HSC Commerce: Marketing and Salesmanship इयत्ता १२ वी - Maharashtra State Board Important Questions for Mathematics and Statistics

State whether the following statement is True or False:  

The degree of a differential equation `"e"^(-("d"y)/("d"x)) = ("d"y)/("d"x) + "c"` is not defined

State whether the following statement is True or False:

Order and degree of differential equation `x ("d"^3y)/("d"x^3) + 6(("d"^2y)/("d"x^2))^2 + y` = 0 is (2, 2)

Find the differential equation by eliminating arbitrary constants from the relation y = (c₁ + c₂x)e^x 

Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`

Find the population of city at any time t given that rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 30000 to 40000.

Solution: Let p be the population at time t.

Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.

∴ `"dp"/"dt" prop "p"`

∴ `"dp"/"dt"` = kp, where k is a constant.

∴ `"dp"/"p"` = k dt

On integrating, we get

`int "dp"/"p" = "k" int "dt"`

∴ log p = kt + c

Initially, i.e. when t = 0, let p = 30000

∴ log 30000 = k × 0 + c       

∴ c = `square`

∴ log p = kt + log 30000

∴ log p - log 30000 = kt

∴ `log("p"/30000)` = kt          .....(1)     

when t = 40, p = 40000

∴ `log (40000/30000) = 40"k"`

∴ k = `square`

∴ equation (1) becomes, `log ("p"/30000)` = `square`

∴ `log ("p"/30000) = "t"/40 log (4/3)`

∴ p = `square`

The order and degree of the differential equation `[1 + ((dy)/(dx))^3]^(2/3) = 8((d^3y)/(dx^3))` are respectively ______.

State whether the following statement is true or false.

The integrating factor of the differential equation `(dy)/(dx) + y/x` = x³ is – x.

y² = (x + c)³ is the general solution of the differential equation ______.

Solve the following differential equation

x²y dx – (x³ + y³)dy = 0

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.

The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.

Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.

`int dx/x = square + c` 

∴ logx = `square`

x = `square` = `square`.e^c

∴ x = `square`.a where a = e^c

At t = 0, x = 800

∴ a = `square`

At t = 5, x = 400

∴ e^–5k = `square`

Now when t = 30 

x = `square` × `square` = 800 × (e^–5k)⁶ = 800 × `square` = `square`.

The mass remaining after 30 days will be `square` mg.

The degree of the differential equation `((d^2y)/dx^2)^2 + (dy/dx)^3` = a^x is 3.

In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

Solution:

Let N be the number of bacteria present at time ‘t’.

Since the rate of increase of N is proportional to N, the differential equation can be written as –

`(dN)/dt αN`

∴ `(dN)/dt` = KN, where K is constant of proportionality

∴ `(dN)/N` = k . dt

∴ `int 1/N dN = K int 1 . dt`

∴ log N = `square` + C   ...(1)

When t = 0, N = N₀ where N₀ is initial number of bacteria.

∴ log N₀ = K × 0 + C

∴ C = log N₀

Also when t = 4, N = 2N₀

∴ log (2 N₀) = K . 4 + `square`   ...[From (1)]

∴ `log((2N_0)/N_0)` = 4K,

∴ log 2 = 4K

∴ K = `square`   ...(2)

Now N = ? when t = 12

From (1) and (2)

log N = `1/4 log 2  . (12) + log N_0`

log N – log N₀ = 3 log 2

∴ `log(N_0/N_0)` = `square`

∴ N = 8 N₀

∴ Bacteria are increased 8 times in 12 hours.

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1,00,000, when will the city have population 4,00,000?

Let ‘p’ be the population at time ‘t’ years.

∴ `("dp")/"dt" prop "p"`

∴ Differential equation can be written as  `("dp")/"dt" = "kp"`

where k is constant of proportionality.

∴ `("dp")/"p" = "k.dt"`

On integrating we get

`square` = kt + c   ...(i)

(i) Where t = 0, p = 1,00,000

∴ from (i)

log 1,00,000 = k(0) + c

∴  c = `square`

∴  log `("p"/(1,00,000)) = "kt"`       ...(ii)

(ii) When t = 25, p = 2,00,000

as population doubles in 25 years

∴ from (ii) log2 = 25k

∴  k = `square`

∴  log`("p"/(1,00,000)) = (1/25log2).t`

(iii) ∴ when p = 4,00,000

`log ((4,00,000)/(1,00,000)) = (1/25log2).t`

∴ `log 4 = (1/25 log2).t`

∴ t = `square ` years

A salesman receives 3% commission on the sales up to ₹ 50,000 and 4% commission on the sales over ₹ 50,000. Find his total income on the sale of ₹ 2,00,000.

A retailer sold a suit for ₹ 8,832 after allowing 8% discount on marked price and further 4% cash discount. If he made 38% profit, find the cost price and the marked price of the suit.

An agent who gives a guarantee to his principal that the party will pay the sale price of goods is called ______.

Fill in the Blank.

A _______ is an agent who brings together the buyer and the seller.

Choose the correct alternative:

The present worth of ₹ 11,660 due 9 months hence is ₹ 11,000. The True discount is ______

State whether the following statement is True or False:

The trade discount is first calculated on the catalogue (list) price