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If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr

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Bunuel
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If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   16 Feb 2021, 21:13
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If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (2m – n)/2

E. b – a ≤ (m + n)/2
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C
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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   16 Feb 2021, 22:03
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Bunuel wrote:
If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (m – n)/2

E. b – a ≤ (m + n)/2


4a + 2b < n and 4b + 2a > m (Multiplying -1 both sides and changing inequality sign)

4a + 2b < n and -4b - 2a < -m

CONCEPT:
Two inequations may be added if their inequality signs are identical


Adding 4a + 2b < n and -4b - 2a < -m

(4a + 2b)+(-4b - 2a) < n-m

i.e. 2a - 2b < n-m

i.e. \(a - b < \frac{(n-m)}{2}\)

i.e. \(b - a > \frac{(m-n)}{2}\)

Answer: Option C

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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   16 Feb 2021, 22:35
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Bunuel wrote:
If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (m – n)/2

E. b – a ≤ (m + n)/2


4a + 2b < n and 4b + 2a > m

Add the same sides of inequality

4a+2b+m<n+4b+2a

Since all options are in terms of m-n, take n on other side
\(m-n<4b+2a-(4a+2b)........m-n<2b-2a........m-n<2(b-a)\)

\(b-a>\frac{m-n}{2}\)

So C is correct.

But C also lies within the range of D : \(b – a ≥ \frac{(m – n)}{2}\). Thus D also MUST be true.
One of C and D requires to be changed as we cannot have two answers.
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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   16 Feb 2021, 22:37
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GMATinsight wrote:
Bunuel wrote:
If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (m – n)/2

E. b – a ≤ (m + n)/2


4a + 2b < n and 4b + 2a > m (Multiplying -1 both sides and changing inequality sign)

4a + 2b < n and -4b - 2a < -m

CONCEPT:
Two inequations may be added if their inequality signs are identical


Adding 4a + 2b < n and -4b - 2a < -m

(4a + 2b)+(-4b - 2a) < n-m

i.e. 2a - 2b < n-m

i.e. \(a - b < \frac{(n-m)}{2}\)

i.e. \(b - a > \frac{(m-n)}{2}\)

Answer: Option C


But why not D, as C is within the range of D?
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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   16 Feb 2021, 23:09
Asked: If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

4b + 2a > m
-2b - 4a > -n
2b -2a > (m-n)
b - a > (m-n)/2

IMO C

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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   18 Feb 2021, 21:15
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Then option D needs correction because I agree D also is part of "must be true". :)

chetan2u wrote:
GMATinsight wrote:
Bunuel wrote:
If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (m – n)/2

E. b – a ≤ (m + n)/2


4a + 2b < n and 4b + 2a > m (Multiplying -1 both sides and changing inequality sign)

4a + 2b < n and -4b - 2a < -m

CONCEPT:
Two inequations may be added if their inequality signs are identical


Adding 4a + 2b < n and -4b - 2a < -m

(4a + 2b)+(-4b - 2a) < n-m

i.e. 2a - 2b < n-m

i.e. \(a - b < \frac{(n-m)}{2}\)

i.e. \(b - a > \frac{(m-n)}{2}\)

Answer: Option C


But why not D, as C is within the range of D?
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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink] New post   18 Feb 2021, 22:40
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chetan2u wrote:
Bunuel wrote:
If 4a + 2b < n and 4b + 2a > m, then which of the following must be true?

A. b – a < (m – n)/2

B. b – a ≤ (m – n)/2

C. b – a > (m – n)/2

D. b – a ≥ (m – n)/2

E. b – a ≤ (m + n)/2


4a + 2b < n and 4b + 2a > m

Add the same sides of inequality

4a+2b+m<n+4b+2a

Since all options are in terms of m-n, take n on other side
\(m-n<4b+2a-(4a+2b)........m-n<2b-2a........m-n<2(b-a)\)

\(b-a>\frac{m-n}{2}\)

So C is correct.

But C also lies within the range of D : \(b – a ≥ \frac{(m – n)}{2}\). Thus D also MUST be true.
One of C and D requires to be changed as we cannot have two answers.

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Edited the typo. Thank you!
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Re: If 4a + 2b < n and 4b + 2a > m, then which of the following must be tr [#permalink]
18 Feb 2021, 22:40
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