Question

The cost of one chair and 1 table is 2000.the cost of 5chair and some tables is 16,200rs .if the no of chair and table interchanged and 5% discount us allowed .U have to pay 11210rs.find the no of table initially .Iproposed to purchase.find the total cost of 7chairs and 6tables

Answer 1

Consider x represents cost of one chair, y represents cost of one table ( costs are in rupees ).

Then, the cost of one chair and 1 table  = x + y

According to the question,

x + y = 2000                                  ...... (1)

Assume cost of 5 chairs and z (say) tables is 16200,

That is, 5x + zy = 16200              ...... (2)

After interchanging number of chairs and table, new cost = zx + 5y

Again, according to the question,

$(100-5)\%\text{ of }(zx+5y)=11210$

         95% of (zx + 5y) = 11210

                 0.95(zx+5y)=11210

                           zx+5y=11800            ...... (3)

Equation (2) - 5 × equation (1)

(z-5)y=16200-10000

(z-5)y=6200

 ⇒ $y=\frac{6200}{z-5}$

Substitute this value in equation (1),

$x+\frac{6200}{z-5}=2000$

          $x=2000-\frac{6200}{z-5}$

             $=\frac{2000z-10000-6200}{z-5}$

             $=\frac{2000z-16200}{z-5}$

Substitute the values of x and y in equation (3),

$z(\frac{2000z-16200}{z-5})+5(\frac{6200}{z-5})=11800$

         $\frac{2000z^2-16200z+31000}{z-5}=11800$

$2000z^2-16200z+31000=11800z-59000$

$2000z^2-28000z+90000=0$

                 $z^2-14z+45=0$

           $z^2-9z-5z+45=0$

         $z(z-9)-5(z-9)=0$

                 $(z-5)(z-9)=0$

                                     $z=5\text{ or }9$

If the value of z is 5, then x and y are not finite.

Thus, z =9.

That is, initially, the number of tables is 9.

Also,

$x=\frac{2000(9)-16200}{9-5}$

  $=\frac{18000-16200}{4}$

  $=\frac{1800}{4}$

  $=450$

$y=\frac{6200}{9-5}$

  $=\frac{6200}{4}$

  $=1550$

And,

$7x+6y=7(450)+6(1550)$

            $=3150+9300$

            $=12,450$

Therefore, the total cost of 7 chairs and 6 tables is rs 12,450.