The cost price of box and pen is 80 the box is sold at 10% profit and the pen is sold at 10% loss of the selling price of the box is Rs 28 more than that of the pen calculate the profit or loss percent in the whole

2.5%

Step-by-step explanation:

Let the cost price of box be x

let the cost price of pen be 80-x

Now for profit of box

s.p =c.p-c.p of p%

s.p = x-x *10/100

s.p= 1.1x

Now for loss of pen

s.p = c.p-c.p of loss%

s.p = (80-x)-10/100 of (80-x)

s.p = 72-0.9x

Now

s.p of box= 28+s.p of pen

1.1x = 28+72-0.9x

20x =100

x =50

total s.p = 1.1x+72-0.9x

By putting the value of x

total s.p = 72+0.2*50

=82

s.p = 82

c.p=80

profit =s.p-c.p

=82-80

=2

Now

p% =p/c.p*100%

=2/80*100

=2.5 profit

2.5%

Step-by-step explanation:

Let the cost price of box be x

let the cost price of pen be 80-x

Now for profit of box

s.p =c.p-c.p of p%

s.p = x-x *10/100

s.p= 1.1x

Now for loss of pen

s.p = c.p-c.p of loss%

s.p = (80-x)-10/100 of (80-x)

s.p = 72-0.9x

Now

s.p of box= 28+s.p of pen

1.1x = 28+72-0.9x

20x =100

x =50

total s.p = 1.1x+72-0.9x

By putting the value of x

total s.p = 72+0.2*50

=82

s.p = 82

c.p=80

profit =s.p-c.p

=82-80

=2

Now

p% =p/c.p*100%

=2/80*100

=2.5 profit

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;