Aptitude Discussion

Q. |
The external length, breadth and height of a closed box are 10 cm, 9 cm and 7 cm respectively. The total inner surface area of the box is 262 sq. cm. If the walls of the box are of uniform thickness \(t\) cm, then \(t\) equals |

✔ A. |
\(1\) cm |

✖ B. |
\(\dfrac{23}{3}\) cm |

✖ C. |
\(1\) or \(\dfrac{23}{3}\) cm |

✖ D. |
None |

**Solution:**

Option(**A**) is correct

The inner dimensions will be \((10-2t), (9-2t)\) and \((7-2t).\)

Thus the inner surface area will be

\(=2 \times [(10 - 2t)(9 - 2t) + (10 - 2t)(7 - 2t) + (9 - 2t)(7 - 2t)]\)

\(= 262\)

This will boil down to a quadratic which can be solved and we will get two values of \(t\). One should not solve the quadratic but make use of the options. Substituting \(t = 1\), we see that it satisfies the above equation.

Obviously \(\dfrac{23}{3}\)is not worth considering as it is greater than 7 and the thickness has to be less than the least

outer dimension, i.e. 7.

**Shreyash Okhade**

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Please some one can explain this doubt??

why we have taken 2t in the solution y we cannot simply take t as it is