2019
June 1 Started business with cash ` 50,000, Bank ` 1,00,000, Goods worth ` 50,000.
5 Purchased goods from Mohan on credit ` 80,000 at 10% Trade Discount.
9 Sold goods to Urmila ` 30,000 at 5% Trade Discount.
12 Paid in to Dena Bank ` 40,000.
15 Goods worth ` 5000 distributed as free sample.
22 Paid for Commission ` 5,000 to Anand.
24 Received ` 28,000 from Urmila in full settlement of her account by Debit Card.
29 Paid for Advertisement ` 9,000/
30 Purchased Laptop for ` 20,000 @28% GST and amount paid by NEFT.
Prepare Cash, Bank, Capital, Purchase and Sales ledger from the above transaction
Answers
Answer:
\large\underline{\sf{Solution-}}
Solution−
Let assume that
\begin{gathered}\rm \: \alpha ,\gamma \: be \: the \: roots \: of {3x}^{2} - 2x - 5 = 0 \\ \end{gathered}
α,γbetherootsof3x
2
−2x−5=0
Consider,
\begin{gathered}\rm \: {3x}^{2} - 2x - 5 = 0 \\ \end{gathered}
3x
2
−2x−5=0
\begin{gathered}\rm \: {3x}^{2} - 5x + 3x - 5 = 0 \\ \end{gathered}
3x
2
−5x+3x−5=0
\begin{gathered}\rm \: x(3x - 5) + 1(3x - 5) = 0 \\ \end{gathered}
x(3x−5)+1(3x−5)=0
\begin{gathered}\rm \: (x + 1)(3x - 5) = 0 \\ \end{gathered}
(x+1)(3x−5)=0
\begin{gathered}\rm\implies \:x = - 1 \: \: or \: \: x = \dfrac{5}{3} \\ \end{gathered}
⟹x=−1orx=
3
5
\begin{gathered}\rm\implies \: \gamma = - 1 \: \: or \: \: \alpha = \dfrac{5}{3} \\ \rm \: or \\ \rm\implies \: \alpha = - 1 \: \: or \: \: \gamma = \dfrac{5}{3} \\\end{gathered}
⟹γ=−1orα=
3
5
or
⟹α=−1orγ=
3
5
Now,
Let assume that,
\begin{gathered}\rm \: \alpha, \beta \: be \: the \: roots \: of \: {2x}^{2} + px - 1 = 0 \\ \end{gathered}
α,βbetherootsof2x
2
+px−1=0
We know,
\begin{gathered}\boxed{\red{\sf Sum\ of\ the\ roots=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \end{gathered}
Sum of the roots=
coefficient of x
2
−coefficient of x
\begin{gathered}\rm\implies \: \alpha + \beta = - \dfrac{p}{2} \\ \end{gathered}
⟹α+β=−
2
p
Also,
\begin{gathered}\boxed{\red{\sf Product\ of\ the\ roots=\frac{Constant}{coefficient\ of\ x^{2}}}} \\ \end{gathered}
Product of the roots=
coefficient of x
2
Constant
\begin{gathered}\rm\implies \: \alpha \beta = - \dfrac{1}{2} \\ \end{gathered}
⟹αβ=−
2
1
It means, we have
\begin{gathered}\rm\implies \: \alpha + \beta = - \dfrac{p}{2} \: \: and \: \: \alpha \beta = - \dfrac{1}{2} \\ \end{gathered}
⟹α+β=−
2
p
andαβ=−
2
1
Case :- 1
\begin{gathered}\rm \: \alpha = - 1 \\\end{gathered}
α=−1
As, we have
\begin{gathered}\rm \: \alpha \beta = - \dfrac{1}{2} \\ \end{gathered}
αβ=−
2
1
\begin{gathered}\rm \: ( - 1) \beta = - \dfrac{1}{2} \\ \end{gathered}
(−1)β=−
2
1
\begin{gathered}\rm \:\beta = \dfrac{1}{2} \\ \end{gathered}
β=
2
1
Now,
\begin{gathered}\rm \: \alpha + \beta = - \dfrac{p}{2} \\ \end{gathered}
α+β=−
2
p
\begin{gathered}\rm \: - 1 + \dfrac{1}{2} = - \dfrac{p}{2} \\ \end{gathered}
−1+
2
1
=−
2
p
\begin{gathered}\rm \: - \dfrac{1}{2} = - \dfrac{p}{2} \\ \end{gathered}
−
2
1
=−
2
p
\begin{gathered}\rm\implies \:p \: = \: 1 \\ \end{gathered}
⟹p=1
Now, Consider Case :- 2
\begin{gathered}\rm \: \alpha = \dfrac{5}{3} \\\end{gathered}
α=
3
5
As, we have
\begin{gathered}\rm \: \alpha \beta = - \dfrac{1}{2} \\ \end{gathered}
αβ=−
2
1
\begin{gathered}\rm \: \frac{5}{3} \times \beta = - \dfrac{1}{2} \\ \end{gathered}
3
5
×β=−
2
1
\begin{gathered}\rm \: \beta = - \dfrac{3}{10} \\ \end{gathered}
β=−
10
3
Now,
\begin{gathered}\rm \: \alpha + \beta = - \dfrac{p}{2} \\ \end{gathered}
α+β=−
2
p
\begin{gathered}\rm \: \dfrac{5}{3} - \dfrac{3}{10} = - \dfrac{p}{2} \\ \end{gathered}
3
5
−
10
3
=−
2
p
\begin{gathered}\rm \: \dfrac{50 - 9}{30} = - \dfrac{p}{2} \\ \end{gathered}
30
50−9
=−
2
p
\begin{gathered}\rm \: \dfrac{41}{30} = - \dfrac{p}{2} \\ \end{gathered}
30
41
=−
2
p
\begin{gathered}\rm\implies \:p \: = \: - \: \dfrac{41}{15}\\ \end{gathered}
⟹p=−
15
41
So,
\begin{gathered}\begin{gathered}\begin{gathered}\bf\: \rm\implies \:\begin{cases} &\bf{p \: = \: 1} \\ \\ &\sf{or}\\ \\ &\bf{p \: = \: - \: \dfrac{41}{15} } \end{cases}\end{gathered}\end{gathered}\end{gathered}
⟹
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎧
p=1
or
p=−
15
41
\rule{190pt}{2pt}
Additional Information :-
Nature of roots :-
Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.
If Discriminant, D > 0, then roots of the equation are real and unequal.
If Discriminant, D = 0, then roots of the equation are real and equal.
If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.
Where,
Discriminant, D = b² - 4ac
Explanation:
anything