Q.11. Three numbers are such that the first is $\frac{1}{3}$ of the second and the third is $\frac{2}{5}$ of the sum of the first and second. If their LCM is 120, then find the third number.

Dear Student,

Please find below the solution to the asked query:

Let , first number  =  x  , So  from given condition we get :

Second number  = $\frac{x}{3}$

And from second condition we get :

Third number = $\frac{2}{5}\left(x+\frac{x}{3}\right) = \frac{2}{5}\left(\frac{3x+x}{3}\right)= \frac{2}{5}\left(\frac{4x}{3}\right) = \frac{\mathbf{8}\mathbf{ }\mathbf{x}}{\mathbf{15}}$

Also given : L.C.M. of these three numbers  = 120 

And we know L.C.M. of fractions  = $\frac{\mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} \mathrm{Numerator}}{\mathrm{H}.\mathrm{C}.\mathrm{F}. \mathrm{of} \mathrm{Denominator}}$

So,

L.C.M. ( x , x and 8 x ) =  8 x 

And

H.C.F ( 1 , 3 and 15 ) =  1 

So,

8x  = 120 ,

x  = 15

Then,

First number = 15  , Second number  = $\frac{15}{3}$ = 5  and Third number  = $\frac{8 \times 15}{15}$ = 8                                   ( Ans )
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