Question

One fashion house has to make $810$ dresses and another one $900$ dresses during the same period time. In the first house the order was ready $3$ days ahead of time and the second house $6$ days ahead time. How many dresses did each fashion house make a day if the second house made $21$ dresses more a day than the first?

Answer 1

Hint: To find the number of dresses made by each house, find the number of dresses made by each house in one day. Then use the given condition that the second house made $21$ dresses more a day than the first to form an equation and solve to find the number of dresses each fashion house made a day.

Complete step-by-step answer:
Find the number of dresses made in a day.
Suppose $F$ be first fashion house and $G$ be second fashion house.
Let $x$ be the same time period
Since house $F$ order was ready $3$ days ahead (three days earlier than time)
$\therefore $ Number of days house $F$ take to make \[810\] dresses is $x - 3$
Since house $G$ order was ready $6$ days ahead (six days earlier than time)
$\therefore $ Number of days house $G$ take to make \[900\] dresses is $x - 6$
Now we find the number of dresses each house can make in a day using a unitary method.
We know if we are given the value of multiple units we can find the value of a single unit by dividing the value of multiple units by the total number of units.
Number of dresses house $F$ make in $x - 3$ days is \[810\].
So, the number of dresses house $F$ make in $1$ day is \[\dfrac{{810}}{{x - 3}}\]. … (i)
Number of dresses the house $G$ make in $x - 6$ days is \[900\].
So, the number of dresses the house $G$ make in $1$ day is \[\dfrac{{900}}{{x - 6}}\]. … (ii)
Now we know from the statement of the question that the second house made $21$ dresses more a day than the first house. So we can write number of dresses made by house $G$ is greater than number of dresses made by house $F$ by $21$
We can write this statement in the form of an equation using equations (i) and (ii).
$ \Rightarrow \dfrac{{900}}{{x - 6}} = \dfrac{{810}}{{x - 3}} + 21$

Solve this equation for $x.$
Shift all the values except the constant to one side of the equation.
$ \Rightarrow \dfrac{{900}}{{x - 6}} - \dfrac{{810}}{{x - 3}} = 21$
Take LCM on the left side of the equation.
\[
   \Rightarrow \dfrac{{900(x - 3) - 810(x - 6)}}{{(x - 6)(x - 3)}} = 21 \\
   \Rightarrow \dfrac{{900x - 2700 - 810x + 4860}}{{x(x - 3) - 6(x - 3)}} = 21 \\
   \Rightarrow \dfrac{{90x + 2160}}{{{x^2} - 3x - 6x + 18}} = 21 \\
 \]
Now we cross multiply the value in denominator of LHS to numerator of RHS
\[
   \Rightarrow 90x + 2160 = 21 \times ({x^2} - 3x - 6x + 18) \\
   \Rightarrow 90x + 2160 = 21 \times ({x^2} - 9x + 18) \\
   \Rightarrow 90x + 2160 = 21{x^2} - 189x + 378 \\
 \]
Shift all constants to one side of the equation.
\[
   \Rightarrow 2160 - 378 = 21{x^2} - 189x - 90x \\
   \Rightarrow 1782 = 21{x^2} - 279x \\
 \]
Shift all values to one side of the equation to make a quadratic equation,
\[ \Rightarrow 21{x^2} - 279x - 1782 = 0\]
Take 3 common from all elements on LHS
\[ \Rightarrow 3(7{x^2} - 93x - 594) = 0\]
Divide both sides of the equation by 3
\[ \Rightarrow \dfrac{{3(7{x^2} - 93x - 594)}}{3} = \dfrac{0}{3}\]
\[ \Rightarrow 7{x^2} - 93x - 594 = 0\] … (iii)
Solve this quadratic equation by using a quadratic formula.
Comparing $7{x^2} - 93x - 594 = 0$ with $a{x^2} + bx + c = 0$ we get:
$a = 7,b = - 93,c = - 594$.
Substituting these values of $a,b,c$ in $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$we get:
 $x = \dfrac{{ - ( - 93) \pm \sqrt {{{( - 93)}^2} - 4(7)( - 594)} }}{{2(7)}}$
$ \Rightarrow x = \dfrac{{93 \pm \sqrt {8649 + 16632} }}{{14}}$
$ \Rightarrow x = \dfrac{{93 \pm \sqrt {25281} }}{{14}}$
$ \Rightarrow x = \dfrac{{93 \pm 159}}{{14}}$
Separate positive and negative signs and write two values of $x.$
$ \Rightarrow x = \dfrac{{93 + 159}}{{14}}$ or $x = \dfrac{{93 - 159}}{{14}}$
$ \Rightarrow x = \dfrac{{252}}{{14}}$ or $x = \dfrac{{ - 66}}{{14}}$
\[ \Rightarrow x = 18\] or $x = \dfrac{{ - 33}}{7}$
Since $x$ is the number of days hence positive.
\[ \Rightarrow x = 18\]
Find the number of dresses made by $F$ and $G.$
$\therefore $ Number of dresses made by house $F = \dfrac{{810}}{{x - 3}}$
                                                                        $
   = \dfrac{{810}}{{18 - 3}} \\
   = \dfrac{{810}}{{15}} \\
   = 54 \\
 $
Number of dresses made by house $G = \dfrac{{900}}{{x - 6}}$
                                                                   $
   = \dfrac{{900}}{{18 - 6}} \\
   = \dfrac{{900}}{{12}} \\
   = 75 \\
 $
Therefore, the first fashion house makes $54$ dresses and the second house makes $75$ dresses.

Note: Students are likely to make mistakes in the calculation part where we shift values from one side of the equation to another, so always keep in mind that the sign changes from positive to negative and vice versa when shifting any constant or variable to the other side.