Aptitude Discussion

Q. |
Ram's age was square of number last year and it will be cube of a number next year. How long must he wait before his age is again a cube of a number? |

✖ A. |
10 year |

✔ B. |
38 year |

✖ C. |
39 year |

✖ D. |
46 year |

**Solution:**

Option(**B**) is correct

Ram's present age = 26.

He will be $4^3=64$

So, required time is,

$=(64-26)$

$= \textbf{38 years}$

**Edit:** Thank you **Deepak** for the detailed explanation in the comments.

**Edit 2:** For an alternative approach to solving the question, check comment by **Sravan Reddy.**

**Sravan Reddy**

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**Subham**

*()
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Not satisfied with the justification

**Deepak**

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Another way of explanation could be as following:

last year $(y-1) = a^2$

next year $(y+1)=b^3$

when will Ram' age be a perfect square AND a perfect cube?

When Ram is 1.

Ram's age | Ram's age squared | Ram's age cubed |
---|---|---|

0 | 0 | 0 |

1 | 1 | 1 |

2 | 4 | 8 |

3 | 9 | 27 |

the only time Ram's AGE will be a square AND cube number all matching is when Ram is 1.

OR....

when will Ram's age MINUS one be ANY square number AND Ram's age PLUS one be any cubed number...

(We are looking for perfect squares and perfect cubes that are a difference of 2)

25, 27

Ram is 26... a yr ago he was $5^2$, next year he will be $3^3$

64 is $8^2$ AND $4^3$ so that is the time when Ram will be cubed & squared.

If Ram is 26, He has **38 years** to wait.

Another kill...

Possible age cubes - 1,8,27,64 (generally exams don't go to 125 as they will be realistic in numbers)

=> so subtraction by 2 is a square - -1, 6, 25, 62 - 25 so 26 is the age, next cube is 64 - so 38 years more