If a b and c

Question

Competition Mode Question If a>b and cc−b B. a+d>b C. b+c>a−d D. b−db^2+c^2

shridhar786 · Accepted Answer

given a > b and c < d a-b > 0 and c-d < 0 from here we know that a-b > c-d a-c > b-d b-d < a-c D is the correct answer

nishthagupta · Answer

Ans: D Because, a>b , c -c>-d . On adding a>b + (-c>-d) = a-c>b-d only d gives this answer choice

eakabuah · Answer

We are given a>b and cb and cb and subtracting a bigger number, d, where d>c from b. Definitely, a-c will always be greater than b-d. In A, a-d>c-b, is not always true. when a=5, b=4, c=1, d=7, a-d(-2)>c-b(-3) however if a=5, b=4, c=6, and d=7; a-d(-2)b is not always true. This is because, we don't know whether a,b, and d are greater than zero or otherwise. when a,b, and d are all greater than 0, then a+d>b. However, if a and b are positive and d is negative, it is possible than a+da-d. This is also not always true. Depending on the magnitudes of a,b,c, and d, b+c>a-d. in other instances, b+ca-d(-2). However, if a=5, b=4, c=-6 and d=-1, then b+c(-2) b^2 + c^2. This is not always true. It is true for a,b,c, and d greater than zero. However, if a,b,c, and d are all negative, then it is not true.

Kinshook · Answer

If a>b and cc−b B. a+d>b C. b+c>a−d D. b−db^2+c^2 a>b d>c a+d>b+c b-d

unraveled · Answer

If a > b and c < d, which of the following MUST be true?

A. a−d > c−b
B. a+d > b
C. b+c > a−d
D. b−d < a−c
E. \(a^2 + d^2 > b^2 + c^2\)

Following cases are possible:

1. b < a < c < d
2. c < d < b < a
3. b < a = c < d
4. c < d = b < a
5. b < c < a = d
6. c < b < a = d
7. b = c < a < d
8. b = c < d < a

Attachment:

File comment: abcd

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Only case 1 and 2 are sufficient.

Answer (D).

exc4libur · Answer

given: a>b and cb-(cb-d…b-d

kris19 · Answer

If a>b and cb --> a-b>0 -- (1) c d>c --> d-c>0 --(2) add (1) + (2) (now, we can add because both (1) & (2) have > 0 on rhs) (a-b)+(d-c)>0 (a+d)-(b+c)>0 (a+d)>(b+c) --(3) A. a−d>c−b Incorrect, after rearranging, we can't get (3) B. a+d>b Incorrect, we can't get (3) C. b+c>a−d Incorrect, we can't get (3) after rearranging D. b−d (a+d)>(b+c) which is (3) E. (a^2 +d^2) >(b^2 +c^2) Incorrect, we can't get (3) after rearranging

snoep · Answer

If a>b and cc−b B. a+d>b C. b+c>a−d D. b−d b^2 + c^2 Eq1 : a > b Eq2 : c -d Add 1 & 2 - a-c > b-d Scan the answer choices that match above expression D it is.

MikeScarn · Answer

Nice question. Manipulate the answer choices to just be addition on both sides. A) a−d>c−b -------> a+b>c+d Clearly not necessarily true. B) a+d>b What if d is a high magnitude negative number? Example: a=1 b=0 d=-100 c=-101 1-100>0.... Not true C) b+c>a−d --------> b+c+d>a Clearly not necessarily true. Make a = a huge number D) b−d b+cb^2+c^2 Trap Answer. What if our small numbers are high magnitude negative numbers? Example: a=0 b=-10 d=0 c=-10 0+0>100+100

woohoo921 · Answer

avigutman For cross multiplying inequalities, is the following correct? Is ad > bc the same as the question stem a/b > c/d? Thank you

avigutman · Answer

woohoo921  you have to think through the arithmetic operations that you're performing here:
You're essentially dividing the inequality by  bd .
Can you do that? 
Well, you can absolutely do that,  as long as you're confident that  bd  > 0 
If  bd  = 0 you can't divide by  bd . If  bd  < 0 you have to flip the direction of the inequality sign.

woohoo921 · Answer

Thank you,  avigutman !

I am gently confirming (assuming that I know which direction to change the signs to, depending on if the variable(s) are negative or positive) that you go from left to right and then right to left?

In other words, assuming all of the variables are positive, I would not do cb > ad... it is ad > cb. I was thinking that this is along the lines of cross-multiplying fractions (e.g., is 2/15 vs 1/13 greater ---> 26 vs 15, so 2/15 is greater).

Thank you again, and I am sorry to bother you.

avigutman · Answer

Correct. 
But, what is "cross-multiplying fractions?"
It's a memorized shortcut procedure which unfortunately causes people to switch off their reasoning.
If we know that x/4 = y/5, for example, we can multiply that equation by 4 and by 5 (so, by 20), and get 5x = 4y.
So, a shortcut of what I just described is to say "cross-multiplying fractions." But, is that memorized shortcut helpful? In my opinion, your question/hesitation,  woohoo921 , show that it is not. It's always better to reason through the steps, particularly when studying for a test that is testing, well, reasoning!

gmatophobia · Answer

Given

1. a > b
2. c < d

Multiplying equation 2 by -1

-c > -d

Adding the resultant equation to 1

a > b
-c > -d
---------------
a - c > b - d

Option D

Rickooreoisb · Answer

If I have to share it by option method, assuming value will take a lot of time. How to select smart values, this is also general question for inequalities which can be solved using option method.

If a > b and c < d , which of the following MUST be true?

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