Solve the Following Question.(5 Marks)

Question 15 Marks

Akash, Sameer, and Sid took a house on rent for one year for ₹ 16,236 . They stayed together for 4 months and then Sid left the house. After 5 more months, Sameer also left the house.
How much rent should each pay?

Answer

Let ' $R$ ' be the rent per month to be paid to the landlord.
Given that, Sid left the house after 4 months
$\therefore$ Rent paid by $\mathrm{Sid}=\frac{R}{3} \times 4=\frac{4 R}{3}$
Sameer left the house after another 5 months,
$\therefore$ Rent paid by Sameer $=\frac{R}{2} \times 5+\frac{R}{3} \times 4$
$=\mathrm{R}\left(\frac{5}{2}+\frac{4}{3}\right)$
$=\frac{23 R}{6}$
Akash stayed in the house for the entire year.
$\therefore$ Rent paid by Akash $=3 \mathrm{R}+\frac{R}{2} \times 5+\frac{R}{3} \times 4$
$=R\left(3+\frac{5}{2}+\frac{4}{3}\right)$
$=\frac{41 R}{6}$
$\therefore$ The rent paid by the three of them, over that period of one year must be in the proportion.
$
\frac{41 \mathrm{R}}{6}: \frac{23 \mathrm{R}}{6}: \frac{4 \mathrm{R}}{3}
$
i.e. in the proportion
$41: 23: 8$.....(multiplying throughout by $\frac{6}{R}$ )
Let $x$ be the constant of proportionality.
Rent to be paid by Akash $=₹ 41 \mathrm{x}$
Rent to be paid by Sameer $=₹ 23 x$
and rent to be paid by Sid $=₹ 8 \mathrm{x}$
The total rent for the house was $₹ 16236$.
$
\begin{aligned}
& \therefore 41 x+23 x+8 x=₹ 16236 \\
& \therefore 72 x=16236 \\
& \therefore x=225.5
\end{aligned}
$
$\therefore$ Akash should pay $41 \mathrm{x}=41 \times 225.5=₹ 9245.5$
Sameer should pay $23 x=23 \times 225.5=₹ 5186.5$
and Sid should pay $8 x=8 \times 225.5=₹ 1804$

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Question 25 Marks

Mr. Dinesh invests $₹ 20,800$ in $6 \% ₹ 100$ shares at $₹ 104$, and $₹ 14,300$ in $10.5 \% ₹ 100$ shares at $₹ 143$. What will be his annual income from the shares?

Answer

For 1 st kind of shares,
Face value of shares (F.V.) $=₹ 100$
Dividend $=6 \%$
$
\begin{aligned}
& \therefore \text { Annual income from one share }=\frac{6}{100} \times 100=₹ 6 \\
& \text { Market value of the share }(M . V .)=₹ 104 \\
& \text { Total amount invested }=₹ 20,800 \\
& \begin{aligned}
\therefore \quad \text { Number of shares } & =\frac{\text { Amount invested }}{\text { Market value }} \\
& =\frac{20,800}{104} \\
& =200
\end{aligned}
\end{aligned}
$
$\therefore$ Total income from 1st kind of shares $=200 \times 6=₹ 1200$
For $2 \mathrm{nd}$ kind of shares,
Face value of shares (F.V.) $=₹ 100$
Dividend $=10.5 \%$
$\therefore$ Annual income from one share $=\frac{10.5}{100} \times 100=₹ 10.5$
Market value of the share $(\mathrm{M} . \mathrm{V})=.₹ 143$
Total amount invested $=₹ 14300$
$
\begin{aligned}
& \therefore \quad \begin{array}{c}
\text { Number of shares }
\end{array}=\frac{\text { Amount invested }}{\text { Market value }} \\
\end{aligned}
$
$\therefore$ Total income from 2 nd kind of shares $=100 \times 10.5=₹ 1050$
$\therefore$ Total annual income of Dinesh from both these shares $=1200+1050=₹ 2250$

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Question 35 Marks

A and B start a business, with A investing the total capital of $₹ 50,000$, on the condition that $B$ pays interest at the rate of $10 \%$ per annum on his half of the capital. A is a working partner and receives ₹ 1,500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?

Answer

Let ' $x$ ' and ' $y$ ' be the profits earned by A and B respectively and let ' $z$ ' be the total profit for the year.
A is the working partner and receives $₹ 1500$ per month from the total profit. i.e. $12 \times 1500=₹ 18,000$ at the end of the year.
The remaining profit is shared between $\mathrm{A}$ and $\mathrm{B}$ equally.
$
\therefore \mathrm{y}=\frac{z-18000}{2} \text {.....(i) }
$
Thus, profit earned by $\mathrm{A}$ at the end of that year is given by
$
\begin{aligned}
& \mathrm{x}=18000+\left(\frac{z-18000}{2}\right) \\
& \therefore \mathrm{x}=\frac{z+18000}{2} \ldots . . . \text { (ii) }
\end{aligned}
$
A invests the entire capital on the condition that B pays A interest at the rate of $10 \%$ per annum on his half of the capital.
$\therefore$ At the end of the first year,
A will receive $\frac{10}{100} \times 25,000$ i.e. $₹ 2500 /$ - over and above his share of profit.
$\therefore$ A's income $=$ Profit of $A+2500=x+2500$
Given that,
income of $\mathrm{A}=$ twice the income of $\mathrm{B}$
$
\therefore \mathrm{x}+2500=2 \mathrm{y} \text {.....(iii) }
$
Using (i) and (ii) in (iii), we get
$
\begin{aligned}
& \frac{z+18000}{2}+2500=2\left(\frac{z-18000}{2}\right) \\
& z+18000+5000=2(z-18000) \\
& z+23000=2 z-36000 \\
& \therefore z=59,000
\end{aligned}
$
$\therefore$ The total profit for the year $=₹ 59,000 /-$

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Question 45 Marks

Mr. Mehta sold his two luxury cars at $₹ 39,10,000$ each. On one he gains $15 \%$ but on the other, he loses $15 \%$. How much does he gain or lose in the whole transaction?

Answer

Let $x, y$ be the C.P. of two cars.
S.P. of both the cars $=39,10,000$ [Given]
$\therefore$ One car is sold at $15 \%$ loss
$\therefore$ S.P. of the first car $=\mathrm{x}-\frac{15}{100} \mathrm{x}$
$
\begin{aligned}
& \therefore \frac{85}{100} \mathrm{x}=39,10,000 \\
& \therefore \mathrm{x}=\frac{39,10,000 \times 100}{85} \\
& \therefore \mathrm{x}=46,000 \times 100 \\
& \therefore \mathrm{x}=46,00,000
\end{aligned}
$
Other car is sold at $15 \%$ gain
$
\begin{aligned}
& \therefore \text { S.P. of second car }=\mathrm{y}+\frac{15}{100} \mathrm{y} \\
& \therefore \mathrm{y}+\frac{15}{100} \mathrm{y}=39,10,000 \\
& \therefore \frac{115}{100} \mathrm{y}=39,10,000 \\
& \therefore \mathrm{y}=\frac{39,10,000 \times 100}{115} \\
& \therefore \mathrm{y}=34,000 \times 100 \\
& \therefore \mathrm{y}=34,00,000 \\
& \mathrm{x}+\mathrm{y}=\text { Total C.P. of two cars } \\
& =46,00,000+34,00,000 \\
& =80,00,000 \\
& \text { Total S.P. }=39,10,000+39,10,000=78,20,000 \\
& \therefore \text { S.P. }<\text { C.P. } \\
& \therefore \text { There is a loss of } ₹(80,00,000-78,20,000)=₹ 1,80,000 \\
& \therefore \text { Loss } \%=\frac{1,80,000}{80,00,000} \times 100 \\
& =\frac{18}{8} \\
& =2.25
\end{aligned}
$
$\therefore$ Mr. Mehta bears a $2.25 \%$ loss in the whole transaction.

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Question 55 Marks

A man sells an article at a profit of $25 \%$. If he had bought it at a $10 \%$ loss and sold it for $₹ 7$ less, he would have gained $35 \%$. Find the cost price of the article.

Answer

Let $₹$ ' $x$ ' be the C.P. of the article
$\therefore$ Article was sold at $25 \%$ profit
$\therefore$ S.P. of the article $=x\left(1+\frac{25}{100}\right)$
$=x\left(1+\frac{1}{4}\right)$
$=1.25 \mathrm{x}$
If the article was bought at $10 \%$ loss
i.e., the new C.P. $=x\left(1-\frac{10}{100}\right)$
$
\begin{aligned}
& =x\left(\frac{9}{10}\right) \\
& =0.9 \mathrm{x} \\
& \text { and sold at } ₹ 7 \text { less } \\
& \therefore \text { New S.P. }=1.25 \mathrm{x}-7
\end{aligned}
$
and sold at $₹ 7$ less
$\therefore$ New S.P. $=1.25 \mathrm{x}-7$
Then, the profit would have been $35 \%$
Using profit percentage $=\frac{\text { S.P.-C.P. }}{\text { C.P. }} \times 100$
$
\begin{array}{ll}
\therefore & 35=\frac{(1.25 x-7)-0.9 x}{0.9 x} \times 100 \\
\therefore & \frac{35}{100}=\frac{0.35 x-7}{0.9 x} \\
\therefore & \frac{7}{20}=\frac{0.35 x-7}{0.9 x} \\
\therefore & 6.3 x=20(0.35 x-7) \\
\therefore & 6.3 x=7 x-140 \\
\therefore & 7 x-6.3 x=140 \\
\therefore & 0.7 x=140 \\
\therefore & x=\frac{140}{0.7} \\
\therefore & x=200
\end{array}
$
$\therefore$ Cost price of the article is $₹ 200$

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Maths (commerce) Solve the Following Question.(5 Marks)

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