-n^2.

Therefore, the answer is E. Answer: E
_________________

When 2 fair dice are tossed, what is the probability that the difference between the 2 numbers that land face up will be 3?

A. 1/6 B. 1/3 C. 1/2 D. 2/3 E. 5/6

=>

6 pairs of numbers with this property can appear on the dice: (1,4), (4,1), (2,5), (5,2), (3,6) and (6,3). The total number of outcomes from rolling two dice is 36. Thus, the probability that the two numbers will have a difference of 3 is 6/36 = 1/6.

Therefore, the answer is A. Answer : A
_________________

When n is divided by 5, the remainder is 3, and when n is divided by 6, the remainder is 3. What is the remainder when the smallest possible integer value of n is divided by 7?

A. 0 B. 1 C. 2 D. 3 E. 4

=> Plugging in numbers is the recommended approach for remainder questions. The integers that have a remainder of 3 when they are divided by 5 are 3, 8, 13, 18, 23, 28, 33, 38, … The integers that have a remainder of 3 when they are divided by 6 are 3, 9, 15, 21, 27, 33, 39, 38, … The smallest integer n that occurs in both lists is 38. Since 38 = 7 ∙5 + 3, this value of n has a remainder of 3 when it is divided by 7. Therefore, the answer is D.

If the distances between consecutive ticks in the above number line are the same, which of the following points represents 2^11?

A. A B. B C. C D. D E. E

=> Let d be the distance between consecutive ticks. Then 2d = 2^10 – 2^9 = 2*2^9 – 2^9 = 2^9 d=2^8. Since 2^11 – 2^10 = 2^32^8 – 2^22^8 = 8(2^8)– 4(2^8)= 4(2^8)= 4d, 2^11 is the fourth point from 2^10, which is D.

Therefore, the answer is D. Answer : D
_________________

Points A, B, C, and D lie on the number line as shown in the figure above. If AC=BD and AB=BC/5, what is the value of C?

A. 7/14 B. 8/14 C. 9/14 D. 10/14 E. 11/14

=>

Suppose d is the distance between A and B. Then CD = d and BC = 5d. Thus, AD = d + 5d + d = 7d. Since A = ½ and D = 2/3, 7d = 2/3 – 1/2 = 4/6 – 3/6 = 1/6, and d = 1/42. So, D – C = 1/42, and C = 2/3 – 1/42 = 28/42 – 1/42 = 27/42 = 9/14.