Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm

Question

Is  4^{\frac12}+4^{\frac13}+4^{\frac14}  greater than 5?

I'm interested in the fastest approach. So, don't forget to start your timer :wink:

chica · Answer

4^1/2 + 4^1/3 + 4^1/4 = 2 + 2*4^2/3 + 2*2 = 2(1 + 4^2/3 + 2) > 5

How much time would be fast enough for the question?

abhijit_sen · Answer

4^1/4 Is not 2*2, it is Sqrt(2).
So your calculation is bit flawed.

And this value will be slightly lesser than 5 and not greater than 5.

abhijit_sen · Answer

You can solve this question by use of Geometric Progress.
In this series a = 4^1/2, r = 4^1/3/4^1/2, and n=3
Then use Sum of GP formula and you have the answer.
This should take less than a minute.

bmwhype2 · Answer

can you elaborate?

abhijit_sen · Answer

Sum of a GP = a(1-r^n)/(1-r)
Where a, r, and n I have already explained.

If GP is infinitely long then sum is = a/(1-r)

GMATBLACKBELT · Answer

I didn't really have a fast approach just an approximation.

We get 2+ 4^1/3 + sqrt2

sqrt2 ~1.41....

So we have 3.41.... + 4^1/3. 4^1/3~ 1.6 would be my guess.

So its greater than 5 slightly.

sreehari · Answer

I don' believe you can use GP formula here... if a,b,c are in geometric progression then r=b/a=c/b

So above series is actually not in GP.

FN · Answer

umm..it cant be greater than 5..

and its very quick to see why

4^1/2=2

4^1/3=sqrt(2)=1.58
4^1/4=sqrt(1.41) ~1.2

2+1.4=3.4+1.2=4.99 which is less than 5

you can also use the GP series too..

terp26 · Answer

it's greater than 5, if you use a calculator you can see how. However only by a small fraction.

abhijit_sen · Answer

Thanks sree for pointing out my mistake. This is not a GP series although it does shows a resemblence to it. But It is still less than 5.

GMATBLACKBELT · Answer

Its not the sqrt of (1.41) but rather the sqrt of 2.

sondenso · Answer

Walker, yesterday I cautht this problem and I chose wrong AC. Can you stimulate your approach?

Thanks,

walker · Answer

Actually, my approach is not so fast and I hoped that anybody would give the best way...

my logic:
let assume that 4^1/2 + 4^1/3 + 4^1/4 > 5
2 + 4^1/3 + 1.4142 > 5
4^1/3 > 1.5858
4 > 1.5858^3

Now I used the formula (a+b)^3=a^3 - 3*a^2*b (at b~0)

4 > (1.6-0.0142)^3 ~ 1.6^3 - 3*1.6^2*0.0142
4 > 0.001*(256*4*4) - 3*2.56*0.0142 ~4.096 - 2.56*0.0426 ~ 4.096 - 0.11 ~ 3.99

So, the assumption was correct and the expression is greater than 5

Any fresh thoughts?

alpha_plus_gamma · Answer

It can not be greater than 5.

sqrt(4) =2

cubert(4) ~ 1.4

4^1/4 < 1.4

even if 4^1/4 is assumed to be 1.4 for simplicity, the sum will be max 4.8.

walker · Answer

the sum is  greater than 5 ! You can use calculator to check it.

GMAT TIGER · Answer

4^(1/2) = 2.000
4^(1/3) = 1.5874
4^(1/4) = 1.41421 
sum = 4^(1/2) + 4^(1/3) + 4^(1/4) 
sum = 2.000 + + 1.5874 + 1.41421 
sum = 5.00161 which is > 5.

jimmyjamesdonkey · Answer

What is the source of this question?

sondenso · Answer

I think this question seems to be designed for you only!

I am lazy and not good at calculating like that. But I have no other way! :lol:

Thanks!

walker · Answer

I found a similar problem in GMATprep but the question was whether the expression is greater than 4. So, I slightly modified the question to warm up our minds.... :roll:

Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm

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