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What is the value of a positive integer k?

1) When k is divided by 5, the remainder is 3.
2) When k is divided by 3, the remainder is 2.

=>Condition 1) & 2)
k = 8, 23, …
Both conditions together are not sufficient either.

Ans: E
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If x, y, and z are positive integers, is x+y divisible by 5?

1) x+z is divisible by 5
2) y+z is divisible by 5

=> Consider 1) & 2)

x = 1, y = 1, z = 4 : x + y is not divisible by 5
x = 5, y = 5, z = 0 : x + y is divisible by 5

Ans: E
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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.

Ans: C
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What is the median of 12, y, and x?

1) x+y=24
2) y=6

=> 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.

Ans: A
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If x^3y^4z^5<0, is xyz>0?

1) y<0
2) x<0


=>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.

Ans: A
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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.

Ans: D
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If |x+5|=|y+5|, what is the value of x+y?

1) xy<0
2) x>3 and y<3

=>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.

Ans: D
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If |x+3|=|y+3|, what is the value of x+y?

1) xy<0
2) x>3 and y<3

=> 1) xy < 0
We can assume x > 0 and y < 0 without loss of generality.
Then we have x + 3 = -y - 3
x + y = -6
This is sufficient.

2) x > 3 and y < 3
Then we have x + 3 = -y - 3
x + y = -6
This is sufficient.

Ans: D
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△ 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.

Ans: B
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Is a positive integer x an odd number?

1) The smallest prime factor of x is 7.
2) The greatest prime factor of x is 7.

=> 1) Since the smallest prime factor of x is 7, it can’t have a prime factor 2.
Thus it should be odd.
This is sufficient.

2) Since the greatest prime factor of x is 7, it may have 2 as a prime factor or it may not.
Thus it may be odd or may be odd.
This is not sufficient.

Ans: A
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[GMAT math practice question]

Is a positive integer x divisible by 12 ?

1) x is divisible by 8
2) x is divisible by 6

=>1) Since 8 is not a multiple of 12, this is not sufficient.

2) Since 6 is not a multiple of 12, this is not sufficient.

1) & 2) Since lcm(8,6) = 24 is a multiple of 12, these are sufficient.

Ans: C
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[GMAT math practice question]

Is x^2+x>0?
1) 0<x<1
2) x^3-x<0

=>

Q: x^2+x>0 ⬄ x(x+1) > 0 ⬄ x < -1 or x > 0

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.


Ans: D
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[GMAT math practice question]

What is the value of x in 10, 13, 15, 17, and x?

1) The median of them is 15
2) The range of them is 7

=>
1) & 2) x = 16, x = 17
We have two solutions. The answer is not unique.
This is not sufficient.

Thus the answer is E.
Ans: E
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[GMAT math practice question]

|x-y|=?

1) x and y are integers
2) xy=5

=>
1) It is not sufficient definitely.

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.

Ans: C
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[GMAT math practice question]


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.

Therefore the answer is D

Ans: D
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[GMAT math practice question]

If x^3y^4z^5<0, is xyz>0?

1) y<0
2) x<0

=>
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.

Ans: A
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[GMAT math practice question]

In the coordinate plane, a triangle ABC has 3 points A(1,0), B(2,0), and C(m,n). What is the area of the triangle ABC?

1) m=1
2) n=2

=>
The area of the triangle is (1/2)*1*|n|. Then the condition 2) is sufficient.

Ans: B
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