If b < x < 0 and w < x < y

Here's a different approach:

I. (w + b)/y < 0
We know that w and b are NEGATIVE, so (w + b) = NEGATIVE
However, we don't know whether y is NEGATIVE or POSITIVE
As such, (w + b)/y can be either POSITIVE or NEGATIVE
So, statement 1 need not be true.

II. (y – b)/b < 0
From the given information, we know that b < y. So, if we subtract b from both sides, we get y - b > 0
In other words, y-b is POSITIVE
Since we also know that b is NEGATIVE, we can see that (y – b)/b = POSITIVE/NEGATIVE = NEGATIVE
So, statement II must be true

III. (b + w) – (x + y) < 0
First rewrite the inequality as: b + w - x - y < 0
The rewrite as: (b - x) + (w - y) < 0
So, statement III can be rewritten as: (b - x) + (w - y) < 0

From the given information, we know that b < x.
If we subtract x from both sides, we get: b - x < 0
In other words, b-x is NEGATIVE

Also, from the given information, we know that w < y.
If we subtract y from both sides, we get: w - y < 0
In other words, w-y is NEGATIVE

This means that (b - x) + (w - y) = NEGATIVE + NEGATIVE = NEGATIVE
So, it must be true that (b - x) + (w - y) < 0

Answer: E

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b < x < 0 --> b and x are negative. b is greater in magnitude than x.

w < x < y --> w and x are negative. w is greater in magnitude than x. We do not know anything about y.

I. (w + b)/y < 0 --> -ve/(Don't know whether y is positive or negative). Not a must be true statement.

II. (y – b)/b < 0 --> If y is negative then b is greater in magnitude than y and y - b will be positive.
If y is positive then y - b will always be positive.
In both the cases (y - b)/b = +ve/-ve = -ve.
Hence its a must be true statement.

III. (b + w) – (x + y) < 0 --> Magnitude wise b + w > x --> b + w - x = -ve
We have to find the sign of -ve - y
If y is negative the b + w - x is greater in magnitude than y --> The expression is negative.
If y is positive then the overall expression is negative.
Hence its a must be true statement.

Answer: E

(1) w+b<0 but y can be >0 or <0
not must be true.

(2) y-b = if y>0 then y-b>0 or if y<0 then y-b>0
and b<0 thus y-b/b<0
must be true

(3)b +w has more magnitude than x+y if y<0
Also if y>0 and more magnitude than b+w then entire equation is more negative value..
thus it must be true for every case
suff

Ans E

b < x < 0 and w < x < y
With the above info, we are pretty sure that b,x,w <0 .
We dont know about y , whether it is >0, 0, or <0.
1. (w+b)/y <0
we know that w+b is <0 , but given equation varies with y. So it must not be true
2. (y-b)/b <0,
we know y>b & b<0.= > y-b >0 & b<0 .
So the given equation always holds & must be true.
3. (b+w) - (x+y)<0
the given equation can be written as (b-x) - (w-y)<0
we know, from given data in the question, that b-x<0 & w-y<0
therefore, subtraction of two negative numbers always be negative.
So, Given equation must be true.

Ans :- E


Regards
Narayana raju

Option E

b<x<0 & w<x<y.

i.e., x, b & w are <0. And, y<0 or >0.

I. \(\frac{{(w+b)}}{y}\)<0 : Not true can be +ve or -ve. Y is unknown.

II. \(\frac{{(y–b)}}{b}\)<0 : -ve. If y is -ve, |b| > |y| & y - b is +ve. If y is +ve, y-b is +ve. True

III. \((b+w)–(x+y)\)<0 : -ve. True

Another approach:

b < x < 0
w < x < y

y can be +ve or -ve

We can pick number here:

b = -4
w = -3
x = -2

Values of y when -ve can be -1
Value of y when +ve can be 10

Statement 1:

1) w+b/y <0

case 1: y is +ve

(-3)+(-4)/10 < 0 - "True"

case 2: y is -ve

(-3)+(-4)/-1 < 0 - "False"

Thus, statement 1 is - "Not always true"

Statement 2:

2) y-b/b <0

simpilfying:

y/b - 1 < 0
y/b < 1

Case 1: y is -ve

(-1)/(-4) < 1 - "True"

Case 2: y is +ve

(10)/(-4) < 1 - "True"

Thus, statement 2 is - "True"

3) (b+w)-(x+y)<0

Case 1: y is +ve

(-4)+(-3)-(-2)-(10)< 0
(-17)+(2)< 0
-15< 0 ------------> "True"

Case 2: y is -ve

(-4)+(-3)-(-2)-(-1)< 0
(-7)+(3)< 0
-4< 0 ------------> "True"

Thus, statement 3 is "True"

Hence, the solution is E.

Given:
1. b < x < 0
2. w < x < y

{b,w}<x<{0,y}

Asked: Which of the following MUST be true?

I. \(\frac{w + b}{y} < 0\)
w+b<0 since w<0 & b<0
y may or may not be >0
NOT NECESSARILY TRUE

II. \(\frac{y – b}{b} < 0\)
y-b>0 & b<0
\(\frac{y-b}{b}<0\)
MUST BE TRUE

III. \((b + w) – (x + y) < 0\)
b+w < x+y
MUST BE TRUE

IMO E

I: \(\frac{w + b}{y} < 0\)
Prompt: b<x<0 and w<x<0, implying that b and w are both NEGATIVE
If y is NEGATIVE, then \(\frac{w+b}{y}\) = \(\frac{NEGATIVE+NEGATIVE}{NEGATIVE}\) = \(POSITIVE\)
Since Statement I does not have to be true, eliminate any answer choice that includes Statement I (C and D).

II: \(\frac{y – b}{b} < 0\)
\(\frac{y}{b} - \frac{b}{b} < 0\)
\(\frac{y}{b} - 1 < 0\)
\(\frac{y}{b} < 1\)
Does it have to be true that \(\frac{y}{b} < 1\)?

Prompt: b < x and x < y, implying that b < y
Since \(b\) is negative, dividing both sides by \(b\) yields the following:
\(1 > \frac{y}{b}\)
\(\frac{y}{b} < 1\)
Since Statement II must be true, eliminate any remaining answer choice that does not include Statement II (B).

Statement III: \((b + w) – (x + y) < 0\)
Prompt: b < x and w < y
Adding together b < x and w < y, we get:
b+w < x+y
(b+w) - (x+y) < 0
Since Statement III must be true, eliminate any remaining answer choice that does not include Statement III (A).


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If b < x < 0 and w < x < y, then which of the following MUST be true?

b, x and w are negative but y could be anything


Case A: w<0, b<0 and y<0, then \(\frac{w + b}{y} > 0\)
Case B: w<0, b<0 and y>0, then \(\frac{w + b}{y} < 0\)
(I) need not be true always

II. \(\frac{y – b}{b} < 0\)
Now b<x<y or b<y, that is y-b>0.
Thus , numerator is positive while the denominator is negative, making the fraction negative.
Always true

III. \((b + w) – (x + y) < 0\)
We know that b<x and w<y.
Add the two equations
b+w<x+y => (b+w)-(x+y)
Always true

E
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