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If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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14 Mar 2017, 01:55
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If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10 ≥ b ≥ –1 , 4 ≥ g ≥ 1, then which of the following expresses the smallest possible value of ag/(bc)? A. –90 B. –20 C. –10 D. 1/90 E. 15/4
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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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14 Mar 2017, 06:30
Bunuel wrote: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10 ≥ b ≥ –1 , 4 ≥ g ≥ 1, then which of the following expresses the smallest possible value of ag/(bc)?
A. –90 B. –20 C. –10 D. 1/90 E. 15/4 smallest value is when numerator is greater ag = 5*4 = 20 denominator is greatest negative value bc = 1*1 = 1 = 20 Ans B



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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27 Apr 2017, 11:37
Can someone provide the OE for this?



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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27 Apr 2017, 11:42
Arvind90 wrote: Can someone provide the OE for this? Arvind90Can u please be little specific???? thanks



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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27 Apr 2017, 11:51
the answer is B. say a=5 g=4 b=1 c=1 we have 20 there is no way to maximize ag.



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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09 Jun 2017, 04:20
rohit8865 wrote: Bunuel wrote: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10 ≥ b ≥ –1 , 4 ≥ g ≥ 1, then which of the following expresses the smallest possible value of ag/(bc)?
A. –90 B. –20 C. –10 D. 1/90 E. 15/4 smallest value is when numerator is greater ag = 5*4 = 20 denominator is greatest negative value bc = 1*1 = 1 = 20 Ans B is there any easy trick to find the answer or do i have to manually calculate all the possibilities



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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09 Jun 2017, 06:42
Imo B This question is little tricky it wants us to make denominator largest and numerator smallest .But if we do so we would be wrong because it has negative numbers so we have to make denominator as 1 and numerator highest highest which being negative will give us least possible value Sent from my ONE E1003 using GMAT Club Forum mobile app
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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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12 Jun 2017, 17:24
We are asked to find \(\frac{ag}{bc}\) So, first ordering the variables in that sequece 5 ≥ a ≥ –2 4 ≥ g ≥ 1
10 ≥ b ≥ –1 3 ≥ c ≥ – 9
Now working through answer choices. A = 90. Not possible, because maximum we can get for numerator a*g is 5*4=20. Eliminate A
B=20. We just checked max for numerator a*g = 20. Now we have to see if we can get a 1 for denominator b*c. We can get bc=1 if b=1 and c=1. So 20 is possible.
Now the questions only asked for smallest possible value. except for 90, 20 is the smallest value amongst answer choices, so we can eliminate all other choices even without calculations.
Answer is B



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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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13 Jun 2017, 11:08
Bunuel wrote: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10 ≥ b ≥ –1 , 4 ≥ g ≥ 1, then which of the following expresses the smallest possible value of ag/(bc)?
A. –90 B. –20 C. –10 D. 1/90 E. 15/4 Because we want the smallest possible value of ag/bc, we can first see that the value of the fraction will have to be negative. Thus, either the product ag will be negative, or bc will be negative. Additionally, we want the absolute value of the entire fraction to be as large as possible, but still in keeping with the given constraints. Let’s start with the product ag: We are given that 5 ≥ a ≥ –2 and 4 ≥ g ≥ 1 If we let a = 5 and g = 4, we have ag = 20. Let’s now move to the denominator bc: We are given that 3 ≥ c ≥ – 9 and 10 ≥ b ≥ –1 If we let c = 1 and b = 1, we have bc = 1. Thus, the smallest value of ag/bc = 20/1 = 20. Answer: B
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Re: If a, b, c and g are nonzero integers, 3 ≥ c ≥ – 9 , 5 ≥ a ≥ –2 , 10
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28 Jul 2018, 12:03
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