Question

rectangular tin box is filled with sugar cubes each 1 cm cubic in volume that in was 3 cm longer than it was wide and 15 cm tall if at most it could hold 140 sugar cubes which equation would give as the possible dimensions of the box
a.5w2+15w <140
b.5w2+15w>_140
c.5w2+15w>140
d.5w2+15w <_140

Answer 1

Solution:

Surface area of cuboid =2(lb+bh+hl)

Therefore,

Surface area of 1 tin box

 =2(30×40+40×50+50×30)=2×(1200+2000+1500)=9400cm2

Surface area of 20 tin box =20×9400=188000cm2=18.8m2

Given that the cost of the tin sheet is R.20 per square meter.

Therefore,

Cost of tin required for 20 tin boxes =20×18.8=Rs.376

Answer 2

5w² + 15w  ≤ 140 is the required inequality representing dimension of the box which can hold at most 140 sugar cubes.

Given : Rectangular tin box is filled with sugar cubes each 1 cm cubic in volume

Length is 3 cm longer than width  and Height is  5 cm

At most it could hold 140 sugar cubes

To Find : Equation for  the possible dimensions of the box

a. 5w²+15w < 140

b. 5w²+15w ≥ 140

c. 5w²+15w >140

d. 5w²+15w ≤ 140

Solution:

Let say width = w cm

Length = w + 3 c m

Height =  5 cm

Volume = Length * width * height

=> Volume = (w + 3)w *  5

Volume of one sugar cube  = 1 cm³

=> Volume of 140  sugar cubes = 140  cm³

at most it could hold 140 sugar cubes Hence Volume is 140 or less than 140

=>  (w + 3)w *  5 ≤ 140

=>   5w² + 15w  ≤ 140

Hence correct option is d )  5w² + 15w  ≤ 140

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