What is the original price of a packet of biscuits bought for Rs230 including 6% tax
1 answer:
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Answer:
The merchant buys 30 shirts originally.
Step-by-step explanation:
Let us assume that the merchant bought x numbers of shirts in $120.
So, the cost for each shirt is $.
Now, if the cost for each shirt is reduced by 1$, then he would have bought 10 shirts more i.e. (x + 10) shirts in $120.
So, we can write the following equation as
⇒(120 - x)(x + 10) = 120x
⇒ 120x - 10x + 1200 - x² = 120x
⇒ x² +10x - 1200 = 0
⇒ x² + 40x - 30x - 1200 = 0
⇒(x + 40)(x - 30) = 0
⇒ x = - 40 or x = 30
But x can not be negative.
Hence, the merchant buys 30 shirts originally. (Answer)
Answer:
restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75
Step-by-step explanation:
to calculate current revenue= $6 x 330 = $1980
suppose x as the number of times the price to be dropped by $0.25
then find new price.. i.e
new price= $(6-0.25x)
and, new sell=330 +15x sandwiches
therefore, the new revenue would be= (6-0.25x)(330 +15x)
in order to maximize the current revenue, simplify the above equation and make it complete square using x
(6-0.25x)(330 +15x)
=1980-82.5x +90x -3.75
=1980 + 7.5x -3.75
=1980-3.75 (-2x+) ----> taking out common
now, to make a complete square lets add and subtract 1 inside the parentheses
=1980-3.75(-1+1-2x+)
=1980 +3.75 -3.75( -2x +1)
=1983.75 -3.75 ---->(1)
as is positive always, minimize the other term in order to maximize the total revenue.
so the minimum possible value of = 0
therefore, x=1
putting x in eq(1) the revenure becomes,
$(1983.75-0)=> $1983.75
therefore, restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75
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as
and be tempted to cancel out the factors of
. But this cancellation is only valid when 
.
When
, you end up with the indeterminate form 
, which is why 
is not a zero.
and be tempted to cancel out the factors of
When