DRDO Multi tasking staff Model paper 9

⇒ Volume of cube = (edge length)‌th =(10)3=1000cm3
Now,
⇒ Volume of a square pyramid = (1/3) × area of square base × perpendicular height = (1/3) × 15 × 15 × 30 = 2250 cm3
When the cube is placed inside the pyramid,
⇒ Remaining volume of pyramid = 2250 – 1000 = 1250 cm3
∴ Required ratio = 1000 ∶ 1250 = 4 ∶ 5