A) 12
B) 20
C) 24
D) 48
Correct Answer: D
Solution :
Let the three numbers are x, y and z. According to the question, \[x=2y\] ?(i) \[\therefore \] \[y=x/2\] ...(ii) and \[\frac{x+y+z}{3}=56\] \[\Rightarrow \] \[x+y+z=56\times 3\] \[\Rightarrow \] \[x+y+z=168\] \[\Rightarrow \] \[x+\frac{x}{2}+z=168\] \[\Rightarrow \] \[\frac{3x+2x}{2}=168\] \[\Rightarrow \] \[3x+2z=168\times 2=336\] \[\Rightarrow \] \[3x+2\times 2x=336\] \[\Rightarrow \] \[7x=336\] \[\therefore \] \[x=48\] \[\therefore \] \[z=2x=2\times 48=96\] \[\therefore \]Difference between first and third numbers \[=96-48=48\]You need to login to perform this action.
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