Question

For the box to satisfy certain requirements , its length must be three meter greater than the

width and its height must be two meter less than the width.

(i) If width is taken as x, which of the following polynomial represent volume of box?

a) x

2

-5x-6 b) x3+x2-6x c)x3

-6x2-6x d)x2+x-6.

(ii) Which is of the following polynomial represent the area of paper sheet used to make box?

a)X

2

-5x-6 b) 6x2+4x-12 c) x3

-6x2-6x d) 6x2+3x-4

(iii)If it must have a volume of 18 unit, what must be its length?

a)6 unit b)3 unit c) 4 unit d) 2 unit

(iv) At a volume of 18 cubic unit, what must be its height?

a)1 unit b) 3 unit c) 2 unit d)4 unit

(v) If a box is made of a paper sheet which cost is 100 rs per square unit, what is the cost of paper?

a) Rs. 5400 b)Rs 10800 c) Rs. 2700 d) Rs. 3400​

Answer 1

Answer:

1. x³+ x² -6x

2. 6 4 x 12

3. 3unit

4. 1 unit

5. rs 5400

Answer 2

Given:

For the box to satisfy certain requirements , its length must be three meter greater than the  width and its height must be two meter less than the width.

To Find:

(i) If width is taken as x, which of the following polynomial represent the volume of the box?

a) x 2 -5x-6

b) x3+x2-6x

c)x3 -6x2-6x

d)x2+x-6.

(ii) Which is of the following polynomial represent the area of the paper sheet used to make the box?

a)x^2-5x-6

b) 6x^2+4x-12

c) x3 -6x2-6x

d) 6x2+3x-4

(iii)If it must have a volume of 18 units, what must be its length?

a)6 unit

b)3 unit

c) 4 unit

d) 2 unit

(iv) At a volume of 18 cubic units, what must be its height?

a)1 unit

b) 3 unit

c) 2 unit

d)4 unit

(v) If a box is made of a paper sheet that cost is Rs100 per square unit, what is the cost of paper?

a) Rs5400

b)Rs10800

c) Rs2700

d) Rs3400​

Solution:

(i) If the width is given as x then the length will be(x+3) and the height will be (x-2)

as we know that the volume will be V=l*b*h putting the values we have

$V=l*b*h\\=(x+3)x(x-2)\\=x(x^2+x-6)\\=x^3+x^2-6x$

Hence, the correct option will be (b).

(ii) To find the total surface area of the box we will use the formula TSA=2(lb*bh*lh) now put the values

$TSA=2(lb*bh*lh)\\=2(x^{2} +3x+x^{2} -2x+x^{2} +x-6)\\=6x^{2} +4x-12$

Hence, the correct option will be (b).

(iii) If its volume is equal to 18 then we need to equate it to the expression we found in question (i)

$18=x^3+x^2-6x\\x^3+x^2-6x-18=0$

Now we don't need to solve the equation but what we can do is put every option value in x and see which will give 0 then when putting x=3

$y=27+9-18-18\\=0$

now we have x=3 so the length will be (x+3)=3+3=6

Hence, the correct option will be (a).

(iv) From the previous question we will have x=3 then its height will be (x-2)=3-2=1

Hence, the correct option will be (a).

(v) using the TSA equation and x=3 we can find the value

$TSA= 6x^2+4x-12\\=6*9+12-12\\=54unit^2$

Now the cost will be

Cost= 54*100= Rs5400

Hence, the correct option will be (a).