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If b, c and d are even integers, and Y is an odd integer, wh
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10 Mar 2020, 05:34
10 Mar 2020, 05:34
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If b, c and d are even integers, and Y is an odd integer, which of the following CANNOT be true?
A) bcd=Y
B) (bc)(d)=Y
C) b+cd=Y
D) \frac{Y – b}{c} = d
E) \frac{b + c + d}{2} = Y
Let's eliminate answer choices that COULD be true.
A) \frac{12}{(2)(2)} = 3. ELIMINATE A
B) (\frac{2}{4})(6) = 3. ELIMINATE B
C) \frac{2 + 4}{2} = 3. ELIMINATE C
D) \frac{Y – b}{c} = d
E) \frac{2 + 2 + 2}{2} = 3. ELIMINATE E
By the process of elimination, the correct answer must be D
ASIDE: Here's why answer choice D is correct
If Y is ODD and b is EVEN, then Y - b = ODD - EVEN = ODD
So, (Y – b)/c = d becomes ODD/c = d
Multiply both sides by c to get: ODD = cd
Since c and d are both EVEN, we get: ODD = (EVEN)(EVEN), which can never be true
Answer: D
Cheers,
Brent
A) bcd=Y
B) (bc)(d)=Y
C) b+cd=Y
D) \frac{Y – b}{c} = d
E) \frac{b + c + d}{2} = Y
Let's eliminate answer choices that COULD be true.
A) \frac{12}{(2)(2)} = 3. ELIMINATE A
B) (\frac{2}{4})(6) = 3. ELIMINATE B
C) \frac{2 + 4}{2} = 3. ELIMINATE C
D) \frac{Y – b}{c} = d
E) \frac{2 + 2 + 2}{2} = 3. ELIMINATE E
By the process of elimination, the correct answer must be D
ASIDE: Here's why answer choice D is correct
If Y is ODD and b is EVEN, then Y - b = ODD - EVEN = ODD
So, (Y – b)/c = d becomes ODD/c = d
Multiply both sides by c to get: ODD = cd
Since c and d are both EVEN, we get: ODD = (EVEN)(EVEN), which can never be true
Answer: D
Cheers,
Brent
Re: If b, c and d are even integers, and Y is an odd integer, wh
[#permalink]
10 Mar 2020, 11:01
10 Mar 2020, 11:01
There are certain rules that we need to remember.
1. Anything multiplied by Even is even.
2. Even + Even = Even
3. \frac{Even}{Even} = Even or Odd. For example, \frac{6}{2} = 3 and \frac{4}{2} = 2
Now, let's see the options.
Option A: \frac{b}{cd} = Y
Assuming value of b = 12 and c = d = 2, we get \frac{12}{2 \times 2} = 3
Option B: (\frac{b}{c})(d) = Y
This is a tricky option as one might think that \frac{Even}{Even} can be Even or Odd and when it is multiplied by Even, it will always result in Even.
Let's disprove this!
Assume value of b = 6 and c = 4 and d = 2, we get \frac{6}{4 } \times 2 = 3
Option B can be true.
Option C: \frac{b + c}{d} = Y
B+C = even + even = even. Now as we have seen in option A, \frac{Even}{Even} can be odd. So, this option is also possible.
Option D: \frac{Y – b}{c} = d
Rearranging the terms, we get Y = cd + b
Now, cd = even \times even = even and B = even. So, EVEN + EVEN can never be ODD. Option D CANNOT be true.
Option E: \frac{b + c + d}{2} = Y
B + C + D = even + even + even = even
Now \frac{Even}{2} can be ODD.
OA, D
1. Anything multiplied by Even is even.
2. Even + Even = Even
3. \frac{Even}{Even} = Even or Odd. For example, \frac{6}{2} = 3 and \frac{4}{2} = 2
Now, let's see the options.
Option A: \frac{b}{cd} = Y
Assuming value of b = 12 and c = d = 2, we get \frac{12}{2 \times 2} = 3
Option B: (\frac{b}{c})(d) = Y
This is a tricky option as one might think that \frac{Even}{Even} can be Even or Odd and when it is multiplied by Even, it will always result in Even.
Let's disprove this!
Assume value of b = 6 and c = 4 and d = 2, we get \frac{6}{4 } \times 2 = 3
Option B can be true.
Option C: \frac{b + c}{d} = Y
B+C = even + even = even. Now as we have seen in option A, \frac{Even}{Even} can be odd. So, this option is also possible.
Option D: \frac{Y – b}{c} = d
Rearranging the terms, we get Y = cd + b
Now, cd = even \times even = even and B = even. So, EVEN + EVEN can never be ODD. Option D CANNOT be true.
Option E: \frac{b + c + d}{2} = Y
B + C + D = even + even + even = even
Now \frac{Even}{2} can be ODD.
OA, D
GreenlightTestPrep wrote:
If b, c and d are even integers, and Y is an odd integer, which of the following CANNOT be true?
A) \frac{b}{cd} = Y
B) (\frac{b}{c})(d) = Y
C) \frac{b + c}{d} = Y
D) \frac{Y – b}{c} = d
E) \frac{b + c + d}{2} = Y
A) \frac{b}{cd} = Y
B) (\frac{b}{c})(d) = Y
C) \frac{b + c}{d} = Y
D) \frac{Y – b}{c} = d
E) \frac{b + c + d}{2} = Y
Re: If b, c and d are even integers, and Y is an odd integer, wh
[#permalink]
23 Jul 2023, 09:39
23 Jul 2023, 09:39
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
