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• ### $450 Tuition Credit & Official CAT Packs FREE February 15, 2019 February 15, 2019 10:00 PM EST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Aug 2009 Posts: 7334 Re: Value of M [#permalink] ### Show Tags 06 Dec 2015, 20:53 2 amithyarli wrote: M = √4 + ∛4 + ∜4, then the value of M is ? A. less than 3 B. equal to 3 C. between 3 and 4 D. equal to 4 E. greater than 4 Hi, the most important point here is that any number >1, if put to any root(100th root or 200th root) will always have a value >1.. now lets see the equation.. M = √4 + ∛4 + ∜4... we know √4=2 and both ∛4 and ∜4 will be >1.. s0 M=2+ something>1 +something>1.. or M is > than 4.. Ans E _________________ 1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html GMAT Expert Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6949 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the [#permalink] ### Show Tags 06 Dec 2015, 23:38 Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. If M=sqrt(4)+cuberoot(4) +sqrt(sqrt(4)) , then the value of M is: A. Less than 3 B. Equal to 3 C. Between 3 and 4 D. Equal to 4 E. Greater than 4 We know that sqrt(4)=2. Since 1^3=1 < (cuberoot(4))^3=4 < 2^3=8, 1<cuberoot(4)<2. Similarly 1^4=4 < (sqrt(sqrt(4)))^4=4 <2^4=16 implies that 1<sqrt(sqrt(4))<2. So 2+1+1<M<2+2+2. M is between 4 and 6. The answer is, therefore, E. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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07 Dec 2015, 09:23
1
Quote:

If M = √4 + ∛4 + ∜4, then the value of M is:

A) less than 3
B) equal to 3
C) between 3 and 4
D) equal to 4
E) greater than 4

√4
√4 = 2

∛4
∛1 = 1
∛8 = 2
So, ∛4 is BETWEEN 1 and 2.
In other words, ∛4 = 1.something

∜4
∜1 = 1
∜16 = 2
So, ∜ is BETWEEN 1 and 2.
In other words, ∜ = 1.something

So, √4 + ∛4 + ∜4 = 2 + 1.something + 1.something
= more than 4
= E

Cheers,
Brent
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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21 Dec 2016, 06:56
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

$$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$

Or, $$M=2+2\sqrt{2}+\sqrt{2}$$

Now, $$2+2\sqrt{2} = 4.xx$$

Hence, the correct answer will always be > 4

Answer will be (E) Greater than 4
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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11 Feb 2017, 21:30
Bunuel wrote:
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. less than 3
B. equal to 3
C. between 3 and 4
D. equal to 4
E. greater than 4

Here is a little trick: any positive integer root from a number more than 1 will be more than 1.

For instance: $$\sqrt[1000]{2}>1$$.

Hence $$\sqrt[3]{4}>1$$ and $$\sqrt[4]{4}>1$$ --> $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}=2+(number \ more \ then \ 1)+(number \ more \ then \ 1)>4$$

Thank you Bunuel!
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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17 Feb 2018, 13:30
shrouded1 wrote:
udaymathapati wrote:

Attachment:
Image2.JPG

$$M=4^{1/2} + 4^{1/3} + 4^{1/4}$$

Now we know that $$4^{1/2} = 2$$

We also know that $$4^{1/4} = \sqrt{2} \approx 1.414 > 1$$

And finally $$4^{1/3} > 4^{1/4} \Rightarrow 4^{1/3}>1$$

So combining all three together $$M > 2+1+1 \Rightarrow M > 4$$

How can it $$4^{1/4}$$ be $$\sqrt{2}$$ what function does exponent 1/4 have

$$\sqrt{2}$$ without 1/4 exponent equals aprox 1.414
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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17 Feb 2018, 14:12
1
dave13 wrote:
shrouded1 wrote:
udaymathapati wrote:

Attachment:
Image2.JPG

$$M=4^{1/2} + 4^{1/3} + 4^{1/4}$$

Now we know that $$4^{1/2} = 2$$

We also know that $$4^{1/4} = \sqrt{2} \approx 1.414 > 1$$

And finally $$4^{1/3} > 4^{1/4} \Rightarrow 4^{1/3}>1$$

So combining all three together $$M > 2+1+1 \Rightarrow M > 4$$

How can it $$4^{1/4}$$ be $$\sqrt{2}$$ what function does exponent 1/4 have

$$\sqrt{2}$$ without 1/4 exponent equals aprox 1.414

$$4^{\frac{1}{4}}=(2^2)^{\frac{1}{4}}=2^{\frac{2}{4}}=2^{\frac{1}{2}}=\sqrt{2}$$
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If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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08 Mar 2018, 15:19
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

Main idea:Approximate RHS by taking the cube root.

Details:

We have M= 4^ (1/3) + 4^ (1/3)+ 4^ (1/3) which is greater than 4

Hence E.
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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12 Mar 2018, 15:35
zz0vlb wrote:
If $$M=\sqrt{4}+\sqrt[3]{4}+\sqrt[4]{4}$$, then the value of M is:

A. Less than 3
B. Equal to 3
C. Between 3 and 4
D. Equal to 4
E. Greater than 4

We are given that M = √4 + ^3√4 + ^4√4. We need to determine the approximate value of M.

Since √4 = 2, we need to determine the value of 2 + ^3√4 + ^4√4

Let’s determine the approximate value of ^3√4. To find this value, we need to find the perfect cube roots just below and just above the cube root of 4.

^3√1 < ^3√4 < ^3√8

1 < ^3√4 < 2

Let’s next determine the approximate value of ^4√4. To find this value, we need to find the perfect fourth roots just below and just above the fourth root of 4.

^4√1 < ^4√4 < ^4√16

1 < ^4√4 < 2

Since both ^3√4 and ^4√4 are greater than 1, so √4 + ^3√4 + ^4√4 > 2 + 1 + 1, and thus, √4 + ^3√4 + ^4√4 > 4.

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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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16 Oct 2018, 05:27
I found this method much easier.
M= 4^1/2+ 4^1/3 + 4^1/4
=4^(1/2+1/3+1/4)
After taking LCM of 4^ the numbers, we get

=4^(36/12)
=4^3 which is greater than 4.

Posted from my mobile device
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the  [#permalink]

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16 Oct 2018, 06:23
topper97 wrote:
I found this method much easier.
M= 4^1/2+ 4^1/3 + 4^1/4
=4^(1/2+1/3+1/4)
After taking LCM of 4^ the numbers, we get

=4^(36/12)
=4^3 which is greater than 4.

Posted from my mobile device

That's totally wrong: $$a^x + a^y \neq a^{x+y}$$
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Re: If M=(root)(4)+(cube root)(4)+(fourth root)(4), then the   [#permalink] 16 Oct 2018, 06:23

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