Aptitude Discussion

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 |

**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 \(\dfrac{7}{72}\)

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

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

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

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

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

\(z=6\)

$⇒x=4×6$ =** 24**

**Sri Devi**

*()
*

**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.

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 :)