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# Is a > c?

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Is a > c? [#permalink]

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13 Aug 2012, 11:06
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Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.
[Reveal] Spoiler: OA

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Thanks Rphardu

Last edited by Bunuel on 14 Aug 2012, 00:14, edited 2 times in total.
Renamed the topic and edited the question.
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13 Aug 2012, 11:33
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

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Re: Is a > c? [#permalink]

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14 Aug 2012, 00:16
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Expert's post
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rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.
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Re: Is a > c? [#permalink]

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14 Aug 2012, 06:45
2
KUDOS
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

When we subtract two inequalities with their signs in opposite directions, we are in fact using addition of two inequalities in the same direction:

$$a>b$$

$$C<D$$ -> this can be rewritten as

$$-C>-D$$

Now we can add the first and the third inequality, because they have the same direction and get $$a-C>b-D.$$
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Re: Is a > c? [#permalink]

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15 Aug 2012, 09:29
1 &2 combo-

a(b^2)-b-(b^2)c+d>0
(b^2)(a-c)-(b-d)>0

note, that 1 states that b>d. in order to make the expression above positive a must be > c
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Posts: 80
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12 Aug 2013, 07:05
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?
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Nikhil

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12 Aug 2013, 09:22
1
KUDOS
nikhil007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I don't understand the solution beyond this part...$$b^2(a-c)>0$$
as per me by dividing both sides of equation by $$b^2$$ we are assuming value of B is not equal 0, else it will be 0/0 which is not defined. so how is this correct?

From (1) and (2) we see $$b^2(a-c)>0$$

Since LHS >0 we must have $$b =! 0$$ and$$a > c$$
as if $$b = 0$$ then LHS = 0 .
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29 Aug 2013, 12:32
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.
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Posts: 43789

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30 Aug 2013, 04:49
Expert's post
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This post was
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swati007 wrote:
EvaJager wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab2 – b > b2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

YES, definitely, you can add two inequalities that have the same direction.

In the above DS question, obviously neither (1) nor (2) alone is sufficient.

(1) and (2) together:
Adding the two inequalities side-by-side we obtain $$ab^2-b+b>b^2c-d+d$$ or $$b^2(a-c)>0$$, which means necessarily $$a-c>0$$ or $$a>c.$$
Sufficient.

I understood till this point - b^2 (a-c) > 0
can someone explain after this step please.

We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.

Hope it's clear.
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30 Aug 2013, 23:14
Quote:

We have $$b^2(a-c)>0$$ ($$b\neq{0}$$). Now, since $$b^2>0$$, then the other multiple must also be greater than 0 --> $$a-c>0$$ --> $$a>c.$$.

Hope it's clear.

Wonderful explanation!!! Thanks Bunuel
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Re: Is a > c? [#permalink]

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16 Jun 2015, 12:39
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d
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Joined: 02 Sep 2009
Posts: 43789
Re: Is a > c? [#permalink]

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16 Jun 2015, 12:50
metskj127 wrote:
Bunuel wrote:
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

Hope it helps.

What if b=0 and d=-1?

In that situation, wouldn't the 2nd equation become:

a(0) > (0)c – d

If b = 0 and d = -1, then ab^2 – b = 0 and b^2c – d = 1. Anyway, what are you trying to say?
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Re: Is a > c? [#permalink]

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16 Jun 2015, 12:58
Never mind- I see my mistake now. Thank you for the help.
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Re: Is a > c? [#permalink]

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21 Jan 2018, 21:48
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

Question asks whether a > c or Is a-c > 0

(1) First statement is obviously not sufficient. Just b>d or b-d > 0 but nothing about a and c is mentioned.

(2) Second statement can be rewritten as:
ab^2 – b^2c > b – d
b^2*(a-c) > (b-d)
Here b^2 cannot be negative but whether a-c is positive or not depends on b-d also (on right hand side). Nothing is mentioned about that so insufficient.

Combining the two statements, from first we know that b-d is positive and since left hand side: b^2*(a-c) is greater than b-d so b^2*(a-c) also must be positive.
Now b^2 cannot be negative. So for b^2*(a-c) to be positive, a-c also must be positive. Which means a-c > 0 or a > c. Sufficient.

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Re: Is a > c? [#permalink]

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23 Jan 2018, 22:47
rphardu wrote:
Is a > c?

(1) b > d

(2) ab^2 – b > b^2c – d

[Reveal] Spoiler:
Can we add (1) and (2) to get the answers.

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 4 variables (a, b, c and d) and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):

b > d
ab^2 - b > b^2c - d
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 > 0 since b > d and b^2 > 0 if b≠0.
⇔ a > c

If b = 0, we have d < 0
ab^2 - b > b^2c - d
⇔ 0 > -d which contradicts b > d
Thus b≠0.

Both conditions together are sufficient.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1):
We don't have any information about a and c from the condition 1) only.
The condition 1) only is not sufficient.

Condition 2):
⇔ ab^2 -b^2c > b - d
⇔ b^2(a-c) > b-d
⇔ a - c > (b-d)/b^2 since b^2 > 0
⇔ a - c > b - d
a = 2, c = 1, b = 1, d = 1 : Yes
a = 1, c = 2, b = 0, d = 1 : No
The condition 2) only is not sufficient.

Therefore, C is the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.
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Re: Is a > c?   [#permalink] 23 Jan 2018, 22:47
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