IN: Practice SET-1
GMAT Logic: Level 1
1.
A list consists of 10 consecutive integers. If -5 is the least integer, what is the sum of the positive integers in the list?
Solution (B): The 10 consecutive integers are:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4.
The positive integers in the list are {1, 2, 3, 4}.
Their sum is \(1 + 2 + 3 + 4 = 10\).
2.
If \(x\) is the sum of four consecutive integers, then \(x\) must be:
Solution (D): Let the integers be \(n, n+1, n+2, n+3\).
Sum \(x = 4n + 6 = 2(2n + 3)\).
Since \(2n+3\) is always odd, the sum \(x\) is 2 times an odd number.
This is always an even number, but it is never divisible by 4.
Example: \(1+2+3+4 = 10\). \(2+3+4+5 = 14\).
3.
If \(a, b,\) and \(c\) are non-zero integers such that \(ab > 0\) and \(bc < 0\), which of the following must be negative?
Solution (A):
1. \(ab > 0\) means \(a\) and \(b\) have the same sign (both positive or both negative).
2. \(bc < 0\) means \(b\) and \(c\) have opposite signs.
Case 1: If \(a, b\) are (+), then \(c\) must be (-). This makes \(a\) (+) and \(c\) (-).
Case 2: If \(a, b\) are (-), then \(c\) must be (+). This makes \(a\) (-) and \(c\) (+).
In both cases, \(a\) and \(c\) have opposite signs, so their product \(ac\) must be negative.
4.
If \(72n\) is the square of an integer, what is the smallest possible positive integer value of \(n\)?
Solution (B):
1. Find the prime factorization of 72: \(72 = 8 \times 9 = 2^3 \times 3^2\).
2. The expression is \( (2^3 \times 3^2) \times n \).
3. For a number to be a perfect square, all exponents in its prime factorization must be even.
4. We have \(3^2\) (even) but \(2^3\) (odd). We need one more factor of 2 to make the exponent \(2^4\).
5. Therefore, the smallest positive integer \(n\) is 2.
5.
If \(y\) is the smallest positive integer such that \(440y\) is the square of an integer, then \(y\) must be:
Solution (E):
1. Find the prime factorization of 440: \(440 = 44 \times 10 = (4 \times 11) \times (2 \times 5) = (2^2 \times 11^1) \times (2^1 \times 5^1) = 2^3 \times 5^1 \times 11^1\).
2. To make all exponents even, we need one more 2 (to get \(2^4\)), one more 5 (to get \(5^2\)), and one more 11 (to get \(11^2\)).
3. The smallest integer \(y\) must be \(2 \times 5 \times 11 = 110\).
6.
If \(x\) and \(n\) are positive integers and \(54x = n^3\), what is the smallest possible integer value of \(x\)?
Solution (C):
1. Find the prime factorization of 54: \(54 = 2 \times 27 = 2^1 \times 3^3\).
2. The expression is \((2^1 \times 3^3) \times x = n^3\).
3. For a number to be a perfect cube, all exponents must be multiples of 3.
4. We have \(3^3\) (multiple of 3) but \(2^1\). We need \(2^3\).
5. We must multiply by \(2^2\) to get \(2^1 \times 2^2 = 2^3\).
6. Therefore, the smallest \(x\) is \(2^2 = 4\).
7.
If \(a\) is an even integer and \(b\) is an odd integer, which of the following expressions must be odd?
Solution (E):
A) \( \text{Even} + \text{Odd} + \text{Odd} = \text{Even} + \text{Even} = \text{Even} \)
B) \( \text{Even} \times \text{Odd} = \text{Even} \)
C) \( \text{Even} + 2(\text{Odd}) = \text{Even} + \text{Even} = \text{Even} \)
D) \( \text{Odd} + 1 = \text{Even} \)
E) \( \text{Even} + \text{Odd} = \text{Odd} \). This must be true.
8.
The sum of three consecutive integers is 18. What is the product of these integers?
Solution (C): Let the integers be \(n, n+1, n+2\).
Sum: \(n + (n+1) + (n+2) = 18\)
\(3n + 3 = 18 \implies 3n = 15 \implies n = 5\).
The integers are 5, 6, and 7.
Their product is \(5 \times 6 \times 7 = 30 \times 7 = 210\).
9.
If k is an integer, then \(x = 10k\). \(x\) must be divisible by which of the following:
I. 2
II. 5
III. 20
I. 2
II. 5
III. 20
Solution (C):
\(x = 10k = 2 \times 5 \times k\).
I. Since \(x\) has a factor of 2, it must be divisible by 2. (True)
II. Since \(x\) has a factor of 5, it must be divisible by 5. (True)
III. If \(k=1\), \(x=10\). 10 is not divisible by 20. (Not must be true)
Therefore, only I and II must be true.
10.
If \(a\) is a negative even integer and \(b\) is a positive odd integer, which of the following is a possible value for \(a/b\)?
Solution (D): We need to test the options.
\(a\) = Negative Even (e.g., -2, -4, -6…)
\(b\) = Positive Odd (e.g., 1, 3, 5…)
The result \(a/b\) must be negative. This eliminates A, B, and E.
(C) -2.5: To get -2.5, we could have \(-5 / 2\). Here, \(a=-5\), which is not even.
(D) -4: To get -4, we could have \(a = -4\) and \(b = 1\). This fits the conditions.
Score: 0 / 10