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# Is r^x < 100?

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Math Expert
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02 Oct 2017, 23:05
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Question Stats:

53% (01:15) correct 47% (01:20) wrong based on 112 sessions

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Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$
[Reveal] Spoiler: OA

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Re: Is r^x < 100? [#permalink]

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02 Oct 2017, 23:22
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Is $$r^x<100$$ ?

(1) $$r∗\sqrt[3]{x}>100$$

if x = 1/8
=> $$r∗\sqrt[3]{\frac{1}{8}}>100$$
=> $$r*\frac{1}{2} > 100$$
=> r > 200
$$r^x<100$$
=>$$300^{\frac{1}{8}}<100$$ True

if x = 8
=> r > 50
$$r^x<100$$
=>$$51^{8}<100$$ False
So here we don't have max and min of x and r, so we cannot calculate value of $$r^x<100$$
Insufficient

(2) x>1
Here we dont know value of r
Insufficient

1+2
$$r∗\sqrt[3]{x}>100$$ and x>1

for x =8>1 => $$r∗\sqrt[3]{x}>100$$
=> $$r∗\sqrt[3]{8}>100$$
=> r >50
so $$r^x<100$$
=> $$51^8<100$$ False

for x=100000000
$$r∗\sqrt[3]{x}>100$$
=> $$r∗\sqrt[3]{100000000}>100$$
=> r* 1000>100
=> r >1/10
so $$r^x<100$$
=> $$1^{1000000000}<100$$ True

So here there is no max limit on x. Hence insufficient

Last edited by Nikkb on 03 Oct 2017, 03:26, edited 4 times in total.

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Re: Is r^x < 100? [#permalink]

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03 Oct 2017, 02:20
Ans : c

1. r * x^1/3 > 100

consider x= positive (eg x= 27 and r =50)

Putting the values 125>100

Substituting x and r values in the given equation , r^x>100

Consider x= negative (eg x= -1 and r = -101)

Putting the values 101>100

Substituting x and r values in the given equation ,
r^x<100

NOt sufficient .

1.x > 1.

Statement alone is not sufficient as r can be any value.

Combining the two and subsituting anyvalue of x>1 and r that satisfies equation 1 ,

r^x is not less than 100 anytime.

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Re: Is r^x < 100? [#permalink]

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03 Oct 2017, 02:50
Bunuel wrote:
Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$

Statement 1: if $$r<0$$, then $$x<0$$ and if $$r>0$$ then $$x>0$$

$$r=-200$$ & $$x=-8$$, then $$r*\sqrt[3]{x}$$ $$=-200*-2=400>100$$

but $$r^x$$ $$= (-200)^{-8} = \frac{1}{(-200)^8}$$$$<100$$

if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$

but $$r^x=200^8>100$$. Hence insufficient

Statement 2: $$x>$$1 but nothing given about $$r$$. hence insufficient

Combining 1 & 2 if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$, then the answer to our question Stem is NO.

but if $$x=0.5$$ and $$r=10^9$$, then $$r*\sqrt[3]{x}>100$$ $$=0.5*10^3=500>100$$ but $$r^x=(0.5)^{10^9}<100$$. So we get a Yes for our question stem.

Hence Insufficient

Option E

Last edited by niks18 on 04 Oct 2017, 06:39, edited 1 time in total.

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Re: Is r^x < 100? [#permalink]

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04 Oct 2017, 06:23
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niks18 wrote:
Bunuel wrote:
Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$

Statement 1: if $$r<0$$, then $$x<0$$ and if $$r>0$$ then $$x>0$$

$$r=-200$$ & $$x=-8$$, then $$r*\sqrt[3]{x}$$ $$=-200*-2=400>100$$

but $$r^x$$ $$= (-200)^{-8} = \frac{1}{(-200)^8}$$$$<100$$

if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$

but $$r^x=200^8>100$$. Hence insufficient

Statement 2: $$x>$$1 but nothing given about $$r$$. hence insufficient

Combining 1 & 2 we know that $$x>0$$, hence $$r>0$$ so $$r^x>100$$. Hence we get a NO for our question stem. Sufficient

Option C

Hi niks18,

When you combine 1 & 2

beside your example, you neglected when x is huge number and 0<r<1

if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$.......Is 200^8 < 100........Answer No

if $$r=0.2$$ and $$x=10^9$$, then $$r*\sqrt[3]{x}>100$$ $$=0.2 *1000=200>100$$.....Is (0.2)^1000000000 < 100...Answer Yes

It should be E

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Re: Is r^x < 100? [#permalink]

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04 Oct 2017, 06:31
Mo2men wrote:
niks18 wrote:
Bunuel wrote:
Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$

Statement 1: if $$r<0$$, then $$x<0$$ and if $$r>0$$ then $$x>0$$

$$r=-200$$ & $$x=-8$$, then $$r*\sqrt[3]{x}$$ $$=-200*-2=400>100$$

but $$r^x$$ $$= (-200)^{-8} = \frac{1}{(-200)^8}$$$$<100$$

if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$

but $$r^x=200^8>100$$. Hence insufficient

Statement 2: $$x>$$1 but nothing given about $$r$$. hence insufficient

Combining 1 & 2 we know that $$x>0$$, hence $$r>0$$ so $$r^x>100$$. Hence we get a NO for our question stem. Sufficient

Option C

Hi niks18,

When you combine 1 & 2

beside your example, you neglected when x is huge number and 0<r<1

if $$r=200$$ and $$x=8$$, then $$r*\sqrt[3]{x}>100$$ $$=200*2=400>100$$.......Is 200^8 < 100........Answer No

if $$r=0.2$$ and $$x=10^9$$, then $$r*\sqrt[3]{x}>100$$ $$=0.2 *1000=200>100$$.....Is (0.2)^1000000000 < 100...Answer Yes

It should be E

Hi Mo2men

Yup agreed . Thanks for highlighting, will edit the solution

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Re: Is r^x < 100? [#permalink]

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06 Nov 2017, 03:53
Bunuel wrote:
Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$

Hello Bunuel, i am getting answer : E ( have shared the solution above)

is OA=C correct ? if so can you please share the solution and also please let me know what mistake i made ?

Thanks

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Re: Is r^x < 100? [#permalink]

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06 Nov 2017, 05:51
Bunuel wrote:
Is $$r^x < 100$$?

(1) $$r*\sqrt[3]{x}>100$$

(2) $$x> 1$$

hi Bunuel...
OA should be E.

combined if r>1, most of the time $$r^x>100$$
if 0<r<1, $$r^x<100$$ always
so insuff

E
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Re: Is r^x < 100?   [#permalink] 06 Nov 2017, 05:51
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