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Is the total profit from the sales of 3 products greater than $6?

1) The least profit from the sales of the 3 products is at least $2.1 2) The 2nd largest profit from the sales of the 3 products is at least $3.1.

=> Condition 1) Since the minimum profit of 3 products is $2.1, the total profit of them is greater than or equal to $2.1 x 3 = $6.3, which is greater than $6. Thus this is sufficient.

Condition 2) Since the 3nd largest profit is at least $3.1, the sum of the first and the second products is at least $6.2, which is greater than $6. Thus this is sufficient too.

We define “A mod n” as the remainder when A is divided by n. What is the value of “A mod 6”?

1) A mod 4 = A mod 12 2) A mod 3 = 2

=> Condition 1) A = 0, 1, 2, 3. This is not sufficient.

Condition 2) A = 2, 5. This is not sufficient.

Condition 1) & 2) From the condition 2), we have four cases as follows. A = 12k + 2 A = 12k + 5 A = 12k + 8 A = 12k + 11 From the condition 1), A should be the case that A = 12k + 2 = 6*(2k) + 2. Thus both conditions together are sufficient.

=> Condition 1) If x = y = 12, then the median of them is 12. If x < y, then the median of them 12, since x < 12 < y. If x > y, then the median of them 12, since y < 12 < x. This is sufficient.

Condition 2) Since we don’t know x, we don’t identify their median. This is not sufficient.

=>The original condition x^3y^4z^5<0 is equivalent to xy < 0 after dividing both sides of the inequality by x^2y^4z^4. Then the question is equivalent to y>0, since xy < 0 from the equivalent condition.

Thus the condition 1) is sufficient, since the question is if y > 0.

The probability that event A occurs is 0.5 and the probability that event B occurs is 0.7. What is the probability that event A occurs but not event B?

1) Event A and event B are independent. 2) The probability that neither event A nor event B occurs is 0.15.

=>Condition 1) We call an event A and an event B independent if P(A∩B) = P(A)*P(B). P(A∩B) = P(A)*P(B) = 0.5 * 0.7 = 0.35. P(A∩BC) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15. This is sufficient.

Condition 2) 1 – P(A∪B) = 0.15. It means P(A∪B) = 0.85. P(A∪B) = P(A) + P(B) - P(A∩B) = 0.5 + 0.7 - P(A∩B) = 0.85. Thus P(A∩B) = 0.15. P(A∩B^C) = P(A) - P(A∩B) = 0.5 – 0.35 = 0.15. This is also sufficient too.

=>Condition 1) |x+5|=|y+5| is equivalent to (x+5)^2=(y+5)^2 or (x+5)^2-(y+5)^2 = 0. (x+5)^2-(y+5)^2 = (x+y+10)(x-y) = 0. Then x = y or x + y + 10 = 0. Since x cannot be equal to y, x + y = -10. This is sufficient.

Condition 2) |x+5|=|y+5| is equivalent to (x+5)^2=(y+5)^2 or (x+5)^2-(y+5)^2 = 0. (x+5)^2-(y+5)^2 = (x+y+10)(x-y) = 0. Then x = y or x + y + 10 = 0. Since x > 3 and y < 3, x is not equal to y. Thus x + y = -10. This is sufficient.

△ is one operation of addition, subtraction, multiplication, or division. Is 1△3=3△1？

1) 1△1=1 2) 2△3=6

=> 1) 1△1=1 △ could be the multiplication or the division. If △ is the multiplication, 1△3 = 3△1. If △ is the division, 1△3 is not equal to 3△1. This is not sufficient.

1) 2△3=6 △ should be the multiplication only. Then we have 1△3 = 3△1. This is sufficient.

1) 0<x<1 Since the range of the question includes the rage of the condition 1), it is sufficient.

2) x^3-x<0 ⬄ x(x^2-1) < 0 ⬄ x(x+1)(x-1) < 0 ⬄ x < -1 or 0 < x < 1 Since the range of the question includes the rage of the condition 2), it is sufficient too.

2) x = 1, y = 2 and x = 2, y = 5/2 We don’t have a unique solution.

1) & 2) We have the following cases. x = 1, y = 5 / x = -1, y = -5 / x = 5, y = 1 / x = 1, y = 5 All of them have |x-y| = 4. Thus we have a unique solution. Both conditions together are sufficient.

If a, b, and c are different positive integers, what is the value of a+b+c?

1) a^2+b^2+c^2=14 2) ab+bc+ca=11

=> Condition 1) We can assume a < b < c without loss of generality. The maximum value of c is 3 and c^2 = 9 a^2 + b^2 = 5. Then we have b = 2 and a = 1. a + b + c = 1 + 2 + 3 = 6

Condition 2) We can assume a < b < c without loss of generality. ab + bc + ca = (a+b)c + ab = 11 Since a + b >= 3, the maximum value of c = 3. If c = 3, ab + 3b + 3a = 11 or ab + 3a + 3b + 9 = 20. We have (a+3)(b+3) = 20. Then a = 1 and b = 2. Thus a + b + c = 1 + 2 + 3 = 6.

=> x^3y^4z^5 < 0 is equivalent to xz < 0. Then the question xyz > 0 is equivalent to y < 0. The condition 1) is sufficient. And the condition 2) is not sufficient.