PSAT questions?

Question

1) If x is 66+(2/3) percent of y, then y is what percent of x?

2) When the positive integer n is divided by 9, the remainder is 7. What is the remainder when 5n is divided by 9?
a) 4 b) 5 c) 6 d) 7 e) 8

3) When an integer is multiplied by itself, it can end in all of the following EXCEPT
a) 1 b) 3 c) 5 d) 6 e) 9

4) If j,k,l, and m are four nonzero numbers, all of the following porportions are equivalent EXCEPT
a) j/k = l/m b) k/m = l/j c) m'l = k/j d) jm/kl = 1 e) l/j = m/k

5) If A is the set of digits and B is the set of even numbers, then the intersection of A and B contains how many elements?

Please EXPLAIN any of the questions you answer. Do not answer questions you are not sure the answer is correct.

Answer 1

1) 66 2/3% = 2/3 (convert percent to fraction)
x is 2/3 of y.
So y is 3/2 of x (reciprocal)
3/2 = 150% (convert fraction to percent)

2) n is of the form (9m+7), where m is an integer, given the remainder is 7.
5n would be 5(9m+7), which is 45m + 35. 45m would have no remainder when divided by 9, since 45 divides by 9, 35's remainder is 8.
(e) 8

3) Try the squares of 1 to 5:
1: 1*1 = 1, ends with 1 (eliminates a)
2: 2*2 = 4, ends with 4
3: 3*3 = 9, ends with 9 (eliminates e)
4: 4*4 =16, ends with 6 (eliminates d)
5: 5*5 = 25, ends with 5 (eliminates c)

That leaves only (b).

4) Cross-multiply each one:
a) j/k = l/m -> jm = lk
b) k/m = l/j -> jk = lm
c) m/l = k/j -> jm = lk
... and already you know that (b) is different.
d) jm/kl = 1 -> jm = kl
e) l/j = m/k -> jm = kl

5) Five (0, 2, 4, 6, 8).

There are only ten digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), and it's pretty easy to weed out the non-even ones.

Answer 2

Hints:

1. x/y=2/3; y/x=1/(2/3)

2. n=9k+7; 5*n=5*(9k+7)=9*(5k)+5*7=9*(5k)+9*3+8

3. Think about how the multiplication algorithm works, then square the numbers 1-10.

4. Assume a., see which of the remaining ones are algebraic manipulations of it.

5. You can interchange "intersection" with "and": to be in AnB, a number must be a digit AND even. The digits are {0,1,2,3,4,5,6,7,8,9}.

Answers:

1. 150%. 1/(2/3)=3/2 (divide by fraction=multiply by reciprocal). Not much more to explain; this is a simple trick.

2. E. 8: From above, you can get 9(5k+3)+8=5n. Since 8 is less than 9, 8 is the remainder.

3. 9 and 1 should be obvious. 36=6*6, 25=5*5, so the only possibility left is 3. Moreover, since the 1's digit is always the product of the 1's digits from the two things being multiplied, you've got your answer.

4. B.: By manipulating A, you can get C,D and E. However, if we try to get B: jm/k=l; m/k=l/j. Note that if you try to flip both sides, you don't get B, but there's nothing left for you to do since you have l/j on one side.

5. 5: of the digits, 0,2,4,6,8 are even. (I don't like the fact that 0 is even, but what are you going to do?)

Answer 3

note
#.3:

1(1) = 1 so 1 works
3(3) = 9 so 9 works

but 3 is prime so it does not work
and 5 is prime so it does not work
and 6 is 2(3) so it does not work

so I think you wrote the question wrong, or more probably
I misunderstood the question