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Q.

Of the three numbers, the first is twice the second and the second is twice the third. The average of the reciprocal of the numbers is (dfrac{7}{72}) .The numbers are:

 A.

16, 8, 4

 B.

20, 10, 5

 C.

24, 12, 6

 D.

36, 18, 9

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Solution:
Option(C) is correct

Let three numbers be $x, y,z$

Given $x=2y⇒x=4z;y=2z;z=z$

The average of reciprocal numbers is \(\dfrac{7}{72}\)

\(\dfrac{1/x+1/y+1/z}{3}=\dfrac{7}{72}\)

\(\dfrac{xy+yz+zy}{3xyz}=\dfrac{7}{72}\)

\(\dfrac{2z^2+4z^2+8z^2}{3×4z×2z×z}=\dfrac{7}{72}\)

\(\dfrac{7}{12z}=\dfrac{7}{72}\)

\(z=\dfrac{504}{84}\)

\(z=6\)

$⇒x=4×6$ = 24


(3) Comment(s)


Sri Devi
 ()

yes...verification through options is the quickest method. however, if anyone wants any other alternative, here is one: :)

f=2s => s=(f/2) ..... (i)

s=2t => t=(s/2) ..... (ii)

(i) & (ii) clearly show that f, f/2, f/4 are the required numbers :)

now,

((1/f)+(2/f)+(4/f))/3=(7/72) (given that avg of reciprocals is 7/72)

7/f = 7/24 => f=24

there is no need to determine the rest of the two numbers as there is only one option with 24 as one of the numbers :)



Vivek
 ()

let

3rd no=1

2nd no=2

1st no= 4

now according to question

=>(1/4x+1/2x/+1/x)/3= 7/72

=>(1+2+4)/4x= 7/24

=>4x=24

=>x=6

now it can be said that 3rd no is 6 2nd is 12 and 1st is 24.



Paramjot
 ()

This is the type of question which can also be simply & fast solve through options.