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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 02:13
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E
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65% (hard)

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46% (01:22) correct 54% (01:06) wrong based on 71 sessions

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Is \((x + y)^3 > (x + y)^2\)?

(1) x > y
(2) x and y are positive.
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E
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 03:02
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Is \((x+y)^3>(x+y)^2\)?

(1) x > y
(2) x and y are positive.

\((x+y)^3 - (x+y)^2 > 0\)
\((x+y)^2( x+y -1) > 0\)
is \( x+y > 1\)?

statement 1:
x> y
x and y can be negative, positive.
not sufficient

statement 2:
x,y > 0
x + y < 1 or x + y > 1

combining both statements,
still x + y < 1 or x + y > 1.
not sufficient

Ans: E
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 03:31
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Answer is supposedly B.

Statement 1: x > y
There are 3 case
1) Both positive (+X&+Y) : (9&7)
2) Both negative (-X&-Y) : (-3&-5)
3) +X and -Y : (4 &-1)
Putting values in the equation (X+Y)^3>(X+Y)^2
Case 2 (Both negative X and Y) is not satisfying the equation ( Not Sufficient)

Statement 2: x and y are positive.
Here X and Y can be (1&1)
When we put the no. In the equation (X+Y)^3>(X+Y)^2
(1+1)^3>(1+1)^2
2^3>2^2
8>4 ( sufficient)

So IOC should be B.

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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. Updated on: 21 Apr 2020, 04:47
Originally posted by freedom128 on 21 Apr 2020, 03:39.
Last edited by freedom128 on 21 Apr 2020, 04:47, edited 1 time in total.
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Q. Is (x+y)^3 > (x+y)^2 ?
--> Is (x+y)^2 (x+y-1) > 0 ?

(1) x > y
- If x=1/2 and y=1/4, (x+y)^2 (x+y-1) <0 (NO!)
- If x=2 and y=1, (x+y)^2 (x+y-1) >0 (YES!)
--> Basing on this information alone, we cannot deduce whether (x+y)^3 > (x+y)^2 or not
NOT SUFFICIENT.

(2) x & y are positive.
- If x=1/2 and y=1/4, (x+y)^2 (x+y-1) <0 (NO!)
- If x=2 and y=1, (x+y)^2 (x+y-1) >0 (YES!)
--> Basing on this information alone, we cannot deduce whether (x+y)^3 > (x+y)^2 or not
NOT SUFFICIENT.

(1)+(2)
- If x=1/2 and y=1/4, (x+y)^2 (x+y-1) <0 (NO!)
- If x=2 and y=1, (x+y)^2 (x+y-1) >0 (YES!)
--> Basing on this information alone, we cannot deduce whether (x+y)^3 > (x+y)^2 or not
NOT SUFFICIENT.

FINAL ANSWER IS (E)

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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 04:47
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Quote:

Is (x+y)^3>(x+y)^2?

(1) x > y
(2) x and y are positive.
Show more

if a>1, then a^3>a^2
if a<1, then a^3<a^2

(1) insufic
(2) insufic

ie. x,y=0.1: (0.2)^3=0.008, (0.2)^2=0.04, a^3<a^2
ie. x,y=2: 4^3=64, 4^2=16, a^3>a^2

(1/2) insufic

Ans (E)
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 04:59
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(1) x > y

If x=0.7, y=0.3, NO
If x=2, y=1, YES

1 is not sufficient

(2) x and y are positive.

If x=0.7, y=0.3, NO
If x=2, y=1, YES

2 is not sufficient

(1)+(2)

Again,

If x=0.7, y=0.3, NO
If x=2, y=1, YES

Not sufficient

Answer is (E)

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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 10:27
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Is (x+y)3>(x+y)2?

(1) x > y
(2) x and y are positive.
given expression can be written as
( x+y)^2 * ( (x+y) -1) >0
#1
x>y
x=-1 , y = -2
and we get no and for x=1 , y=2 we get yes
insufficient
#2
x and y are positive.
( x+y)^2 * ( (x+y) -1) >0
sufficient
OPTION B
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 10:50
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Is \((x+y)^3 > (x+y)^2\) ?
\((x+y)^3 - (x+y)^2 > 0\)
\((x+y)^2(x+y -1) > 0\)
Since \((x+y)^2 > 0\), x + y > 1 ?

(1) x > y
We don't know signs.
x = 3, y = 2 \(5^3 > 5^2\) YES
x = -2, y = -3 \((-5)^3 > (-5)^2\) NO

INSUFFICIENT.

(2) x and y are positive.
x = 3, y = 2 \(5^3 > 5^2\) YES
\(x = \frac{1}{2}\), \(y = \frac{1}{3}\) \((\frac{5}{6})^3 > (\frac{5}{6})^2\) NO

INSUFFICIENT.

Together 1 and 2
Still both YES and NO cases are possible.

INSUFFICIENT.

Answer E.
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 11:07
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Statement 1: x>y

Let x = 3 and y = 2. Then, (5)^3 > (5)^2 —-> 125>25 —— (Yes)
Now, let x = -2 and y = -3. Then, (-5)^3 > (-5)^2 ———> -125> 25 —- (No)

Hence, statement 1 is insufficient.

Statement 2: x and y are positive

If x and y are positive, then (positive + positive)^3 > (positive + positive)^2 —-> (positive)^3 > (positive)^2 ALWAYS YES.

Statement 2 is sufficient.

Answer (B)

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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 21 Apr 2020, 11:17
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Is (x+y)3>(x+y)2?
(1) x > y
(2) x and y are positive.
stem:(x+y)^2 will always be positive
1)
if x=-1, y=-2, it fails
if x=2, y=1, it satisfies
Not sufficient
2)
can be decimals.
so not sufficient
Ans E
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Is (x + y)^3 > (x + y)^2? (1) x > y (2) x and y are positive. 24 Oct 2023, 05:22
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