HSC Commerce (English Medium) 12th Standard Board Exam - Maharashtra State Board Question Bank Solutions

If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^–3 = 0.0497.

P[X = x] = `square`

Since P[X = 2] = P[X = 3]

`square` = `square`

`m^2/2 = m^3/6` 

∴ m = `square`

Now, P[X ≥ 2] = 1 – P[x < 2]

= 1 – {P[X = 0] + P[X = 1]

= `1 - {square/(0!) + square/(1!)}`

= 1 – e^–3[1 + 3]

= 1 – `square` = `square`

`sqrt((sump_1q_0)/(sump_0q_0)) xx sqrt((sump_1q_1)/(sump_0q_1)) xx 100`

Conditions under which the object function is to be maximum or minimum are called ______.

A department store has four workers to pack goods. The times (in minutes) required for each worker to complete the packings per item sold is given below. How should the manager of the store assign the jobs to the workers, so as to minimize the total time of packing?

If X – P(m) with P(X = 1) = P(X = 2), then find the mean and P(X = 2) given that e^–2 = 0.1353

Solution: Since P(X = 1) = P(X = 2)

`(e^-mm^1)/(1!) = square`

∴ m = `square`

∴ mean = `square` = `square`

Then P(X = 2) = `square` = `square`

Find the equation of tangent and normal to the curve y = x² + 5 where the tangent is parallel to the line 4x – y + 1 = 0.

Laspeyre’s Price Index Number uses current year’s quantities as weights.

An agent places insurance for ₹ 4,00,000 on life of a person. The premium is to be paid annually at the rate of ₹ 35 per thousand per annum. Find the agent’s commission at 15% on the premium.

An agent was paid ₹ 88,000 as a commission on the sales of computers at the rate of 12.5%. If the price of each computer was ₹ 32,000, how many computers did he sell?

Calculate Marshall – Edgeworth’s price index number for the following data:

Evaluate `int1/(x(x-1)) dx`

Evaluate `int 1/(x(x-1))dx`

Evaluate `int(5x^2 - 6x+3)/(2x- 3 )dx`

If f' (x) = 4x³ - 3x² + 2x + k, f(O) = 1 and f(1) = finf (x).

If f'(x) - 4x³ - 3x² + 2x + k, f(0) = 1 and f(1) = 4, find f(x).

If ∑ p₀q₀ = 120, ∑ p₀q₁ = 160, ∑ p₁q₁ = 140, ∑ p₁q_o = 200, find Laspeyre’s, Paasche’s, Dorbish-Bowley’s and Marshall-Edgeworth’s Price Index Numbers.

Five wagons are available at stations 1, 2, 3, 4 and 5. These are required at 5 stations I, II, III, IV and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?

In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.

(Given that e^-0.2 = 0.8187)

Solution:

Here, m = `square` and X − P(m) with parameter m.

The p.m.f. X is:

P(X = x) = `(e^(−m).m^x)/(x!), x = 0, 1, 2,...`

P(X ≥ 3) = 1 − P(X < 3)

= 1 − [`square + square + square`]

= `1 − [(e^(− 0.2)(0.2)^0)/(0!) + (e^(−0.2)(0.2)^1)/(1!) + (e^(−0.2)(0.2)^2)/(2!)]`

= 1 − [0.8187(1 + 0.2 + 0.02)]

= 1 − `square`

= `square`

In the following table, Laspeyre's and Paasche's Price Index Numbers are equal. Complete the following activity to find x :

Solution: P₀₁(L) = P₀₁(P)

`(sum "p"_1"q"_0)/(sum "p"_0"q"_0) xx 100 = square/(sum "p"_0"q"_1) xx 100`

`(20 + 5x)/square xx 100 = square/14 xx 100`

∴ x = `square`

A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table :

Complete the following activity to find the optimal assignment to minimize the total processing cost.

Solution:

Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :

Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.

Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:

Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.

Examine the rows one by one starting with the first row with exactly one zero is found. Mark the zero by enclosing it in (`square`), indicating assignment of the job. Cross all the zeros in the same column. This step is shown in the following table :

Step 5: It is observed that all the zeros are assigned and each row and each column contains exactly one assignment. Hence, the optimal (minimum) assignment schedule is :

Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`