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Is abc at least 4?

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Is abc at least 4?  [#permalink]

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New post 08 Feb 2019, 08:02
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GMATH practice exercise (Quant Class 14)

Is \(abc\,\, \ge \,\,4\) ?

\(\left( 1 \right)\,\,b + c \ge 2\)
\(\left( 2 \right)\,\,ab \ge ac \ge 4\)

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Is abc at least 4?  [#permalink]

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New post 08 Feb 2019, 08:21
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is \(abc\,\, \ge \,\,4\) ?

\(\left( 1 \right)\,\,b + c \ge 2\)
\(\left( 2 \right)\,\,ab \ge ac \ge 4\)


Key word: abc can be +ive or -ive numbers.
Even one can take fractions for satisfying the condition.

said that
from 1, we dont get anything about a, a can be +ive or -ive

from 2, we get two cases
C1 when a, b and c, are +ive 4*1 >= 4*1 >= 4
C2 when a,b and c are -ive -4*-1 > = -4 * -1 >=4
C3 when a,b and c are fractions 1/2*8 >= 1/2*8 >=4

So only when we combine, we get a unique value

as we cannot take b and c as -ive numbers

C
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Re: Is abc at least 4?  [#permalink]

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New post 08 Feb 2019, 14:12
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fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is \(abc\,\, \ge \,\,4\) ?

\(\left( 1 \right)\,\,b + c \ge 2\)
\(\left( 2 \right)\,\,ab \ge ac \ge 4\)


Nice problem, Fabio!

Statement 1:
Case 1: a=4, b=1 and c=1, with the result that b+c≥2
In this case, abc = 4, so the answer to the question stem is YES.
Case 2: a=0, b=1 and c=1, with the result that b+c≥2
In this case, abc = 0, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2:
Case 1: a=4, b=1 and c=1, with the result that ab=4 and ac=4
In this case, abc = 4, so the answer to the question stem is YES.
Case 2: a=-4, b=-1 and c=-1, with the result that ab=4 and ac=4
In this case, abc = -4, so the answer to the question stem is NO.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statements combined:
ab ≥ ac ≥ 4 requires that a, b and c have the SAME SIGN.
Since b+c≥2, and b and c have the same sign, b and c must both be POSITIVE.
Implication:
a, b and c are ALL positive.

Thus:
ab ≥ ac
(ab)/a ≥ (ac)/a
b ≥ c

Adding together b ≥ c and b+c ≥ 2, we get:
b + b + c ≥ c + 2
2b ≥ 2
b ≥ 1

Inequalities constrained to positive values can be MULTIPLIED.
Multiplying b≥1 and ac≥4, we get:
abc ≥ 4
Thus, the answer to the question stem is YES.
SUFFICIENT.


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Is abc at least 4?  [#permalink]

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New post 08 Feb 2019, 14:26
fskilnik wrote:
GMATH practice exercise (Quant Class 14)

Is \(abc\,\, \ge \,\,4\) ?

\(\left( 1 \right)\,\,b + c \ge 2\)
\(\left( 2 \right)\,\,ab \ge ac \ge 4\)

Hi, Mitch! Thanks for the words and for your beautiful contribution!

\(abc\,\,\mathop \ge \limits^? \,\,4\)

\(\left( 1 \right)\,\,b + c\,\, \ge 2\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {1,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {4,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,ab \ge ac \ge 4\,\,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {a,b,c} \right) = \left( {4,1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b,c} \right) = \left( { - 2, - 2, - 2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( {1 + 2} \right)\,\,a \ne 0\,\,\,\,\,\left( {ac \ne 0} \right)\,\,\,\,::\,\,\,\,\left\{ \matrix{
\,a < 0\,\,\,\, \Rightarrow \,\,\,\,b < 0\,\,\,\,\left( {ab > 0} \right)\,\,\,\,\,and\,\,\,\,\,c < 0\,\,\,\left( {ac > 0} \right)\,\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,{\rm{impossible}} \hfill \cr
\,a > 0\,\,\,\, \Rightarrow \,\,\,\,b > 0\,\,\,\,\left( {ab > 0} \right)\,\,\,\,\,and\,\,\,\,\,c > 0\,\,\,\left( {ac > 0} \right) \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,a,b,c\,\,\, > 0\,\,\,\,\,\left( * \right)\)

\(\left. \matrix{
ab \ge 4\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,c\,\,\left( * \right)} \,\,\,abc \ge 4c\,\,\, \hfill \cr
ac \ge 4\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,b\,\,\left( * \right)} \,\,\,abc \ge 4b \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( + \right)} \,\,\,\,\,\,\,2abc \ge 4\left( {b + c} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,2} \,\,\,\,\,\,\,abc \ge 2\left( {b + c} \right)\,\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


The correct answer is therefore (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Is abc at least 4?   [#permalink] 08 Feb 2019, 14:26
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