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

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

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!

Ans: D
Because, a>b , c<d. In inequalities, opposite signs can be subtracted. Or, c<d --> -c>-d . On adding a>b
+ (-c>-d)
= a-c>b-d
only d gives this answer choice

We are given a>b and c<d, we are to determine which of the answer choices must be true.
D is the answer. This is because a>b and c<d, hence b-d will always be less than a-c. We are subtracting a smaller number, c, from a, where a>b 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)<c-b(3).

Similarly, in B, a+d>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+d<b. Hence B is not always true.

C b+c>a-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+c<a-d. To illustrate, let a=5, b=4, c=6 and d=7 then b+c(10)>a-d(-2). However, if a=5, b=4, c=-6 and d=-1, then b+c(-2)<a-d(6)

E a^2 + d^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.

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. \(a^2+d^2>b^2+c^2\)

a>b
d>c
a+d>b+c
b-d<a-c

IMO D

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.

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



Only case 1 and 2 are sufficient.

Answer (D).
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given: a>b and c<d have different signs, we can subtract them, maintaining the former sign.
a>b-(c<d)=a-c>b-d…b-d<a-c

Answer (D)

If a>b and c<d, which of the following MUST be true?
a>b --> a-b>0 -- (1)
c<d --> 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−c
Correct, after rearranging, (b+c)<(a+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

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. \(a^2\)+\(d^2\)>\(b^2\)+\(c^2\)

Eq1 : a > b
Eq2 : c<d ...Multiply both sides by -1, -c > -d
Add 1 & 2 - a-c > b-d

Scan the answer choices that match above expression

D it is.

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) \(b−d<a−c\) --------> b+c<a+d True. Both sides are pure addition and the larger numbers are on the same side.

E) \(a^2+d^2>b^2+c^2\) 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

avigutman

For cross multiplying inequalities, is the following correct?

Is ad > bc the same as the question stem a/b > c/d?

Thank you

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.

@avitgutman

You are the best! Thank you for all of your time and help and for pushing me to critically reason

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
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