SSC MTS 2 Aug 2019 Shift 2 Paper

Question : 27 of 100

Marks: +1, -0

The length, breadth and height of a cuboid are 15 cm, 12 cm and 11 cm respectively. The length is decreased by

6{2/3}%

, the breadth increased by

8{1/3}%

, whereas the height is kept unchanged. What is the change in the total area of the four side faces (considering the rectangle contained by the length and breadth as the base) of the cuboid?

Decrease by 22 cm $^2$
Increase by 22 cm $^2$
Increase by 33 cm $^2$
No change

Validate

Solution:

Length = 15 cm, Breadth = 12 cm, Height = 11 cm
⇒ The total area of the four side faces = 2h(l + b)
= 22 × 27 = 594 cm

^2

⇒

6{2/3}%=1/15,8{1/3}%=1/12

⇒ Length = 15 × 14/15 = 14 cm
⇒ Breadth = 12 × 13/12 = 13 cm
⇒ Height = 11 cm
⇒ The total area of the four side faces after change in length and breadth = 2 × 11(14 + 13)
⇒ The total area of the four side faces after change in length and breadth = 594 cm

^2

⇒ Change in total area = 594 – 594 = 0
∴ No change in the total area of the four side faces.

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