# Difficult Geometry & Mensuration Solved QuestionAptitude 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.

## (3) Comment(s)

()

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

Saikiran
()

Please some one can explain this doubt??

Shreya
()

We took 2t in the solution because we have surface area of a cuboid i.e in sq. Cm that's why we took it as 2t