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If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac

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Bunuel
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If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   17 Jun 2018, 04:40
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If a, b, and c are positive, is a + b > c?

(1) a/c > b > 1

(2) ab > ac
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D
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mrinalsharma1990
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If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   17 Jun 2018, 04:58
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Bunuel wrote:
If a, b, and c are positive, is a + b > c?

(1) a/c > b > 1

(2) ab > ac


Option 1
Since a/c>b a is>c hence a+b>c- sufficient

Option 2
ab>ac which means b>c hence a+b>c- sufficient

Ans- D
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baru
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 04:39
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(1) a/c > b > 1
=> a>bc>c
=> a>c
=> a + ( some thing positive) is always > c ==> Sufficient

(2) ab > ac
if we consider some fractions here it will fail .
a = 1/4 , b = 3/4 , c = 1 ==> No
a = 2 , b = 3 , c = 1 ==> Yes ====> Insufficient

Option A
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If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 05:24
B imo.

1) gives us a>bc but this is not sufficent to answer the question. If a=2, b=\(\frac{1}{2}\) and c=3 we get 2>\(\frac{3}{2}\). If we plug in the same values in the equation from the stem we receive:

2+\(\frac{1}{2}\)<3. This means that the inital statement would be incorrect.
However, if we choose a=6, b=3 and c=1 and plug those values into the initial statement we receive:

6+3>9. This means that the initial statement would be correct.
In conclusion, statement 1 is not sufficent.


2) can be converted to a>c so a+b>c will always be true. Hence, this statement is sufficient.
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 09:52
Masterscorp wrote:
B imo.

1) gives us a>bc but this is not sufficent to answer the question. If a=2, b=\(\frac{1}{2}\) and c=3 we get 2>\(\frac{3}{2}\). If we plug in the same values in the equation from the stem we receive:

2+\(\frac{1}{2}\)<3. This means that the inital statement would be incorrect.
However, if we choose a=6, b=3 and c=1 and plug those values into the initial statement we receive:

6+3>9. This means that the initial statement would be correct.
In conclusion, statement 1 is not sufficent.


2) can be converted to a>c so a+b>c will always be true. Hence, this statement is sufficient.



I think the first example doesn't qualify here.

Since all the numbers are positive, the first expression can be written as

a>bc>c

This implies that a>c. The first example that you chose doesn't meet this criteria. If you try examples where a>c then you will see that this is always sufficient.
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 10:05
rajudantuluri wrote:
Masterscorp wrote:
B imo.

1) gives us a>bc but this is not sufficent to answer the question. If a=2, b=\(\frac{1}{2}\) and c=3 we get 2>\(\frac{3}{2}\). If we plug in the same values in the equation from the stem we receive:

2+\(\frac{1}{2}\)<3. This means that the inital statement would be incorrect.
However, if we choose a=6, b=3 and c=1 and plug those values into the initial statement we receive:

6+3>9. This means that the initial statement would be correct.
In conclusion, statement 1 is not sufficent.


2) can be converted to a>c so a+b>c will always be true. Hence, this statement is sufficient.



I think the first example doesn't qualify here.

Since all the numbers are positive, the first expression can be written as

a>bc>c

This implies that a>c. The first example that you chose doesn't meet this criteria. If you try examples where a>c then you will see that this is always sufficient.

Can you please explain why the first example doesn't qualify in your opinion?
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 10:21
Since all are positive:
(1) a/c > 1; a > c
(1) b > c

Answer D. Each statement by itself is sufficient.
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 13:27
Masterscorp wrote:
rajudantuluri wrote:
Masterscorp wrote:
B imo.

1) gives us a>bc but this is not sufficent to answer the question. If a=2, b=\(\frac{1}{2}\) and c=3 we get 2>\(\frac{3}{2}\). If we plug in the same values in the equation from the stem we receive:

2+\(\frac{1}{2}\)<3. This means that the inital statement would be incorrect.
However, if we choose a=6, b=3 and c=1 and plug those values into the initial statement we receive:

6+3>9. This means that the initial statement would be correct.
In conclusion, statement 1 is not sufficent.


2) can be converted to a>c so a+b>c will always be true. Hence, this statement is sufficient.


As stated before we must choose examples where a>c but the values you’ve chosen for a and c are 2, 3 respectively.
So it doesn’t meet the criteria.


I think the first example doesn't qualify here.

Since all the numbers are positive, the first expression can be written as

a>bc>c

This implies that a>c. The first example that you chose doesn't meet this criteria. If you try examples where a>c then you will see that this is always sufficient.

Can you please explain why the first example doesn't qualify in your opinion?



As stated before, you need to choose values for a and c such that a is greater than c. You’ve chosen 2 and 3 respectively for a and c so it doesn’t meet the criteria
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   18 Jun 2018, 21:12
Bunuel wrote:
If a, b, and c are positive, is a + b > c?

(1) a/c > b > 1

(2) ab > ac


(1) Since a/c > 1 and both a/c are positive, we can write a > c. Now b is a positive number, so adding b to a will further increase the value of a. So definitely a+b > c. Sufficient.

(2) ab > ac. Since a is positive we can divide both sides by a to get b > c. Now a is a positive number, so adding a to be will further increase the value of b. So definitely a+b > c. Sufficient.

Hence D answer
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If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   23 Jun 2018, 22:10
Hello, @baru

(1) a/c > b > 1
=> a>bc>c
=> a>c
=> a + ( some thing positive) is always > c ==> Sufficient

(2) ab > ac
if we consider some fractions here it will fail .
a = 1/4 , b = 3/4 , c = 1 ==> No
a = 2 , b = 3 , c = 1 ==> Yes ====> Insufficient

Option A[/quote]

When we take statement-2:
ab>ac
if we omit the common item "a", then we get
b>c
As per your reasoning of the first option,
b + something(i.e. a) will always be greater than "c".

So, we can find a solution from both the options. And the answer is D.
Hope it helps.
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Doubt: [#permalink] New post   12 Jul 2018, 04:51
Hi,

I have a doubt with the question below, please help:

Please refer to the attached screenshot. As per me the answer should be -8, however the answer as per the solution offered for this question is -800.

Please help me understand where I am going wrong.

Thanks in advance.
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink] New post   26 Jun 2019, 22:40
Bunuel wrote:
If a, b, and c are positive, is a + b > c?

(1) a/c > b > 1

(2) ab > ac


Start with Easy Statement 2.
b>c, as a>0
now definitely b+a>c if b>c. Sufficient.

Statement 1.
1 < b < a/c
Let's try to satisfy a + b > c
1 < 2 < 7/2, Okay.

Now Let's try to negate a + b > c, or satisfy c>a+b
1 < 2 < a/5, a has to be more than 10, not possible.
So, 1 < b < a/c tells us a + b > c is the only possible inference.

D. Each statement independently sufficient.
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Re: If a, b, and c are positive, is a + b > c? (1) a/c > b > 1 (2) ab > ac [#permalink]
26 Jun 2019, 22:40
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