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Exercise 3.1 · Q1

Q.If the sides of a cubic box are increased by 1,2,31, 2, 3 units respectively to form a cuboid, then the volume is increased by 5252 cubic units. Find the volume of the cuboid.

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Step 1. Set up the equation. Let the cube's side be xx. The cuboid's dimensions are x+1, x+2, x+3x+1,\ x+2,\ x+3, so its volume is (x+1)(x+2)(x+3)=x3+6x2+11x+6(x+1)(x+2)(x+3)=x^3+6x^2+11x+6.

Step 2. Use the given volume increase. The increase over the cube's volume x3x^3 is 6x2+11x+66x^2+11x+6, and this equals 5252:

6x2+11x+6=52  ⟹  6x2+11x−46=0.6x^2+11x+6=52 \implies 6x^2+11x-46=0.

Step 3. Solve the quadratic. Δ=112−4(6)(−46)=121+1104=1225=352\Delta=11^2-4(6)(-46)=121+1104=1225=35^2. So x=−11±3512x=\dfrac{-11\pm35}{12}, giving x=2x=2 or x=−4612x=-\dfrac{46}{12} (rejected, since a side length must be positive).

Step 4. Compute the cuboid's volume. With x=2x=2: dimensions 3,4,53,4,5, so volume =3×4×5=60=3\times4\times5=60. Check: cube volume =8=8; increase =60−8=52=60-8=52 ✓.

✓Final answer

The volume of the cuboid is 60\boxed{60} cubic units.

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