A company produces computers with 7 different speeds, and the ratio of

Question

A company produces computers with 7 different speeds, and the ratio of each speed to the next one is a fixed ratio. If the fastest computer is 5.4 GHz, while slowest speed is 0.2 GHz, what is the speed of the second slowest computer?

A. \(\sqrt{3}*0.2\)

B. \(\sqrt[7]{27}*0.2\)

C. \(\sqrt[7]{5^2}+ 0.2\)

D. \(0.2^2\)

E. \(0.8\)

A.  \sqrt{3}*0.2

B.  \sqrt {27}*0.2

C.  \sqrt {5^2}+ 0.2

D.  0.2^2

E.  0.8

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CareerGeek · Answer

The given speeds are in Geometric Progression
Let the slowest = first term (a) = 0.2
--> Value of 7th term,  a*r^6 = 5.4 
-->  0.2*r^6 = 5.4 
-->  r^6 = \frac{5.4}{0.2} = 27 = 3^3 
-->  r = (3^3)^{\frac{1}{6}} = 3^{\frac{1}{2}} = \sqrt{3}

Second slowest = 2nd term =  a*r = 0.2*\sqrt{3}

Option A

exc4libur · Answer

geometric progression: ar, ar^1, ar^2, ar^3… ar^(n-1)
a=0.2, n=7
ar^(n-1)=5.4
0.2r^6=5.4…r^6=27…r=3^(3*(1/6))=3^(1/2)
ar^2=0.2(3^(1/2))=0.2√3

Ans (A)

cpn · Answer

Let us say the speeds are S1 S2....S7. Also let us assume that S2=k *S1, S3=k*S2....S7=k*S6 as the ratio of the speeds is said to be a constant. Here we have assumed S1 to be the slowest and S7 the fastest.
So (S2/S1)*(S3/S2)*(S4/S3)*(S5/S4)*(S6/S5)*(S7/S6)=k^6
That is S7/S1= k^6 as others will cancel off..leaving S1/S7=k^6
So 0.54/0.2=k^6 or 27=k^6 which means k=√3 therefore S2 the second slowest is √3S1= √3 (0.2)

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cpn · Answer

Let the speeds be S1, S2,....S7 with S1 the slowest and S7 the fastest...since the ratio of the speeds is a constant, S2/S1=S3/S2=....S7/S6=k
Therefore S2/S1*S3/S2*S4/S3*S5/S4*S6/S5*S7/S6=k^6
So S7/S1=k^6 as others will cancel. That means 5.4/0.2=k^6 or k=√3
Second slowest is S2= k S1= √3(0.2) . So A

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CrackverbalGMAT · Answer

The first statement of the question clearly tells us that the speeds are in a Geometric sequence since they increase in a fixed ratio. We also come to know that there are seven terms in the sequence. Let the terms be  t_1 ,  t_2 ,  t_3 ……  t_7 .

From the question,  t_2  =  t_1  * r,  t_3 = t_1*r^2  and so on. Therefore,  t_7 = t_1 * r^6 . Substituting the value of  t_7  = 5.4 and  t_1  = 0.2, we get  r^6  =  \frac{t_7 }{ t_1}  = 27.

r^6  = ( 3^3 ) which means r =  3^{\frac{1}{2}} .

Therefore,  t_2  =  t_1  * r = 0.2 *  3^{\frac{1}{2}}  = 0.2 * √3.
 The correct answer option is B .

Bunuel , why is this question categorized under Geometry? Is it possibly because of the keywords matching (Geometry and Geometrical Progression)?

Bunuel · Answer

_______________________
Removed the tag. Thank you.

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