Question

The following table gives the monthly incomes of 12
families in a town:
S. No.

1 2 3 4. 5 6 7 8 9 10 11 12
Monthly
income (₹)
280 180 96 98 104 | 75 80 94 100 75 600 200
Calculate the median of the above incomes.
​

Answer 1

Answer:

₹ 1,982

Step-by-step explanation:

QUESTION:If α, β and γ are the roots of equation x³ - px + q = 0, then the cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)] will be?SOLUTION:Given:α, β and γ are the roots of equation x³ - px + q = 0To Find:Cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)].Answer:Option A : (p + q - 1)x³ - (2p + 3q)x² + (p + 3q)x - q = 0 is correct.Step By Step Explanation:

We know that for a cubic equation, ax³ + bx² + cx + d = 0,

Sum of zeroes = -b/aSum of product of zeroes taken two at a time = c/aProduct of zeroes = -d/a

So,

Sum of zeroes = α + β + γ = -b/a = -(0)/1 = 0 ————(1)Sum of product of zeroes taken two at a time = αβ + βγ + γα = c/a = (-p)/1 = -p ————(2)Product of zeroes = αβγ = -d/a = -(q)/1 = -q ————(3)

\begin{gathered}\\\end{gathered}

Now, we need to find the cubic equation whose roots are [α/(1+α)], [β/(1+β)] and [γ/(1+γ)].

For this, we will first find the sum of zeroes, sum of product of zeroes taken two at a time and product of zeroes for the required cubic equation.

\begin{gathered}\\\end{gathered}

First, we will find sum of zeroes.

So,

\implies\text{Sum of zeroes =} \dfrac{\alpha}{1+\alpha}+\dfrac{\beta}{1+\beta}+\dfrac{\gamma}{1+\gamma}⟹Sum of zeroes =1+αα+1+ββ+1+γγ

\implies\text{Sum of zeroes =}\dfrac{\alpha(1+\beta)(1+\gamma)+ \beta(1+\alpha)(1+\gamma)+ \gamma(1+\beta)(1+\alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}⟹Sum of zeroes =(1+α)(1+β)(1+γ)α(1+β)(1+γ)+β(1+α)(1+γ)+γ(1+β)(1+α)

\implies\text{Sum of zeroes =}\dfrac{(\alpha+\alpha\gamma+\beta\alpha+\alpha\beta\gamma)+(\beta+\beta\gamma+\beta\alpha+\alpha\beta\gamma)+(\gamma+\alpha\gamma+\beta\gamma+\alpha\beta\gamma)}{1+\alpha+\beta+\alpha\beta+\gamma+\alpha\gamma+\beta\gamma+\alpha\beta\gamma}⟹Sum of zeroes =1+α+β+αβ+γ+αγ+βγ+αβγ(α+αγ+βα+αβγ)+(β+βγ+βα+αβγ)+(γ+αγ+βγ+αβγ)

Grouping the terms,

\implies\text{Sum of zeroes =}\dfrac{(\alpha+\beta+\gamma)+2(\alpha\beta+\beta\gamma+\gamma\alpha)+3(\alpha\beta\gamma)}{1+(\alpha+\beta+\gamma)+(\alpha\beta+\alpha\gamma+\beta\gamma)+(\alpha\beta\gamma)}⟹Sum of zeroes =1+(α+β+γ)+(αβ+αγ+βγ)+(αβγ)(α+β+γ)+2(αβ+βγ+γα)+3(αβγ)

Using (1), (2) and (3),

\implies\text{Sum of zeroes =}\dfrac{(0)+2(-p)+3(-q)}{1+(0)+(-p)+(-q)}⟹Sum of zeroes =1+(0)+(−p)+(−q)(0)+2(−p)+3(−q)

\implies\text{Sum of zeroes =}\dfrac{-2p-3q}{1-p-q}=\dfrac{2p+3q}{p+q-1}⟹Sum of zeroes =1−p−q−2p−3q=p+q−12p+3q

Hence,

\implies\text{Sum of zeroes =}\dfrac{2p+3q}{p+q-1}- - - -(4)⟹Sum of zeroes =p+q−12p+3q−−−−(4)

\begin{gathered}\\\end{gathered}

Now, we will find sum of product of zeroes taken two at a time.

So,

\implies\text{Sum of product of zeroes taken two at a time =} \left(\dfrac{\alpha}{1+\alpha}\times\dfrac{\beta}{1+\beta}\right)+ \left(\dfrac{\alpha}{1+\alpha}\times\dfrac{\gamma}{1+\gamma}\right)+ \left(\dfrac{\gamma}{1+\gamma}\times\dfrac{\beta}{1+\beta}\right)⟹Sum of product of zeroes taken two at a time =(1+αα×1+ββ)+(1+αα×1+γγ)+(1+γγ×1+ββ)

\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta}{(1+\alpha)(1+\beta)}+ \dfrac{\alpha\gamma}{(1+\alpha)(1+\gamma)} + \dfrac{\gamma\beta}{(1+\gamma)(1+\beta)}⟹Sum of product of zeroes taken two at a time =(1+α)(1+β)αβ+(1+α)(1+γ)αγ+(1+γ)(1+β)γβ

\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta(1+\gamma)+\alpha\gamma(1+\beta)+ \gamma\beta(1+\alpha)}{(1+\alpha)(1+\beta)(1+\gamma)}⟹Sum of product of zeroes taken two at a time =(1+α)(1+β)(1+γ)αβ(1+γ)+αγ(1+β)+γβ(1+α)

\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{\alpha\beta+\alpha\beta\gamma+ \alpha\gamma+\alpha\beta\gamma+ \gamma\beta+\alpha\beta\gamma}{1-p-q}⟹Sum of product of zeroes taken two at a time =1−p−qαβ+αβγ+αγ+αβγ+γβ+αβγ

We had calculated the value of (1+α)(1+β)(1+γ) to be (1-p-q), that’s why we put its value directly.

Grouping the terms,

\implies\text{Sum of product of zeroes taken two at a time =}\dfrac{(\alpha\beta+\alpha\gamma+\gamma\beta)+3(\alpha\beta\gamma)}{1-p-q}⟹Sum of product of zeroes taken two at a time =1−p−q(αβ+αγ+γβ)+3(αβγ)

Using (2) and (3),

Answer 2

Answer of this question is 99