A) Three numbers are in the ratio 1:2:3 and the sum of their cubes is 4500. The numbers will be 5, 10, 15.
B) The digit in the units place for the cube of a four digit number of the form \[xyz8\] is 2.
C) The smallest number by which 3600 be divided to make it a perfect cube is 450.
D) None of these
Correct Answer: D
Solution :
[a] Let the numbers be x, 2x and 3x Sum of cubes of the numbers\[=4500\] \[\therefore {{(x)}^{3}}+{{(2x)}^{3}}+{{(3x)}^{3}}=4500\] \[\Rightarrow {{x}^{3}}+8{{x}^{3}}+27{{x}^{3}}=4500\] \[\Rightarrow 36{{x}^{3}}=4500\Rightarrow {{x}^{3}}=\frac{4500}{36}=125\] \[\Rightarrow x=\sqrt[3]{125}\] Now, \[125=\underline{5\times 5\times 5}={{5}^{3}}\] \[\Rightarrow \sqrt[3]{125}=5\] \[\therefore x=\sqrt[3]{125}=5\] \[\therefore \] Required numbers are 5, 10 and 15. [b] Since the unit?s place digit of xyz8 is 8 \[\therefore \] Unit?s place digit of cube of \[xyz8\] is 2. [c] \[3600={{2}^{3}}\times {{5}^{2}}\times {{3}^{2}}\times 2\] To make it a perfect cube, it must be divided by \[{{5}^{2}}\times {{3}^{2}}=2\] i.e., 450.
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