Question

If the sides of a cubic box are increased by 1, 2, 3 units respectively to form a cuboid, then the volume is increased by 52 cubic units. Find the volume of the cuboid.

Answer 1

Answer:

Volume of the cuboid is 60 cubic units

Step-by-step explanation:

Let the length of the side of the cubic be x.

volume of the cubic box $=x^3$ cubic units

when the sides of the cubic box are increased by 1, 2 and 3 units, we get a cuboid of dimensions (x+1), (x+2) and (x+3)

Now,

Volume of the cuboid $=(x+1)(x+2)(x+3)$

As per given data,

$x^3+52=(x+1)(x+2)(x+3)$

$x^3+52=x^3+6x^2+11x+6$

$52=6x^2+11x+6$

$6x^2+11x+6-52=0$

$6x^2+11x-46=0$

$(x-2)(6x+23)=0$

$x=2,\frac{-23}{6}$

But x cannot be negative

$\therefore\:x=2$

Volume of the cuboid $=(x+1)(x+2)(x+3)$

Volume of the cuboid $=(2+1)(2+2)(2+3)$

Volume of the cuboid $=(3)(4)(5)$

Volume of the cuboid $=60$ cubic units

Answer 2

Given:

V2 - V1 =52

(a+1)(a+2)(a+3) -a^3 = 52

(a^2 +3a+2)(a+3)-a^3 =52

a^3 +3a^2 +2a+3a^2 +9a+6 -a^3 = 52

6a^2 +11a+6 = 52

6a^2 +11a -46 =0

6a^2 +23a-12a-46 =0

a(6a+23)-2(6a+23) =0

(6a+23)(a-2) =0

=)6a= -23 , a=2

=)a= -23/6 (not possible)

So we choose a=2

Volume of cube = a^3 = 8cubic units

Volume of cuboid = (a+1)(a+2)(a+3)

= 3*4*5

= 60 cubic units

V of cube = 60 cu. units

V of cuboid = 8 cu. units