Question : A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side be x units, make a quadratic equation from the above information. [1 Mark]
(ii) Write the corresponding quadratic equation in standard form. [1 Mark]
(iii) (a) Find the value of x, the length of side of a tile by factorisation. [2 Marks]
OR
(b) Solve the quadratic equation for x, using quadratic formula. [2 Marks]
CBSE Outside Delhi 2024 (30/3/1)
Solution :
This is a case study question from Quadratic Equations.
Here, it should be understood that the area covered by 200 square shaped tiles of each side 'x' units must be equal to the area covered by 128 different square shaped tiles of each area (x+1) units.
(i) Assuming the original length of each side be x units, make a quadratic equation from the above information.
Ans :
200×Area of each tiles of side x units = 128×Area of each tiles of side (x+1) units.
200×(x)²=128(x+1)²
(ii) Write the corresponding quadratic equation in standard form.
Ans :
200×(x)²=128(x+1)²
Dividing both sides by 8
25x²=16(x²+2x+1)
25x²=16x²+32x+16
25x²-16x²-32x-16=0
9x²-32x-16=0
which is the required quadratic equation.
(iii) (a) Find the value of x, the length of side of a tile by factorisation.
Ans :
9x²-32x-16=0
9x²-(36-4)x-16=0
[By Splitting the middle term]
9x²-36x+4x-16=0
9x(x-4)+4(x-4)=0
(x-4)=(9x+4)=0
(x-4)=0 or (9x+4)=0
x=4 or 9x=-4
x=4 or x=-4/9
But Side of a tile can't be negative. So, x≠0.
Hence, x=4
OR
(b) Solve the quadratic equation for x, using quadratic formula.
Ans :
9x²-32x-16=0
Here
a=9
b=-32
c=-16
D=b²-4ac
D=(-32)²-4(9)(-16)
D=1024+576
D=1600
D>0
∴ Roots are real and equal.
Using Quadratic Formula
x=[-b±√D]/2a
x=[-(-32)±√1600]/2(9)
x=[32±40]/18
x=[32±40]/18
x=[32+40]/18 OR x=[32-40]/18
x=[72]/18 OR x=[-8]/18
x=4 OR x=-4/9
But Side of a tile can't be negative. So, x≠0.
Hence, x=4